[{"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "Simplify, 16x plus 14 minus the entire expression 3x squared plus x minus 9. 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."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. But now I'm going to distribute the negative sign here. So negative 1 times 3x squared is negative 3x squared."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. Negative 1 times negative 9, remember you have to consider this negative right over there. That is part of the term."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. Negative times a negative is a positive. So then we have positive 9."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. So what's our highest degree term here? I like to write it in that order."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. We only have one of those, so let me write it over here. Negative 3x squared."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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? 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."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Example 3 Subtracting polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 14 plus 9, they're both constant terms, or they're both being multiplied by x to the 0. 14 plus 9 is 23."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "We're told to study the growth of bacteria. A scientist measures the area in square millimeters occupied by a sample population. The growth of the population can be modeled by f of t is equal to 24 times e to the 0.4 times t power, where t is the number of hours since the experiment began. Here's the graph of f. So I guess f is going to be, the output of this function is going to be the number of square millimeters after t hours. All right, so here we have the graph. We see how as time goes on, the square millimeters of our little bacterial population keeps growing, and it clearly is growing, or it looks like it's growing exponentially. In fact, we know it's growing exponential because it's an exponential function right over here."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Here's the graph of f. So I guess f is going to be, the output of this function is going to be the number of square millimeters after t hours. All right, so here we have the graph. We see how as time goes on, the square millimeters of our little bacterial population keeps growing, and it clearly is growing, or it looks like it's growing exponentially. In fact, we know it's growing exponential because it's an exponential function right over here. And they say, when does the population first occupy an area of 400 square millimeters? So pause this video and try to figure that out. All right, and this is a screenshot from the Khan Academy exercise."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "In fact, we know it's growing exponential because it's an exponential function right over here. And they say, when does the population first occupy an area of 400 square millimeters? So pause this video and try to figure that out. All right, and this is a screenshot from the Khan Academy exercise. So we wanna say, when does the population first occupy an area of 400 square millimeters? Let's see, 400 square millimeters is right over there. And so it looks like after seven hours that we are going to be 400 square millimeters or larger."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "All right, and this is a screenshot from the Khan Academy exercise. So we wanna say, when does the population first occupy an area of 400 square millimeters? Let's see, 400 square millimeters is right over there. And so it looks like after seven hours that we are going to be 400 square millimeters or larger. So it first hits it after seven hours. So seven hours, just like that. Now let's do the next few examples that build on this."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "And so it looks like after seven hours that we are going to be 400 square millimeters or larger. So it first hits it after seven hours. So seven hours, just like that. Now let's do the next few examples that build on this. So if I go back up to the top, and now we're told the same thing. We're using square millimeters to study the growth. This is the function."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Now let's do the next few examples that build on this. So if I go back up to the top, and now we're told the same thing. We're using square millimeters to study the growth. This is the function. But then they add this next line. Here is the graph. Here is the graph of F, and the graph of the line Y equals 600."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "This is the function. But then they add this next line. Here is the graph. Here is the graph of F, and the graph of the line Y equals 600. So they added that graph there. And then they say, which statement represents the meaning of the intersection point of the graphs? All right, so let's look at the choices here."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Here is the graph of F, and the graph of the line Y equals 600. So they added that graph there. And then they say, which statement represents the meaning of the intersection point of the graphs? All right, so let's look at the choices here. So, and it says choose all that apply. So pause this video and see if you can answer that. All right, so choice A says it describes the time when the population occupies 600 square millimeters."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "All right, so let's look at the choices here. So, and it says choose all that apply. So pause this video and see if you can answer that. All right, so choice A says it describes the time when the population occupies 600 square millimeters. So which statement represents the meaning of the intersection point of the graph? So they're talking about, they're talking about this point right over there. So does that describe the time when the population occupies 600 square millimeters?"}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "All right, so choice A says it describes the time when the population occupies 600 square millimeters. So which statement represents the meaning of the intersection point of the graph? So they're talking about, they're talking about this point right over there. So does that describe the time when the population occupies 600 square millimeters? So that is the time when the population has indeed reached 600 square millimeters, because that's the line Y is equal to 600 square millimeters. So I like that choice. I will select it."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "So does that describe the time when the population occupies 600 square millimeters? So that is the time when the population has indeed reached 600 square millimeters, because that's the line Y is equal to 600 square millimeters. So I like that choice. I will select it. The next choice. It gives the solution to the equation 24 times E to the 0.4T is equal to 600. Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "I will select it. The next choice. It gives the solution to the equation 24 times E to the 0.4T is equal to 600. Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power. This is Y is equal to 600. So the T value at which these two graphs equal, that means that they're both equal to the same Y value. Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Well, if you think about it, this right over here in blue, we've already talked about it, that is Y is equal to 24 times E to the 0.4T power. This is Y is equal to 600. So the T value at which these two graphs equal, that means that they're both equal to the same Y value. Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600. So I like this too. It gives a T value where this is true. So that's the solution to that equation."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Or another way to think about it, it means that that is equal to that, or that 24 times E to the 0.4T power is indeed equal to 600. So I like this too. It gives a T value where this is true. So that's the solution to that equation. It describes the situation where the area the population occupies is equal to the number of hours. That's definitely not the case, because the area here is 600 square millimeters. The hours looks like it's a little bit after eight."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "So that's the solution to that equation. It describes the situation where the area the population occupies is equal to the number of hours. That's definitely not the case, because the area here is 600 square millimeters. The hours looks like it's a little bit after eight. So they're definitely not equal. It describes the area the population occupies after 600 hours. No, we don't have to look up there."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "The hours looks like it's a little bit after eight. So they're definitely not equal. It describes the area the population occupies after 600 hours. No, we don't have to look up there. This T axis doesn't even go to 600 hours. So we wouldn't select that as well. Now let's keep building and go to the next part of this."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "No, we don't have to look up there. This T axis doesn't even go to 600 hours. So we wouldn't select that as well. Now let's keep building and go to the next part of this. And it says, it says, so once again, we measure the area in square millimeters to figure out the growth of the population. The growth of, oh, so here, we have two populations here. It says the growth of population A can be modeled by F of T is equal to that."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Now let's keep building and go to the next part of this. And it says, it says, so once again, we measure the area in square millimeters to figure out the growth of the population. The growth of, oh, so here, we have two populations here. It says the growth of population A can be modeled by F of T is equal to that. We've seen that already. But now they are introducing another population. The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "It says the growth of population A can be modeled by F of T is equal to that. We've seen that already. But now they are introducing another population. The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began. Here are the graphs of F and G. So now we have two populations. They're both growing exponentially, but at different rates. And then it says, when do the populations occupy the same area?"}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "The growth of population B can be modeled by G of T is equal to this, where T is the number of hours since the experiment began. Here are the graphs of F and G. So now we have two populations. They're both growing exponentially, but at different rates. And then it says, when do the populations occupy the same area? It says, round your answer to the nearest integer. And you could pause this video and try to think about that if you like. Well, you can see very clearly that it looks like they intersect right around there."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "And then it says, when do the populations occupy the same area? It says, round your answer to the nearest integer. And you could pause this video and try to think about that if you like. Well, you can see very clearly that it looks like they intersect right around there. So that's the point at which they're going to occupy the same area. It looks like it's about 175 square millimeters. But they're not asking about the area."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Well, you can see very clearly that it looks like they intersect right around there. So that's the point at which they're going to occupy the same area. It looks like it's about 175 square millimeters. But they're not asking about the area. They're saying, when does it happen? And it looks like it happens after about five hours. So round to the nearest integer."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "But they're not asking about the area. They're saying, when does it happen? And it looks like it happens after about five hours. So round to the nearest integer. Let's say five hours. Now let's do the last part. So it's the same setup, but now they are asking us a different question."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "So round to the nearest integer. Let's say five hours. Now let's do the last part. So it's the same setup, but now they are asking us a different question. They are asking us, which statements represent the meaning of the intersection points of the graphs? All right, so choice A says, and then pause the video again and try to answer these on your own. All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "So it's the same setup, but now they are asking us a different question. They are asking us, which statements represent the meaning of the intersection points of the graphs? All right, so choice A says, and then pause the video again and try to answer these on your own. All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time. So let's see this. That looks about right. I had estimated 175, but we could call that 180."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "All right, choice A says, it means that the populations both occupied about 180 square millimeters at the same time. So let's see this. That looks about right. I had estimated 175, but we could call that 180. And it looks like that does roughly happen at around the fifth hour. So it looks like they're occupying the same area at around the same time. So I like that choice."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "I had estimated 175, but we could call that 180. And it looks like that does roughly happen at around the fifth hour. So it looks like they're occupying the same area at around the same time. So I like that choice. It means that at the beginning, population A was larger than population B. Well, the point of intersection doesn't tell us what population was larger to begin with. We could try to answer it by looking over here."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "So I like that choice. It means that at the beginning, population A was larger than population B. Well, the point of intersection doesn't tell us what population was larger to begin with. We could try to answer it by looking over here. When time t equals zero, when time t equals zero, population A is the blue curve. It is F. And so it does look like population A was larger than population B at time t equals zero at the beginning. But that's not what the point of intersection tells us."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "We could try to answer it by looking over here. When time t equals zero, when time t equals zero, population A is the blue curve. It is F. And so it does look like population A was larger than population B at time t equals zero at the beginning. But that's not what the point of intersection tells us. So they're not just asking us for true statements. They're saying which statements represent the meaning, the meaning of the intersection point of the graphs. But that doesn't tell us about what the starting situation was."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "But that's not what the point of intersection tells us. So they're not just asking us for true statements. They're saying which statements represent the meaning, the meaning of the intersection point of the graphs. But that doesn't tell us about what the starting situation was. It gives a solution to the equation 24 times e to the 0.4t is equal to nine times e to the 0.6t. Well, we already talked about that in the last example where we only had one curve. And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "But that doesn't tell us about what the starting situation was. It gives a solution to the equation 24 times e to the 0.4t is equal to nine times e to the 0.6t. Well, we already talked about that in the last example where we only had one curve. And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A. And then y is equal to nine times e to the 0.6t is the curve for population B. And so the point at which these two curves intersect, that's the point at which both this, we're at a t value that gives the same so that this expression will give you the same y value as this expression. Or another way to say it is we're at the t value where this is equal to this."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "And that actually is the case because y is equal to 24 times e to the 0.4t is the curve for population A. And then y is equal to nine times e to the 0.6t is the curve for population B. And so the point at which these two curves intersect, that's the point at which both this, we're at a t value that gives the same so that this expression will give you the same y value as this expression. Or another way to say it is we're at the t value where this is equal to this. So it does indeed give the solution to the equation. And then the last choice is it gives a solution to the equation 24 times e to the 0.4t is equal to zero? No, that would happen, if you wanna know when it's equal to zero, you would look at the curve y equals zero."}, {"video_title": "Solving equations by graphing word problems Algebra 2 Khan Academy.mp3", "Sentence": "Or another way to say it is we're at the t value where this is equal to this. So it does indeed give the solution to the equation. And then the last choice is it gives a solution to the equation 24 times e to the 0.4t is equal to zero? No, that would happen, if you wanna know when it's equal to zero, you would look at the curve y equals zero. I'll do that in a different color, which is right over here. And see where it intersects the function f, which is equal to 24 times e to the 0.4t. But that's not what this point of intersection represents."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "All right, let's tackle this one in purple first. And you might first notice that on both sides of the equation I have different bases, so it would be nice to have a common base. And when you look at it, you're like, well, 32 is not a power of eight, or at least it's not an integer power of eight, but they are both powers of two. 32 is the same thing as two to the fifth power, two to the fifth power, and eight is the same thing as two to the third power, two to the third power. So I can rewrite our original equation as instead of writing 32, I could write it as two to the fifth, and then that's going to be raised to the x over three power, x over three power, is equal to, instead of writing eight, I could write two to the third power, two to the third power, and I'm raising that to the x minus 12, x minus 12. Now if I raise something to a power and then raise that to a power, I could just multiply these exponents. So I could rewrite the left-hand side as two to the five x over three, five x over three power."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "32 is the same thing as two to the fifth power, two to the fifth power, and eight is the same thing as two to the third power, two to the third power. So I can rewrite our original equation as instead of writing 32, I could write it as two to the fifth, and then that's going to be raised to the x over three power, x over three power, is equal to, instead of writing eight, I could write two to the third power, two to the third power, and I'm raising that to the x minus 12, x minus 12. Now if I raise something to a power and then raise that to a power, I could just multiply these exponents. So I could rewrite the left-hand side as two to the five x over three, five x over three power. I just multiply these exponents, and that's going to be equal to two to the, and now I just multiply the three times x minus 12. So two to the three x minus 36, and now things have simplified nicely. I have two to this power is equal to two to that power, so these two exponents must be equal to each other."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So I could rewrite the left-hand side as two to the five x over three, five x over three power. I just multiply these exponents, and that's going to be equal to two to the, and now I just multiply the three times x minus 12. So two to the three x minus 36, and now things have simplified nicely. I have two to this power is equal to two to that power, so these two exponents must be equal to each other. Five x over three must be equal to three x minus 36. So let's set them equal to each other and solve for x. So five x over three is equal to three x minus 36."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "I have two to this power is equal to two to that power, so these two exponents must be equal to each other. Five x over three must be equal to three x minus 36. So let's set them equal to each other and solve for x. So five x over three is equal to three x minus 36. Let's see, we could multiply, we could multiply everything by three. Let's do that. So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So five x over three is equal to three x minus 36. Let's see, we could multiply, we could multiply everything by three. Let's do that. So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108. We're in the home stretch here. Divide, whoops, sorry about that. We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So if we multiply everything times three, you are going to get five x is equal to nine x minus, what is this, nine x minus 108, and now we can subtract nine x from both sides, and so we will get five x minus nine x is going to be negative four x is equal to negative 108. We're in the home stretch here. Divide, whoops, sorry about that. We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27. X is equal to 27, and we are all done. We're all done, and if you substituted x back in there, you would get 32 to the 27 divided by three, so 32 to the ninth power is the same thing as eight to the 27 minus 12th power, so eight to the 15th, yeah, 27 minus 12, eight to the 15th power. So anyway, that was fun."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "We could divide both sides by negative four, negative four, and we are left with x is equal to, what is this going to be, 27. X is equal to 27, and we are all done. We're all done, and if you substituted x back in there, you would get 32 to the 27 divided by three, so 32 to the ninth power is the same thing as eight to the 27 minus 12th power, so eight to the 15th, yeah, 27 minus 12, eight to the 15th power. So anyway, that was fun. Let's do the next one now. So this one looks interesting in other ways. We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So anyway, that was fun. Let's do the next one now. So this one looks interesting in other ways. We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five. We know that 25 is the same thing as five squared, so we can rewrite this as five to the four x plus three over, instead of 25, I could rewrite that as five squared, and then I'm gonna raise that to the nine minus x, to the nine minus x, and that, of course, is going to be equal to five to the two x plus five. Now, five to the second, and then that to the nine minus x, I can just multiply these exponents, so this is going to be five to the four x plus three over five to the, two times nine is 18, two times negative x is negative two x, and that is going to be equal to, that is going to be equal to five to the two x plus five. And now, let's see, there's multiple ways that we could tackle it."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "We have rational expressions, we have an exponential up here, exponential down here, and the key realization here is, well, the first thing I'd like to do, let me write this 25 in terms of five. We know that 25 is the same thing as five squared, so we can rewrite this as five to the four x plus three over, instead of 25, I could rewrite that as five squared, and then I'm gonna raise that to the nine minus x, to the nine minus x, and that, of course, is going to be equal to five to the two x plus five. Now, five to the second, and then that to the nine minus x, I can just multiply these exponents, so this is going to be five to the four x plus three over five to the, two times nine is 18, two times negative x is negative two x, and that is going to be equal to, that is going to be equal to five to the two x plus five. And now, let's see, there's multiple ways that we could tackle it. We could multiply both sides of this equation by five to the 18 minus two x, that's one way to do it, or we could say, hey, look, I have five to some exponent divided by five to some other exponent, so I could just subtract this blue exponent from this yellow one, so the left-hand side will simplify to five to the four x plus three minus, let me do this with a minus and a neutral color, minus 18 minus two x, 18 minus two x, and that, of course, is going to be equal to what we've had on the right-hand side, five to the two x plus five. Now we just have to simplify a little bit. Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "And now, let's see, there's multiple ways that we could tackle it. We could multiply both sides of this equation by five to the 18 minus two x, that's one way to do it, or we could say, hey, look, I have five to some exponent divided by five to some other exponent, so I could just subtract this blue exponent from this yellow one, so the left-hand side will simplify to five to the four x plus three minus, let me do this with a minus and a neutral color, minus 18 minus two x, 18 minus two x, and that, of course, is going to be equal to what we've had on the right-hand side, five to the two x plus five. Now we just have to simplify a little bit. Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool. Whoops. All right. So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "Let's see, this is going to be, in fact, we could just say, look, I'm having trouble with my little pen tool. Whoops. All right. So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five. So there's a bunch of different things we could do here. One, we could subtract two x from both sides. That'll clean it up a little bit."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So now we could say this exponent needs to be equal to that exponent because we have the same base, and so what we have here on the left-hand side, that I could rewrite as four x plus three minus 18 plus two x. I'm just multiplying the negative times both of these terms, so plus two x is going to be equal to two x plus five, two x plus five. So there's a bunch of different things we could do here. One, we could subtract two x from both sides. That'll clean it up a little bit. Two x from both sides. We could also subtract five from both sides. So let's just do that."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "That'll clean it up a little bit. Two x from both sides. We could also subtract five from both sides. So let's just do that. So, well, let me just subtract it. Subtract five from both sides. I'm skipping some steps here, but I figure you're at this point reasonably comfortable with linear equations."}, {"video_title": "Solving exponential equations using exponent properties (advanced) High School Math Khan Academy.mp3", "Sentence": "So let's just do that. So, well, let me just subtract it. Subtract five from both sides. I'm skipping some steps here, but I figure you're at this point reasonably comfortable with linear equations. So then, on the left-hand side, we are going to have four x, and then you have three minus 18 minus five. Three minus 18 is negative 15, minus five is negative 20, is going to be equal to zero. And then, because those cancel out, and so add 20 to both sides, you get four x is equal to 20."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So let's say this is the year, and we're gonna think about how much we have at the beginning of the year, and then this is the dollars in our account. And let's say that the bank is always willing to give us 5% per year, which is pretty good. It's very hard to find a bank account that will actually give you 5% growth per year. So that means if you put $100 in at the end of a year or exactly a year later, it'd be $105. If you put $1,000 in a year later, it'd be 1,050. It'd be 5% larger. And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So that means if you put $100 in at the end of a year or exactly a year later, it'd be $105. If you put $1,000 in a year later, it'd be 1,050. It'd be 5% larger. And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account. That's pretty straightforward. But then what happens in year two? I'm going to deposit $1,000, but then that original $1,000 that I have would have grown."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "And so let's say that we wanna put $1,000 in per year, and I wanna think about, well, what is going to be my balance at the beginning of year one, at the beginning of year two, at the beginning of year three, and then see if we can come up with a general expression for the beginning of year n. So year one, right at the beginning of the year, I put in $1,000 in the account. That's pretty straightforward. But then what happens in year two? I'm going to deposit $1,000, but then that original $1,000 that I have would have grown. So I'm going to deposit $1,000, and then the original $1,000 that I put at the beginning of year one, that has now grown by 5%. Growing by 5% is the same thing as multiplying by 1.05. So this is now going to be plus $1,000 times 1.05."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "I'm going to deposit $1,000, but then that original $1,000 that I have would have grown. So I'm going to deposit $1,000, and then the original $1,000 that I put at the beginning of year one, that has now grown by 5%. Growing by 5% is the same thing as multiplying by 1.05. So this is now going to be plus $1,000 times 1.05. Fairly straightforward. Now what about the beginning of year three? How much would I have in the bank account right when I've made that first, that year three deposit?"}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So this is now going to be plus $1,000 times 1.05. Fairly straightforward. Now what about the beginning of year three? How much would I have in the bank account right when I've made that first, that year three deposit? Pause this video, see if you can figure that out. Well, just like at the beginning of year two and the beginning of year one, we're going to make a $1,000 deposit. But now the money from year two has grown by 5%."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "How much would I have in the bank account right when I've made that first, that year three deposit? Pause this video, see if you can figure that out. Well, just like at the beginning of year two and the beginning of year one, we're going to make a $1,000 deposit. But now the money from year two has grown by 5%. So this is now going to be $1,000 times 1.05. And then that money that we originally deposited from year one, that was 1,000 times 1.05 in year two, that's going to grow by another 5%. And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "But now the money from year two has grown by 5%. So this is now going to be $1,000 times 1.05. And then that money that we originally deposited from year one, that was 1,000 times 1.05 in year two, that's going to grow by another 5%. And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%. Well, we could just rewrite this part right over here as 1.05 squared. So do you see a general pattern that's going to happen here? Well, as you go to year n, in fact, pause the video again and see if you could write a general expression."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "And so this is going to be plus 1,000 times 1.05 times 1.05, we're growing by another 5%. Well, we could just rewrite this part right over here as 1.05 squared. So do you see a general pattern that's going to happen here? Well, as you go to year n, in fact, pause the video again and see if you could write a general expression. You're gonna have to do a little bit of this dot, dot, dot action in order to do it. But see if you could write a general expression for year n. Well, for year n, you're going to make that original $1,000 at the beginning of year n, and then you're going to have 1,000 times 1.05 for that $1,000 that you deposited at the beginning of year n minus one. And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "Well, as you go to year n, in fact, pause the video again and see if you could write a general expression. You're gonna have to do a little bit of this dot, dot, dot action in order to do it. But see if you could write a general expression for year n. Well, for year n, you're going to make that original $1,000 at the beginning of year n, and then you're going to have 1,000 times 1.05 for that $1,000 that you deposited at the beginning of year n minus one. And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding. So you could view this $1,000 as the one that you put in year one, and then how many years has it compounded? Well, when you go from one to two, you've compounded one year. When you go from one to three, you've compounded two years."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "And then this is just going to keep going, and it's going to go all the way to plus $1,000 to times 1.05 to the power of the number of years you've been compounding. So you could view this $1,000 as the one that you put in year one, and then how many years has it compounded? Well, when you go from one to two, you've compounded one year. When you go from one to three, you've compounded two years. So when we're talking about the beginning of year n, you go up to the exponent that is one less than that. And so this is going to be to the n minus one power. So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three?"}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "When you go from one to three, you've compounded two years. So when we're talking about the beginning of year n, you go up to the exponent that is one less than that. And so this is going to be to the n minus one power. So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three? Or how much do we have in our bank account at the beginning of year n? These are geometric series, and I'll write that word down, geometric series. Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So what we just did here is we've just constructed each one of these when we're saying, okay, how much money do we have in our bank account at the beginning of year three? Or how much do we have in our bank account at the beginning of year n? These are geometric series, and I'll write that word down, geometric series. Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences. Sequences, and let me go down a little bit so that you can, so we have a little bit more space. A sequence is an ordered list of numbers. A sequence might be something like, well, let's say we have a geometric sequence."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "Now, just as a little bit of a review, or it might not be a review, it might be a primer, series are related to sequences, and you can really view series as sums of sequences. Sequences, and let me go down a little bit so that you can, so we have a little bit more space. A sequence is an ordered list of numbers. A sequence might be something like, well, let's say we have a geometric sequence. In a geometric sequence, each successive term is the previous term times a fixed number. So let's say we start at two, and every time we multiply by three. So we'll go from two, two times three is six, six times three is 18, 18 times three is 54."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "A sequence might be something like, well, let's say we have a geometric sequence. In a geometric sequence, each successive term is the previous term times a fixed number. So let's say we start at two, and every time we multiply by three. So we'll go from two, two times three is six, six times three is 18, 18 times three is 54. This is a geometric sequence, ordered list of numbers. Now, if we wanna think about the geometric series, or the one that's analogous to this, is that we would sum the terms here. So this would be two plus six plus 18 plus 54."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So we'll go from two, two times three is six, six times three is 18, 18 times three is 54. This is a geometric sequence, ordered list of numbers. Now, if we wanna think about the geometric series, or the one that's analogous to this, is that we would sum the terms here. So this would be two plus six plus 18 plus 54. Or we could even write it, and this will look similar to what we had just done with our little savings example, is this is two plus two times three plus two times three squared plus two times three to the third power. And so with a geometric series, you're going to have a sum where each successive term in the expression is equal to, if you put them all in order, is going to be equal to the term before it times a fixed amount. So the second term is equal to the first term times three, and we're summing them."}, {"video_title": "Geometric series introduction Algebra 2 Khan Academy.mp3", "Sentence": "So this would be two plus six plus 18 plus 54. Or we could even write it, and this will look similar to what we had just done with our little savings example, is this is two plus two times three plus two times three squared plus two times three to the third power. And so with a geometric series, you're going to have a sum where each successive term in the expression is equal to, if you put them all in order, is going to be equal to the term before it times a fixed amount. So the second term is equal to the first term times three, and we're summing them. In a sequence, you're just looking at it. It's an ordered list, so to speak, but here you are actually adding up the ordered list. So what we just saw in this example is why a, one, what a geometric series is, but also a famous example of how it's useful, and this is just scratching the surface."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "So for example, if I had the polynomial, and this would be a quadratic polynomial, let's say x squared plus three x plus two, and I wanted to divide it by x plus one. Pause this video and think about what would that be? What would I have to multiply x plus one by to get x squared plus three x plus two? Well, one way to approach it is we could try to factor x squared plus three x plus two, and we've done that multiple times in our lives. We think about, well, what two numbers add up to three? You know, if I were to multiply them, I'd get two, and the ones that might jump out at you are two and one, and so we could express x squared plus three x plus two as x plus two times x plus one, and then all of that is going to be over x plus one. And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be?"}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "Well, one way to approach it is we could try to factor x squared plus three x plus two, and we've done that multiple times in our lives. We think about, well, what two numbers add up to three? You know, if I were to multiply them, I'd get two, and the ones that might jump out at you are two and one, and so we could express x squared plus three x plus two as x plus two times x plus one, and then all of that is going to be over x plus one. And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? Well, you're just going to be left with an x plus two. This is going, I don't even have to put parentheses. This is going to be an x plus two."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "And so if you were to take x plus two times x plus one, and then divide that by x plus one, what is that going to be? Well, you're just going to be left with an x plus two. This is going, I don't even have to put parentheses. This is going to be an x plus two. And if we wanna be really mathematically precise, we would say, hey, this would be true as long as x does not equal negative one, because if x equals negative one in this expression or this expression, we're going to be dividing by zero, and we know that leads to all sorts of mathematical problems. But as we see, for any other x, as long as we're not dividing by zero here, this expression is going to be the same thing as x plus two, and that's because x plus two times x plus one is equal to what we have in this numerator here. Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "This is going to be an x plus two. And if we wanna be really mathematically precise, we would say, hey, this would be true as long as x does not equal negative one, because if x equals negative one in this expression or this expression, we're going to be dividing by zero, and we know that leads to all sorts of mathematical problems. But as we see, for any other x, as long as we're not dividing by zero here, this expression is going to be the same thing as x plus two, and that's because x plus two times x plus one is equal to what we have in this numerator here. Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division. And if it sounds familiar, because you first learned about long division in fourth or fifth grade, it's because it's a very similar process, where you would take your x plus one, and you would try to divide it into your x squared plus three x plus two. And you do something very, and I'm gonna do a very quick example right over here, but we're gonna do much more detailed examples in future videos. But you look at the highest degree terms."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "Now, as we go deeper into polynomial division, we're going to approach things that aren't as easy to do just purely through factoring, and that's where we're going to have a technique called polynomial long division, polynomial long division, sometimes known as algebraic long division. And if it sounds familiar, because you first learned about long division in fourth or fifth grade, it's because it's a very similar process, where you would take your x plus one, and you would try to divide it into your x squared plus three x plus two. And you do something very, and I'm gonna do a very quick example right over here, but we're gonna do much more detailed examples in future videos. But you look at the highest degree terms. You say, okay, I have a first degree term and a second degree term here. How many times does x go into x squared? Well, it goes x times."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "But you look at the highest degree terms. You say, okay, I have a first degree term and a second degree term here. How many times does x go into x squared? Well, it goes x times. So you put the x in the first degree column, and then you multiply your x times x plus one. X times x is x squared. X times one is x."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "Well, it goes x times. So you put the x in the first degree column, and then you multiply your x times x plus one. X times x is x squared. X times one is x. And then you subtract this from that. So you might already start to see some parallels with the long division that you first learned in school many years ago. So when you do that, these cancel out."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "X times one is x. And then you subtract this from that. So you might already start to see some parallels with the long division that you first learned in school many years ago. So when you do that, these cancel out. Three x minus x, we are left with a two x. And then you bring down that two. So two x plus two."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "So when you do that, these cancel out. Three x minus x, we are left with a two x. And then you bring down that two. So two x plus two. And you say, how many times does x go into two x? Well, it goes two times. So you have a plus two here."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "So two x plus two. And you say, how many times does x go into two x? Well, it goes two times. So you have a plus two here. Two times x plus one, two times x is two x. Two times one is two. You can subtract these."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "So you have a plus two here. Two times x plus one, two times x is two x. Two times one is two. You can subtract these. And then you are going to be left with nothing. Two minus two is zero. Two x minus two x is zero."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "You can subtract these. And then you are going to be left with nothing. Two minus two is zero. Two x minus two x is zero. So in this situation, it divided cleanly into it, and we got x plus two, which is exactly what we had over there. Now, an interesting scenario that we're also going to approach in the next few videos is, what if things don't divide cleanly? For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "Two x minus two x is zero. So in this situation, it divided cleanly into it, and we got x plus two, which is exactly what we had over there. Now, an interesting scenario that we're also going to approach in the next few videos is, what if things don't divide cleanly? For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three. And if I were to try to divide that by x plus one, well, it's not going to divide cleanly anymore. You could do it either approach. One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "For example, if I were to add one to x squared plus three x plus two, I would get x squared plus three x plus three. And if I were to try to divide that by x plus one, well, it's not going to divide cleanly anymore. You could do it either approach. One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one. And then all of that's going to be over x plus one. And then you could say, hey, this is the same thing as x squared plus three x plus two over x plus one, x plus one over x plus one plus one over x plus one, plus one over x plus one. And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two."}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "One way to think about it, if we know we can factor x squared plus three x plus two, is say, hey, this is the same thing as x squared plus three x plus two plus one. And then all of that's going to be over x plus one. And then you could say, hey, this is the same thing as x squared plus three x plus two over x plus one, x plus one over x plus one plus one over x plus one, plus one over x plus one. And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. So this is going to be equal to x plus two, but then we have that one that we weren't able to divide x plus one into, so we're just left with a one over x plus one. And we'll study that in a lot more detail in other videos. What does this remainder mean?"}, {"video_title": "Polynomial division introduction Algebra 2 Khan Academy.mp3", "Sentence": "And we already figured out that this expression on the left, as long as x does not equal negative one, this is going to be equal to x plus two. So this is going to be equal to x plus two, but then we have that one that we weren't able to divide x plus one into, so we're just left with a one over x plus one. And we'll study that in a lot more detail in other videos. What does this remainder mean? And how do we calculate it if we can't factor a part of what we have in the numerator? And as we do our polynomial long division, we'll see that the remainder will show up at the end when we are done dividing. We'll see those examples in future videos."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So at time equals zero, she is, looks like about two, what is this, this would be one and a half, so it looks like she's about two meters off the ground. And then as time increases, she gets as high as, it looks like this is close to 30, maybe 34 meters, and then she comes back down, back to, looks like two meters, and up to 34 meters again. So let's read the question. So the question asks us, approximately how long does it take Divya to complete one revolution on the Ferris wheel? All right, so this is interesting. So this is when she's at the bottom of the Ferris wheel. So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the question asks us, approximately how long does it take Divya to complete one revolution on the Ferris wheel? All right, so this is interesting. So this is when she's at the bottom of the Ferris wheel. So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again. So it took her 60, and t is in terms of seconds, so it took her 60 seconds to go from the bottom to the bottom again. Another 60 seconds, she would have completed another revolution. And so, let me fill that in."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So then she gets to the top of the Ferris wheel, and then she keeps rotating until she gets back to the bottom of the Ferris wheel again. So it took her 60, and t is in terms of seconds, so it took her 60 seconds to go from the bottom to the bottom again. Another 60 seconds, she would have completed another revolution. And so, let me fill that in. It is going to take her 60 seconds. 60 seconds, and we of course can check our answer if we like. Let's do another one of these."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so, let me fill that in. It is going to take her 60 seconds. 60 seconds, and we of course can check our answer if we like. Let's do another one of these. So here we have a doctor observes the electrical activity of Finn's heart over a period of time. The electrical activity of Finn's heart is cyclical, as we hope it would be, and peaks every 0.9 seconds. Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts?"}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's do another one of these. So here we have a doctor observes the electrical activity of Finn's heart over a period of time. The electrical activity of Finn's heart is cyclical, as we hope it would be, and peaks every 0.9 seconds. Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? Well over here it looks like we peaked at zero seconds, and here we're peaking a little bit more than one. This looks like maybe at 1.1, and maybe at 2.2 and 3.3. This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "Which of the following graphs could model a situation if t stands for time in seconds, and e stands for the electrical activity of Finn's heart in volts? Well over here it looks like we peaked at zero seconds, and here we're peaking a little bit more than one. This looks like maybe at 1.1, and maybe at 2.2 and 3.3. This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a. This one is peaking, it looks like the interval between peaks is less than a second, but it looks like a good bit less than a second. It looks like maybe every 3 quarters of a second, or maybe every 4 fifths of a second. Not quite 9 tenths."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "This looks like it's peaking a little bit more than every one second, so like maybe every 1.1 seconds, not every 0.9 seconds, so I'd rule out a. This one is peaking, it looks like the interval between peaks is less than a second, but it looks like a good bit less than a second. It looks like maybe every 3 quarters of a second, or maybe every 4 fifths of a second. Not quite 9 tenths. 9 tenths, this first peak would be a little bit closer to one, but this one is close. Choice C is looking good. The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "Not quite 9 tenths. 9 tenths, this first peak would be a little bit closer to one, but this one is close. Choice C is looking good. The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one. It looks like a tenth less than one. So I like choice C. Now choice D, it looks like we're peaking every half second, so it's definitely not that. So this looks like a peak of every 0.9 seconds."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "The first, we're at zero, then the first peak, this looks pretty close to one, but it's less than one. It looks like a tenth less than one. So I like choice C. Now choice D, it looks like we're peaking every half second, so it's definitely not that. So this looks like a peak of every 0.9 seconds. This is the best representation that I, this is the best representation that I can think of. And you can actually verify that. If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this looks like a peak of every 0.9 seconds. This is the best representation that I, this is the best representation that I can think of. And you can actually verify that. If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds. So one, two, three, four. This looks like it's at 3.6. Over here, you have one, two, three, four."}, {"video_title": "Periodicity of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "If you have a peak every 0.9 seconds, you're gonna have 4 peaks in 3.6 seconds. So one, two, three, four. This looks like it's at 3.6. Over here, you have one, two, three, four. You've had 4 peaks in less than 3 seconds. So this definitely one, this one definitely isn't 0.9. So instead of just even forcing yourself to eyeball just between this peak and that peak, you can say, well, if we're every 0.9 seconds, how long would 3 peaks take, or 4 peaks?"}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work through this together. And we can see that all of the choices are expressed as a polynomial in factored form. And factored form is useful when we're thinking about the roots of a polynomial, the x values that make that polynomial equal to zero. The roots are also evident when we look at this graph here. We have a root at x equals negative four, a root at x equals negative 1 1 1 1 1, or negative 3 1 1 1 1 1, and a root at x is equal to one. So really what we have to do is say, which of these factors are consistent with the roots that we see? So let's go root by root."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "The roots are also evident when we look at this graph here. We have a root at x equals negative four, a root at x equals negative 1 1 1 1 1, or negative 3 1 1 1 1 1, and a root at x is equal to one. So really what we have to do is say, which of these factors are consistent with the roots that we see? So let's go root by root. So here on the left, we have a root at x equals negative four. In order for this polynomial to be zero when x is equal to negative four, that means that x plus four must be a factor, or some multiple, or some constant times x plus four must be a factor of our polynomial. Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let's go root by root. So here on the left, we have a root at x equals negative four. In order for this polynomial to be zero when x is equal to negative four, that means that x plus four must be a factor, or some multiple, or some constant times x plus four must be a factor of our polynomial. Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them. The first one has a two as an exponent, it's being squared, while the others have a one as the exponent. Now what we've talked about in other videos when we talked about multiplicity, we said, hey, if we see a sign change around a root, like we're seeing right over here around x equals negative four, that means that we are going to see an odd exponent on the corresponding factor. But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Now we can see in the choices that we have a bunch of x plus fours, but they have different exponents on them. The first one has a two as an exponent, it's being squared, while the others have a one as the exponent. Now what we've talked about in other videos when we talked about multiplicity, we said, hey, if we see a sign change around a root, like we're seeing right over here around x equals negative four, that means that we are going to see an odd exponent on the corresponding factor. But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent. Now we clearly see the sign change, so we would expect an odd exponent, and of course, one is an odd number and two isn't. So if you just have a straight up x plus four, you would have a sign change around x equals negative four. So I can rule out this first choice."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "But if we didn't see a sign change, as we see in this other root over here, that means we would see an even exponent. Now we clearly see the sign change, so we would expect an odd exponent, and of course, one is an odd number and two isn't. So if you just have a straight up x plus four, you would have a sign change around x equals negative four. So I can rule out this first choice. These other three choices are still looking good based on just the x plus four factor. Now let's move on to the next factor right over here, so, or the next root. The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So I can rule out this first choice. These other three choices are still looking good based on just the x plus four factor. Now let's move on to the next factor right over here, so, or the next root. The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant. Now when we look at the choices or the remaining choices, we don't see x plus 3 1 2, but we do see something that involves a two and a three, and so one way to think about it is, hey, if I just multiply this times the constant two, so that would get us two x plus three, well, I do see that right over here, and then the next question is what should be the exponent? Well, once again, we have a sign change around x equals negative 3 1 2, so we would expect an odd exponent there, and you can see out of the choices, only two of them have an exponent of one, which is an odd number, while the other one has an even exponent there, so we can rule that one out as well, and then we go to this last root. I will do this in an orange color."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "The next root is at x is equal to negative 3 1 2, and so one way to think about it is you could have a factor that looks like x plus 3 1 2, or this times some constant. Now when we look at the choices or the remaining choices, we don't see x plus 3 1 2, but we do see something that involves a two and a three, and so one way to think about it is, hey, if I just multiply this times the constant two, so that would get us two x plus three, well, I do see that right over here, and then the next question is what should be the exponent? Well, once again, we have a sign change around x equals negative 3 1 2, so we would expect an odd exponent there, and you can see out of the choices, only two of them have an exponent of one, which is an odd number, while the other one has an even exponent there, so we can rule that one out as well, and then we go to this last root. I will do this in an orange color. We have a root at x equals one, so we would expect x minus one, or this multiplied times some constant, to be one of the factors, and what's interesting here is we don't see a sign change around x equals one, so we would expect an even exponent, and so out of the remaining choices, we see an x minus one in both of them, but only choice C has the even exponent that we would expect, so choice C is looking good. If we were to go with choice D, where this is to the first power, we would expect a sign change around x is equal to one, so this would be a situation where the curve would keep going down, something like that, so we like choice C. Let's do another example. So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "I will do this in an orange color. We have a root at x equals one, so we would expect x minus one, or this multiplied times some constant, to be one of the factors, and what's interesting here is we don't see a sign change around x equals one, so we would expect an even exponent, and so out of the remaining choices, we see an x minus one in both of them, but only choice C has the even exponent that we would expect, so choice C is looking good. If we were to go with choice D, where this is to the first power, we would expect a sign change around x is equal to one, so this would be a situation where the curve would keep going down, something like that, so we like choice C. Let's do another example. So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that. All right, we're gonna do the same idea. Let's go to this first root right over here. We have a root at x equals negative three, so we would expect some multiple of x plus three to be one of the factors."}, {"video_title": "Zeros of polynomials (multiplicity) Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So once again, we are asked what could be the equation of P, and we're given a graph, so again, pause this video and try to work through that. All right, we're gonna do the same idea. Let's go to this first root right over here. We have a root at x equals negative three, so we would expect some multiple of x plus three to be one of the factors. We also have a sign change around x equals negative three, so we would expect an odd multiplicity, and we would expect an odd exponent on the x plus three factor. When we look at all the choices, C and D have an even exponent, so if we had x plus three to the fourth, then you wouldn't have a sign change here. You would just touch the x-axis and then go back to where it was coming from, so we can rule out these choices, and now let's look at the second root right over here at x is equal to two, so we would expect x minus two to be one of the factors or a multiple of this, and because we don't have a sign change around x equals two, the graph just touches the x-axis and then goes back to where it was coming from."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "18x to the fourth minus 3x squared plus 6x minus 4. All of that over 6x. So there's a couple of ways to think about them. They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decompose this numerator up here."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "They're all really equivalent. You can really just view this up here as being the exact same thing as 18x to the fourth over 6x plus negative 3x squared over 6x, or you could say minus 3x squared over 6x, plus 6x over 6x minus 4 over 6x. Now, there's a couple of ways to think about it. One is I just kind of decompose this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a over d plus b over d plus c over d. Or maybe not so clearly, but hopefully that helps clarify. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "One is I just kind of decompose this numerator up here. If I just had a bunch of stuff, a plus b plus c over d, that's clearly equal to a over d plus b over d plus c over d. Or maybe not so clearly, but hopefully that helps clarify. Another way to think about it is kind of like you're distributing the division. If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. The other way to think about it is that we're multiplying this entire expression, so this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If I divide a whole expression by something, that's equivalent to dividing each of the terms by that something. The other way to think about it is that we're multiplying this entire expression, so this is the same thing as 1 over 6x times this entire thing, times 18x to the fourth minus 3x squared plus 6x minus 4. And so here, this would just be the straight distributive property to get to this. Whatever seems to make sense for you. They're all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Whatever seems to make sense for you. They're all equivalent. They're all logical, good things to do to simplify this thing. Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here we can just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now, once you have it here, now we just have a bunch of monomials that we're just dividing by 6x. And here we can just use exponent properties. This first one over here, we can take the coefficients and divide them. 18 divided by 6 is 3. And then you have x to the fourth divided by x to the... Well, they don't tell us, but if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the four minus one power, or x to the third power."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "18 divided by 6 is 3. And then you have x to the fourth divided by x to the... Well, they don't tell us, but if it's just an x, that's the same thing as x to the first power. So it's x to the fourth divided by x to the first. That's going to be x to the four minus one power, or x to the third power. Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's going to be x to the four minus one power, or x to the third power. Then we have this coefficient over here, or these coefficients. We have negative 3 divided by 6. So I'm going to do this part next. Negative 3 divided by 6 is negative 1 half. And then you have x squared divided by x. We already know that x is the same thing as x to the first."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So I'm going to do this part next. Negative 3 divided by 6 is negative 1 half. And then you have x squared divided by x. We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We already know that x is the same thing as x to the first. So that's going to be x to the 2 minus 1 power, which is just 1. Or I could just leave it as an x right there. Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just... Well, I'll write it. I could write a 1 here."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Then we have these coefficients, 6 divided by 6. Well, that's just 1. So I could just... Well, I'll write it. I could write a 1 here. Let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. It gives you two ways."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I could write a 1 here. Let me just write the 1 here, because we said 2 minus 1 is 1. And then x divided by x is x to the first over x to the first. It gives you two ways. Anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, which is x to the 0, which is also equal to 1. Either way."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It gives you two ways. Anything divided by anything is just 1. Or you could view it as x to the 1 divided by x to the 1 is going to be x to the 1 minus 1, which is x to the 0, which is also equal to 1. Either way. You knew how to do this before you even learned that exponent property, because x divided by x is 1. And then assuming x does not equal 0. And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Either way. You knew how to do this before you even learned that exponent property, because x divided by x is 1. And then assuming x does not equal 0. And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0. And then finally we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is both 4..."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then finally, and we kind of have to assume x doesn't equal 0 in this whole thing, otherwise we would be dividing by 0. And then finally we have 4 over 6x. And there's a couple of ways to think about it. So the simplest way is both 4... Negative 4... Let me do it. Negative 4 over 6 is the same thing as negative 2 thirds. Just simplify that fraction."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So the simplest way is both 4... Negative 4... Let me do it. Negative 4 over 6 is the same thing as negative 2 thirds. Just simplify that fraction. And we're multiplying that times 1 over x. And so we could multiply that times 1 over x. So we could do this 4 times 1 over x."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Just simplify that fraction. And we're multiplying that times 1 over x. And so we could multiply that times 1 over x. So we could do this 4 times 1 over x. Another way to think about it is, you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the 1st power. And then when you tried to simplify it using your exponent properties, you would have... Well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we could do this 4 times 1 over x. Another way to think about it is, you could have viewed this 4 as being multiplied by x to the 0 power, and this being x to the 1st power. And then when you tried to simplify it using your exponent properties, you would have... Well, that would be x to the 0 minus 1 power, which is x to the negative 1 power. So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. It's going to be 3x to the 3rd minus 1 half x plus 1. Because this thing right here is just 1."}, {"video_title": "Polynomial divided by monomial Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we could have written an x to the negative 1 here, but x to the negative 1 is the exact same thing as 1 over x. And so let's just write our answer completely simplified. It's going to be 3x to the 3rd minus 1 half x plus 1. Because this thing right here is just 1. So plus 1. And then minus 2 times 1 in the numerator over 3 times x in the denominator. And we are done."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "So we have negative 30x to the fifth is equal to negative 10x to the third times f. And I encourage you to pause this video and see if you can figure out what f is going to be. Well, the way that we can tackle it, we could just isolate the f on the right-hand side here if we divide by negative 10x to the third. So we might say, well, we want to divide this side by negative 10x to the third. But if we want the equality to be true, if we want the left side to stay being the same as the right side, we have to do, whatever we do to the right side, we have to do to the left side as well. So we have to divide the left side by negative 10x to the third. And then what does that leave us with? Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "But if we want the equality to be true, if we want the left side to stay being the same as the right side, we have to do, whatever we do to the right side, we have to do to the left side as well. So we have to divide the left side by negative 10x to the third. And then what does that leave us with? Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third. Well, multiplying by something and then dividing by that same thing is the same thing as just multiplying by one, or you could, one way to think about it, they just cancel out. So we are just going to be left with an f. We're going to be left with an f on the right-hand side. There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "Well, on the right-hand side, up top, we're multiplying by negative 10x to the third, and then we're dividing by negative 10x to the third. Well, multiplying by something and then dividing by that same thing is the same thing as just multiplying by one, or you could, one way to think about it, they just cancel out. So we are just going to be left with an f. We're going to be left with an f on the right-hand side. There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients. We could say negative 30 divided by, negative 30 divided by negative 10 is positive three. So that's going to be three. And then x to the, x to the fifth power divided by x to the third power."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "There's the whole point, we wanted to solve for f. And on the left-hand side, we see, we can first look at the coefficients. We could say negative 30 divided by, negative 30 divided by negative 10 is positive three. So that's going to be three. And then x to the, x to the fifth power divided by x to the third power. Well, that's going to be x squared. X squared. You could either think of it in terms of our exponent properties."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "And then x to the, x to the fifth power divided by x to the third power. Well, that's going to be x squared. X squared. You could either think of it in terms of our exponent properties. We would subtract these two exponents, x to the five minus three, which is x squared, or you could say, hey, up on top, that's x times x times x times x times x. Did I say that right? Five x's."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "You could either think of it in terms of our exponent properties. We would subtract these two exponents, x to the five minus three, which is x squared, or you could say, hey, up on top, that's x times x times x times x times x. Did I say that right? Five x's. You could view it as x, let me do it in that, on top, you have x to the fifth, which is this. I always like to remind myself why the exponent properties even work. And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "Five x's. You could view it as x, let me do it in that, on top, you have x to the fifth, which is this. I always like to remind myself why the exponent properties even work. And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared. So you get f is equal to three x squared. So you could write, we could write that negative 30 x to the fifth is equal to, is equal to negative 10 x to the third times f. And now we know that f is three x squared. Three x squared."}, {"video_title": "Worked example finding the missing monomial factor High School Math Khan Academy.mp3", "Sentence": "And then on the denominator, on the denominator, you have x times x times x, and these three x's are going to cancel, and you're just going to be left with x times x, which is just x squared. So you get f is equal to three x squared. So you could write, we could write that negative 30 x to the fifth is equal to, is equal to negative 10 x to the third times f. And now we know that f is three x squared. Three x squared. And so another way to describe what's going on in this equation, we could say that negative 30 x to the fifth is divisible by either one of these factors, that negative 30 x to the fifth is divisible by negative 10 x to the third, or we could say negative 30 x to the fifth is divisible by three x squared, or we could say that three x squared is a factor of negative 30 x to the fifth. And the way that we can make these claims about factor and divisibility is we're dealing with, we're dealing with non-fractional coefficients right over here, and we're also dealing with non-fractional exponents right over here. So that's why we're saying, hey, these are factors, this yellow thing and this magenta thing, factors of this blue thing, or this blue thing is divisible by either one of these."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's first think about what an even function is. One way to think about an even function is that if you were to flip it over the y-axis, that the function looks the same. So here's a classic example of an even function. It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It would be this right over here, your classic parabola where your vertex is on the y-axis. This is an even function. So this one is maybe the graph of f of x is equal to x squared. And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions?"}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And notice, if you were to flip it over the y-axis, you're going to get the exact same graph. Now a way that we can talk about that mathematically, and we have talked about this when we introduced the idea of reflection, to say that a function is equal to its reflection over the y-axis, that's just saying that f of x is equal to f of negative x. Because if you were to replace your x's with a negative x, that flips your function over the y-axis. Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now what about odd functions? So odd functions, you get the same function if you flip over the y and the x-axes. So let me draw a classic example of an odd function. Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Our classic example would be f of x is equal to x to the third, is equal to x to the third, and it looks something like this. So notice, if you were to flip first over the y-axis, you would get something that looks like this. So I'll do it as a dotted line. If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "If you were to flip just over the y-axis, it would look like this. And then if you were to flip that over the x-axis, well, then you're going to get the same function again. Now how would we write this down mathematically? Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, that means that our function is equivalent to not only flipping it over the y-axis, which would be f of negative x, but then flipping that over the x-axis, which is just taking the negative of that. So this is doing two flips. So some of you might be noticing a pattern or think you might be on the verge of seeing a pattern that connects the words even and odd with the notions that we know from earlier in our mathematical lives. I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "I've just shown you an even function where the exponent is an even number. And I've just showed you an odd function where the exponent is an odd number. Now I encourage you to try out many, many more polynomials and try out the exponents, but it turns out that if you just have f of x is equal to, if you just have f of x is equal to x to the n, then this is going to be an even function if n is even, and it's going to be an odd function if n is odd. So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So that's one connection. Now some of you are thinking, wait, but there seem to be a lot of functions that are neither even nor odd, and that is indeed the case. For example, if you just had the graph x squared plus two, this right over here is still going to be even because if you flip it over, you have the symmetry around the y-axis, you're going to get back to itself. But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "But if you had x minus two squared, which looks like this, x minus two, that would shift two to the right, it'll look like that, that is no longer even because notice, if you flip it over the y-axis, you're no longer getting the same function. So it's not just the exponent, it also matters on the structure of the expression itself. If you have something very simple like just x to the n, well then, that could be or that would be even or odd depending on what your n is. Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd?"}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Similarly, if we were to shift this f of x, if we were to even shift it up, it's no longer, it is no longer, so if this is x to the third, let's say plus three, this is no longer odd because you flip it over once, you get right over there, but then you flip it again, you're going to get this. You're going to get something like this. So you're no longer back to your original function. Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer."}, {"video_title": "Function symmetry introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now an interesting thing to think about, can you imagine a function that is both even and odd? So I encourage you to pause that video or pause the video and try to think about it. Is there a function where f of x is equal to f of negative x and f of x is equal to the negative of f of negative x? Well, I'll give you a hint or actually I'll just give you the answer. Imagine if f of x is just equal to the constant zero. Notice, this thing is just a horizontal line, just like that at y is equal to zero, and if you flip it over the y-axis, you get back to where it was before, then if you flip it over the x-axis again, then you're still back to where you were before. So this over here is both even and odd, very interesting case."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy (2).mp3", "Sentence": "45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "Carlos has taken an initial dose of a prescription medication. The relationship between the elapsed time, t, in hours since he took the first dose, and the amount of medication, m of t, in milligrams in his bloodstream is modeled by the following function. All right. In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "In how many hours will Carlos have one milligram of medication remaining in his bloodstream? So m of what t is equal to, so we need to essentially solve for m of t is equal to one milligram, because m of t outputs, whatever value it outputs is going to be in milligram. So let's just solve that. So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So m of t is, they give us a definition. Its model is an exponential function. 20 times e to the negative 0.8 t is equal to one. So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So let's see, we can divide both sides by 20, and so we will get e to the negative 0.8 t is equal to one over 20, one over 20, which we could write as 0.05, 0.05, I have a feeling we're gonna have to deal with decimals here regardless. And so how do we solve this? Well, one way to think about it, one way to think about it, what happens if we took the natural log of both sides? And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "And just remember, a reminder, the natural log is the logarithm base e. So actually let me write this a little bit differently. This is zero, that is 0.05. So I'm gonna take the natural log of both sides. So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "So ln, ln. So the natural log, this says, what power do I have to raise e to to get to e to the negative 0.8 t? Well, I've got to raise e to the, this simplifies to negative 0.8 t. Once again, natural log, this thing, let me clarify, ln of e to the negative 0.8 t, this is equivalent to if I were to write log base e of e to the negative 0.8 t. What power do I have to raise e to to get to e to the negative 0.8 t? We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "We'll have to raise it to the negative 0.8 t power. So that's why the left-hand side's simplified to this, and that's going to be equal to the natural log, actually I'll just leave it in those terms, the natural log of 0.05, natural log of 0.05, all of that, and now we can divide both sides by negative 0.8 to solve for t. So let's do that. So we divide by negative 0.8, divide by negative 0.8, and so t is going to be equal to all of this business. On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74."}, {"video_title": "Exponential model word problem medication dissolve High School Math Khan Academy.mp3", "Sentence": "On the left-hand side now we just have a t, and on the right-hand side we have all of this business, which I think a calculator will be valuable for. So let me get a calculator out, clear it out, and let's start with 0.05. Let's take the natural log, that's that button right over there, the natural log, we get that value, and we want to divide it by negative 0.8. So divide it by, divide it by 0.8 negative, so we're going to divide by 0.8 negative, is equal to, let's see, they want us to round to the nearest hundredth, so 3.74. So it'll take 3.74, 7.4 hours for his dosage to go down to one milligram, where it actually started at 20 milligrams. When t equals zero, it's 20. After 3.74 hours, he's down in his bloodstream to one milligram, I guess his body has metabolized the rest of it in some way."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. The first thing we can look at is let's look at the coefficients."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. You could do it by looking at a prime factorization."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. Prime factorization of six is just two times three."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. Three is divisible into all of them."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. We can't say a three and a five, because five isn't divisible into 12 or six."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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? 12, you could have said, okay, I can get 12 by saying one times 12, or two times six, or three times four."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And then 15, you could have said, well, one times 15, or three times five."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. So when we're looking at the greatest common monomial factor, the coefficient is going to be three."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. We have x to the fourth, x to the third, and x squared."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. X squared is divisible into x to the fourth and x to the third, and of course, x squared itself."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. So if this is three x squared, we can then figure out what the width is."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. Notice, three x squared times four x squared is 12 x to the fourth."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And then last but not least, we have 15 divided by three is going to be five."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. So once again, the length, we figured that out, it was the greatest common monomial factor of these terms."}, {"video_title": "Factoring polynomials common factor area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. We could write, we could write that, excuse me, I wanna see the original thing."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "We're told a roller coaster has C cars, each containing 20 seats, and it completes R rides a day. Assuming that no one can ride it more than once a day, the maximum number of people that can ride the roller coaster in a single day is P. Write an equation that relates P, C, and R. Pause this video and see if you can do that. All right, before I even look at the variables, I'm just gonna try to think it out in plain language. So what we wanna think about is what is the max number of people per day? People per day. And so that's going to be equal to the number of cars in our roller coaster. So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "So what we wanna think about is what is the max number of people per day? People per day. And so that's going to be equal to the number of cars in our roller coaster. So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car. So this would just tell you the maximum number of people per ride. So then we have to multiply it times the number of rides per day. So times, we'll do this in a new color, times number of rides per day."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "So number of cars times the maximum number of people per car times the maximum number of people per car times the maximum number per car. So this would just tell you the maximum number of people per ride. So then we have to multiply it times the number of rides per day. So times, we'll do this in a new color, times number of rides per day. So what are each of these things? They would have either given us numbers or variables for each of them. The max number of people per day, that's what we're trying to set on one side of the equation."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "So times, we'll do this in a new color, times number of rides per day. So what are each of these things? They would have either given us numbers or variables for each of them. The max number of people per day, that's what we're trying to set on one side of the equation. That is this variable P right over here. So we'll say capital P is equal to what's the number of cars per coaster, I guess you could say. Let me write it this way, per coaster, per roller coaster."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "The max number of people per day, that's what we're trying to set on one side of the equation. That is this variable P right over here. So we'll say capital P is equal to what's the number of cars per coaster, I guess you could say. Let me write it this way, per coaster, per roller coaster. So they give us that right over here. A roller coaster has C cars. So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car?"}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "Let me write it this way, per coaster, per roller coaster. So they give us that right over here. A roller coaster has C cars. So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car? Well, they say each containing 20 seats. So I'd multiply that times 20 for this part. And then I want to multiply that times the number of rides per day for the entire roller coaster."}, {"video_title": "Modeling with multiple variables Roller coaster Modeling Algebra II Khan Academy.mp3", "Sentence": "So that's going to be this variable here in orange or this part of it, that C. Now what's the maximum number of people per car? Well, they say each containing 20 seats. So I'd multiply that times 20 for this part. And then I want to multiply that times the number of rides per day for the entire roller coaster. So that's going to be times R. And we're done, we could rearrange this a little bit. We could write this as P is equal to 20 times CR. 20 times CR."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And as we'll see it a little, it may feel a little magical at first, but in future videos we will prove it and we'll see, well, like many things in mathematics, when you actually think it through, maybe it's not so much magic. So what is the polynomial remainder theorem? Well it tells us that if we start with some polynomial, f of x, so this right over here is a polynomial, polynomial, and we divide it, we divide, divide by x minus a, then the remainder, then the remainder from that, essentially, polynomial long division is going to be f of a. It is going to be, it is going to be f of a. f of a, I know this might seem a little bit abstract right now, I'm talking about f of x's and x minus a's, let's make it a little bit more concrete. So let's say that f of x, f of x is equal to, I'm just gonna make up a, let's say a second degree polynomial. This would be true for any polynomial though. So three x squared minus four x plus seven, and let's say that a is, I don't know, a is one."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It is going to be, it is going to be f of a. f of a, I know this might seem a little bit abstract right now, I'm talking about f of x's and x minus a's, let's make it a little bit more concrete. So let's say that f of x, f of x is equal to, I'm just gonna make up a, let's say a second degree polynomial. This would be true for any polynomial though. So three x squared minus four x plus seven, and let's say that a is, I don't know, a is one. So we're gonna divide by, we're gonna divide that, we're gonna divide by x minus, x minus one. So a, in this case, is equal to one. So let's just do the polynomial long division, and I encourage you to pause the video, and if you're unfamiliar with polynomial long division, I encourage you to watch that before watching this video because I will assume you know how to do a polynomial long division."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So three x squared minus four x plus seven, and let's say that a is, I don't know, a is one. So we're gonna divide by, we're gonna divide that, we're gonna divide by x minus, x minus one. So a, in this case, is equal to one. So let's just do the polynomial long division, and I encourage you to pause the video, and if you're unfamiliar with polynomial long division, I encourage you to watch that before watching this video because I will assume you know how to do a polynomial long division. So divide three x squared minus four x plus seven, divide it by x minus one, see what you get as your remainder, and see if that remainder really is f of one. So assuming you've had a go at it, so let's work through it together. So let's divide x minus one, x minus one into three x squared, into three x squared minus four x plus seven."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's just do the polynomial long division, and I encourage you to pause the video, and if you're unfamiliar with polynomial long division, I encourage you to watch that before watching this video because I will assume you know how to do a polynomial long division. So divide three x squared minus four x plus seven, divide it by x minus one, see what you get as your remainder, and see if that remainder really is f of one. So assuming you've had a go at it, so let's work through it together. So let's divide x minus one, x minus one into three x squared, into three x squared minus four x plus seven. All right, a little bit of polynomial long division is never a bad way to start your morning. It's morning for me, I don't know what it is for you. All right, so, how many, I'll look at, I always look at the x term here, the highest degree term, and then I'll start with the highest degree term here."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's divide x minus one, x minus one into three x squared, into three x squared minus four x plus seven. All right, a little bit of polynomial long division is never a bad way to start your morning. It's morning for me, I don't know what it is for you. All right, so, how many, I'll look at, I always look at the x term here, the highest degree term, and then I'll start with the highest degree term here. So how many times is x going to three x squared? Well, it goes three x times. Three x times x is three x squared, so I'll write three x over here."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All right, so, how many, I'll look at, I always look at the x term here, the highest degree term, and then I'll start with the highest degree term here. So how many times is x going to three x squared? Well, it goes three x times. Three x times x is three x squared, so I'll write three x over here. I'm writing it in the, I guess you could say the first degree place. Three x times x is three x squared. Three x times negative one is negative three x."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Three x times x is three x squared, so I'll write three x over here. I'm writing it in the, I guess you could say the first degree place. Three x times x is three x squared. Three x times negative one is negative three x. And now we want to subtract, we want to subtract this thing. And this is just the way that you do traditional long division. And so, what do we get?"}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Three x times negative one is negative three x. And now we want to subtract, we want to subtract this thing. And this is just the way that you do traditional long division. And so, what do we get? Well, three x squared minus three x squared, that's just going to be zero. So these just add up to zero. And this negative four x, this is going to be plus three x, right?"}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And so, what do we get? Well, three x squared minus three x squared, that's just going to be zero. So these just add up to zero. And this negative four x, this is going to be plus three x, right? A negative of a negative. Negative four x plus three x is going to be negative x. Let me do this in a new color."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And this negative four x, this is going to be plus three x, right? A negative of a negative. Negative four x plus three x is going to be negative x. Let me do this in a new color. So it's going to be negative negative x. And then we can bring down the seven. Complete analogy to how you first learn long division in maybe, I don't know, third or fourth grade."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let me do this in a new color. So it's going to be negative negative x. And then we can bring down the seven. Complete analogy to how you first learn long division in maybe, I don't know, third or fourth grade. So all I did is I multiplied three x times this, you get three x squared minus three x. And then I subtracted that from three x squared minus four x to get this right over here. Or you could say I subtracted it from this whole, from the whole polynomial, and then I got negative x plus seven."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Complete analogy to how you first learn long division in maybe, I don't know, third or fourth grade. So all I did is I multiplied three x times this, you get three x squared minus three x. And then I subtracted that from three x squared minus four x to get this right over here. Or you could say I subtracted it from this whole, from the whole polynomial, and then I got negative x plus seven. So now, how many times does x minus one go into negative x plus seven? Well, x goes into negative x negative one times. Negative one times."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Or you could say I subtracted it from this whole, from the whole polynomial, and then I got negative x plus seven. So now, how many times does x minus one go into negative x plus seven? Well, x goes into negative x negative one times. Negative one times. Negative one times x is negative x. Negative one times negative one is positive one. But then we're going to want to subtract this thing."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Negative one times. Negative one times x is negative x. Negative one times negative one is positive one. But then we're going to want to subtract this thing. We're going to want to subtract this thing, and this is going to give us our remainder. So negative x minus negative x, that's the same thing as negative x plus x. So these are just going to add up to zero."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "But then we're going to want to subtract this thing. We're going to want to subtract this thing, and this is going to give us our remainder. So negative x minus negative x, that's the same thing as negative x plus x. So these are just going to add up to zero. And then you have seven. This isn't going to be seven plus one. Remember, we have this negative out."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So these are just going to add up to zero. And then you have seven. This isn't going to be seven plus one. Remember, we have this negative out. So if you distribute the negative, this is going to be a negative one. Seven minus one is six. So your remainder here is six."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Remember, we have this negative out. So if you distribute the negative, this is going to be a negative one. Seven minus one is six. So your remainder here is six. One way to think about it, you could say, you could say that, you could say, well, actually, I'll save that for a future video. This right over here is the remainder. Remainder."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So your remainder here is six. One way to think about it, you could say, you could say that, you could say, well, actually, I'll save that for a future video. This right over here is the remainder. Remainder. And you know when you got to the remainder, and this is just all a review of polynomial long division, is when you get something that has a lower degree. This is a, I guess you could call this a zero degree polynomial. This has a lower degree than what you are actually dividing into, or then the x minus one, than your divisor."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Remainder. And you know when you got to the remainder, and this is just all a review of polynomial long division, is when you get something that has a lower degree. This is a, I guess you could call this a zero degree polynomial. This has a lower degree than what you are actually dividing into, or then the x minus one, than your divisor. So this is a lower degree, so this is the remainder. You can't take this into this any more times. Now, by the polynomial remainder theorem, if it's true, and I just picked a random example here, this is by no means a proof, but just a kind of a way to make it tangible of what the polynomial remainder theorem is telling us."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This has a lower degree than what you are actually dividing into, or then the x minus one, than your divisor. So this is a lower degree, so this is the remainder. You can't take this into this any more times. Now, by the polynomial remainder theorem, if it's true, and I just picked a random example here, this is by no means a proof, but just a kind of a way to make it tangible of what the polynomial remainder theorem is telling us. If the polynomial remainder theorem is true, it's telling us that f of a, in this case one, f of one should be equal to six. It should be equal to this remainder. Now let's verify that."}, {"video_title": "Polynomial remainder theorem Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now, by the polynomial remainder theorem, if it's true, and I just picked a random example here, this is by no means a proof, but just a kind of a way to make it tangible of what the polynomial remainder theorem is telling us. If the polynomial remainder theorem is true, it's telling us that f of a, in this case one, f of one should be equal to six. It should be equal to this remainder. Now let's verify that. This is going to be equal to three times one squared, which is going to be three, minus four times one, so that's just going to be minus four, plus seven. Three minus four is negative one, plus seven is indeed, we deserve a minor drum roll, is indeed equal to six. So, this is just kind of, at least for this particular case, we're saying, okay, it seems like the polynomial remainder theorem worked, but the utility of it is if someone said, hey, what's the remainder if I were to divide three x squared minus four x plus seven by x minus one, if all they care about is the remainder, they don't care about the actual quotient, all they care about is the remainder, you could say, hey, look, I can just take that, you know, in this case a is one, I can throw that in, I can evaluate f of one, and I'm going to get six."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "There are many videos on Khan Academy where we talk about factoring polynomials, but what we're going to do in this video is do a few more examples of factoring higher degree polynomials. So let's start with a little bit of a warmup. Let's say that we wanted to factor six x squared plus nine x times x squared minus four x plus four. Pause this video and see if you can factor this into the product of even more expressions. All right, now let's do this together. And the way that this might be a little bit different than what you've seen before is this is already partially factored. This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can factor this into the product of even more expressions. All right, now let's do this together. And the way that this might be a little bit different than what you've seen before is this is already partially factored. This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions. But as you might be able to tell, we can factor this further. For example, six x squared plus nine x, both six x squared and nine x are divisible by three x. So let's factor out a three x here."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "This polynomial, this higher degree polynomial, is already expressed as the product of two quadratic expressions. But as you might be able to tell, we can factor this further. For example, six x squared plus nine x, both six x squared and nine x are divisible by three x. So let's factor out a three x here. So this is the same thing as three x times. Three x times what is six x squared? Well, three times two is six, and x times x is x squared."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So let's factor out a three x here. So this is the same thing as three x times. Three x times what is six x squared? Well, three times two is six, and x times x is x squared. And then three x times what is nine x? Well, three x times three is nine x. And you can verify that if we were to distribute this three x you would get six x squared plus nine x."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, three times two is six, and x times x is x squared. And then three x times what is nine x? Well, three x times three is nine x. And you can verify that if we were to distribute this three x you would get six x squared plus nine x. And then what about this second expression right over here? Can we factor this? Well, you might recognize this as a perfect square."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And you can verify that if we were to distribute this three x you would get six x squared plus nine x. And then what about this second expression right over here? Can we factor this? Well, you might recognize this as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four. And you might say, hey, that's negative two and negative two. And so this would be x minus two."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, you might recognize this as a perfect square. Some of you might have said, hey, I need to come up with two numbers whose product is four and whose sum is negative four. And you might say, hey, that's negative two and negative two. And so this would be x minus two. We could write x minus two squared or we could write it as x minus two times x minus two. If what I just did is unfamiliar, I encourage you to go back and watch videos on factoring perfect square quadratics and things like that. But there you have it."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And so this would be x minus two. We could write x minus two squared or we could write it as x minus two times x minus two. If what I just did is unfamiliar, I encourage you to go back and watch videos on factoring perfect square quadratics and things like that. But there you have it. I think we have factored this as far as we could go. So now let's do a slightly trickier higher degree polynomial. So let's say we wanted to factor x to the third minus four x squared plus six x minus 24."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "But there you have it. I think we have factored this as far as we could go. So now let's do a slightly trickier higher degree polynomial. So let's say we wanted to factor x to the third minus four x squared plus six x minus 24. And just like always, pause this video and see if you can have a go at it. And I'll give you a little bit of a hint. You can factor in this case by grouping."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So let's say we wanted to factor x to the third minus four x squared plus six x minus 24. And just like always, pause this video and see if you can have a go at it. And I'll give you a little bit of a hint. You can factor in this case by grouping. And in some ways, it's a little bit easier than what we've done in the past. Historically, when we've learned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadratic, and we broke it up so that we had four terms. Here, we already have four terms."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "You can factor in this case by grouping. And in some ways, it's a little bit easier than what we've done in the past. Historically, when we've learned factoring by grouping, we've looked at a quadratic and then we looked at the middle term, the x term of the quadratic, and we broke it up so that we had four terms. Here, we already have four terms. And see if you can have a go at that. All right, now let's do it together. So you can't always factor a third degree polynomial by grouping, but sometimes you can, so it's good to look for it."}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Here, we already have four terms. And see if you can have a go at that. All right, now let's do it together. So you can't always factor a third degree polynomial by grouping, but sometimes you can, so it's good to look for it. So when we see it written like this, we say, okay, x to the third minus four x squared, is there a common factor here? Well, yeah, both x to the third and negative four x squared are divisible by x squared. So what happens if we factor out an x squared?"}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So you can't always factor a third degree polynomial by grouping, but sometimes you can, so it's good to look for it. So when we see it written like this, we say, okay, x to the third minus four x squared, is there a common factor here? Well, yeah, both x to the third and negative four x squared are divisible by x squared. So what happens if we factor out an x squared? So that's x squared times x minus four. And what about these second two terms? Is there a common factor between six x and negative 24?"}, {"video_title": "Factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So what happens if we factor out an x squared? So that's x squared times x minus four. And what about these second two terms? Is there a common factor between six x and negative 24? Yeah, they're both divisible by six. So let's factor out a six here. So plus six times x minus four."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We are asked, what could be the equation of p? And we have the graph of our polynomial p right over here. You could view this as the graph of y is equal to p of x. So pause this video and see if you can figure that out. All right, now let's work on this together. And you can see that all the choices have p of x in factored form, where it's very easy to identify the zeros, or the x values that would make our polynomial equal to zero. And we could also look at this graph, and we can see what the zeros are."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure that out. All right, now let's work on this together. And you can see that all the choices have p of x in factored form, where it's very easy to identify the zeros, or the x values that would make our polynomial equal to zero. And we could also look at this graph, and we can see what the zeros are. This is where we're going to intersect the x-axis, also known as the x-intercepts. So you can see when x is equal to negative four, we have a zero, because our polynomial is zero there. So we know p of negative four is equal to zero."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And we could also look at this graph, and we can see what the zeros are. This is where we're going to intersect the x-axis, also known as the x-intercepts. So you can see when x is equal to negative four, we have a zero, because our polynomial is zero there. So we know p of negative four is equal to zero. We also know that p of, it looks like 1 1\u20442, or I could say 3\u20442, p of 3\u20442 is equal to zero. And we also know that p of three is equal to zero. So let's look for an expression where that is true."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So we know p of negative four is equal to zero. We also know that p of, it looks like 1 1\u20442, or I could say 3\u20442, p of 3\u20442 is equal to zero. And we also know that p of three is equal to zero. So let's look for an expression where that is true. And because it's in factored form, each of the parts of the product will probably make our polynomial zero for one of these zeros. So let's see, if in order for our polynomial to be equal to zero when x is equal to negative four, we probably want to have a term that has an x plus four in it. Or we wanna have, I should say, a product that has an x plus four in it, because x plus four is equal to zero when x is equal to negative four."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let's look for an expression where that is true. And because it's in factored form, each of the parts of the product will probably make our polynomial zero for one of these zeros. So let's see, if in order for our polynomial to be equal to zero when x is equal to negative four, we probably want to have a term that has an x plus four in it. Or we wanna have, I should say, a product that has an x plus four in it, because x plus four is equal to zero when x is equal to negative four. Well, we have an x plus four there, and we have an x plus four there. So I'm liking choices B and D so far. Now, for this second root, we have p of 3\u20442 is equal to zero."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Or we wanna have, I should say, a product that has an x plus four in it, because x plus four is equal to zero when x is equal to negative four. Well, we have an x plus four there, and we have an x plus four there. So I'm liking choices B and D so far. Now, for this second root, we have p of 3\u20442 is equal to zero. So I would look for something like x minus 3\u20442 in our product. I don't see an x minus 3\u20442 here. But as we've mentioned in other videos, you can also multiply these times constants."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Now, for this second root, we have p of 3\u20442 is equal to zero. So I would look for something like x minus 3\u20442 in our product. I don't see an x minus 3\u20442 here. But as we've mentioned in other videos, you can also multiply these times constants. So if I were to multiply, let's see, to get rid of this fraction here, if I multiply by two, this would be the same thing as, let me scroll down a little bit, same thing as 2x minus three. And you could test that out. 2x minus three is equal to zero when x is equal to 3\u20442."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "But as we've mentioned in other videos, you can also multiply these times constants. So if I were to multiply, let's see, to get rid of this fraction here, if I multiply by two, this would be the same thing as, let me scroll down a little bit, same thing as 2x minus three. And you could test that out. 2x minus three is equal to zero when x is equal to 3\u20442. And let's see, we have a 2x minus three right over there. So choice D is looking awfully good, but let's just verify it with this last one. For p of three to be equal to zero, we could have an expression like x minus three in the product because this is equal to zero when x is equal to three."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "2x minus three is equal to zero when x is equal to 3\u20442. And let's see, we have a 2x minus three right over there. So choice D is looking awfully good, but let's just verify it with this last one. For p of three to be equal to zero, we could have an expression like x minus three in the product because this is equal to zero when x is equal to three. And we indeed have that right over there. So choice D is looking very good. When x is equal to negative four, this part of our product is equal to zero, which makes the whole thing equal to zero."}, {"video_title": "Zeros of polynomials matching equation to graph Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "For p of three to be equal to zero, we could have an expression like x minus three in the product because this is equal to zero when x is equal to three. And we indeed have that right over there. So choice D is looking very good. When x is equal to negative four, this part of our product is equal to zero, which makes the whole thing equal to zero. When x is equal to 3\u20442, 2x minus three is equal to zero, which makes the entire product equal to zero. And when an x is equal to three, it makes x minus three equal to zero. Zero times something times something is going to be equal to zero."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "All right, so if we're saying what is this top expression divided by this bottom expression, another way to think about it is what do I have to multiply? So I'm going to multiply something, I'll put that in parentheses, if I multiply that something times x, I should get x to the fourth minus two x to the third plus five x. Now how do I approach that? Well, there's two ways that I could tackle it. One way is I could just rewrite this expression as being, and I will just make this x in yellow so I can keep track of it. I could just rewrite this as one over x times, times x to the fourth minus two x to the third plus five x, and then I can distribute the one over x. And so what is that going to be equal to?"}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "Well, there's two ways that I could tackle it. One way is I could just rewrite this expression as being, and I will just make this x in yellow so I can keep track of it. I could just rewrite this as one over x times, times x to the fourth minus two x to the third plus five x, and then I can distribute the one over x. And so what is that going to be equal to? Well, it's going to be equal to x to the fourth. Let me do this. X to the fourth over x minus two x to the third over x plus five x, plus five x over x."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "And so what is that going to be equal to? Well, it's going to be equal to x to the fourth. Let me do this. X to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of these going to be equal to? X to the fourth divided by x, if I have four x's that I'm multiplying together and then I divide by x, that's going to be equivalent to x to the third power. So this right over here is equal to x to the third."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "X to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of these going to be equal to? X to the fourth divided by x, if I have four x's that I'm multiplying together and then I divide by x, that's going to be equivalent to x to the third power. So this right over here is equal to x to the third. You could also get there from your exponent properties. In the denominator, you have an x to the first power, and so you would subtract the exponents. You have the same base here, so that's x to the third."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "So this right over here is equal to x to the third. You could also get there from your exponent properties. In the denominator, you have an x to the first power, and so you would subtract the exponents. You have the same base here, so that's x to the third. And then this part right over here, what would that equal to? Well, it's going to be minus two x to the third divided by x to the first. Well, by the same property, that's going to be x squared."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "You have the same base here, so that's x to the third. And then this part right over here, what would that equal to? Well, it's going to be minus two x to the third divided by x to the first. Well, by the same property, that's going to be x squared. And then last but not least, if you take five x's and then you divide by x, you are just going to be left with five. And you can verify that this indeed, if I were to multiply it by x, I'm going to get x to the fourth minus two x to the third plus five x. Let me do that."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "Well, by the same property, that's going to be x squared. And then last but not least, if you take five x's and then you divide by x, you are just going to be left with five. And you can verify that this indeed, if I were to multiply it by x, I'm going to get x to the fourth minus two x to the third plus five x. Let me do that. If I put x to the third minus two x squared plus five times x, what I can do is distribute the x. X times x to the third is x to the fourth. X times negative two x squared is negative two x to the third. X times five is five x."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "Let me do that. If I put x to the third minus two x squared plus five times x, what I can do is distribute the x. X times x to the third is x to the fourth. X times negative two x squared is negative two x to the third. X times five is five x. Now I mentioned there's two ways that I could do it. Another way that I could try to tackle it is I could look at this numerator and try to factor an x out. I would try to factor out whatever I see in the denominator."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "X times five is five x. Now I mentioned there's two ways that I could do it. Another way that I could try to tackle it is I could look at this numerator and try to factor an x out. I would try to factor out whatever I see in the denominator. So if I do that, actually let me just rewrite the numerator. So I can rewrite x to the fourth as x times x to the third. And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "I would try to factor out whatever I see in the denominator. So if I do that, actually let me just rewrite the numerator. So I can rewrite x to the fourth as x times x to the third. And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could write this five x as being equal to plus x times five. And then I'm gonna divide everything by x. Divide everything by x. I just rewrote the numerator here, but for each of those terms, I factored out an x. And now I can factor out x out of the whole thing."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "And then I can rewrite the minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could write this five x as being equal to plus x times five. And then I'm gonna divide everything by x. Divide everything by x. I just rewrote the numerator here, but for each of those terms, I factored out an x. And now I can factor out x out of the whole thing. So I sometimes think of factoring out an x out of the whole thing is reverse distributive property. So if I factor out this x out of every term, what am I left with? I'm left with an x times x to the third minus two x squared plus five."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "And now I can factor out x out of the whole thing. So I sometimes think of factoring out an x out of the whole thing is reverse distributive property. So if I factor out this x out of every term, what am I left with? I'm left with an x times x to the third minus two x squared plus five. I ended up doing that in the same color, but hopefully you're following. Plus five. And then all of that is divided by x."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy.mp3", "Sentence": "I'm left with an x times x to the third minus two x squared plus five. I ended up doing that in the same color, but hopefully you're following. Plus five. And then all of that is divided by x. And as long as x does not equal zero, x divided by x is going to be equal to one, and we're left with what we had to begin with, or the answer that we had to begin with. So these are two different approaches. Nothing super, super sophisticated here."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "we're asked to solve the equation 3 plus the principal square root of 5x plus 6 is equal to 12. And so the general strategy to solve this type of equation is to isolate the radical sign on one side of the equation and then you can square it to essentially get the radical sign to go away. But you have to be very careful there because when you square radical signs you actually lose the information that you're taking the principal square root, not the negative square root or not the plus or minus square root. You're only taking the positive square root. And so when we get our final answer we do have to check and make sure that it gels with taking the principal square root. So let's try, let's see what I'm talking about. So the first thing I want to do is I want to isolate this on one side of the equation."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "You're only taking the positive square root. And so when we get our final answer we do have to check and make sure that it gels with taking the principal square root. So let's try, let's see what I'm talking about. So the first thing I want to do is I want to isolate this on one side of the equation. The best way to isolate that is to get rid of this 3. And the best way to get rid of the 3 is to subtract 3 from the left-hand side. And of course if I do it on the left-hand side I also have to do it on the right-hand side."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the first thing I want to do is I want to isolate this on one side of the equation. The best way to isolate that is to get rid of this 3. And the best way to get rid of the 3 is to subtract 3 from the left-hand side. And of course if I do it on the left-hand side I also have to do it on the right-hand side. Otherwise I would lose the ability to say that they're equal. And so the left-hand side right over here simplifies to the principal square root of 5x plus 6. And this is equal to 12 minus 3."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And of course if I do it on the left-hand side I also have to do it on the right-hand side. Otherwise I would lose the ability to say that they're equal. And so the left-hand side right over here simplifies to the principal square root of 5x plus 6. And this is equal to 12 minus 3. This is equal to 9. And now we can square both sides of this equation. So we could square the principal square root of 5x plus 6 and we can square 9."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And this is equal to 12 minus 3. This is equal to 9. And now we can square both sides of this equation. So we could square the principal square root of 5x plus 6 and we can square 9. When you do this, when you square this you get 5x plus 6. If you square the square root of 5x plus 6 you're going to get 5x plus 6. And this is where we actually lost some information."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we could square the principal square root of 5x plus 6 and we can square 9. When you do this, when you square this you get 5x plus 6. If you square the square root of 5x plus 6 you're going to get 5x plus 6. And this is where we actually lost some information. Because we would have also gotten this if we squared the negative square root of 5x plus 6. And so that's why we have to be careful with the answers we get. And actually make sure it works when the original equation was the principal square root."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And this is where we actually lost some information. Because we would have also gotten this if we squared the negative square root of 5x plus 6. And so that's why we have to be careful with the answers we get. And actually make sure it works when the original equation was the principal square root. So we get 5x plus 6 on the left-hand side and on the right-hand side we get 81. And now this is just a straight up linear equation. We want to isolate the x terms."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And actually make sure it works when the original equation was the principal square root. So we get 5x plus 6 on the left-hand side and on the right-hand side we get 81. And now this is just a straight up linear equation. We want to isolate the x terms. Let's subtract 6 from both sides. On the left-hand side we have 5x and on the right-hand side we have 75. And then we can divide both sides by 5."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We want to isolate the x terms. Let's subtract 6 from both sides. On the left-hand side we have 5x and on the right-hand side we have 75. And then we can divide both sides by 5. Divide both sides by 5. We get x is equal to 15. 5 times 10 is 50."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And then we can divide both sides by 5. Divide both sides by 5. We get x is equal to 15. 5 times 10 is 50. 5 times 5 is 25. This is 75. So we get x is equal to 15."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "5 times 10 is 50. 5 times 5 is 25. This is 75. So we get x is equal to 15. But we need to make sure that this actually works for our original equation. Maybe this would have worked if this was the negative square root. So we need to make sure it actually works for the positive square root, for the principal square root."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we get x is equal to 15. But we need to make sure that this actually works for our original equation. Maybe this would have worked if this was the negative square root. So we need to make sure it actually works for the positive square root, for the principal square root. So let's apply it to our original equation. So we get 3 plus the principal square root of 5 times 15. So 75 plus 6."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we need to make sure it actually works for the positive square root, for the principal square root. So let's apply it to our original equation. So we get 3 plus the principal square root of 5 times 15. So 75 plus 6. So I just took 5 times 15 over here. I put our solution in. It should be equal to 12."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So 75 plus 6. So I just took 5 times 15 over here. I put our solution in. It should be equal to 12. Or we get 3 plus the square root of 75 plus 6 is 81. It needs to be equal to 12. And this is the principal root of 81."}, {"video_title": "Solving radical equations Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "It should be equal to 12. Or we get 3 plus the square root of 75 plus 6 is 81. It needs to be equal to 12. And this is the principal root of 81. So it's positive 9. So it's 3 plus 9 needs to be equal to 12, which is absolutely true. So we can feel pretty good about this answer."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "But just as a bit of a refresher, it's the idea that you do a bunch of legitimate algebraic operations, you get a solution or some solutions at the end, but then when you test it in the original equation, it doesn't satisfy the original equation. And so the key of this video is why do extraneous solutions even occur? And it all is due to the notion of reversibility. There's certain operations in algebra that you can do in one direction, but you can't, and it'll always be true in one direction, but it isn't always true in the other direction. And I'll show you those two operations. One is squaring, and the other is multiplying both sides by a variable expression. So let's see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "There's certain operations in algebra that you can do in one direction, but you can't, and it'll always be true in one direction, but it isn't always true in the other direction. And I'll show you those two operations. One is squaring, and the other is multiplying both sides by a variable expression. So let's see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution. So we know, for example, that if a is equal to b, I could square both sides, and then a squared is going to be equal to b squared. But the other way is not true. For example, if a squared is equal to b squared, it is not always the case that a is equal to b."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution. So we know, for example, that if a is equal to b, I could square both sides, and then a squared is going to be equal to b squared. But the other way is not true. For example, if a squared is equal to b squared, it is not always the case that a is equal to b. What's an example that shows that this is not always the case, but actually pause the video, try to think about it. Well, negative two squared is indeed equal to two squared, but negative two is not equal to two. So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "For example, if a squared is equal to b squared, it is not always the case that a is equal to b. What's an example that shows that this is not always the case, but actually pause the video, try to think about it. Well, negative two squared is indeed equal to two squared, but negative two is not equal to two. So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true. Another non-reversible operation sometimes is multiplying both sides by a variable expression. So multiplying both sides, actually, that got us confused, it looks like an x. Multiply both sides by variable."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true. Another non-reversible operation sometimes is multiplying both sides by a variable expression. So multiplying both sides, actually, that got us confused, it looks like an x. Multiply both sides by variable. I'll just write variable, but it could be a variable expression as well. For example, we know that if a is equal to b, that if we multiply both sides by a variable, that's still going to be true. Xa is going to be equal to xb."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Multiply both sides by variable. I'll just write variable, but it could be a variable expression as well. For example, we know that if a is equal to b, that if we multiply both sides by a variable, that's still going to be true. Xa is going to be equal to xb. But the other, the reverse, isn't always the case. If xa is equal to xb, is it always the case that a is equal to b? Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Xa is going to be equal to xb. But the other, the reverse, isn't always the case. If xa is equal to xb, is it always the case that a is equal to b? Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work. Well, what if a was two and b is three and the variable x just happened to take on the value zero? So we know that zero times two is indeed equal to zero times three, but two is not equal to three. Now, how does all of this connect to the extraneous solutions you've seen when you're solving radical equations or when you're solving some rational or equations with rational expressions on both sides?"}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Well, the simple answer is no, and I always encourage you, pause this video and see if you can find an example where this doesn't work. Well, what if a was two and b is three and the variable x just happened to take on the value zero? So we know that zero times two is indeed equal to zero times three, but two is not equal to three. Now, how does all of this connect to the extraneous solutions you've seen when you're solving radical equations or when you're solving some rational or equations with rational expressions on both sides? Well, let's look at an example. Let's solve a radical equation. If I wanted to solve the equation, the square root of five x minus four is equal to x minus two."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Now, how does all of this connect to the extraneous solutions you've seen when you're solving radical equations or when you're solving some rational or equations with rational expressions on both sides? Well, let's look at an example. Let's solve a radical equation. If I wanted to solve the equation, the square root of five x minus four is equal to x minus two. A typical first step is, hey, let's get rid of this radical by squaring both sides. So I am going to square both sides, and then I'm going to get five x minus four is equal to x squared minus four x plus four. Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "If I wanted to solve the equation, the square root of five x minus four is equal to x minus two. A typical first step is, hey, let's get rid of this radical by squaring both sides. So I am going to square both sides, and then I'm going to get five x minus four is equal to x squared minus four x plus four. Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations. And let's see, we can subtract five x from both sides. We can add four to both sides. I'm just trying to get a zero on the left-hand side."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations. And let's see, we can subtract five x from both sides. We can add four to both sides. I'm just trying to get a zero on the left-hand side. And so I'm going to be left with zero is equal to x squared minus nine x plus eight, or zero is equal to x minus eight times x minus one, or we could say that x minus one is equal to zero, or x minus eight is equal to zero. We get x equals one or x equals eight. So let's test these solutions."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "I'm just trying to get a zero on the left-hand side. And so I'm going to be left with zero is equal to x squared minus nine x plus eight, or zero is equal to x minus eight times x minus one, or we could say that x minus one is equal to zero, or x minus eight is equal to zero. We get x equals one or x equals eight. So let's test these solutions. If x equals eight, we would get, and I'll color-code it a little bit. For x equals eight, if I test it in the original equation, I get the square root of 36 is equal to six, which is absolutely true, so that one works. But what about x equals one?"}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's test these solutions. If x equals eight, we would get, and I'll color-code it a little bit. For x equals eight, if I test it in the original equation, I get the square root of 36 is equal to six, which is absolutely true, so that one works. But what about x equals one? I get the square root of five times one minus four is one is equal to one minus two, which is equal to negative one. That did not work. This right over here is an extraneous solution."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "But what about x equals one? I get the square root of five times one minus four is one is equal to one minus two, which is equal to negative one. That did not work. This right over here is an extraneous solution. If someone said, what are all the x values that satisfy this equation, you would not say x equals one, even though you got there with legitimate algebraic steps. And the reason that is true is, actually, pause this video. Look back."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "This right over here is an extraneous solution. If someone said, what are all the x values that satisfy this equation, you would not say x equals one, even though you got there with legitimate algebraic steps. And the reason that is true is, actually, pause this video. Look back. For which of these steps does x equal one still work, and what step does it not work? Well, you'll see that x equals one works for all of these equations below this purple line. It just doesn't work for the square root of five minus four x is equal to x minus two."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Look back. For which of these steps does x equal one still work, and what step does it not work? Well, you'll see that x equals one works for all of these equations below this purple line. It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one, and then you could deduce all the way up to this line here. But the issue here is that squaring is not a reversible operation. This is analogous to saying, hey, we know that a squared is equal to b squared."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one, and then you could deduce all the way up to this line here. But the issue here is that squaring is not a reversible operation. This is analogous to saying, hey, we know that a squared is equal to b squared. We know that this is equal to this, but then that doesn't mean that a is necessarily equal to b for x equals one. And we could do the same thing with a rational or an equation that deals with rational expressions. So, for example, we might have to deal with, and let me make sure I have some space here."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "This is analogous to saying, hey, we know that a squared is equal to b squared. We know that this is equal to this, but then that doesn't mean that a is necessarily equal to b for x equals one. And we could do the same thing with a rational or an equation that deals with rational expressions. So, for example, we might have to deal with, and let me make sure I have some space here. If I had to solve x squared over x minus one is equal to one over x minus one, the first thing I might wanna do is multiply both sides by x minus one. So multiply, multiply by x minus one. Now, notice, I'm multiplying both sides by a variable expression, so we have to be a little bit conscientious now, but if I multiply both sides by x minus one, I'm going to get x squared is equal to one, or I could say that x equals one, or x is equal to negative one."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So, for example, we might have to deal with, and let me make sure I have some space here. If I had to solve x squared over x minus one is equal to one over x minus one, the first thing I might wanna do is multiply both sides by x minus one. So multiply, multiply by x minus one. Now, notice, I'm multiplying both sides by a variable expression, so we have to be a little bit conscientious now, but if I multiply both sides by x minus one, I'm going to get x squared is equal to one, or I could say that x equals one, or x is equal to negative one. But we could test these for x equals one. If I go up here, I'm dividing by zero on both sides, so this is an extraneous solution. The key here is that we multiplied both sides by a variable expression."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Now, notice, I'm multiplying both sides by a variable expression, so we have to be a little bit conscientious now, but if I multiply both sides by x minus one, I'm going to get x squared is equal to one, or I could say that x equals one, or x is equal to negative one. But we could test these for x equals one. If I go up here, I'm dividing by zero on both sides, so this is an extraneous solution. The key here is that we multiplied both sides by a variable expression. In this case, we multiplied both sides by x minus one. You can do that. You can multiply both sides by a variable expression, and it is a legitimate algebraic operation."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "The key here is that we multiplied both sides by a variable expression. In this case, we multiplied both sides by x minus one. You can do that. You can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because zero times two is equal to zero times three does not mean that two is equal to three. It's completely analogous because we multiplied by a variable expression that actually takes on the value zero when x is equal to one."}, {"video_title": "Extraneous solutions Equations Algebra 2 Khan Academy.mp3", "Sentence": "You can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because zero times two is equal to zero times three does not mean that two is equal to three. It's completely analogous because we multiplied by a variable expression that actually takes on the value zero when x is equal to one. So the big takeaway here is, hopefully you understand why extraneous solutions happen a little bit more. When you square, when you multiply both sides by a variable expression, completely legitimate as long as you do it properly, but it's not always the case that the reverse is true. You could add or subtract anything from both sides of an equation, and that's always going to be reversible, and so that's not going to lead to extraneous solutions."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "So we have an interesting equation here. Let's see if we can solve for k. And we're going to assume that m is greater than zero. Like always, pause the video, try it out on your own, and then I will do it with you. All right, let's work on this a little bit. So you can imagine that the key to this is to simplify it using our knowledge of exponent properties and there's a couple of ways to think about it. First, we can look at this rational expression here, m to the 7 9ths power divided by m to the 1 3rd power. And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "All right, let's work on this a little bit. So you can imagine that the key to this is to simplify it using our knowledge of exponent properties and there's a couple of ways to think about it. First, we can look at this rational expression here, m to the 7 9ths power divided by m to the 1 3rd power. And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. And it actually comes straight out of the notion that x to the a over x to the b, x to the a over x to the b, is the same thing as x to the a times one over x to the b, which is the same thing as x to the a times one over x to the b. That's the same thing as x to the negative b, which is going to be the same thing as if I have a base to one exponent times the same base to another exponent. That's the same thing as that base to the sum of the exponents, a plus negative b, which is just going to be a minus b."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "And the key realization here is that if I have x to the a over x to the b, that this is going to be equal to x to the a minus b power. And it actually comes straight out of the notion that x to the a over x to the b, x to the a over x to the b, is the same thing as x to the a times one over x to the b, which is the same thing as x to the a times one over x to the b. That's the same thing as x to the negative b, which is going to be the same thing as if I have a base to one exponent times the same base to another exponent. That's the same thing as that base to the sum of the exponents, a plus negative b, which is just going to be a minus b. So we got to the same place. So we can rewrite this as, so we can rewrite this part as being equal to m to the 7 9ths power minus 1 3rd power is equal to, is equal to m to the k over 9. And I think you see where this is going."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "That's the same thing as that base to the sum of the exponents, a plus negative b, which is just going to be a minus b. So we got to the same place. So we can rewrite this as, so we can rewrite this part as being equal to m to the 7 9ths power minus 1 3rd power is equal to, is equal to m to the k over 9. And I think you see where this is going. What is 7 9ths minus 1 3rd? Well, 1 3rd is the same thing we want to have a common denominator. 1 3rd is the same thing as 3 9ths."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "And I think you see where this is going. What is 7 9ths minus 1 3rd? Well, 1 3rd is the same thing we want to have a common denominator. 1 3rd is the same thing as 3 9ths. So I can rewrite this as 3 9ths. So 7 9ths minus 3 9ths is going to be 4 9ths. So this is the same thing as m to the, m to the 4 9ths power is going to be equal to m to the k 9ths power."}, {"video_title": "Simplifying quotient of powers (rational exponents) Algebra I High School Math Khan Academy.mp3", "Sentence": "1 3rd is the same thing as 3 9ths. So I can rewrite this as 3 9ths. So 7 9ths minus 3 9ths is going to be 4 9ths. So this is the same thing as m to the, m to the 4 9ths power is going to be equal to m to the k 9ths power. So 4 9ths must be the same thing as k 9ths. So we can say 4 9ths is equal to k 9ths. 4 over 9 is equal to k over 9, which tells us that k must be equal to 4."}, {"video_title": "Plotting complex numbers on the complex plane Precalculus Khan Academy.mp3", "Sentence": "So we have a complex number here. It has a real part, negative 2. It has an imaginary part. You have 2 times i. And what you see here is we're going to plot it on this kind of two-dimensional grid, but it's not our traditional coordinate axes. In our traditional coordinate axis, you're plotting a real x value versus a real y coordinate. Here, on the horizontal axis, that's going to be the real part of our complex number, and our vertical axis is going to be the imaginary part."}, {"video_title": "Plotting complex numbers on the complex plane Precalculus Khan Academy.mp3", "Sentence": "You have 2 times i. And what you see here is we're going to plot it on this kind of two-dimensional grid, but it's not our traditional coordinate axes. In our traditional coordinate axis, you're plotting a real x value versus a real y coordinate. Here, on the horizontal axis, that's going to be the real part of our complex number, and our vertical axis is going to be the imaginary part. So in this example, this complex number, our real part is the negative 2, and then our imaginary part is a positive 2. And so that right over there in the complex plane is the point negative 2 plus 2i. Let's do a few more of these."}, {"video_title": "Plotting complex numbers on the complex plane Precalculus Khan Academy.mp3", "Sentence": "Here, on the horizontal axis, that's going to be the real part of our complex number, and our vertical axis is going to be the imaginary part. So in this example, this complex number, our real part is the negative 2, and then our imaginary part is a positive 2. And so that right over there in the complex plane is the point negative 2 plus 2i. Let's do a few more of these. So 5 plus 2i, once again, real part is 5, imaginary part is 2, and we're done. Let's do two more of these. 1 plus 5i."}, {"video_title": "Plotting complex numbers on the complex plane Precalculus Khan Academy.mp3", "Sentence": "Let's do a few more of these. So 5 plus 2i, once again, real part is 5, imaginary part is 2, and we're done. Let's do two more of these. 1 plus 5i. 1, that's the real part, plus 5i right over that im. All right, let's do one more of these. 4 minus 4i."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "In this video, we're going to learn to divide polynomials. And sometimes this is called algebraic long division. But you'll see what I'm talking about when we do a few examples. Let's say I just want to divide 2x plus 4 and divide it by 2. And we're not really changing the value. We're just changing how we're going to express the value. So we already know how to simplify this."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's say I just want to divide 2x plus 4 and divide it by 2. And we're not really changing the value. We're just changing how we're going to express the value. So we already know how to simplify this. We've done this in the past. We could divide the numerator and the denominator by 2. And this would be equal to what?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we already know how to simplify this. We've done this in the past. We could divide the numerator and the denominator by 2. And this would be equal to what? This would be equal to x plus 2. Let me write it this way. It would be equal to, if you divide this by 2, it becomes an x."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And this would be equal to what? This would be equal to x plus 2. Let me write it this way. It would be equal to, if you divide this by 2, it becomes an x. You divide the 4 by 2, it becomes a 2. If you divide the 2 by 2, you get a 1. So this is equal to x plus 2, which is pretty straightforward, I think."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It would be equal to, if you divide this by 2, it becomes an x. You divide the 4 by 2, it becomes a 2. If you divide the 2 by 2, you get a 1. So this is equal to x plus 2, which is pretty straightforward, I think. The other way is you could have factored a 2 out of here, and then those would have canceled out. But I'll also show you how to do it using algebraic long division, which is a bit of overkill for this problem. But I just want to show you that it's not fundamentally anything new."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is equal to x plus 2, which is pretty straightforward, I think. The other way is you could have factored a 2 out of here, and then those would have canceled out. But I'll also show you how to do it using algebraic long division, which is a bit of overkill for this problem. But I just want to show you that it's not fundamentally anything new. It's just a different way of doing things. But it's useful for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "But I just want to show you that it's not fundamentally anything new. It's just a different way of doing things. But it's useful for more complicated problems. So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2, you'd always start with the highest degree term. 2 goes into the highest degree term."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So you could have also written this as 2 goes into 2x plus 4 how many times? And you would perform this the same way you would do traditional long division. You'd say 2, you'd always start with the highest degree term. 2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "2 goes into the highest degree term. You would ignore the 4. 2 goes into 2x how many times? Well, it goes into 2x x times. And you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, it goes into 2x x times. And you put the x in the x place. x times 2 is 2x. And just like traditional long division, you now subtract. You now subtract. So 2x plus 4 minus 2x is what? It's 4, right?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And just like traditional long division, you now subtract. You now subtract. So 2x plus 4 minus 2x is what? It's 4, right? And then 2 goes into 4 how many times? It goes into 2 times, a positive 2 times. Put that in the constants place."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's 4, right? And then 2 goes into 4 how many times? It goes into 2 times, a positive 2 times. Put that in the constants place. 2 times 2 is 4. You subtract. Remainder is 0."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Put that in the constants place. 2 times 2 is 4. You subtract. Remainder is 0. So this might seem overkill for what was probably a problem that you already knew how to do and do it in a few steps. But we're now going to see that this is a very generalizable process. You can do this really for any degree polynomial, dividing into any other degree polynomial."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Remainder is 0. So this might seem overkill for what was probably a problem that you already knew how to do and do it in a few steps. But we're now going to see that this is a very generalizable process. You can do this really for any degree polynomial, dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You can do this really for any degree polynomial, dividing into any other degree polynomial. Let me show you what I'm talking about. So let's say we wanted to divide x plus 1 into x squared plus 3x plus 6. So what do we do here? So you look at the highest degree term here, which is an x. And you look at the highest degree term here, which is an x squared. So you can ignore everything else."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So what do we do here? So you look at the highest degree term here, which is an x. And you look at the highest degree term here, which is an x squared. So you can ignore everything else. And that really simplifies the process. And you say x goes into x squared how many times? Well, x squared divided by x is just x, right?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So you can ignore everything else. And that really simplifies the process. And you say x goes into x squared how many times? Well, x squared divided by x is just x, right? x goes into x squared x times. And you put it in the x place. This is the x place right here, or the x to the first power place."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, x squared divided by x is just x, right? x goes into x squared x times. And you put it in the x place. This is the x place right here, or the x to the first power place. So x times x plus 1 is what? x times x is x squared. x times 1 is x."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This is the x place right here, or the x to the first power place. So x times x plus 1 is what? x times x is x squared. x times 1 is x. So it's x squared plus x. And just like we did over here, we now subtract. And what do we get?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "x times 1 is x. So it's x squared plus x. And just like we did over here, we now subtract. And what do we get? x squared plus 3x plus 6 minus x squared. Let me be very careful. This is minus x squared plus x."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And what do we get? x squared plus 3x plus 6 minus x squared. Let me be very careful. This is minus x squared plus x. Don't want to make sure that negative sign only applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This is minus x squared plus x. Don't want to make sure that negative sign only applies to this whole thing. So x squared minus x squared, those cancel out. 3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "3x, this is going to be a minus x. Let me put that sign there. So this is minus x squared minus x, just to be clear. We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We're subtracting the whole thing. 3x minus x is 2x. And then you bring down the 6, or 6 minus 0 is nothing. So 2x plus 6. Now you look at the highest degree term, an x and a 2x. How many times does x go into 2x? It goes into it two times."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So 2x plus 6. Now you look at the highest degree term, an x and a 2x. How many times does x go into 2x? It goes into it two times. 2 times x is 2x. 2 times 1 is 2. 2 times 1 is 2."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It goes into it two times. 2 times x is 2x. 2 times 1 is 2. 2 times 1 is 2. So we get 2 times x plus 1 is 2x plus 2. But we're going to want to subtract this from this up here, so we're going to subtract it. Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "2 times 1 is 2. So we get 2 times x plus 1 is 2x plus 2. But we're going to want to subtract this from this up here, so we're going to subtract it. Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them. These guys cancel out. 6 minus 2 is 4. And how many times does x go into 4?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Instead of writing 2x plus 2, we could just write negative 2x minus 2 and then add them. These guys cancel out. 6 minus 2 is 4. And how many times does x go into 4? Well, we could just say that 0 times, or we could say that 4 is the remainder. So if we wanted to rewrite x squared plus 3x plus 6 over x plus 1, notice this is the same thing as x squared plus 3x plus 6 divided by x plus 1. This thing divided by this."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And how many times does x go into 4? Well, we could just say that 0 times, or we could say that 4 is the remainder. So if we wanted to rewrite x squared plus 3x plus 6 over x plus 1, notice this is the same thing as x squared plus 3x plus 6 divided by x plus 1. This thing divided by this. We can now say that this is equal to x plus 2. It is equal to x plus 2 plus the remainder divided by x plus 1 plus 4 over x plus 1. This right here and this right here are equivalent."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This thing divided by this. We can now say that this is equal to x plus 2. It is equal to x plus 2 plus the remainder divided by x plus 1 plus 4 over x plus 1. This right here and this right here are equivalent. And if you wanted to check that, if you wanted to go from this back to that, what you could do is multiply this by x plus 1 over x plus 1 and then add the 2. So this is the same thing as x plus 2. I'm just going to multiply that times x plus 1 over x plus 1, that's just multiplying it by 1."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This right here and this right here are equivalent. And if you wanted to check that, if you wanted to go from this back to that, what you could do is multiply this by x plus 1 over x plus 1 and then add the 2. So this is the same thing as x plus 2. I'm just going to multiply that times x plus 1 over x plus 1, that's just multiplying it by 1. And then to that, add 4 over x plus 1. I did that so I have the same common denominator. And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here, you should get x squared plus 3x plus 6."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I'm just going to multiply that times x plus 1 over x plus 1, that's just multiplying it by 1. And then to that, add 4 over x plus 1. I did that so I have the same common denominator. And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here, you should get x squared plus 3x plus 6. Let's do another one of these. They're kind of fun. So let's say that we want to simplify x squared plus 5x plus 4 over x plus 4."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And when you perform this addition right here, when you multiply these two binomials and then add the 4 up here, you should get x squared plus 3x plus 6. Let's do another one of these. They're kind of fun. So let's say that we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's say that we want to simplify x squared plus 5x plus 4 over x plus 4. So once again, we can do our algebraic long division. We can divide x plus 4 into x squared plus 5x plus 4. And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And once again, same exact process. Look at the highest degree terms in both of them. x goes into x squared how many times? It goes into it x times. Put in the x place. This is our x place right here. x times x is x squared."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It goes into it x times. Put in the x place. This is our x place right here. x times x is x squared. x times 4 is 4x. And then of course, we're going to want to subtract these from there. So let me just put a negative sign there."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "x times x is x squared. x times 4 is 4x. And then of course, we're going to want to subtract these from there. So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and you could even see this coming."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let me just put a negative sign there. And then these cancel out. 5x minus 4x is x. 4 minus 0 is plus 4. x plus 4, and you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time. Or you could just look at, if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "4 minus 0 is plus 4. x plus 4, and you could even see this coming. You could say x plus 4 goes into x plus 4 obviously one time. Or you could just look at, if you were not looking at the constant terms, you would completely just say, well, x goes into x how many times? Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, one time. Plus 1. 1 times x is x. 1 times 4 is 4. We're going to subtract them from up here. So it cancels out. So we have no remainder."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "1 times 4 is 4. We're going to subtract them from up here. So it cancels out. So we have no remainder. So this right here simplifies to, this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we have no remainder. So this right here simplifies to, this is equal to x plus 1. And there's other ways you could have done this. We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 times x plus 1."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We could have tried to factor this numerator. x squared plus 5x plus 4 over x plus 4. This is the same thing as what? We could have factored this numerator as x plus 4 times x plus 1. 4 times 1 is 4. 4 plus 1 is 5. All of that over x plus 4."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We could have factored this numerator as x plus 4 times x plus 1. 4 times 1 is 4. 4 plus 1 is 5. All of that over x plus 4. That cancels out, and you're left just with x plus 1. Either way would have worked. But the algebraic long division will always work, even if you can't cancel out factors like that."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All of that over x plus 4. That cancels out, and you're left just with x plus 1. Either way would have worked. But the algebraic long division will always work, even if you can't cancel out factors like that. Even if you did have remainder in this situation, you didn't. So this was equal to x plus 1. Let's do another one of these."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "But the algebraic long division will always work, even if you can't cancel out factors like that. Even if you did have remainder in this situation, you didn't. So this was equal to x plus 1. Let's do another one of these. Just to make sure that you really have, because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared. Let me just change it up."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's do another one of these. Just to make sure that you really have, because this is actually a very, very useful skill to have in your toolkit. So let's say we have x squared. Let me just change it up. Let's say we had 2x squared. I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let me just change it up. Let's say we had 2x squared. I could really make these numbers up on the fly. 2x squared minus 20x plus 12 divided by. Actually, let's make it really interesting. Just to show you that it'll always work. I want to go above quadratic."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "2x squared minus 20x plus 12 divided by. Actually, let's make it really interesting. Just to show you that it'll always work. I want to go above quadratic. So let's say we have 3x to the third minus 2x squared plus 7x minus 4. And we want to divide that. We want to divide that by x squared plus 1."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I want to go above quadratic. So let's say we have 3x to the third minus 2x squared plus 7x minus 4. And we want to divide that. We want to divide that by x squared plus 1. I just made this up. But we can just do the algebraic long division to figure out what this is going to be, or what this is. We simplify it, what it'll be."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We want to divide that by x squared plus 1. I just made this up. But we can just do the algebraic long division to figure out what this is going to be, or what this is. We simplify it, what it'll be. x squared plus 1 divided into this thing up here. 3x to the third minus 2x squared plus 7x minus 4. Once again, look at the highest degree term."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We simplify it, what it'll be. x squared plus 1 divided into this thing up here. 3x to the third minus 2x squared plus 7x minus 4. Once again, look at the highest degree term. x squared goes into 3x to the third how many times? Well, it's going to go into it 3x times. You multiply 3x times this, you get 3x to the third."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Once again, look at the highest degree term. x squared goes into 3x to the third how many times? Well, it's going to go into it 3x times. You multiply 3x times this, you get 3x to the third. So it's going to go into it 3x times. So you have to write the 3x over here in the x term. So it's going to go into it 3x times, just like that."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You multiply 3x times this, you get 3x to the third. So it's going to go into it 3x times. So you have to write the 3x over here in the x term. So it's going to go into it 3x times, just like that. Now let's multiply. 3x times x squared is 3x to the third plus 3x times 1. So we have a 3x over here."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it's going to go into it 3x times, just like that. Now let's multiply. 3x times x squared is 3x to the third plus 3x times 1. So we have a 3x over here. I'm making sure to put it in the x place. And we're going to want to subtract them. And what do we have when we do that?"}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we have a 3x over here. I'm making sure to put it in the x place. And we're going to want to subtract them. And what do we have when we do that? These cancel out. We have a minus 2x squared. And then 7x minus 3x is plus 4x."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And what do we have when we do that? These cancel out. We have a minus 2x squared. And then 7x minus 3x is plus 4x. And we have a minus 4. Once again, look at the highest degree term. x squared and a negative 2x squared."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then 7x minus 3x is plus 4x. And we have a minus 4. Once again, look at the highest degree term. x squared and a negative 2x squared. So x squared goes into negative 2x squared negative 2 times. Put it in the constants place. Negative 2 times x squared is negative 2x squared."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "x squared and a negative 2x squared. So x squared goes into negative 2x squared negative 2 times. Put it in the constants place. Negative 2 times x squared is negative 2x squared. Negative 2 times 1 is negative 2. Now we're going to want to subtract these from there. So let's multiply them by negative 1, or those become a positive."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Negative 2 times x squared is negative 2x squared. Negative 2 times 1 is negative 2. Now we're going to want to subtract these from there. So let's multiply them by negative 1, or those become a positive. These two guys cancel out. 4x minus 0 is 4x minus. Let me switch colors."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So let's multiply them by negative 1, or those become a positive. These two guys cancel out. 4x minus 0 is 4x minus. Let me switch colors. 4x minus 0 is 4x. Negative 4 minus negative 2, or negative 4 plus 2, is equal to negative 2. And then x squared now has a higher degree than 4x, the highest degree here."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let me switch colors. 4x minus 0 is 4x. Negative 4 minus negative 2, or negative 4 plus 2, is equal to negative 2. And then x squared now has a higher degree than 4x, the highest degree here. So we view this as the remainder. So this expression we could rewrite it as being equal to 3x minus 2. That's the 3x minus 2."}, {"video_title": "Polynomial division Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then x squared now has a higher degree than 4x, the highest degree here. So we view this as the remainder. So this expression we could rewrite it as being equal to 3x minus 2. That's the 3x minus 2. Plus our remainder, 4x minus 2. All of that over x squared plus 1. x squared plus 1. Hopefully you found that as fun as I did."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "And we know that because three times three is equal to nine. This is equivalent to saying, what is the principal root of nine? Well, that is equal to three. But what would happen if I took nine to the negative 1 1\u20442 power? Now we have a negative fractional exponent. And the key to this is to just not to get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "But what would happen if I took nine to the negative 1 1\u20442 power? Now we have a negative fractional exponent. And the key to this is to just not to get too worried or intimidated by this, but just think about it step by step. Just ignore for the second that this is a fraction. And just look at this negative first. Just breathe slowly and realize, OK, I got a negative. Negative exponent."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Just ignore for the second that this is a fraction. And just look at this negative first. Just breathe slowly and realize, OK, I got a negative. Negative exponent. That means that this is just going to be 1 over 9 to the 1 1\u20442. That's what that negative is the cue for. This is 1 over 9 to the 1 1\u20442."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "Negative exponent. That means that this is just going to be 1 over 9 to the 1 1\u20442. That's what that negative is the cue for. This is 1 over 9 to the 1 1\u20442. And we know that 9 to the 1 1\u20442 is equal to 3. So this is just going to be equal to 1. This is just going to be equal to 1 3rd."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "This is 1 over 9 to the 1 1\u20442. And we know that 9 to the 1 1\u20442 is equal to 3. So this is just going to be equal to 1. This is just going to be equal to 1 3rd. Let's take things a little bit further. What would this evaluate to? And I encourage you to pause the video after trying it."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "This is just going to be equal to 1 3rd. Let's take things a little bit further. What would this evaluate to? And I encourage you to pause the video after trying it. Or pause the video to try it. Negative 27 to the negative 1 3rd power. So I encourage you to pause the video and think about what this would evaluate to."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "And I encourage you to pause the video after trying it. Or pause the video to try it. Negative 27 to the negative 1 3rd power. So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1 3rd power."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and think about what this would evaluate to. So remember, just take a deep breath. You can always get rid of this negative in the exponent by taking the reciprocal and raising it to the positive. So this is going to be equal to 1 over negative 27 to the positive 1 3rd power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 over negative 27 to the positive 1 3rd power. And I know what you're saying. Hey, I still can't breathe easily. I have this negative number to this fractional exponent. But this is just saying, what number, if I were to multiply it three times, so if I have that number, so whatever the number this is, I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I have to use here to get negative 27? Well, we already know that 3 to the 3rd, which is equal to 3 times 3 times 3 is equal to positive 27. So that's a pretty good clue."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "I have this negative number to this fractional exponent. But this is just saying, what number, if I were to multiply it three times, so if I have that number, so whatever the number this is, I were to multiply it, if I took three of them and I multiply them together, if I multiplied 1 by that number three times, what number would I have to use here to get negative 27? Well, we already know that 3 to the 3rd, which is equal to 3 times 3 times 3 is equal to positive 27. So that's a pretty good clue. What would negative 3 to the 3rd power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9 times negative 3 is negative 27. So we just found this number, this question mark."}, {"video_title": "Negative fractional exponent examples Algebra I Khan Academy.mp3", "Sentence": "So that's a pretty good clue. What would negative 3 to the 3rd power be? Well, that's negative 3 times negative 3 times negative 3, which is negative 3 times negative 3 is positive 9 times negative 3 is negative 27. So we just found this number, this question mark. Negative 3 times negative 3 times negative 3 is equal to negative 27. So negative 27 to the 1 3rd, this part right over here, is equal to negative 3. So this is going to be equal to 1 over negative 3, which is the same thing as negative 1 3rd."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And it first bears mentioning how this widget works. So this point right over here, it helps you define the midline, the thing that you could imagine your sine or cosine function oscillates around. And then you also define a neighboring extreme point, either a maximum or a minimum point, to graph your function. So let's think about how we would do this. And like always, I encourage you to pause this video and think about how you would do it yourself. But the first way I like to think about it is what would a regular just... If this just said y is equal to sine of x, how would I graph that?"}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So let's think about how we would do this. And like always, I encourage you to pause this video and think about how you would do it yourself. But the first way I like to think about it is what would a regular just... If this just said y is equal to sine of x, how would I graph that? Well, sine of 0 is 0. Sine of pi over 2 is 1. And then sine of pi is 0 again."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "If this just said y is equal to sine of x, how would I graph that? Well, sine of 0 is 0. Sine of pi over 2 is 1. And then sine of pi is 0 again. And so this is what just regular sine of x would look like. But let's think about how this is different. Well, first of all, it's not just sine of x."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And then sine of pi is 0 again. And so this is what just regular sine of x would look like. But let's think about how this is different. Well, first of all, it's not just sine of x. It's sine of 1 half x. So what would be the graph of just sine of 1 half x? Well, one way to think about it, there's actually two ways to think about it, is a coefficient right over here on your x term that tells you how fast the thing that's being inputted into sine is growing."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, first of all, it's not just sine of x. It's sine of 1 half x. So what would be the graph of just sine of 1 half x? Well, one way to think about it, there's actually two ways to think about it, is a coefficient right over here on your x term that tells you how fast the thing that's being inputted into sine is growing. And now it's going to grow half as fast. And so one way to think about it is your period is now going to be twice as long. So one way to think about it is instead of getting to this next maximum point at pi over 2, you're going to get there at pi."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Well, one way to think about it, there's actually two ways to think about it, is a coefficient right over here on your x term that tells you how fast the thing that's being inputted into sine is growing. And now it's going to grow half as fast. And so one way to think about it is your period is now going to be twice as long. So one way to think about it is instead of getting to this next maximum point at pi over 2, you're going to get there at pi. And you could test that. If you...when x is equal to pi, this will be 1 half pi. Sine of 1 half pi is indeed equal to 1."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So one way to think about it is instead of getting to this next maximum point at pi over 2, you're going to get there at pi. And you could test that. If you...when x is equal to pi, this will be 1 half pi. Sine of 1 half pi is indeed equal to 1. Another way to think about it is you might be familiar with the formula, although I always like you to think about where these formulas come from, that to figure out the period of a sine or cosine function, you take 2 pi and you divide it by whatever this coefficient is. So 2 pi divided by 1 half is going to be 4 pi. And you can see the period here, we go up, down, and back to where we were over 4 pi."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Sine of 1 half pi is indeed equal to 1. Another way to think about it is you might be familiar with the formula, although I always like you to think about where these formulas come from, that to figure out the period of a sine or cosine function, you take 2 pi and you divide it by whatever this coefficient is. So 2 pi divided by 1 half is going to be 4 pi. And you can see the period here, we go up, down, and back to where we were over 4 pi. And that makes sense because if you just had a 1 coefficient here, your period would be 2 pi. 2 pi radians, you make one circle around the unit circle is one way to think about it. So right here we have the graph of sine of 1 half x."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And you can see the period here, we go up, down, and back to where we were over 4 pi. And that makes sense because if you just had a 1 coefficient here, your period would be 2 pi. 2 pi radians, you make one circle around the unit circle is one way to think about it. So right here we have the graph of sine of 1 half x. Now what if we wanted to instead think about 3 times the graph of sine of 1 half x, or 3 sine 1 half x? Well, then our amplitude is just going to be 3 times as much. And so instead of our maximum point going from... instead of our maximum point being at 1, it will now be at 3."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So right here we have the graph of sine of 1 half x. Now what if we wanted to instead think about 3 times the graph of sine of 1 half x, or 3 sine 1 half x? Well, then our amplitude is just going to be 3 times as much. And so instead of our maximum point going from... instead of our maximum point being at 1, it will now be at 3. Or another way to think about it is we're going 3 above the midline and 3 below the midline. So this right over here is the graph of 3 sine of 1 half x. Now we have one thing left to do, and this is this minus 2."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And so instead of our maximum point going from... instead of our maximum point being at 1, it will now be at 3. Or another way to think about it is we're going 3 above the midline and 3 below the midline. So this right over here is the graph of 3 sine of 1 half x. Now we have one thing left to do, and this is this minus 2. So this minus 2 is just going to shift everything down by 2. So we just have to shift everything down. So let me shift this one down by 2, and let me shift this one down by 2."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Now we have one thing left to do, and this is this minus 2. So this minus 2 is just going to shift everything down by 2. So we just have to shift everything down. So let me shift this one down by 2, and let me shift this one down by 2. And so there you have it. Notice our period is still 4 pi. Our amplitude, how much we oscillate above or below the midline, is still 3."}, {"video_title": "Example Graphing y=3\u22c5sin(\u00bd\u22c5x)-2 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So let me shift this one down by 2, and let me shift this one down by 2. And so there you have it. Notice our period is still 4 pi. Our amplitude, how much we oscillate above or below the midline, is still 3. And now we have this minus 2. Another way to think about it, when x is equal to 0, this whole first term is going to be 0, and y should be equal to negative 2. And we're done."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "That looks like a rectangle. And if this side's length is L, then this side's length is also going to be L. And if this width is W, then this width up here is W. And the perimeter is just how, what is the distance if you were to go around this rectangle. And so that distance is going to be this W plus this L plus this W, or that width, plus this length. And if you have one W and you add it to another W, that's going to give you two Ws. So that's two Ws. And then if you have one L and then you have another L, that's going to give you, if you add them together, that's going to give you two Ls. So the perimeter is going to be two Ls plus two Ws."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "And if you have one W and you add it to another W, that's going to give you two Ws. So that's two Ws. And then if you have one L and then you have another L, that's going to give you, if you add them together, that's going to give you two Ls. So the perimeter is going to be two Ls plus two Ws. They just wrote it in a different order than the way I wrote it. But the same thing. So hopefully that makes sense."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "So the perimeter is going to be two Ls plus two Ws. They just wrote it in a different order than the way I wrote it. But the same thing. So hopefully that makes sense. Now their question is, rewrite the formula so that it solves for width. So the formula, the way it's written now, it says P is equal to something. They want us to write it, so it's this W right here, they want it to be W is equal to a bunch of stuff with Ls and Ps in it and maybe some numbers there."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "So hopefully that makes sense. Now their question is, rewrite the formula so that it solves for width. So the formula, the way it's written now, it says P is equal to something. They want us to write it, so it's this W right here, they want it to be W is equal to a bunch of stuff with Ls and Ps in it and maybe some numbers there. So let's think about how we can do this. So they tell us that P is equal to 2 times L plus 2 times W. We want to solve for W. Well, a good starting point might be to get rid of the L on this side of the equation. And to get rid of it on that side of the equation, we could subtract the 2L from both sides of the equation."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "They want us to write it, so it's this W right here, they want it to be W is equal to a bunch of stuff with Ls and Ps in it and maybe some numbers there. So let's think about how we can do this. So they tell us that P is equal to 2 times L plus 2 times W. We want to solve for W. Well, a good starting point might be to get rid of the L on this side of the equation. And to get rid of it on that side of the equation, we could subtract the 2L from both sides of the equation. So let's do it this way. So you subtract 2L over here, minus 2L. You're also going to have to do that on the left-hand side."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "And to get rid of it on that side of the equation, we could subtract the 2L from both sides of the equation. So let's do it this way. So you subtract 2L over here, minus 2L. You're also going to have to do that on the left-hand side. So you're going to have minus 2L. We're doing it on both sides of the equation. Remember, an equation says P is equal to that."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "You're also going to have to do that on the left-hand side. So you're going to have minus 2L. We're doing it on both sides of the equation. Remember, an equation says P is equal to that. So if you do anything to that, you have to do it to P. So if you subtract 2L from this, you're going to have to subtract 2L from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2L. And then that is going to be equal to 2L minus 2L."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "Remember, an equation says P is equal to that. So if you do anything to that, you have to do it to P. So if you subtract 2L from this, you're going to have to subtract 2L from P in order for the equality to keep being true. So the left-hand side is going to be P minus 2L. And then that is going to be equal to 2L minus 2L. The whole reason why we subtract the 2L is because these are going to cancel out. So these cancel out, and you're just left with a 2W here. We're almost there."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "And then that is going to be equal to 2L minus 2L. The whole reason why we subtract the 2L is because these are going to cancel out. So these cancel out, and you're just left with a 2W here. We're almost there. We've almost solved for W. To finish it up, we just have to divide both sides of this equation by 2. And the whole reason why I'm dividing both sides of this equation by 2 is to get rid of this 2 coefficient. This 2 that's multiplying W. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "We're almost there. We've almost solved for W. To finish it up, we just have to divide both sides of this equation by 2. And the whole reason why I'm dividing both sides of this equation by 2 is to get rid of this 2 coefficient. This 2 that's multiplying W. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything. So this is just going to be a W. And then we have our left-hand side. So we're done."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "This 2 that's multiplying W. So if you divide both sides of this equation by 2, once again, if you do something to one side of the equation, you do it to the other side. The whole reason why I divided the right-hand side by 2 is 2 times anything divided by 2 is just going to be that anything. So this is just going to be a W. And then we have our left-hand side. So we're done. If we flip these two sides, we have our W will be equal to this thing over here. Equals P minus 2L all of that over 2. Now, this is the correct answer."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we're done. If we flip these two sides, we have our W will be equal to this thing over here. Equals P minus 2L all of that over 2. Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this. So let me square this off because this is completely the correct answer."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now, this is the correct answer. There's other ways to write it, though. You might want to rewrite this. So let me square this off because this is completely the correct answer. This is the correct answer. But there's other ways that you might have been able to get this answer or other expressions for this answer. You might have also another completely legitimate way to do this problem."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "So let me square this off because this is completely the correct answer. This is the correct answer. But there's other ways that you might have been able to get this answer or other expressions for this answer. You might have also another completely legitimate way to do this problem. Let me write it this way. So our original problem is P is equal to 2L plus 2W. Is on this right-hand side."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "You might have also another completely legitimate way to do this problem. Let me write it this way. So our original problem is P is equal to 2L plus 2W. Is on this right-hand side. What if we factor out a 2? So let me make this clear. You have a 2 here and you have a 2 here."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "Is on this right-hand side. What if we factor out a 2? So let me make this clear. You have a 2 here and you have a 2 here. So you can imagine undistributing the 2. So we would get P is equal to 2 times L plus W. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2 so that we get rid of this 2 on the right-hand side."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "You have a 2 here and you have a 2 here. So you can imagine undistributing the 2. So we would get P is equal to 2 times L plus W. This is an equally legitimate way to do this problem. Now, we can divide both sides of this equation by 2 so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out. 2 times anything divided by 2 is just going to be the anything. It's equal to P over 2."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "Now, we can divide both sides of this equation by 2 so that we get rid of this 2 on the right-hand side. So if you divide both sides of this equation by 2, these 2's are going to cancel out. 2 times anything divided by 2 is just going to be the anything. It's equal to P over 2. So let me just rewrite this over here. Let me just rewrite this. So we will get P over 2 is going to be equal to L plus W. And then if we want to solve for W, we just subtract L from both sides."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "It's equal to P over 2. So let me just rewrite this over here. Let me just rewrite this. So we will get P over 2 is going to be equal to L plus W. And then if we want to solve for W, we just subtract L from both sides. And sometimes you can write it in a separate line like this. Sometimes you can just write it like this. You can say, I'm going to subtract an L on that side."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we will get P over 2 is going to be equal to L plus W. And then if we want to solve for W, we just subtract L from both sides. And sometimes you can write it in a separate line like this. Sometimes you can just write it like this. You can say, I'm going to subtract an L on that side. If I do it on that side, I have to do it on this side too. That's the same thing as adding a negative L. And so the right-hand side, you're just left with a W. And then the left-hand side, it could be a negative L plus P over 2, or you could just change the order. And you can write this as P over 2 minus L. And this is also an equally legitimate answer."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "You can say, I'm going to subtract an L on that side. If I do it on that side, I have to do it on this side too. That's the same thing as adding a negative L. And so the right-hand side, you're just left with a W. And then the left-hand side, it could be a negative L plus P over 2, or you could just change the order. And you can write this as P over 2 minus L. And this is also an equally legitimate answer. And you're probably saying, hey Sal, wait, these things look different. P minus 2L over 2, that looks different than P over 2 minus L. And they're not. Think about this."}, {"video_title": "Example Solving for a variable Linear equations Algebra I Khan Academy.mp3", "Sentence": "And you can write this as P over 2 minus L. And this is also an equally legitimate answer. And you're probably saying, hey Sal, wait, these things look different. P minus 2L over 2, that looks different than P over 2 minus L. And they're not. Think about this. We could rewrite this as P over 2 minus 2L over 2. If I have A minus B and they're both being divided by 2, I can just separate. You can imagine I'm distributing the division by 2 right over here."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "The two things I'm going to graph are y is equal to 2 to the x power and y is equal to the log base 2 of x. I encourage you to pause the video, make a table for each of them and try to graph them on the same graph paper and see how they are related. If you see how they are related, think about why they are related that way. Let's first start with y equals 2 to the x power. I'll make a little table here, different x values and the corresponding y values. x and y, we can start with negative 2, negative 1, 0, 1, 2, 3. In each case, y is going to be 2 raised to this power. 2 to the negative 2 power is going to be 1 fourth."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "I'll make a little table here, different x values and the corresponding y values. x and y, we can start with negative 2, negative 1, 0, 1, 2, 3. In each case, y is going to be 2 raised to this power. 2 to the negative 2 power is going to be 1 fourth. 2 to the negative 1 power is 1 half. 2 to the 0 power is 1. 2 to the first power is 2."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the negative 2 power is going to be 1 fourth. 2 to the negative 1 power is 1 half. 2 to the 0 power is 1. 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. Let's graph that."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. Let's graph that. 2 to the third power is 8. 2 to the third power is 8. 2 to the second power is 4."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Let's graph that. 2 to the third power is 8. 2 to the third power is 8. 2 to the second power is 4. 2 to the first power is 2. 2 to the zeroth power is 1. 2 to the negative 1 power is 1 half."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the second power is 4. 2 to the first power is 2. 2 to the zeroth power is 1. 2 to the negative 1 power is 1 half. 2 to the negative 2 power is 1 fourth. 2 to the negative third power is 1 eighth. It's going to look something like this."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the negative 1 power is 1 half. 2 to the negative 2 power is 1 fourth. 2 to the negative third power is 1 eighth. It's going to look something like this. The graph is going to look something like this right over here. It's kind of your classic, sometimes this will be called your exponential hockey stick. It kind of looks like a hockey stick."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "It's going to look something like this. The graph is going to look something like this right over here. It's kind of your classic, sometimes this will be called your exponential hockey stick. It kind of looks like a hockey stick. It just starts kind of slow and just shoots straight up. Notice, as we go to the left, as x becomes more and more negative, our value approaches 0 but never quite gets there. If we have 2 to the negative 1 millionth power, it's going to be a very, very small number, very, very close to 0."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "It kind of looks like a hockey stick. It just starts kind of slow and just shoots straight up. Notice, as we go to the left, as x becomes more and more negative, our value approaches 0 but never quite gets there. If we have 2 to the negative 1 millionth power, it's going to be a very, very small number, very, very close to 0. But it's not going to be quite 0. We're going to have a horizontal asymptote at y is equal to 0, or the x-axis is a horizontal asymptote. Fair enough."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "If we have 2 to the negative 1 millionth power, it's going to be a very, very small number, very, very close to 0. But it's not going to be quite 0. We're going to have a horizontal asymptote at y is equal to 0, or the x-axis is a horizontal asymptote. Fair enough. Now let's graph y is equal to log base 2 of x. Before I graph that, let's just think about another way of representing it. This literally says for any x, what power, what exponent y, if I raise 2 to that, would give me x?"}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Fair enough. Now let's graph y is equal to log base 2 of x. Before I graph that, let's just think about another way of representing it. This literally says for any x, what power, what exponent y, if I raise 2 to that, would give me x? This is an equivalent statement as saying 2 to the y power is equal to x. If you notice what we've done here, between these two things, you're essentially just switching the x's and the y's. Here it's 2 to the x power is equal to y."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "This literally says for any x, what power, what exponent y, if I raise 2 to that, would give me x? This is an equivalent statement as saying 2 to the y power is equal to x. If you notice what we've done here, between these two things, you're essentially just switching the x's and the y's. Here it's 2 to the x power is equal to y. Here it's 2 to the y power is equal to x. Really, this and this, you've swapped the x's and the y's. You've swapped the x's and the y's."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Here it's 2 to the x power is equal to y. Here it's 2 to the y power is equal to x. Really, this and this, you've swapped the x's and the y's. You've swapped the x's and the y's. What we will see is that we can essentially swap these two columns, x and y. Let me just do 1 fourth, 1 half, 1, 1, 2, 4, and 8. Here now we're saying if x is 1 fourth, what power do we have to raise 2 to to get to 1 fourth?"}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "You've swapped the x's and the y's. What we will see is that we can essentially swap these two columns, x and y. Let me just do 1 fourth, 1 half, 1, 1, 2, 4, and 8. Here now we're saying if x is 1 fourth, what power do we have to raise 2 to to get to 1 fourth? We have to raise it to the negative 2 power. 2 to the negative 1 power is equal to 1 half. 2 to the 0 power is equal to 1."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Here now we're saying if x is 1 fourth, what power do we have to raise 2 to to get to 1 fourth? We have to raise it to the negative 2 power. 2 to the negative 1 power is equal to 1 half. 2 to the 0 power is equal to 1. 2 to the first power is equal to 2. 2 to the second power is equal to 4. 2 to the third power is equal to 8."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the 0 power is equal to 1. 2 to the first power is equal to 2. 2 to the second power is equal to 4. 2 to the third power is equal to 8. Notice all we did is we essentially swapped these two columns. Let's graph this. When x is equal to 1 fourth, y is equal to negative 2."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "2 to the third power is equal to 8. Notice all we did is we essentially swapped these two columns. Let's graph this. When x is equal to 1 fourth, y is equal to negative 2. When x is 1 half, y is equal to negative 1. When x is 1, y is 0. When x is 2, y is 1."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "When x is equal to 1 fourth, y is equal to negative 2. When x is 1 half, y is equal to negative 1. When x is 1, y is 0. When x is 2, y is 1. When x is 4, y is 2. When x is 8, y is 3. It's going to look like this."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "When x is 2, y is 1. When x is 4, y is 2. When x is 8, y is 3. It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two?"}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "It's going to look like this. Notice, I think you might already be seeing a pattern right over here. These two graphs are essentially the reflections of each other. What would you have to reflect about to get these two? You'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think about it, if you swap the axis, you would get the other graph, which is essentially what we're doing. Notice it's symmetric about that line."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "What would you have to reflect about to get these two? You'd have to reflect about y is equal to x. If you swap the x's and the y's, another way to think about it, if you swap the axis, you would get the other graph, which is essentially what we're doing. Notice it's symmetric about that line. That's because these are essentially the inverse functions of each other. One way to think about it is we swapped the x's and y's. Just as x becomes more and more negative, you see y approaching 0."}, {"video_title": "Comparing exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Notice it's symmetric about that line. That's because these are essentially the inverse functions of each other. One way to think about it is we swapped the x's and y's. Just as x becomes more and more negative, you see y approaching 0. Here you see as y is becoming more and more negative, x is approaching 0. Or you could say as x approaches 0, y becomes more and more negative. The whole point of this is just to give you an appreciation for the relationship between an exponential function and a logarithmic function."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So this red curve is the graph of f of x, and this blue curve is the graph of g of x. And I want to try to express g of x in terms of f of x. And so let's see how they're related. So we pick any x, and we could start right here at the vertex of f of x. And we see that at least at that point, g of x is exactly one higher than that. So g of 2, I could write this down, g of 2 is equal to f of 2 plus 1. Let's see if that's true for any x."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So we pick any x, and we could start right here at the vertex of f of x. And we see that at least at that point, g of x is exactly one higher than that. So g of 2, I could write this down, g of 2 is equal to f of 2 plus 1. Let's see if that's true for any x. So then we can just sample over here. Well, let's see. f of 4 is right over here."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "Let's see if that's true for any x. So then we can just sample over here. Well, let's see. f of 4 is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is one more than that."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "f of 4 is right over here. g of 4 is one more than that. f of 6 is right here. g of 6 is one more than that. So it looks like if we pick any point over here, even though there's a little bit of an optical illusion, it looks like they get closer together. They do if you try to find the closest distance between the two. But if you look at vertical distance, you see that it stays a constant."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "g of 6 is one more than that. So it looks like if we pick any point over here, even though there's a little bit of an optical illusion, it looks like they get closer together. They do if you try to find the closest distance between the two. But if you look at vertical distance, you see that it stays a constant. It stays a constant 1. So we can actually generalize this. This is true for any x. g of x is equal to f of x plus 1."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "But if you look at vertical distance, you see that it stays a constant. It stays a constant 1. So we can actually generalize this. This is true for any x. g of x is equal to f of x plus 1. Let's do a few more examples of this. So right over here, here is f of x in red again. And here is g of x."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "This is true for any x. g of x is equal to f of x plus 1. Let's do a few more examples of this. So right over here, here is f of x in red again. And here is g of x. And so let's say we picked x equals negative 4. This is f of negative 4. And we see g of negative 4 is 2 less than that."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "And here is g of x. And so let's say we picked x equals negative 4. This is f of negative 4. And we see g of negative 4 is 2 less than that. And we see whatever f of x is, g of x, no matter what x we pick, g of x seems to be exactly 2 less. g of x is exactly 2 less. So in this case, very similar to the other one, g of x is going to be equal to f of x."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "And we see g of negative 4 is 2 less than that. And we see whatever f of x is, g of x, no matter what x we pick, g of x seems to be exactly 2 less. g of x is exactly 2 less. So in this case, very similar to the other one, g of x is going to be equal to f of x. But instead of adding, we're going to subtract 2 from f of x. f of x minus 2. Let's do a few more examples. So here we have f of x in red again."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So in this case, very similar to the other one, g of x is going to be equal to f of x. But instead of adding, we're going to subtract 2 from f of x. f of x minus 2. Let's do a few more examples. So here we have f of x in red again. I'll relabel it. f of x. And here is g of x."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So here we have f of x in red again. I'll relabel it. f of x. And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of 3."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "And here is g of x. So let's think about it a little bit. Let's pick an arbitrary point here. Let's say we have in red here, this point right over there is the value of f of 3. So that, or f of negative 3, I should say. This is negative 3. This is the point negative 3, f of 3."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "Let's say we have in red here, this point right over there is the value of f of 3. So that, or f of negative 3, I should say. This is negative 3. This is the point negative 3, f of 3. So negative 3, f of 3. Now g hits that same value when x is equal to negative 1. When x is equal to negative 1."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "This is the point negative 3, f of 3. So negative 3, f of 3. Now g hits that same value when x is equal to negative 1. When x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. f of negative 3 is equal to f of negative 3. And we could do that with a bunch of points."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "When x is equal to negative 1. So let's think about this. g of negative 1 is equal to f of negative 3. f of negative 3 is equal to f of negative 3. And we could do that with a bunch of points. We could see that g of 0, which is right there. Let me do it in a color you can see. g of 0 is equivalent to f of negative 2."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "And we could do that with a bunch of points. We could see that g of 0, which is right there. Let me do it in a color you can see. g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "g of 0 is equivalent to f of negative 2. So let me write that down. g of 0 is equal to f of negative 2. We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "We could keep doing that. We could say g of 1, which is right over here. This is 1. g of 1 is equal to f of negative 1. g of 1 is equal to f of negative 1. So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of, well, it's going to be 2 less than x. So f of x minus 2."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So I think you see the pattern here. g of whatever is equal to the function evaluated at 2 less than whatever is here. So we could say that g of x is equal to f of, well, it's going to be 2 less than x. So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. It's important to realize here, when I did f of x minus 2 here, and remember, the function is being evaluated."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So f of x minus 2. So this is the relationship. g of x is equal to f of x minus 2. It's important to realize here, when I did f of x minus 2 here, and remember, the function is being evaluated. This is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counterintuitive unless you go through this exercise right over here."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "It's important to realize here, when I did f of x minus 2 here, and remember, the function is being evaluated. This is the input. x minus 2 is the input. When I subtract the 2, this is shifting the function to the right, which is a little bit counterintuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2, we would have actually shifted f to the left. Now let's think about this one."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "When I subtract the 2, this is shifting the function to the right, which is a little bit counterintuitive unless you go through this exercise right over here. So g of x is equal to f of x minus 2. If it was f of x plus 2, we would have actually shifted f to the left. Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image, but it looks like it's been flattened out. So let's think of it this way."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "Now let's think about this one. This one seems kind of wacky. So first of all, g of x, it almost looks like a mirror image, but it looks like it's been flattened out. So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So let's think of it this way. Let's take the mirror image of what g of x is. So I'm going to try my best to take the mirror image of it. So let's see. It gets to about 2 there. Then it gets pretty close to 1 right over there. And then it gets about right over there."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "So let's see. It gets to about 2 there. Then it gets pretty close to 1 right over there. And then it gets about right over there. So if I were to take its mirror image, it looks something like this. Its mirror image, if I were to reflect it across the x-axis, it looks something like this. It looks something like this."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "And then it gets about right over there. So if I were to take its mirror image, it looks something like this. Its mirror image, if I were to reflect it across the x-axis, it looks something like this. It looks something like this. So this right over here, we would call. So if this is g of x, when we flip it that way, this is the negative g of x. When x equals 4, g of x looks like it's, I don't know, about negative 3 and 1 half."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "It looks something like this. So this right over here, we would call. So if this is g of x, when we flip it that way, this is the negative g of x. When x equals 4, g of x looks like it's, I don't know, about negative 3 and 1 half. You take the negative of that, you get positive. I guess it should be closer. You get positive 3 and 1 half if you were to take the exact mirror image."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "When x equals 4, g of x looks like it's, I don't know, about negative 3 and 1 half. You take the negative of that, you get positive. I guess it should be closer. You get positive 3 and 1 half if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us there. It looks like we actually have to triple this value for any point."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "You get positive 3 and 1 half if you were to take the exact mirror image. So that's negative g of x. But that still doesn't get us there. It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3."}, {"video_title": "Shifting & reflecting functions Algebra II High School Math Khan Academy.mp3", "Sentence": "It looks like we actually have to triple this value for any point. And you see it here. This gets to 2, but we need to get to 6. This gets to 1, but we need to get to 3. So it looks like this red graph right over here is 3 times this graph. So this is 3 times negative g of x, which is equal to negative 3 g of x. So here we have f of x is equal to negative 3 times g of x."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's learn a little bit about the wonderful world of logarithms. So, we already know how to take exponents. If I were to say 2 to the 4th power, what does that mean? Well, that means 2 times 2 times 2 times 2. 2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, times 2 is 16. But what if we think about things in another way?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that means 2 times 2 times 2 times 2. 2 multiplied, or repeatedly multiplied, 4 times. And so this is going to be 2 times 2 is 4, times 2 is 8, times 2 is 16. But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is. So, for example, let's say that I start with 2, and I say I'm raising it to some power. What does that power have to be to get 16?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "But what if we think about things in another way? What if we're essentially, we know that we get to 16 when we raise 2 to some power, and we want to know what that power is. So, for example, let's say that I start with 2, and I say I'm raising it to some power. What does that power have to be to get 16? Well, we just figure that out. X would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "What does that power have to be to get 16? Well, we just figure that out. X would have to be 4. And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful. Log base 2, so I'll do this 2 in blue, log base 2 of 16 is equal to what? Or is equal in this case, since we have the x there, is equal to x."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this is what logarithms are fundamentally about, figuring out what power you have to raise to to get another number. Now, the way that we would denote this with logarithm notation is we would say log base, actually let me make it a little bit more colorful. Log base 2, so I'll do this 2 in blue, log base 2 of 16 is equal to what? Or is equal in this case, since we have the x there, is equal to x. This and this are completely equivalent statements. This is saying, hey, well, if I take 2 to some x power, I get 16. This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or is equal in this case, since we have the x there, is equal to x. This and this are completely equivalent statements. This is saying, hey, well, if I take 2 to some x power, I get 16. This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you've got to raise it to the fourth power. Once again, x is equal to 4. So with that out of the way, let's try more examples of evaluating logarithmic expressions."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is saying, what power do I need to raise 2 to to get 16, and I'm going to set that to be equal to x. And you would say, well, you've got to raise it to the fourth power. Once again, x is equal to 4. So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log base 3 of 81. What would this evaluate to? Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So with that out of the way, let's try more examples of evaluating logarithmic expressions. So let's say you had log base 3 of 81. What would this evaluate to? Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting properties later on."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, just as a reminder, this evaluates to the power we have to raise 3 to to get to 81. So if you want to, you could set this to be equal to an x, set that to be equal to an x, and you can restate this equation as 3 to the x power is equal to 81. Why is a logarithm useful? And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81. You had to use algebra here. Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "And you'll see that it has very interesting properties later on. But you didn't necessarily have to use algebra to do it this way, to say that the x is the power that you raise 3 to to get to 81. You had to use algebra here. Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra. We didn't have to set it equal to x. We could just say this evaluates to the power I need to raise 3 to to get to 81. Well, what power do you have to raise 3 to to get to 81?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, with just a straight-up logarithmic expression, you didn't really have to use any algebra. We didn't have to set it equal to x. We could just say this evaluates to the power I need to raise 3 to to get to 81. Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit. So 3 to the first power is just 3. 3 to the second power is 9."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, what power do you have to raise 3 to to get to 81? Well, let's experiment a little bit. So 3 to the first power is just 3. 3 to the second power is 9. 3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 to the second power is 9. 3 to the third power is 27. 3 to the fourth power, 27 times 3, is equal to 81. 3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 to the fourth power is equal to 81. x is equal to 4. So we could say log base 3 of 81 is equal to 4. Let's do several more of these examples. And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number. Let's say we want to take log base 6 of 216. What will this evaluate to?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I really encourage you to give a shot on your own, and you'll hopefully get the hang of it. So let's try a little larger number. Let's say we want to take log base 6 of 216. What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to the second power is 36."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "What will this evaluate to? Well, we're asking ourselves, what power do we have to raise 6 to to get to 216? 6 to the first power is 6. 6 to the second power is 36. 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "6 to the second power is 36. 36 times 6 is 216. This is equal to 216. So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is 6 to the third power is equal to 216. So if someone says, what power do I have to raise 6 to, this base here, to get to 216? Well, that's just going to be equal to 3. 6 to the third power is equal to 216. Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "6 to the third power is equal to 216. Let's try another one. Let's say I had log base 2 of 64. So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to raise 2 to to get to 64."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "So what does this evaluate to? Well, once again, we're asking ourselves, or this will evaluate to the exponent that I have to raise this base to. And you do this as a little subscript right here. The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "The exponent that I have to raise 2 to to get to 64. So 2 to the first power is 2. 2 to the second power is 4. 8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the sixth power."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "8, 16, 32, 64. So this right over here is 2 to the sixth power is equal to 64. So when you evaluate this expression, you say, what power do I have to raise 2 to to get to 64? Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending on how you view it. What is log base 100 of 1?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, I have to raise it to the sixth power. Let's do a slightly more straightforward one. Or maybe this will be less straightforward, depending on how you view it. What is log base 100 of 1? Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about it."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "What is log base 100 of 1? Let me think about that for a second. So the 100 is a subscript, and then it's log base 100 of 1. That's one way to think about it. I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1?"}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's one way to think about it. I could put a parentheses around the 1. What does this evaluate to? Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation. So if I set this to be equal to x, this is literally saying 100 to what power is equal to 1? Well, anything to the zeroth power is equal to 1."}, {"video_title": "Logarithms Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, this is asking ourselves, or we would evaluate this as, what power do I have to raise 102 to get to 1? So let me write this down as an equation. So if I set this to be equal to x, this is literally saying 100 to what power is equal to 1? Well, anything to the zeroth power is equal to 1. So in this case, x is equal to 0. So log base 100 of 1 is going to be equal to 0. Log base anything of 1 is going to be equal to 0, because anything to the zeroth power, and we're not talking about 0 here, anything to the zeroth power that's not 0 is going to be equal to 1."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "All right, complete the following sentence about the daily percent change in the mass of the sunfish. Every day there is a blank percent addition or removal from the mass of the sunfish. So one thing that we know almost from the get-go, we know that the sunfish gains weight. And we also see that as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one, m of t is going to grow. So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "And we also see that as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one, m of t is going to grow. So I already know it's going to be about addition to the mass of the sunfish. But let's think about how much is added every day. Well, let's think about it. Well, let's see if we can rewrite this. This is, I'm gonna just focus on the right-hand side of this expression."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "But let's think about how much is added every day. Well, let's think about it. Well, let's see if we can rewrite this. This is, I'm gonna just focus on the right-hand side of this expression. So 1.35 to the t over six plus five, that's the same thing as 1.35 to the fifth power times 1.35 to the t over sixth power. And that's going to be equal to 1.35 to the fifth power times 1.35, and I can separate this t over six as 1 sixth times t. So 1.35 to the 1 sixth power, and then that being raised to the t power. So let's think about it."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "This is, I'm gonna just focus on the right-hand side of this expression. So 1.35 to the t over six plus five, that's the same thing as 1.35 to the fifth power times 1.35 to the t over sixth power. And that's going to be equal to 1.35 to the fifth power times 1.35, and I can separate this t over six as 1 sixth times t. So 1.35 to the 1 sixth power, and then that being raised to the t power. So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass and multiply it by this common ratio. The common ratio here isn't, the way I've written it, isn't 1.35. It's 1.35 to the 1 sixth power."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass and multiply it by this common ratio. The common ratio here isn't, the way I've written it, isn't 1.35. It's 1.35 to the 1 sixth power. Let me draw a little table here to make that really, really clear. And all of that algebraic manipulation I just did is just so I could simplify this so I have some common ratio to the t power. So t and m of t. So based on how I've just written it, when t is zero, well, if t is zero, this is one, so then we just have our initial mass is going to be 1.35 to the fifth power."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "It's 1.35 to the 1 sixth power. Let me draw a little table here to make that really, really clear. And all of that algebraic manipulation I just did is just so I could simplify this so I have some common ratio to the t power. So t and m of t. So based on how I've just written it, when t is zero, well, if t is zero, this is one, so then we just have our initial mass is going to be 1.35 to the fifth power. And then when t is equal to one, when t is equal to one, it's gonna be our initial mass, 1.35 to the fifth power times our common ratio, times 1.35 to the 1 sixth power. When t equals two, we're just gonna multiply what we had at t equals one, and we're just going to multiply that times 1.35 to the 1 sixth again. And so every day, well, let me get, every day we are growing, every day we are growing by our common ratio, 1.35 to the 1 sixth power."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "So t and m of t. So based on how I've just written it, when t is zero, well, if t is zero, this is one, so then we just have our initial mass is going to be 1.35 to the fifth power. And then when t is equal to one, when t is equal to one, it's gonna be our initial mass, 1.35 to the fifth power times our common ratio, times 1.35 to the 1 sixth power. When t equals two, we're just gonna multiply what we had at t equals one, and we're just going to multiply that times 1.35 to the 1 sixth again. And so every day, well, let me get, every day we are growing, every day we are growing by our common ratio, 1.35 to the 1 sixth power. Actually, let me get a calculator out. We're allowed to use calculators in this exercise. So 1.35 to the, to the, open parentheses, one divided by six, close parentheses, power is equal to 1.05, I'll say 1.051 approximately."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "And so every day, well, let me get, every day we are growing, every day we are growing by our common ratio, 1.35 to the 1 sixth power. Actually, let me get a calculator out. We're allowed to use calculators in this exercise. So 1.35 to the, to the, open parentheses, one divided by six, close parentheses, power is equal to 1.05, I'll say 1.051 approximately. So this is approximately 1.051. So we could say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing by a factor of 1.051."}, {"video_title": "Interpreting change in exponential models with manipulation High School Math Khan Academy.mp3", "Sentence": "So 1.35 to the, to the, open parentheses, one divided by six, close parentheses, power is equal to 1.05, I'll say 1.051 approximately. So this is approximately 1.051. So we could say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing by a factor of 1.051. Well, growing by a factor of 1.051 means that you're adding a little bit more than 5%. You're adding 0.51 every day of your mass. So that's, you're adding 5.1%."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "She observed that the population loses 5.6% of its size every 2.8 months. The population of narwhals can be modeled by a function n, which depends on the amount of time t in months. When Chepi began the study, she observed that there were 89,000 narwhals in the Arctic Ocean. Write a function that models the population of narwhals t months since the beginning of Chepi's study. Like always, pause the video and see if you can do it on your own before we work through it together. So now let's work through it together. And to get a sense of what this function needs to do, it's always valuable to create a table for some interesting inputs for the function and seeing how the function should behave."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "Write a function that models the population of narwhals t months since the beginning of Chepi's study. Like always, pause the video and see if you can do it on your own before we work through it together. So now let's work through it together. And to get a sense of what this function needs to do, it's always valuable to create a table for some interesting inputs for the function and seeing how the function should behave. So first of all, if t is in months and n of t is, and n is the, that models, n is the number of narwhals, or narwhales. So when t is equal to zero, what is n of zero? Well, we know at t equals zero, there are 89,000 narwhals in the ocean."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "And to get a sense of what this function needs to do, it's always valuable to create a table for some interesting inputs for the function and seeing how the function should behave. So first of all, if t is in months and n of t is, and n is the, that models, n is the number of narwhals, or narwhales. So when t is equal to zero, what is n of zero? Well, we know at t equals zero, there are 89,000 narwhals in the ocean. So 89,000. And now what's another interesting one? Well, t is in months, and we know that the population decreases 5.6% every 2.8 months."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we know at t equals zero, there are 89,000 narwhals in the ocean. So 89,000. And now what's another interesting one? Well, t is in months, and we know that the population decreases 5.6% every 2.8 months. So let's think about when t is 2.8, 2.8 months. Well, then the population should have gone down 5.6%. So going down 5.6% is the same thing as retaining what?"}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, t is in months, and we know that the population decreases 5.6% every 2.8 months. So let's think about when t is 2.8, 2.8 months. Well, then the population should have gone down 5.6%. So going down 5.6% is the same thing as retaining what? What's one minus 5.6%? Retaining 94.4%. Let me be clear, 100%."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "So going down 5.6% is the same thing as retaining what? What's one minus 5.6%? Retaining 94.4%. Let me be clear, 100%. If you lose 5.6%, you are going to be left with 94.4%. The.6 plus.4 gets you to 95, plus another five gets you 100. So another way of saying this sentence, that the population loses 5.6% of its size every 2.8 months, is to say that the population is 94.4% of its size every 2.8 months, or shrinks to 94.4% of its original size every, or let me phrase this clearly."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let me be clear, 100%. If you lose 5.6%, you are going to be left with 94.4%. The.6 plus.4 gets you to 95, plus another five gets you 100. So another way of saying this sentence, that the population loses 5.6% of its size every 2.8 months, is to say that the population is 94.4% of its size every 2.8 months, or shrinks to 94.4% of its original size every, or let me phrase this clearly. After every 2.8 months, the population, you could either say it shrinks 5.6%, or you could say it's gone from, it's 94.4% of the population at the beginning of those 2.8 months. So after 2.8 months, the population should be 89,000 times, I could write times 94.4%, or I could write times 0.944. Now if we go another 2.8 months, so two times 2.8, I obviously could just write that as, I could write that as 5.6 months, but let me just write this as 2.8 months."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "So another way of saying this sentence, that the population loses 5.6% of its size every 2.8 months, is to say that the population is 94.4% of its size every 2.8 months, or shrinks to 94.4% of its original size every, or let me phrase this clearly. After every 2.8 months, the population, you could either say it shrinks 5.6%, or you could say it's gone from, it's 94.4% of the population at the beginning of those 2.8 months. So after 2.8 months, the population should be 89,000 times, I could write times 94.4%, or I could write times 0.944. Now if we go another 2.8 months, so two times 2.8, I obviously could just write that as, I could write that as 5.6 months, but let me just write this as 2.8 months. Where are we gonna be? We're gonna be at 89,000 times 0.944, this is where we were before at the beginning of this period, and we're gonna be 94.4% of that. So we're gonna multiply by 94.4% again, or 0.944 again, or we can just say times 0.944 squared."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now if we go another 2.8 months, so two times 2.8, I obviously could just write that as, I could write that as 5.6 months, but let me just write this as 2.8 months. Where are we gonna be? We're gonna be at 89,000 times 0.944, this is where we were before at the beginning of this period, and we're gonna be 94.4% of that. So we're gonna multiply by 94.4% again, or 0.944 again, or we can just say times 0.944 squared. And after three of these periods, well, we're gonna be times 0.944 again, so it's gonna be 89,000 times 0.944 squared times 0.944, which is gonna be 0.944 to the third power. And I think you might see what's going on here. We have an exponential function."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're gonna multiply by 94.4% again, or 0.944 again, or we can just say times 0.944 squared. And after three of these periods, well, we're gonna be times 0.944 again, so it's gonna be 89,000 times 0.944 squared times 0.944, which is gonna be 0.944 to the third power. And I think you might see what's going on here. We have an exponential function. Between every 2.8 months, we are multiplying by this common ratio of 0.944. And so we can write our function, N of t, our initial value is 89,000, 89,000 times 0.944 to the power of however many of these 2.8 month periods we've gone so far. So if we take the number of months, and we divide by 2.8, that's how many 2.8 month periods we have gone."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have an exponential function. Between every 2.8 months, we are multiplying by this common ratio of 0.944. And so we can write our function, N of t, our initial value is 89,000, 89,000 times 0.944 to the power of however many of these 2.8 month periods we've gone so far. So if we take the number of months, and we divide by 2.8, that's how many 2.8 month periods we have gone. And so notice, when t equals zero, all of this turns into one. You raise something to the zero power, it just becomes one, you have 89,000. When t is equal to 2.8, this exponent is one, and we're gonna multiply by 0.944 once."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if we take the number of months, and we divide by 2.8, that's how many 2.8 month periods we have gone. And so notice, when t equals zero, all of this turns into one. You raise something to the zero power, it just becomes one, you have 89,000. When t is equal to 2.8, this exponent is one, and we're gonna multiply by 0.944 once. When t is 5.6, the exponent is going to be two, and we're gonna multiply by 0.944 twice. And I'm just doing the values that make the exponent integers, but it's going to work for the ones in between. I encourage you to graph it, or to try those values on a calculator if you like."}, {"video_title": "Constructing exponential models percent change Mathematics II High School Math Khan Academy.mp3", "Sentence": "When t is equal to 2.8, this exponent is one, and we're gonna multiply by 0.944 once. When t is 5.6, the exponent is going to be two, and we're gonna multiply by 0.944 twice. And I'm just doing the values that make the exponent integers, but it's going to work for the ones in between. I encourage you to graph it, or to try those values on a calculator if you like. But there you have it, we're done. We have modeled our narwhals. So let me just underline that, we're done."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if we want to write the same information, really, in logarithmic form, we could say that the power, the power that I need to raise 10 to, to get to 100 is equal to two, or log base 10, log base 10 of 100 is equal to two. Notice, these are equivalent statements. This is just an exponential form. This is in logarithmic form. This is saying the power that I need to raise 10 to, to get to 100, is equal to two, which is the same thing as saying that 10 to the second power is 100. 10 to the second power is 100. And the way that I specified the base is by doing this underscore right over here."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is in logarithmic form. This is saying the power that I need to raise 10 to, to get to 100, is equal to two, which is the same thing as saying that 10 to the second power is 100. 10 to the second power is 100. And the way that I specified the base is by doing this underscore right over here. So underscore 10, log base 10 of 100 is equal to two. Now they, here they ask us to rewrite the following equation in exponential form. So this is log base five of one over 125 is equal to negative three."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the way that I specified the base is by doing this underscore right over here. So underscore 10, log base 10 of 100 is equal to two. Now they, here they ask us to rewrite the following equation in exponential form. So this is log base five of one over 125 is equal to negative three. This is, one way to think about it is saying the power that I need to raise five to, to get to one over 125 is equal to negative three, or that five to the negative three power, five to the negative three power is equal to one over 125. And we can verify that this has formatted it the right way. Five to the negative three power is one over 125."}, {"video_title": "Writing in logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base five of one over 125 is equal to negative three. This is, one way to think about it is saying the power that I need to raise five to, to get to one over 125 is equal to negative three, or that five to the negative three power, five to the negative three power is equal to one over 125. And we can verify that this has formatted it the right way. Five to the negative three power is one over 125. The exact same truth about the universe, just in different forms, logarithmic form and exponential form. So let me check my answer, make sure I got it right. And I did."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And what we're going to start off doing is just graph a plain vanilla function, f of x is equal to x squared. That looks as we would expect it to look. But now let's think about how we could shift it up or down. Well, one thought is, well, to shift it up, we just have to make the value of f of x higher so we could add a value. And that does look like it shifted it up by one. Whatever f of x was before, we're now adding one to it, so it shifts the graph up by one. That's pretty intuitive."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, one thought is, well, to shift it up, we just have to make the value of f of x higher so we could add a value. And that does look like it shifted it up by one. Whatever f of x was before, we're now adding one to it, so it shifts the graph up by one. That's pretty intuitive. If we subtract one, or actually let's subtract three, notice it shifted it down. The vertex was right over here at zero, zero. Now it is at zero, negative three, so it shifted it down."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's pretty intuitive. If we subtract one, or actually let's subtract three, notice it shifted it down. The vertex was right over here at zero, zero. Now it is at zero, negative three, so it shifted it down. And we can set up a slider here to make that a little bit clearer. So if I just replace this with, if I just replace this with the variable k, then, let me delete this little thing here, that little subscript thing that happened. Then we can add a slider k here."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now it is at zero, negative three, so it shifted it down. And we can set up a slider here to make that a little bit clearer. So if I just replace this with, if I just replace this with the variable k, then, let me delete this little thing here, that little subscript thing that happened. Then we can add a slider k here. And this is just allowing us to set what k is equal to. So here k is equal to one. So this is x squared plus one."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Then we can add a slider k here. And this is just allowing us to set what k is equal to. So here k is equal to one. So this is x squared plus one. And notice, we have shifted up. And if we increase the value of k, notice how it shifts the graph up. And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is x squared plus one. And notice, we have shifted up. And if we increase the value of k, notice how it shifts the graph up. And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin. And then as we decrease the value of k, it shifts our graph down. And that's pretty intuitive, because we're adding or subtracting that amount to x squared so it changes, we could say, the y value. It shifts it up or down."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And as we decrease the value of k, if k is zero, we're back where our vertex is right at the origin. And then as we decrease the value of k, it shifts our graph down. And that's pretty intuitive, because we're adding or subtracting that amount to x squared so it changes, we could say, the y value. It shifts it up or down. But how do we shift to the left or to the right? So what's interesting here is, to shift to the left or to the right, we can replace our x with an x minus something. So let's see how that might work."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It shifts it up or down. But how do we shift to the left or to the right? So what's interesting here is, to shift to the left or to the right, we can replace our x with an x minus something. So let's see how that might work. So I'm going to replace our x with an x minus, let's replace it with an x minus one. What do you think's going to happen? Do you think that's going to shift it one to the right or one to the left?"}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's see how that might work. So I'm going to replace our x with an x minus, let's replace it with an x minus one. What do you think's going to happen? Do you think that's going to shift it one to the right or one to the left? So let's just put the one in. Well, that's interesting. Before, our vertex was at zero, zero."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Do you think that's going to shift it one to the right or one to the left? So let's just put the one in. Well, that's interesting. Before, our vertex was at zero, zero. Now, our vertex is at one, zero. So by replacing our x with an x minus one, we actually shifted one to the right. Now, why does that make sense?"}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Before, our vertex was at zero, zero. Now, our vertex is at one, zero. So by replacing our x with an x minus one, we actually shifted one to the right. Now, why does that make sense? Well, one way to think about it, before we put this x, before we replaced our x with an x minus one, the vertex was when we were squaring zero. Now, in order to square zero, squaring zero happens when x is equal to one. When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, why does that make sense? Well, one way to think about it, before we put this x, before we replaced our x with an x minus one, the vertex was when we were squaring zero. Now, in order to square zero, squaring zero happens when x is equal to one. When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero. So it makes sense that you have a similar behavior of the graph at the vertex now when x equals one as before you had when x equals zero. And to see how this can be generalized, let's put another variable here and let's add a slider for h, and then we can see that when h is zero and k is zero, our function is really then just x squared. And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to one, you do one minus one, you get zero, and then that's when you are squaring zero. So it makes sense that you have a similar behavior of the graph at the vertex now when x equals one as before you had when x equals zero. And to see how this can be generalized, let's put another variable here and let's add a slider for h, and then we can see that when h is zero and k is zero, our function is really then just x squared. And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left. Now, right here, h is equal to negative five. You typically won't see x minus negative five. You would see that written as x plus five."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then if h increases, we're replacing our x with x minus a larger value, that's shifting to the right, and then as h decreases, as it becomes negative, that shifts to the left. Now, right here, h is equal to negative five. You typically won't see x minus negative five. You would see that written as x plus five. So if you replace your x's with an x plus five, that actually shifts everything five units to the left. And of course, we can shift both of them together like this. So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You would see that written as x plus five. So if you replace your x's with an x plus five, that actually shifts everything five units to the left. And of course, we can shift both of them together like this. So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that. And everything we did just now is with the x squared function as our core function, but you could do it with all sorts of functions. You could do it with an absolute value function. Let's do absolute value."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So here, we're shifting it up, and then we can get back to our neutral horizontal shift, and then we can shift it to the right like that. And everything we did just now is with the x squared function as our core function, but you could do it with all sorts of functions. You could do it with an absolute value function. Let's do absolute value. That's always a fun one. So instead of squaring all this business, let's have an absolute value here. So I'm gonna put an absolute, whoops, absolute value, and there you have it."}, {"video_title": "Shifting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do absolute value. That's always a fun one. So instead of squaring all this business, let's have an absolute value here. So I'm gonna put an absolute, whoops, absolute value, and there you have it. You can start at, let me make both of these variables equal to zero. So that would just be the graph of f of x is equal to the absolute value of x. But let's say you wanted to shift it so that this point right over here that's at the origin is at the point negative five, negative five, which is right over there."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Fractional exponents can be a little daunting at first, so it never hurts to do as many examples as possible. So let's do a few. What if we had 25 over 9, and we wanted to raise it to the 1 half power? So we're essentially just saying, well, what is the principal square root of 25 over 9? So what number times itself is going to be 25 over 9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5 over 3?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So we're essentially just saying, well, what is the principal square root of 25 over 9? So what number times itself is going to be 25 over 9? Well, we know 5 times 5 is 25, and 3 times 3 is 9. So why don't we just go with 5 over 3? Because notice, if you have 5 over 3 times 5 over 3, that is going to be 25 over 9. Or another way of saying this, that 5 over 3 squared is equal to 25 over 9. So 25 over 9 to the 1 half is going to be equal to 5 thirds."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So why don't we just go with 5 over 3? Because notice, if you have 5 over 3 times 5 over 3, that is going to be 25 over 9. Or another way of saying this, that 5 over 3 squared is equal to 25 over 9. So 25 over 9 to the 1 half is going to be equal to 5 thirds. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81 over 256 to the negative 1 fourth power."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So 25 over 9 to the 1 half is going to be equal to 5 thirds. Now let's escalate things a little bit. Let's take a really hairy one. Let's raise 81 over 256 to the negative 1 fourth power. I encourage you to pause this and try this on your own. So what's going on here? This negative, the first thing I always like to do is I want to get rid of this negative in the exponent."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Let's raise 81 over 256 to the negative 1 fourth power. I encourage you to pause this and try this on your own. So what's going on here? This negative, the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256 over 81 to the 1 fourth power. And so now I can say, well, what number times itself is going to be equal to 256?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "This negative, the first thing I always like to do is I want to get rid of this negative in the exponent. So let me just take the reciprocal of this and raise it to the positive. So I could just say that this is equal to 256 over 81 to the 1 fourth power. And so now I can say, well, what number times itself is going to be equal to 256? And what number times itself times itself times itself? Did I say that four times? Well, what number, if I take four of them and multiply, do I get 81?"}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "And so now I can say, well, what number times itself is going to be equal to 256? And what number times itself times itself times itself? Did I say that four times? Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing. And we'll talk about this in more depth later on when we talk about exponent properties. But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Well, what number, if I take four of them and multiply, do I get 81? And one way to think about it, this is going to be the same thing. And we'll talk about this in more depth later on when we talk about exponent properties. But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9, or 25 to the 1 half over 9 to the 1 half. So we're just doing that over here."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "But this is going to be the exact same thing as 256 to the 1 fourth over 81 to the 1 fourth. You, in fact, saw it over here. This over here was the same thing as the square root of 25 over the square root of 9, or 25 to the 1 half over 9 to the 1 half. So we're just doing that over here. So we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "So we're just doing that over here. So we still have to think about what number this is. And this is a little bit of, there's no easy way to do this. You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "You kind of have to just play around a little bit to come up with it. But 4 might jump out at you if you recognize that 16 times 16 is 256. We know that 4 to the fourth power, or you're about to know this, is 4 times 4 times 4 times 4. And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256. Or we could say 4 is equal to 256 to the 1 fourth power. Fair enough."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "And 4 times 4 is 16, times 4 is 64, times 4 is equal to 256. So 4 to the fourth is 256. Or we could say 4 is equal to 256 to the 1 fourth power. Fair enough. Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81."}, {"video_title": "Evaluating fractional exponents fractional base Algebra I Khan Academy.mp3", "Sentence": "Fair enough. Now what about 81? Well, 3 might jump out at you. We know that 3 to the fourth power is equal to 3 times 3 times 3 times 3, which is equal to 81. So 3 is equal to 81 to the 1 fourth. So this top number, 256 to the 1 fourth, that's just 4. 81 to the 1 fourth, that is just 3."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "Two to the x squared minus three power is equal to one over the cube root of x. Pause this video and see if you can solve this. Well, you probably realize that this is not so easy to solve. The way that I would at least attempt to tackle it is you would say this is two to the x squared minus three is equal to x to the, I could rewrite this, this is one over x to the 1 3rd, so this is x to the negative 1 3rd power. Maybe I can simplify it by raising both sides to the negative three power. And so then I would get, if I raise something to an exponent and then raise that to an exponent, I can just multiply the exponents. So it would be two to the negative three x squared plus nine power, I just multiplied both of these terms times negative three, is equal to x to the negative 1 3rd to the negative three."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "The way that I would at least attempt to tackle it is you would say this is two to the x squared minus three is equal to x to the, I could rewrite this, this is one over x to the 1 3rd, so this is x to the negative 1 3rd power. Maybe I can simplify it by raising both sides to the negative three power. And so then I would get, if I raise something to an exponent and then raise that to an exponent, I can just multiply the exponents. So it would be two to the negative three x squared plus nine power, I just multiplied both of these terms times negative three, is equal to x to the negative 1 3rd to the negative three. Negative 1 3rd times negative three is just one, so that's just going to be equal to x. So it looks a little bit simpler, but still not so easy. I could try to take log base two of both sides and I'd get negative three x squared plus nine is equal to log base two of x."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "So it would be two to the negative three x squared plus nine power, I just multiplied both of these terms times negative three, is equal to x to the negative 1 3rd to the negative three. Negative 1 3rd times negative three is just one, so that's just going to be equal to x. So it looks a little bit simpler, but still not so easy. I could try to take log base two of both sides and I'd get negative three x squared plus nine is equal to log base two of x. But once again, not having an easy time to solve this. And the reason why I gave you this equation is to appreciate that some equations are not so easy to solve algebraically. But we have other tools."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "I could try to take log base two of both sides and I'd get negative three x squared plus nine is equal to log base two of x. But once again, not having an easy time to solve this. And the reason why I gave you this equation is to appreciate that some equations are not so easy to solve algebraically. But we have other tools. We have things like computers. We can graph things and they can at least get us really close to knowing what the solution is. And the way that we can do that is we could say, hey, well, what if I had one function or one equation that was y is equal to two x, two to the x squared minus three, I should say."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "But we have other tools. We have things like computers. We can graph things and they can at least get us really close to knowing what the solution is. And the way that we can do that is we could say, hey, well, what if I had one function or one equation that was y is equal to two x, two to the x squared minus three, I should say. And then you had another that was y is equal to one over the cube root of x. And then you could graph each of these and then you could see where they intersect because where they intersect, that means two to the x squared minus three is giving you the same y as one over the cube root of x. Or another way to think about it is they're going to intersect at an x value where these two expressions are equal to each other."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "And the way that we can do that is we could say, hey, well, what if I had one function or one equation that was y is equal to two x, two to the x squared minus three, I should say. And then you had another that was y is equal to one over the cube root of x. And then you could graph each of these and then you could see where they intersect because where they intersect, that means two to the x squared minus three is giving you the same y as one over the cube root of x. Or another way to think about it is they're going to intersect at an x value where these two expressions are equal to each other. And so what we could do is we could go to a graphing calculator or we could go to a site like Desmos and graph it and at least try to approximate what the point of intersection is. And so let's do that. So I graphed this ahead of time on Desmos."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "Or another way to think about it is they're going to intersect at an x value where these two expressions are equal to each other. And so what we could do is we could go to a graphing calculator or we could go to a site like Desmos and graph it and at least try to approximate what the point of intersection is. And so let's do that. So I graphed this ahead of time on Desmos. So you can see here, this is our two sides of our equation, but now we've expressed each of them as a function. Right here in blue, we have two, we have f of x, or I could even say this is y is equal to f of x, which is equal to two to the x squared minus three. And then in this yellowish color, I have y is equal to g of x, which is equal to one over the cube root of x."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "So I graphed this ahead of time on Desmos. So you can see here, this is our two sides of our equation, but now we've expressed each of them as a function. Right here in blue, we have two, we have f of x, or I could even say this is y is equal to f of x, which is equal to two to the x squared minus three. And then in this yellowish color, I have y is equal to g of x, which is equal to one over the cube root of x. And we can see where they intersect. They intersect right over there. And we're not going to get an exact answer, but even at this level of zoom, and on a tool like Desmos, you can keep zooming in in order to get a more and more precise answer."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "And then in this yellowish color, I have y is equal to g of x, which is equal to one over the cube root of x. And we can see where they intersect. They intersect right over there. And we're not going to get an exact answer, but even at this level of zoom, and on a tool like Desmos, you can keep zooming in in order to get a more and more precise answer. In fact, you can even scroll over this and it'll even tell you where they intersect. But even if we're trying to approximate just looking at the graph, we can see that the x value right over here, it looks like it is happening at around, let's see, this is 1.5, and each of these is a tenth, so this is 1.6, and then it looks like it's about 2 1\u20443 of the way to the next one. So this looks like it's about, I'll say this is approximately 1.66."}, {"video_title": "Solving equations by graphing Algebra 2 Khan academy.mp3", "Sentence": "And we're not going to get an exact answer, but even at this level of zoom, and on a tool like Desmos, you can keep zooming in in order to get a more and more precise answer. In fact, you can even scroll over this and it'll even tell you where they intersect. But even if we're trying to approximate just looking at the graph, we can see that the x value right over here, it looks like it is happening at around, let's see, this is 1.5, and each of these is a tenth, so this is 1.6, and then it looks like it's about 2 1\u20443 of the way to the next one. So this looks like it's about, I'll say this is approximately 1.66. And if you were to actually find the exact solution, you'd actually find this awfully close to 1.66. So the whole point here is, is that even when it's algebraically difficult to solve something, you could set up or restate your problem or reframe your problem in a way that makes it easier to solve. You can set this up as, hey, let's make two functions, and then let's graph them and see where they intersect, and the x value where they intersect, well, that would be a solution to that equation, and that's exactly what we did right there."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "But we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's have the equation the square root of x is equal to 2 times x minus 6. Now one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that. I have the square root of x isolated on the left-hand side, then we square both sides of the equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that. I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared. So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to this squared is going to be 2x squared, which is 4x squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "If that is equal to that, then that squared should also be equal to that squared. So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x, and it's going to be twice that. So minus 24x."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And we get x is equal to this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x, and it's going to be twice that. So minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually the special case where we square binomials. But the general view is this squared, which is that."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually the special case where we square binomials. But the general view is this squared, which is that. And then you have minus 2 times the product of these two. The product of those two is minus 12x, or negative 12x. 2 times that is negative 24x."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "But the general view is this squared, which is that. And then you have minus 2 times the product of these two. The product of those two is minus 12x, or negative 12x. 2 times that is negative 24x. And then that squared. So this is what our equation has, I guess we could say, simplified to. And let's see what happens if we subtract x from both sides of this equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "2 times that is negative 24x. And then that squared. So this is what our equation has, I guess we could say, simplified to. And let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes 0. And the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation has simplified to just a standard quadratic equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes 0. And the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation has simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and group it and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this x can be negative b, negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So this radical equation has simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and group it and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this x can be negative b, negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625 minus 4 times a, which is 4, times c, which is 36. All of that over 2 times 4. All of that over 8."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625 minus 4 times a, which is 4, times c, which is 36. All of that over 2 times 4. All of that over 8. So let's get our calculator out to figure out what this is over here. So let's say we have 625 minus, let's see, this is going to be 16 times 36. 16 times 36 is equal to 49."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "All of that over 8. So let's get our calculator out to figure out what this is over here. So let's say we have 625 minus, let's see, this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice. It's a nice, perfect square. We know what the square root of 49 is."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "16 times 36 is equal to 49. That's nice. It's a nice, perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7. All of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7. All of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32. 32 over 8, which is equal to 4. And then our other solution, let me do that in a different color, x is equal to 25 minus 7, which is 18 over 8. 8 goes into 18 two times, remainder 2."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32. 32 over 8, which is equal to 4. And then our other solution, let me do that in a different color, x is equal to 25 minus 7, which is 18 over 8. 8 goes into 18 two times, remainder 2. So this is equal to 2 and 2 eighths, or 2 and 1 fourth, or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "8 goes into 18 two times, remainder 2. So this is equal to 2 and 2 eighths, or 2 and 1 fourth, or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. So if x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. So if x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So if x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.25 should be equal to 2 times 2.25 minus 6."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out that the square root of 2.25 is 1.5."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "According to this, we should be able to take the square root, the principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out that the square root of 2.25 is 1.5. But let me just use a calculator to verify that for you. So 2.25, take the square root, it's 1.5. The principal root is 1.5."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And then from that, you might be able to figure out that the square root of 2.25 is 1.5. But let me just use a calculator to verify that for you. So 2.25, take the square root, it's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then according to this, this should be equal to 2 times 2.25 is 4.5 minus 6."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And then according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is extraneous."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "This is not true. 2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is extraneous. Now, here's the conundrum. Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So 2.25 is extraneous. Now, here's the conundrum. Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here that something we did gave us this solution that doesn't quite apply over here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here that something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So there's something here that something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here, you're going to see that it works."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here, you're going to see that it works. Put in 4 for x here, and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Put in 2.25 for x here, you're going to see that it works. Put in 4 for x here, and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation. And the answer is, there's two ways you could think about it. To go back from this equation to that equation, we take the square root."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation. And the answer is, there's two ways you could think about it. To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root. Going from this right here, let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here, only the valid one satisfies the original problem. So let me write the equation that both of them satisfy."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Notice, this is only taking the principal square root. Going from this right here, let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here, only the valid one satisfies the original problem. So let me write the equation that both of them satisfy. Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why, when you square both sides, you are, to some degree, you can either think of it as losing or gaining some information."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So let me write the equation that both of them satisfy. Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why, when you square both sides, you are, to some degree, you can either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "And why, when you square both sides, you are, to some degree, you can either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to, you multiply negative 1 times that, you get negative 2x plus 6, or 6 minus 2x squared. This and this are two ways of writing that."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to, you multiply negative 1 times that, you get negative 2x plus 6, or 6 minus 2x squared. This and this are two ways of writing that. Now, when we took our square root, or when we, I guess there's two ways you can think about it, when we squared it, we're assuming that this was the only interpretation. But this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "This and this are two ways of writing that. Now, when we took our square root, or when we, I guess there's two ways you can think about it, when we squared it, we're assuming that this was the only interpretation. But this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here. And I hope you get what I'm saying, because we're kind of only taking, you can kind of imagine it, the positive square root. We're not considering the negative square root of this. Because when you take the square root of both sides to get here, we're only taking the principal root."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So we found two solutions to this, but only 4 satisfies this interpretation right here. And I hope you get what I'm saying, because we're kind of only taking, you can kind of imagine it, the positive square root. We're not considering the negative square root of this. Because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it, let me rewrite the original equation. Let me rewrite the original equation. We had the square root of x is equal to 2x minus 6."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it, let me rewrite the original equation. Let me rewrite the original equation. We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would have been a solution if we said both of the square roots of x is equal to 2x minus 6."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would have been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out, and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25. So that is equal to 4.5 minus 6, which is negative 1.5."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "2.25 would have been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out, and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25. So that is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So that is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions. And if you square this, maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all that."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "So that's why we got two solutions. And if you square this, maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all that. My intention is not to confuse you. The simple thing to think about when you're solving radical equations is, look, isolate radicals, square, keep on solving, you might get more than one answer. Plug your answers back in."}, {"video_title": "Extraneous solutions to radical equations Algebra I Khan Academy.mp3", "Sentence": "Now, you might have found that a little bit confusing and all that. My intention is not to confuse you. The simple thing to think about when you're solving radical equations is, look, isolate radicals, square, keep on solving, you might get more than one answer. Plug your answers back in. Answers that don't work, they're extraneous solutions. But most of my explanation in this video is really why does that extraneous solution pop up? And hopefully I gave you some intuition."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's say we have the radical equation two x minus one is equal to the square root of eight minus x. So we already have the radical isolated on one side of the equation, so we might say, well, let's just get rid of the radical. Let's square both sides of this equation. So we might say that this is the same thing as two x minus one squared is equal to the square root of eight minus x, eight minus x squared. And then we would get, let's see, two x minus one squared is four x squared minus four x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel like, hey, we did legitimate operations."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we might say that this is the same thing as two x minus one squared is equal to the square root of eight minus x, eight minus x squared. And then we would get, let's see, two x minus one squared is four x squared minus four x plus one is equal to eight minus x. Now we have to be very, very, very careful here. We might feel like, hey, we did legitimate operations. We did the same thing to both sides, that these are equivalent equations, but they aren't quite equivalent because when you're squaring something, one way to think about it is when you're squaring it, you're losing information. So for example, this would be true even if the original equation were two x, let me do this in a different color, even if the original equation were two x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there because the negative squared would be equal to a positive."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "We might feel like, hey, we did legitimate operations. We did the same thing to both sides, that these are equivalent equations, but they aren't quite equivalent because when you're squaring something, one way to think about it is when you're squaring it, you're losing information. So for example, this would be true even if the original equation were two x, let me do this in a different color, even if the original equation were two x minus one is equal to the negative of the square root of eight minus x. Because if you squared both sides of this, you would also get, you would also get that right over there because the negative squared would be equal to a positive. So when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here. If it's a solution to this right-hand side one and not the yellow one, then we would call that an extraneous solution. So let's see if we can solve this."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "Because if you squared both sides of this, you would also get, you would also get that right over there because the negative squared would be equal to a positive. So when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here. If it's a solution to this right-hand side one and not the yellow one, then we would call that an extraneous solution. So let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's see if we can solve this. So let's write this as kind of a standard quadratic. Let's subtract eight from both sides. So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So plus x, plus x, and we are going to get, we are going to get four x squared minus three x minus seven, minus seven is equal to, is equal to zero. And let's see, we would want to factor this right over here, and let's see, maybe I could do this by, if I do it by, well, I'll just use a quadratic formula here. So the solutions are going to be x is going to be equal to negative b, so three, plus or minus the square root of b squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract eight from both sides to get rid of this eight over here, and let's add x to both sides. So plus x, plus x, and we are going to get, we are going to get four x squared minus three x minus seven, minus seven is equal to, is equal to zero. And let's see, we would want to factor this right over here, and let's see, maybe I could do this by, if I do it by, well, I'll just use a quadratic formula here. So the solutions are going to be x is going to be equal to negative b, so three, plus or minus the square root of b squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times, well, I'll just write, I'll write a seven here, then that negative is going to make this a positive. All of that over two a, so two times four is eight. So this is going to be three plus or minus the square root of, let's see, four times four is 16 times seven."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So the solutions are going to be x is going to be equal to negative b, so three, plus or minus the square root of b squared, so negative three squared is nine, minus four times a, which is four, times c, which is negative seven. So I could just say times, well, I'll just write, I'll write a seven here, then that negative is going to make this a positive. All of that over two a, so two times four is eight. So this is going to be three plus or minus the square root of, let's see, four times four is 16 times seven. 16 times seven is going to be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven, two, four, so it's 112, plus nine, so 121, that worked out nicely."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this is going to be three plus or minus the square root of, let's see, four times four is 16 times seven. 16 times seven is going to be 70 plus 42. Let me make sure I'm doing this right. So 16 times seven, two, four, so it's 112, plus nine, so 121, that worked out nicely. So plus or minus the square root of 121, all of that over eight, well, that is equal to three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14 eighths, or if we subtract 11, three minus 11 is negative eight. Negative eight divided by eight is negative one."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "So 16 times seven, two, four, so it's 112, plus nine, so 121, that worked out nicely. So plus or minus the square root of 121, all of that over eight, well, that is equal to three plus or minus 11, all of that over eight. So that is equal to, if we add 11, that is 14 eighths, or if we subtract 11, three minus 11 is negative eight. Negative eight divided by eight is negative one. So we have to think about, you might say, okay, I found two solutions to the radical equation, but remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure that they're legitimate or maybe one of these is an extraneous solution. In fact, one is very likely a solution to this radical equation, which wasn't our original goal."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "Negative eight divided by eight is negative one. So we have to think about, you might say, okay, I found two solutions to the radical equation, but remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides. We have to make sure that they're legitimate or maybe one of these is an extraneous solution. In fact, one is very likely a solution to this radical equation, which wasn't our original goal. So let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "In fact, one is very likely a solution to this radical equation, which wasn't our original goal. So let's see. Let's try out x equals negative one. If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one. So that would be negative two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine, this is the positive square root."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "If x equals negative one, we would have two times negative one minus one is equal to the square root of eight minus negative one. So that would be negative two minus one is equal to the square root of, is equal to the square root of nine. And so we'd have negative three is equal to the square root of nine. The principle root of nine, this is the positive square root. This is not true. So this right over here, that is an extraneous solution. Extraneous."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "The principle root of nine, this is the positive square root. This is not true. So this right over here, that is an extraneous solution. Extraneous. Extraneous solution. It is a solution to this one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "Extraneous. Extraneous solution. It is a solution to this one right over here. Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three. So it checks out for this one. So this one right over here is the extraneous solution."}, {"video_title": "Extraneous solutions of radical equations Mathematics III High School Math Khan Academy.mp3", "Sentence": "Because notice, for that one, if you substitute two times negative one minus one is equal to the negative of eight minus negative one. So this is negative three is equal to the negative of three. So it checks out for this one. So this one right over here is the extraneous solution. This one right over here is going to be the actual solution for our original equation. And you can test it out on your own. In fact, I encourage you to do so."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So let me try to roughly graph these two equations. So my y-axis, this is my x-axis, this is my x-axis. This right over here, x squared plus y squared is equal to 25. That's going to be a circle centered at 0 with radius 5. You don't have to know that to solve this problem, but it helps to visualize it. So if this is 5, this is 5, 5, 5. This right over here is negative 5."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "That's going to be a circle centered at 0 with radius 5. You don't have to know that to solve this problem, but it helps to visualize it. So if this is 5, this is 5, 5, 5. This right over here is negative 5. This right over here is negative 5. This equation would be represented by this set of points. Or this is a set of points that satisfy this equation."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "This right over here is negative 5. This right over here is negative 5. This equation would be represented by this set of points. Or this is a set of points that satisfy this equation. So let me, there you go. Trying to draw it as close to a perfect circle as I can. And then y equals x plus 1 is a line of slope 1 with a 1y intercept."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "Or this is a set of points that satisfy this equation. So let me, there you go. Trying to draw it as close to a perfect circle as I can. And then y equals x plus 1 is a line of slope 1 with a 1y intercept. So this is 1, 2, 3, 4. y intercept is there. It has a slope of 1. So it looks something like this."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "And then y equals x plus 1 is a line of slope 1 with a 1y intercept. So this is 1, 2, 3, 4. y intercept is there. It has a slope of 1. So it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So it looks something like this. So when we're looking for the solutions, we're looking for the points that satisfy both. The points that satisfy both are the points that sit on both. So it's that point. Let me do it in green. It's this point, and it's this point right over here. So how do we actually figure that out?"}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So it's that point. Let me do it in green. It's this point, and it's this point right over here. So how do we actually figure that out? Well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So how do we actually figure that out? Well, sometimes the easiest way is to substitute one of these constraints into the other constraint. And since they've already solved for y here, we can substitute y in the blue equation with x plus 1 with this constraint right over here. So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. And now we can attempt to solve for x. So we get x squared plus."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So instead of saying x squared plus y squared equals 25, we can say x squared plus, and instead of writing a y, we're adding the constraint that y must be x plus 1. So x squared plus x plus 1 squared must be equal to 25. And now we can attempt to solve for x. So we get x squared plus. Now we square this. We'll get, I'm going to write it in magenta. We'll get x squared plus 2x plus 1."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So we get x squared plus. Now we square this. We'll get, I'm going to write it in magenta. We'll get x squared plus 2x plus 1. And that must be equal to 25. We have 2x squared. Now I'm just combining these two terms."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "We'll get x squared plus 2x plus 1. And that must be equal to 25. We have 2x squared. Now I'm just combining these two terms. 2x squared plus 2x plus 1 is equal to 25. Now we could just use the quadratic formula to find the, well, we have to be careful. We have to set this equal to 0 and then use the quadratic formula."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "Now I'm just combining these two terms. 2x squared plus 2x plus 1 is equal to 25. Now we could just use the quadratic formula to find the, well, we have to be careful. We have to set this equal to 0 and then use the quadratic formula. So let's subtract 25 from both sides. And you could get 2x squared plus 2x minus 24 is equal to 0. And actually, just to simplify this, let's divide both sides by 2."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "We have to set this equal to 0 and then use the quadratic formula. So let's subtract 25 from both sides. And you could get 2x squared plus 2x minus 24 is equal to 0. And actually, just to simplify this, let's divide both sides by 2. And you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use a quadratic formula. We can factor this right over here."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "And actually, just to simplify this, let's divide both sides by 2. And you get x squared plus x minus 12 is equal to 0. And actually, we don't even have to use a quadratic formula. We can factor this right over here. What are two numbers that when we take their product, we get negative 12? And when we add them, we get positive 1. Well, positive 4 and negative 3 would do the trick."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "We can factor this right over here. What are two numbers that when we take their product, we get negative 12? And when we add them, we get positive 1. Well, positive 4 and negative 3 would do the trick. So we have x plus 4 times x minus 3 is equal to 0. So x could be equal to, well, if this is if x plus 4 is 0, then that would make this whole thing true. So x could be equal to negative 4, or x could be equal to positive 3."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "Well, positive 4 and negative 3 would do the trick. So we have x plus 4 times x minus 3 is equal to 0. So x could be equal to, well, if this is if x plus 4 is 0, then that would make this whole thing true. So x could be equal to negative 4, or x could be equal to positive 3. So this right over here is a situation where x is negative 4. This right over here is a situation where x is 3. So we're almost done."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So x could be equal to negative 4, or x could be equal to positive 3. So this right over here is a situation where x is negative 4. This right over here is a situation where x is 3. So we're almost done. We just have to find the corresponding y's. And for that, we can just resort to the simplest equation right over here. y is x plus 1."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "So we're almost done. We just have to find the corresponding y's. And for that, we can just resort to the simplest equation right over here. y is x plus 1. So in this situation, when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative 4 comma negative 3."}, {"video_title": "Systems of nonlinear equations 3 Systems of equations and inequalities Algebra II Khan Academy.mp3", "Sentence": "y is x plus 1. So in this situation, when x is negative 4, y is going to be that plus 1. So y is going to be negative 3. This is the point negative 4 comma negative 3. Likewise, when x is 3, y is going to be equal to 4. So this is the point 3 comma 4. These are the two solutions to this nonlinear system of equations."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "When we go from one term to the next, what are we doing? Well, we're multiplying by 10 11ths. To go from one to 10 11ths, you multiply by 10 over 11ths, then you multiply by 10 over 11ths, 10 over 11 again, and we keep doing this. So we wanna find the first 50 terms of it. So we can apply the formula we derived for the sum of a finite geometric series. And that tells us that the sum of, let's say in this case, the first 50 terms, actually, let me do it down here. So the sum, the sum of the first 50 terms is going to be equal to the first term, which is one."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So we wanna find the first 50 terms of it. So we can apply the formula we derived for the sum of a finite geometric series. And that tells us that the sum of, let's say in this case, the first 50 terms, actually, let me do it down here. So the sum, the sum of the first 50 terms is going to be equal to the first term, which is one. So it's going to be one times one minus, let me do that in a different color, one times one minus the common ratio. So the common ratio here is 10 11ths, 10 11ths to the 50th power, to the power of how many terms we have. All of that over one minus our common ratio."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So the sum, the sum of the first 50 terms is going to be equal to the first term, which is one. So it's going to be one times one minus, let me do that in a different color, one times one minus the common ratio. So the common ratio here is 10 11ths, 10 11ths to the 50th power, to the power of how many terms we have. All of that over one minus our common ratio. And so I'm not going to solve it completely, but we can simplify this a little bit. This is going to be one minus, and let me put parentheses here just to make sure we're not just taking the 10 to the 50th power. So one minus 10 11ths to the 50th power over, this is 11 11ths minus 10 11ths is over one over 11."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "All of that over one minus our common ratio. And so I'm not going to solve it completely, but we can simplify this a little bit. This is going to be one minus, and let me put parentheses here just to make sure we're not just taking the 10 to the 50th power. So one minus 10 11ths to the 50th power over, this is 11 11ths minus 10 11ths is over one over 11. And so this is the same thing as multiplying the numerator by 11. So this is going to be equal to 11 times one minus 10 11ths to the 50th power. And you could try to simplify this even more, but this gets us pretty far."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So one minus 10 11ths to the 50th power over, this is 11 11ths minus 10 11ths is over one over 11. And so this is the same thing as multiplying the numerator by 11. So this is going to be equal to 11 times one minus 10 11ths to the 50th power. And you could try to simplify this even more, but this gets us pretty far. At this point, it is just arithmetic. Let's do another one of these. This is kind of fun."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "And you could try to simplify this even more, but this gets us pretty far. At this point, it is just arithmetic. Let's do another one of these. This is kind of fun. So this is more clearly a geometric series. And let's just first think about how many terms we're going to take the sum of. You might be tempted to say, okay, I'm going to take it to the 79th power."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "This is kind of fun. So this is more clearly a geometric series. And let's just first think about how many terms we're going to take the sum of. You might be tempted to say, okay, I'm going to take it to the 79th power. There must be 79 terms here. But be very careful, because the first term is when we're taking things to the zeroth power. We're taking 0.99 to the zeroth power."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "You might be tempted to say, okay, I'm going to take it to the 79th power. There must be 79 terms here. But be very careful, because the first term is when we're taking things to the zeroth power. We're taking 0.99 to the zeroth power. The second term is where we're taking it to the first power. The third term is where we're taking it to the second power. The fourth term is where we're taking it to the third power."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "We're taking 0.99 to the zeroth power. The second term is where we're taking it to the first power. The third term is where we're taking it to the second power. The fourth term is where we're taking it to the third power. So on and so forth. So this right over here is the 80th, the 80th term, 80th term. So we want to find S sub 80."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "The fourth term is where we're taking it to the third power. So on and so forth. So this right over here is the 80th, the 80th term, 80th term. So we want to find S sub 80. And so this is going to be equal to our first, our first term is going to be one times one minus our common ratio to the 80th power, to the 80th power, all over, and I'm leaving a blank because we still need to figure out our common ratio, all over one minus our common ratio. So at first, you might say, well, maybe the common ratio here is 0.99. But notice, we have a change in sign here."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So we want to find S sub 80. And so this is going to be equal to our first, our first term is going to be one times one minus our common ratio to the 80th power, to the 80th power, all over, and I'm leaving a blank because we still need to figure out our common ratio, all over one minus our common ratio. So at first, you might say, well, maybe the common ratio here is 0.99. But notice, we have a change in sign here. And the key thing is to say, well, to go from one term to the next, what are we multiplying by? Well, to go from the first term to the second term, we multiply by negative 0.99. And then, so we're multiplying by negative 0.99."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "But notice, we have a change in sign here. And the key thing is to say, well, to go from one term to the next, what are we multiplying by? Well, to go from the first term to the second term, we multiply by negative 0.99. And then, so we're multiplying by negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99. So the common ratio is not positive 0.99, but negative 0, negative 0.99. So I'm gonna write that negative 0.99."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "And then, so we're multiplying by negative 0.99. Now to go to the next term, we're again multiplying by negative 0.99. So the common ratio is not positive 0.99, but negative 0, negative 0.99. So I'm gonna write that negative 0.99. And of course, that is going to be to the 80th power all over one minus negative 0.99. And so we could simplify this a little bit. This is all going to be equal to, well, that one, we don't have to worry too much about that."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So I'm gonna write that negative 0.99. And of course, that is going to be to the 80th power all over one minus negative 0.99. And so we could simplify this a little bit. This is all going to be equal to, well, that one, we don't have to worry too much about that. And so this is gonna be one minus, so negative 0.99 to the 80th power. Actually, I put parentheses there to make sure we are taking the negative 0.99 to the 80th power. Well, we're taking it to an even power, so it's going to be positive."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "This is all going to be equal to, well, that one, we don't have to worry too much about that. And so this is gonna be one minus, so negative 0.99 to the 80th power. Actually, I put parentheses there to make sure we are taking the negative 0.99 to the 80th power. Well, we're taking it to an even power, so it's going to be positive. So that's going to be the same thing as 0.99 to the 80th power. And all of that over, well, subtracting a negative, that's just going to be adding the positive. So all of that over 1.99."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "Well, we're taking it to an even power, so it's going to be positive. So that's going to be the same thing as 0.99 to the 80th power. And all of that over, well, subtracting a negative, that's just going to be adding the positive. So all of that over 1.99. And we could attempt to simplify it more, but if we had a calculator, we could actually find this exact value, or close value, actually. Most calculators don't give you the exact value when you take something to the 80th power. But this is what that sum is going to be."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So all of that over 1.99. And we could attempt to simplify it more, but if we had a calculator, we could actually find this exact value, or close value, actually. Most calculators don't give you the exact value when you take something to the 80th power. But this is what that sum is going to be. Let's do one more of these. All right, so here we have a series defined recursively. And so it's useful to just think about what it would actually look like."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "But this is what that sum is going to be. Let's do one more of these. All right, so here we have a series defined recursively. And so it's useful to just think about what it would actually look like. So the first term is 10. And then the next term, so the second term, a sub two, is equal to a sub one times 9 tenths. Right, so the next term is gonna be the previous term times 9 tenths."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "And so it's useful to just think about what it would actually look like. So the first term is 10. And then the next term, so the second term, a sub two, is equal to a sub one times 9 tenths. Right, so the next term is gonna be the previous term times 9 tenths. So it's gonna be 10 times nine over 10. And then the next term is going to be that times, it's gonna be the second term. The third term is the second term times 9 tenths."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "Right, so the next term is gonna be the previous term times 9 tenths. So it's gonna be 10 times nine over 10. And then the next term is going to be that times, it's gonna be the second term. The third term is the second term times 9 tenths. So 10 times nine over 10, nine over 10 squared. And the way it's written right now, we don't, I haven't written it as a finite geometric series. So let's say we want to take the sum, let's say we want the sum of first, first, I don't know, 30 terms."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "The third term is the second term times 9 tenths. So 10 times nine over 10, nine over 10 squared. And the way it's written right now, we don't, I haven't written it as a finite geometric series. So let's say we want to take the sum, let's say we want the sum of first, first, I don't know, 30 terms. Sum of first 30 terms. So what will this be? Well, we're gonna take s sub one, s sub 30, I don't know why I wrote 10, s sub 30, the sum of the first 30 terms is going to be equal to the first term."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "So let's say we want to take the sum, let's say we want the sum of first, first, I don't know, 30 terms. Sum of first 30 terms. So what will this be? Well, we're gonna take s sub one, s sub 30, I don't know why I wrote 10, s sub 30, the sum of the first 30 terms is going to be equal to the first term. We've done this before, the first term times one minus the common ratio, one minus the common ratio to the 30th power. All of that over one minus the common ratio. And let's see, we could, one minus 9 tenths, this is 1 tenth right over here."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "Well, we're gonna take s sub one, s sub 30, I don't know why I wrote 10, s sub 30, the sum of the first 30 terms is going to be equal to the first term. We've done this before, the first term times one minus the common ratio, one minus the common ratio to the 30th power. All of that over one minus the common ratio. And let's see, we could, one minus 9 tenths, this is 1 tenth right over here. You divide by 1 tenth, this is the same thing as multiplying by 100. So this is gonna be 100 times one minus 9 tenths to the, well, let me write it this way. 9 tenths to the 30th power."}, {"video_title": "Worked examples finite geometric series High School Math Khan Academy.mp3", "Sentence": "And let's see, we could, one minus 9 tenths, this is 1 tenth right over here. You divide by 1 tenth, this is the same thing as multiplying by 100. So this is gonna be 100 times one minus 9 tenths to the, well, let me write it this way. 9 tenths to the 30th power. And actually these parentheses, you always wanna, so I can put parentheses there just to make sure we see that we're taking both the nine and the 10, or the 9 tenths, the whole thing to the 30th power, not just the nine. So there, there you go. Yep, there you go, we're done."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Let's solve for a, and I encourage you right now to pause this video and try it on your own. Well if you have a fifth root right over here, one thing that you might be tempted to do to undo the fifth root is to raise it to the fifth power. And of course we can't just raise one side of an equation to the fifth power. Whatever we do to one side, we have to do to the other side if we want this to still be equal. So let's raise both sides of this equation to the fifth power. Now this left-hand side, we just have to remember a little bit of our exponent properties. 3a to the fifth power, and if we want to just remind ourselves where that comes from, it's the same thing as 3 to the a times 3 to the a times 3 to the a times 3 to the a times 3 to the a."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Whatever we do to one side, we have to do to the other side if we want this to still be equal. So let's raise both sides of this equation to the fifth power. Now this left-hand side, we just have to remember a little bit of our exponent properties. 3a to the fifth power, and if we want to just remind ourselves where that comes from, it's the same thing as 3 to the a times 3 to the a times 3 to the a times 3 to the a times 3 to the a. Well what's that going to be equal to? That's going to be 3 to the a plus a plus a plus a plus a power, which is the same thing as 3 to the 5a power. So the exponent property here is if you raise a base to some exponent and then raise that whole thing to another exponent, that's the equivalent of raising the base to an exponent that is the product of these two exponents."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "3a to the fifth power, and if we want to just remind ourselves where that comes from, it's the same thing as 3 to the a times 3 to the a times 3 to the a times 3 to the a times 3 to the a. Well what's that going to be equal to? That's going to be 3 to the a plus a plus a plus a plus a power, which is the same thing as 3 to the 5a power. So the exponent property here is if you raise a base to some exponent and then raise that whole thing to another exponent, that's the equivalent of raising the base to an exponent that is the product of these two exponents. So we could rewrite this left-hand side as 3 to the 5a power is going to be equal to, well if you take something that's a fifth root and you raise it to the fifth power, then you're just left with what you had under the radical. That's going to be equal to 3 squared. So now things become a lot clearer."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So the exponent property here is if you raise a base to some exponent and then raise that whole thing to another exponent, that's the equivalent of raising the base to an exponent that is the product of these two exponents. So we could rewrite this left-hand side as 3 to the 5a power is going to be equal to, well if you take something that's a fifth root and you raise it to the fifth power, then you're just left with what you had under the radical. That's going to be equal to 3 squared. So now things become a lot clearer. 3 to the 5a needs to be equal to 3 squared. Or another way of thinking about it, we have the same base on both sides, so this exponent needs to be equal to this exponent right over there. Or we could write that 5 times a needs to be equal to 2."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So now things become a lot clearer. 3 to the 5a needs to be equal to 3 squared. Or another way of thinking about it, we have the same base on both sides, so this exponent needs to be equal to this exponent right over there. Or we could write that 5 times a needs to be equal to 2. And of course now we can just divide both sides by 5, and we get a is equal to 2 fifths. And this is an interesting result. What's neat about this example, it kind of shows you the motivation for how we define rational exponents."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "Or we could write that 5 times a needs to be equal to 2. And of course now we can just divide both sides by 5, and we get a is equal to 2 fifths. And this is an interesting result. What's neat about this example, it kind of shows you the motivation for how we define rational exponents. So let's just put this back into the original expression. We've just said, or we've just solved for a, and we've gotten that 3 to the 2 fifths power to the 2, and actually let me color code it a little bit, because I think that will be interesting. 3 to the 2 over 5 power is equal to the fifth root."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "What's neat about this example, it kind of shows you the motivation for how we define rational exponents. So let's just put this back into the original expression. We've just said, or we've just solved for a, and we've gotten that 3 to the 2 fifths power to the 2, and actually let me color code it a little bit, because I think that will be interesting. 3 to the 2 over 5 power is equal to the fifth root. Notice the fifth root. So the denominator here, that's the root. So the fifth root of 3 squared."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "3 to the 2 over 5 power is equal to the fifth root. Notice the fifth root. So the denominator here, that's the root. So the fifth root of 3 squared. So if you take this base 3, you square it, but then you take the fifth root of that, that's the same thing as raising it to the 2 fifths power. Notice, take this 3, take it to the second power, and then you find the fifth root of it. Or we could, if you use this property that we just saw right over here, you could rewrite this."}, {"video_title": "Exponential equation with rational answer Exponent expressions Algebra I Khan Academy.mp3", "Sentence": "So the fifth root of 3 squared. So if you take this base 3, you square it, but then you take the fifth root of that, that's the same thing as raising it to the 2 fifths power. Notice, take this 3, take it to the second power, and then you find the fifth root of it. Or we could, if you use this property that we just saw right over here, you could rewrite this. This is the same thing as 3 squared, and then you raise that to the 1 fifth power. We saw that property at play over here. You could just multiply these two exponents."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Let's say that we are dealing with a geometric series and there are some things that we know about this geometric series. For example, we know that the first term of our geometric series is a. So that is our first term. We also know the common ratio of our geometric series and we're gonna call that r. So this is the common ratio. And we also know that it's a finite geometric series. So let me write this. It is finite."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "We also know the common ratio of our geometric series and we're gonna call that r. So this is the common ratio. And we also know that it's a finite geometric series. So let me write this. It is finite. So it has a finite number of terms. And let's say that n is equal to the number of terms. The number of terms."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "It is finite. So it has a finite number of terms. And let's say that n is equal to the number of terms. The number of terms. And we're gonna use a notation. We're going to use a notation s sub n to denote the sum. The sum."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "The number of terms. And we're gonna use a notation. We're going to use a notation s sub n to denote the sum. The sum. The sum of first. First. Of first n terms."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "The sum. The sum of first. First. Of first n terms. And the goal of this whole video is using this information, coming up with a general formula for the sum of the first n terms. A formula for evaluating a geometric series. So let's write out s sub n. Just get a feeling for what it would look like."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Of first n terms. And the goal of this whole video is using this information, coming up with a general formula for the sum of the first n terms. A formula for evaluating a geometric series. So let's write out s sub n. Just get a feeling for what it would look like. So s sub n is going to be equal to, well you have your first term here, which is an a. And then what's our second term going to be? Well it's going to be, this is a geometric series."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So let's write out s sub n. Just get a feeling for what it would look like. So s sub n is going to be equal to, well you have your first term here, which is an a. And then what's our second term going to be? Well it's going to be, this is a geometric series. So it's going to be a times the common ratio. So it's going to be the first term times the common ratio. So the first term times r. Now what's the third term going to be?"}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Well it's going to be, this is a geometric series. So it's going to be a times the common ratio. So it's going to be the first term times the common ratio. So the first term times r. Now what's the third term going to be? Well it's going to be the second term times our common ratio again. So it's going to be ar times r. Or ar squared. Ar squared."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So the first term times r. Now what's the third term going to be? Well it's going to be the second term times our common ratio again. So it's going to be ar times r. Or ar squared. Ar squared. And we could go all the way to our nth term. So we're going to go all the way to the nth term. And you might be tempted to say it's going to be a times r to the nth power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Ar squared. And we could go all the way to our nth term. So we're going to go all the way to the nth term. And you might be tempted to say it's going to be a times r to the nth power. But we have to be careful here. Because notice, our first term is really ar to the zeroth power. Our second term is ar to the first power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And you might be tempted to say it's going to be a times r to the nth power. But we have to be careful here. Because notice, our first term is really ar to the zeroth power. Our second term is ar to the first power. Our third term is ar to the second power. So whatever term we're on, the exponent is that term number minus one. So if we're on the nth term, it's going to be ar to the n minus oneth power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Our second term is ar to the first power. Our third term is ar to the second power. So whatever term we're on, the exponent is that term number minus one. So if we're on the nth term, it's going to be ar to the n minus oneth power. So we want to come up with a nice, clean formula for evaluating this. And we're going to use a little trick to do it. To do it, we're going to think about what r times the sum is."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So if we're on the nth term, it's going to be ar to the n minus oneth power. So we want to come up with a nice, clean formula for evaluating this. And we're going to use a little trick to do it. To do it, we're going to think about what r times the sum is. And we're going to subtract that out. So we're going to take the r times that sum. R times the sum of the first nth term."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "To do it, we're going to think about what r times the sum is. And we're going to subtract that out. So we're going to take the r times that sum. R times the sum of the first nth term. Actually, let's just multiply negative r. Negative r times the sum. And then we can just add these two things. And you'll see that it cleans this thing up nicely."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "R times the sum of the first nth term. Actually, let's just multiply negative r. Negative r times the sum. And then we can just add these two things. And you'll see that it cleans this thing up nicely. So what is this going to be equal to? This is going to be equal to, well, if you multiply, if we multiply a times negative r, we will get negative ar. And I'm just going to write it right underneath this one."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And you'll see that it cleans this thing up nicely. So what is this going to be equal to? This is going to be equal to, well, if you multiply, if we multiply a times negative r, we will get negative ar. And I'm just going to write it right underneath this one. So if you multiply this times negative r, I'm just going to multiply every one of these terms by negative r. That's the equivalent of multiplying negative r times the sum. Distributing the negative r. So if I multiply it times this term, a times negative r, that's going to be negative, that's going to be negative ar. And if I multiply ar times negative r, that's going to be negative ar squared."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And I'm just going to write it right underneath this one. So if you multiply this times negative r, I'm just going to multiply every one of these terms by negative r. That's the equivalent of multiplying negative r times the sum. Distributing the negative r. So if I multiply it times this term, a times negative r, that's going to be negative, that's going to be negative ar. And if I multiply ar times negative r, that's going to be negative ar squared. Negative ar squared. You might see where this is going. And just to be clear what's going on, that's that term times negative r. Times negative r. This is that term times negative r. And we would keep going all the way to the term, the term, this, the term before this times negative r. So if I, the term before this times negative r is going to be a, is going to be negative."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And if I multiply ar times negative r, that's going to be negative ar squared. Negative ar squared. You might see where this is going. And just to be clear what's going on, that's that term times negative r. Times negative r. This is that term times negative r. And we would keep going all the way to the term, the term, this, the term before this times negative r. So if I, the term before this times negative r is going to be a, is going to be negative. Actually, let me put subtraction signs. It's going to be negative a times r to the n minus one power. r to the n minus one power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And just to be clear what's going on, that's that term times negative r. Times negative r. This is that term times negative r. And we would keep going all the way to the term, the term, this, the term before this times negative r. So if I, the term before this times negative r is going to be a, is going to be negative. Actually, let me put subtraction signs. It's going to be negative a times r to the n minus one power. r to the n minus one power. That was the term right before this. That was a times r to the n minus two times negative r is going to give us this. So it's going to get us right over there."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "r to the n minus one power. That was the term right before this. That was a times r to the n minus two times negative r is going to give us this. So it's going to get us right over there. And then finally we take this last term and you multiply it by negative r, what do you get? You get negative a, negative a, and then times r to the n. r to the n. You multiply this times the negative, you get the negative a. And then r to the n minus one times r, or times r to the first, well this is going to be r to the n. And now what's interesting here is this we can add up the left side and we can add up the right hand side."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So it's going to get us right over there. And then finally we take this last term and you multiply it by negative r, what do you get? You get negative a, negative a, and then times r to the n. r to the n. You multiply this times the negative, you get the negative a. And then r to the n minus one times r, or times r to the first, well this is going to be r to the n. And now what's interesting here is this we can add up the left side and we can add up the right hand side. So let's do that. So let's do that. On the left hand side, we get s sub n minus r, minus r, minus r times s sub n, s sub n. And on the right hand side, we have something very cool happening."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And then r to the n minus one times r, or times r to the first, well this is going to be r to the n. And now what's interesting here is this we can add up the left side and we can add up the right hand side. So let's do that. So let's do that. On the left hand side, we get s sub n minus r, minus r, minus r times s sub n, s sub n. And on the right hand side, we have something very cool happening. Notice this a, we still have that. The a sits there. But everything else, except for this last thing, is going to cancel out."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "On the left hand side, we get s sub n minus r, minus r, minus r times s sub n, s sub n. And on the right hand side, we have something very cool happening. Notice this a, we still have that. The a sits there. But everything else, except for this last thing, is going to cancel out. So these two are gonna cancel out. These two are gonna cancel out. Let me do that a little neater."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "But everything else, except for this last thing, is going to cancel out. So these two are gonna cancel out. These two are gonna cancel out. Let me do that a little neater. These two are gonna cancel out. These two are gonna cancel out. And all we're gonna have left with is a negative a r to the n. So it's gonna be a minus a times r to the nth power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Let me do that a little neater. These two are gonna cancel out. These two are gonna cancel out. And all we're gonna have left with is a negative a r to the n. So it's gonna be a minus a times r to the nth power. And now we can just solve for s sub n and we have our formula, what we were looking for. So let's see, we can factor out an s sub n on the left hand side. So you get an s sub n, the sum of our first n terms."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And all we're gonna have left with is a negative a r to the n. So it's gonna be a minus a times r to the nth power. And now we can just solve for s sub n and we have our formula, what we were looking for. So let's see, we can factor out an s sub n on the left hand side. So you get an s sub n, the sum of our first n terms. You factor that out. So it's gonna be times, times one minus r is going to be equal to, and on the right hand side we can actually factor out an a. So it's gonna be a times one minus r to the n. And so to solve for s sub n, the sum of our first n terms, we deserve a little bit of a drum roll here."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So you get an s sub n, the sum of our first n terms. You factor that out. So it's gonna be times, times one minus r is going to be equal to, and on the right hand side we can actually factor out an a. So it's gonna be a times one minus r to the n. And so to solve for s sub n, the sum of our first n terms, we deserve a little bit of a drum roll here. S sub n is going to be equal to this divided by one minus r. So it's going to be a times one minus r to the n one minus r to the n over one minus r. And we're done. We have figured out our formula for the sum or for the sum of a finite geometric series. And so in the next few videos or in future videos we will apply this."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So it's gonna be a times one minus r to the n. And so to solve for s sub n, the sum of our first n terms, we deserve a little bit of a drum roll here. S sub n is going to be equal to this divided by one minus r. So it's going to be a times one minus r to the n one minus r to the n over one minus r. And we're done. We have figured out our formula for the sum or for the sum of a finite geometric series. And so in the next few videos or in future videos we will apply this. And I encourage you, whenever you use this formula, it's very important now that you know where it came from that you really keep close track of how many terms you're actually summing up. Sometimes you might have a sigma notation that starts, it might start its index at zero and then goes up to a number, in which case you're gonna have that number plus one term. So you have to be very careful."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And so in the next few videos or in future videos we will apply this. And I encourage you, whenever you use this formula, it's very important now that you know where it came from that you really keep close track of how many terms you're actually summing up. Sometimes you might have a sigma notation that starts, it might start its index at zero and then goes up to a number, in which case you're gonna have that number plus one term. So you have to be very careful. This is the number of terms. This is the first term here, we define it up here. N is the number of terms, the first n terms."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And frankly, this is already quite simple, but I'm assuming they want us to use some logarithm properties and manipulate this in some way, maybe to actually make it a little bit more complicated. But let's give our best shot at it. So the logarithm property that jumps out at me, because this right over here, we're saying what power do we have to raise three to to get 27x? 27x is the same thing as 27 times x. And so the logarithm property, it seems like they want us to use, is log base, let me write it, log base b of a times c. I'll write it this way. Log base b of a times c. This is equal to the logarithm base b of a plus the logarithm base b of c. And this comes straight out of the exponent properties, that if you have two exponents with the same base, you can add the exponents. So let me make that a little bit clearer to you."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "27x is the same thing as 27 times x. And so the logarithm property, it seems like they want us to use, is log base, let me write it, log base b of a times c. I'll write it this way. Log base b of a times c. This is equal to the logarithm base b of a plus the logarithm base b of c. And this comes straight out of the exponent properties, that if you have two exponents with the same base, you can add the exponents. So let me make that a little bit clearer to you. And this part is a little confusing. The important part for this example is that you know how to apply this. But it's even better if you know the intuition."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let me make that a little bit clearer to you. And this part is a little confusing. The important part for this example is that you know how to apply this. But it's even better if you know the intuition. So let's say that log base b of a times c is equal to x. So this thing right over here evaluates to x. Let's say that this thing right over here evaluates to y."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "But it's even better if you know the intuition. So let's say that log base b of a times c is equal to x. So this thing right over here evaluates to x. Let's say that this thing right over here evaluates to y. So log base b of a is equal to y. And let's say that this thing over here evaluates to z. So log base b of c is equal to z."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's say that this thing right over here evaluates to y. So log base b of a is equal to y. And let's say that this thing over here evaluates to z. So log base b of c is equal to z. Now, what we know is this thing right over here, this thing right over here, or this thing right over here tells us, tells us that b to the x power is equal to a times c. Now, this right over here is telling us that b to the y power is equal to a. b to the y power is equal to a. And this over here is telling us that b to the z power is equal to c. Let me do that in that same green. So I'm just writing the same truth I'm writing as an exponential function or exponential equation instead of a logarithmic equation."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So log base b of c is equal to z. Now, what we know is this thing right over here, this thing right over here, or this thing right over here tells us, tells us that b to the x power is equal to a times c. Now, this right over here is telling us that b to the y power is equal to a. b to the y power is equal to a. And this over here is telling us that b to the z power is equal to c. Let me do that in that same green. So I'm just writing the same truth I'm writing as an exponential function or exponential equation instead of a logarithmic equation. So b to the z power is equal to c. This is the same statement. This is the same statement. Or the same truth said in a different way."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm just writing the same truth I'm writing as an exponential function or exponential equation instead of a logarithmic equation. So b to the z power is equal to c. This is the same statement. This is the same statement. Or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, we can, and c, and c is equal to bz, then we can write b to the x power is equal to b to the y power, b to the y power, that's what a is, we know that already, times b to the z power, times b to the z power. And we know from our exponent properties, we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the, I'll do it in a neutral color, b to the y plus z power."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or the same truth said in a different way. And this is the same truth said in a different way. Well, if we know that a is equal to this, is equal to b to the y, we can, and c, and c is equal to bz, then we can write b to the x power is equal to b to the y power, b to the y power, that's what a is, we know that already, times b to the z power, times b to the z power. And we know from our exponent properties, we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the, I'll do it in a neutral color, b to the y plus z power. This comes straight out of our exponent properties. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. X must be equal to y plus z. If this is confusing to you, don't worry about it too much."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we know from our exponent properties, we know from our exponent properties that if we take b to the y times b to the z, this is the same thing as b to the, I'll do it in a neutral color, b to the y plus z power. This comes straight out of our exponent properties. And so if b to the y plus z power is the same thing as b to the x power, that tells us that x must be equal to y plus z. X must be equal to y plus z. If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing is that you know how to apply it, and then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents. And I know when people first tell me that, I was like, well, what does that mean?"}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "If this is confusing to you, don't worry about it too much. The important thing, or at least the first important thing is that you know how to apply it, and then you can think about this a little bit more, and you can even try it out with some numbers. You just have to realize that logarithms are really just exponents. And I know when people first tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base three of 27 times x, I'll write it that way, is equal to log base three of 27 plus log base three of x. And then this right over here we can evaluate."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I know when people first tell me that, I was like, well, what does that mean? But when you evaluate a logarithm, you're getting an exponent that you would have to raise b to to get to a times c. But let's just apply this property right over here. So if we apply it to this one, we know that log base three of 27 times x, I'll write it that way, is equal to log base three of 27 plus log base three of x. And then this right over here we can evaluate. This tells us what power do I have to raise three to to get to 27? You could view it as this way. Three to the question mark is equal to 27."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "And then this right over here we can evaluate. This tells us what power do I have to raise three to to get to 27? You could view it as this way. Three to the question mark is equal to 27. Well, three to the third power is equal to 27. Three times three is nine, times three is 27. So this right over here evaluates to three."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "Three to the question mark is equal to 27. Well, three to the third power is equal to 27. Three times three is nine, times three is 27. So this right over here evaluates to three. So if we were to simplify, or I guess I wouldn't even call it simplifying it, I would just call it expanding it out or using this property because we now have two terms where we started off with one term. Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes three."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right over here evaluates to three. So if we were to simplify, or I guess I wouldn't even call it simplifying it, I would just call it expanding it out or using this property because we now have two terms where we started off with one term. Actually, if we started with this, I'd say that this is the more simple version of it. But when we rewrite it, this first term becomes three. So this first term becomes three. And then we're left with plus log base three of x. So this is just an alternate way of writing this original statement."}, {"video_title": "Sum of logarithms with same base Logarithms Algebra II Khan Academy.mp3", "Sentence": "But when we rewrite it, this first term becomes three. So this first term becomes three. And then we're left with plus log base three of x. So this is just an alternate way of writing this original statement. Log base three of 27x. So once again, not clear that this is simpler than this right over here. It's just another way of writing it using logarithm properties."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "But let's just start with a little bit of review. If I were to ask you, what is a plus b squared, what would that be? Pause the video and try to figure it out. Well, some of you might immediately know what a binomial like this squared is, but I'll work it out. So this is the same thing as a plus b times a plus b. And then we can multiply this a times that a. So that's going to give us a squared."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "Well, some of you might immediately know what a binomial like this squared is, but I'll work it out. So this is the same thing as a plus b times a plus b. And then we can multiply this a times that a. So that's going to give us a squared. And then I can multiply that a times that b, and that's going to give us a b. Then I could multiply this b times that a. I could write that as b a or a b, so I'll just write it as a b again. And then I multiply this b times that b, so plus b squared."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "So that's going to give us a squared. And then I can multiply that a times that b, and that's going to give us a b. Then I could multiply this b times that a. I could write that as b a or a b, so I'll just write it as a b again. And then I multiply this b times that b, so plus b squared. And what I really just did is apply the distributive property twice. We go into a lot of detail in previous videos. Some people also like to call it the FOIL method."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "And then I multiply this b times that b, so plus b squared. And what I really just did is apply the distributive property twice. We go into a lot of detail in previous videos. Some people also like to call it the FOIL method. Either way, this should all be a review. If it's not, I encourage you to look at those introductory videos. But this is going to simplify to a squared plus we have an a b and another a b, so you add those together, you get two a b plus b squared."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "Some people also like to call it the FOIL method. Either way, this should all be a review. If it's not, I encourage you to look at those introductory videos. But this is going to simplify to a squared plus we have an a b and another a b, so you add those together, you get two a b plus b squared. Now, why did I go through this review? Well, now we can use this idea that a plus b squared is equal to a squared plus two a b plus b squared to tackle things that at least look a little bit more involved. So if I were to ask you, what is five x to the sixth plus four squared?"}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "But this is going to simplify to a squared plus we have an a b and another a b, so you add those together, you get two a b plus b squared. Now, why did I go through this review? Well, now we can use this idea that a plus b squared is equal to a squared plus two a b plus b squared to tackle things that at least look a little bit more involved. So if I were to ask you, what is five x to the sixth plus four squared? Pause this video and try to figure it out and try to keep this and this in mind. Well, there's several ways you could approach this. You could just expand this out the way we just did, or you could recognize this pattern that we just established, that if I have a plus b and a squared, it's going to be this."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "So if I were to ask you, what is five x to the sixth plus four squared? Pause this video and try to figure it out and try to keep this and this in mind. Well, there's several ways you could approach this. You could just expand this out the way we just did, or you could recognize this pattern that we just established, that if I have a plus b and a squared, it's going to be this. And so what you might notice is the role of a is being played by five x to the sixth right over there, and the role of b is being played by four right over there. So we could say, hey, this is going to be equal to a squared, we have our a squared there, so what is a squared? Well, five x to the sixth squared is going to be 25 x to the 12th power."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "You could just expand this out the way we just did, or you could recognize this pattern that we just established, that if I have a plus b and a squared, it's going to be this. And so what you might notice is the role of a is being played by five x to the sixth right over there, and the role of b is being played by four right over there. So we could say, hey, this is going to be equal to a squared, we have our a squared there, so what is a squared? Well, five x to the sixth squared is going to be 25 x to the 12th power. And then it's going to be plus two times a times b, so plus two times five x to the sixth times four. Actually, let me just write it out just so we don't confuse ourselves. Two times five x to the, I'll color code it too."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "Well, five x to the sixth squared is going to be 25 x to the 12th power. And then it's going to be plus two times a times b, so plus two times five x to the sixth times four. Actually, let me just write it out just so we don't confuse ourselves. Two times five x to the, I'll color code it too. Two times five x to the sixth times four, times four, plus b squared. So plus four squared, so that's going to be plus 16. And then we can simplify this."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "Two times five x to the, I'll color code it too. Two times five x to the sixth times four, times four, plus b squared. So plus four squared, so that's going to be plus 16. And then we can simplify this. So this is going to be equal to 25 x to the 12th. Two times five times four is 40. Two times five is 10, times four is 40."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "And then we can simplify this. So this is going to be equal to 25 x to the 12th. Two times five times four is 40. Two times five is 10, times four is 40. So plus 40 x to the sixth plus 16. Let's do another example. And I'll do this one even a little bit faster just because we're getting, I think, pretty good at this."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "Two times five is 10, times four is 40. So plus 40 x to the sixth plus 16. Let's do another example. And I'll do this one even a little bit faster just because we're getting, I think, pretty good at this. So let's say we're trying to determine what three t squared minus seven t to the sixth power squared is. Pause the video and try to figure it out. All right, we're going to do it together now."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "And I'll do this one even a little bit faster just because we're getting, I think, pretty good at this. So let's say we're trying to determine what three t squared minus seven t to the sixth power squared is. Pause the video and try to figure it out. All right, we're going to do it together now. So this is our a, and our b now, we should view as negative seven t to the sixth because this says plus b. So you could view this as plus negative seven t to the sixth. We could even write that if we want."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "All right, we're going to do it together now. So this is our a, and our b now, we should view as negative seven t to the sixth because this says plus b. So you could view this as plus negative seven t to the sixth. We could even write that if we want. We could write this plus negative seven t to the sixth if it helps us recognize this whole thing as b. So this is going to be equal to a squared, which is nine t to the fourth, plus two times this times this, two times a times b. So two times three t squared is going to be six t squared times negative seven t to the sixth."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "We could even write that if we want. We could write this plus negative seven t to the sixth if it helps us recognize this whole thing as b. So this is going to be equal to a squared, which is nine t to the fourth, plus two times this times this, two times a times b. So two times three t squared is going to be six t squared times negative seven t to the sixth. Actually, let me write this out. This is getting a little bit complicated. So this is going to be plus two times three t squared times negative seven t to the sixth power."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "So two times three t squared is going to be six t squared times negative seven t to the sixth. Actually, let me write this out. This is getting a little bit complicated. So this is going to be plus two times three t squared times negative seven t to the sixth power. And then last but not least, we're going to square negative seven t to the sixth. So that's going to be negative seven squared is positive 49. And t to the sixth squared is t to the 12th, t to the 12th power."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "So this is going to be plus two times three t squared times negative seven t to the sixth power. And then last but not least, we're going to square negative seven t to the sixth. So that's going to be negative seven squared is positive 49. And t to the sixth squared is t to the 12th, t to the 12th power. And so this is going to be equal to nine t to the fourth. And let's see, two times three is six times negative seven is negative 42. And t squared times t to the sixth, we add the exponents, we have the same base, so it's going to be t to the eighth."}, {"video_title": "Polynomial special products perfect square Algebra 2 Khan Academy.mp3", "Sentence": "And t to the sixth squared is t to the 12th, t to the 12th power. And so this is going to be equal to nine t to the fourth. And let's see, two times three is six times negative seven is negative 42. And t squared times t to the sixth, we add the exponents, we have the same base, so it's going to be t to the eighth. And then we have plus 49 t to the 12th power. So it looks like we did something really fancy. We have this higher degree polynomial."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So if you've been watching these videos, you know that we have a lot of scenarios where people seem to be walking up to us on the street and asking us to do math problems. And I guess this will be no different. So let's say someone walks up to you on the street and says, quick, you x squared plus five x plus eight over x plus two. What can this be simplified to? Or what is x squared plus five x plus eight divided by x plus two? Pause this video and see if you can work through that. So there's two ways that we can approach this."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "What can this be simplified to? Or what is x squared plus five x plus eight divided by x plus two? Pause this video and see if you can work through that. So there's two ways that we can approach this. We can try to factor our numerator and see if we have a common factor there. Or we can try to use algebraic long division. So let's first try to factor this numerator."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So there's two ways that we can approach this. We can try to factor our numerator and see if we have a common factor there. Or we can try to use algebraic long division. So let's first try to factor this numerator. And we would ideally want x plus two to be one of the factors. So let's see, what two numbers can add up to five? And when I multiply them, I get to eight."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So let's first try to factor this numerator. And we would ideally want x plus two to be one of the factors. So let's see, what two numbers can add up to five? And when I multiply them, I get to eight. And ideally, two is one of them. So I could think of two and three. But two times three is going to be equal to six, not eight."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "And when I multiply them, I get to eight. And ideally, two is one of them. So I could think of two and three. But two times three is going to be equal to six, not eight. And I can't think of anything else. But that still gives us some progress. Because what if we did say, all right, let's rewrite part of it."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "But two times three is going to be equal to six, not eight. And I can't think of anything else. But that still gives us some progress. Because what if we did say, all right, let's rewrite part of it. What if we were to write x squared plus five x? And we wanna write a plus six because that actually would be divisible by x plus two. So I'm gonna write a plus six."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "Because what if we did say, all right, let's rewrite part of it. What if we were to write x squared plus five x? And we wanna write a plus six because that actually would be divisible by x plus two. So I'm gonna write a plus six. But of course, we have an eight here. So then we're going to have an extra two right over there. And then all of that is divisible by x plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So I'm gonna write a plus six. But of course, we have an eight here. So then we're going to have an extra two right over there. And then all of that is divisible by x plus two. And now I can rewrite this part up here in orange. That is x plus two times x plus three. So let me write it here."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "And then all of that is divisible by x plus two. And now I can rewrite this part up here in orange. That is x plus two times x plus three. So let me write it here. X plus two times x plus three. I still have that plus two sitting out there in the numerator. Plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So let me write it here. X plus two times x plus three. I still have that plus two sitting out there in the numerator. Plus two. And then all of that over x plus two. Or I could write this as being over x plus two. And this being over x plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "Plus two. And then all of that over x plus two. Or I could write this as being over x plus two. And this being over x plus two. All I did is I said, hey, if I have something plus something else over x plus two, that could be the first something over x plus two plus the second something over x plus two. And then here, we could say, hey, look, this first part, as long as x does not equal negative two, because then we would be changing the domain, then these two would cancel out. You could say, hey, I'm just dividing the numerator and the denominator by x plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "And this being over x plus two. All I did is I said, hey, if I have something plus something else over x plus two, that could be the first something over x plus two plus the second something over x plus two. And then here, we could say, hey, look, this first part, as long as x does not equal negative two, because then we would be changing the domain, then these two would cancel out. You could say, hey, I'm just dividing the numerator and the denominator by x plus two. And so this would be equal to x plus three plus, and I don't necessarily even have to put parentheses there, plus two over x plus two. And I would have to constrain the domain. So this is for x does not equal negative two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "You could say, hey, I'm just dividing the numerator and the denominator by x plus two. And so this would be equal to x plus three plus, and I don't necessarily even have to put parentheses there, plus two over x plus two. And I would have to constrain the domain. So this is for x does not equal negative two. So in this situation, we had a remainder. And people will refer to the two as the remainder. We divide as far as we can, but we still have, it still remains to be done to divide the two by x plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So this is for x does not equal negative two. So in this situation, we had a remainder. And people will refer to the two as the remainder. We divide as far as we can, but we still have, it still remains to be done to divide the two by x plus two. And so we would refer to the two as the remainder. Now that wasn't too difficult, but it also wasn't too straightforward. And we'll see that in this is a situation where the algebraic long division is actually a little bit more straightforward."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "We divide as far as we can, but we still have, it still remains to be done to divide the two by x plus two. And so we would refer to the two as the remainder. Now that wasn't too difficult, but it also wasn't too straightforward. And we'll see that in this is a situation where the algebraic long division is actually a little bit more straightforward. So let's try that out. And once again, pause this video and see if you can figure out what this is through algebraic long division. So we're trying to take x plus two and divide it into x squared plus five x plus eight."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "And we'll see that in this is a situation where the algebraic long division is actually a little bit more straightforward. So let's try that out. And once again, pause this video and see if you can figure out what this is through algebraic long division. So we're trying to take x plus two and divide it into x squared plus five x plus eight. Look at the highest degree terms, the x and the x squared. X goes into x squared x times. Put it in the first degree column."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So we're trying to take x plus two and divide it into x squared plus five x plus eight. Look at the highest degree terms, the x and the x squared. X goes into x squared x times. Put it in the first degree column. X times two is two x. X times x is x squared. Subtract these from x squared plus five x. And we get five x minus two x is three x. X squared minus x squared."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "Put it in the first degree column. X times two is two x. X times x is x squared. Subtract these from x squared plus five x. And we get five x minus two x is three x. X squared minus x squared. That's just zero. Bring down that eight. Look at the highest degree term."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "And we get five x minus two x is three x. X squared minus x squared. That's just zero. Bring down that eight. Look at the highest degree term. And we get x goes into three x three times. Put that in the constant column or the zeroth degree column. So plus three."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "Look at the highest degree term. And we get x goes into three x three times. Put that in the constant column or the zeroth degree column. So plus three. Three times two is six. Three times x is three x. Subtract these. And we're left with, let me scroll down a little bit."}, {"video_title": "Dividing quadratics by linear expressions with remainders Algebra 2 Khan Academy.mp3", "Sentence": "So plus three. Three times two is six. Three times x is three x. Subtract these. And we're left with, let me scroll down a little bit. You're left with, those cancel out, and you're left with eight minus six which is indeed equal to two. And we could say, hey, we don't really know how to divide x plus two into two for an arbitrary x. So we will say, hey, this is going to be equal to x plus three with a remainder, with a remainder of two."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Her height above the ground in meters is modeled by h of t, where t is the time in seconds, and we can see that right over here. Now what I wanna focus on this video is some features of this graph, and the features we're going to focus on, actually the first of them, is going to be the midline. So pause this video and see if you can figure out the midline of this graph, or the midline of this function, and then we're gonna think about what it actually represents. Well, Alexa starts off at five meters above the ground, and then she goes higher and higher and higher, gets as high as 25 meters, and then goes back as low as five meters above the ground, then as high as 25 meters, and what we can view the midline as is the midpoint between these extremes, or the average of these extremes. Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Well, Alexa starts off at five meters above the ground, and then she goes higher and higher and higher, gets as high as 25 meters, and then goes back as low as five meters above the ground, then as high as 25 meters, and what we can view the midline as is the midpoint between these extremes, or the average of these extremes. Well, the extremes are, she goes as low as five and as high as 25. So what's the average of five and 25? Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Well, that would be 15. So the midline would look something like this, and I'm actually gonna keep going off the graph, and the reason is is to help us think about what does that midline even represent? And one way to think about it is it represents the center of our rotation in this situation, or how high above the ground is the center of our Ferris wheel? And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "And to help us visualize that, let me draw a Ferris wheel. So I'm going to draw a circle with this as the center, and so the Ferris wheel would look something like, would look something like this, and it has some type of maybe support structure. So the Ferris wheel might look something like that, and this height above the ground, that is 15 meters, that is what the midline is representing. Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Now, the next feature I want to explore is the amplitude. Pause this video and think about what is the amplitude of this oscillating function right over here, and then we'll think about what does that represent in the real world, or where does it come from in the real world? Well, the amplitude is the maximum difference or the maximum magnitude away from that midline, and you can see it right over here, actually right when Alexa starts, we have starting 10 meters below the midline, 10 meters below the center, and this is when Alexa is right over here. She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "She is 10 meters below the midline, and then after a, looks like 10 seconds, she is right at the midline, so that means that she's right over here. Maybe the Ferris wheel is going this way, at least in my imagination, it's going clockwise, and then after another 10 seconds, she is at 25 meters, so she is right over there, and you can see that. She is right over there. I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that."}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "I drew that circle intentionally of that size, and so we see the amplitude in full effect, 10 meters below to begin the midline and 10 meters above, and so it's the maximum displacement or the maximum change from that midline, and so over here, it really represents the radius of our Ferris wheel, 10 meters, and then from this part, she starts going back down again, and then over here, she's back to where she started. Now, the last feature I want to explore is the notion of a period. What is the period of this periodic function? Pause this video and think about that. Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation?"}, {"video_title": "Interpreting trigonometric graphs in context Trigonometry Algebra Khan Academy.mp3", "Sentence": "Pause this video and think about that. Well, the period is how much time does it take to complete one cycle? So here, she's starting at the bottom, and let's see, after 10 seconds, not at the bottom yet, after 20 seconds, not at the bottom yet, after 30 seconds, not at the bottom yet, and then here she is, after 40 seconds, she's back at the bottom and about to head up again, and so this time right over here, that 40 seconds, that is the period, and if you think about what's going on over here, she starts over here, five meters above the ground, after 10 seconds, she is right over here, and that corresponds to this point right over here, after 10 more seconds, she's right over there, that corresponds to that point, after 10 more seconds, she's over here, that corresponds to that, and after 10 more seconds, or a total of 40 seconds, she is back to where she started. So the period in this example shows how long does it take to complete one full rotation? Now we have to be careful sometimes when we're trying to visually inspect the period, because sometimes it might be tempting to say, start right over here and say, okay, we're 15 meters above the ground, all right, let's see, we're going down, now we're going up again, and look, we're 15 meters above the ground, maybe this 20 seconds is a period, but when you look at it over here, it's clear that that is not the case. This point represents this point at being 15 meters above the ground, going down, that's getting us to this point, and then after another 10 seconds, we get back over here. Notice, all this is measuring is half of a cycle, going halfway around."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is a screenshot of Desmos. It's an online graphing calculator. And what we're going to do is use it to understand how we can go about scaling functions. And I encourage you to go to Desmos and try it on your own, either during this video or after. So let's start with a nice, interesting function. Let's say f of x is equal to the absolute value of x. So that's pretty straightforward."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And I encourage you to go to Desmos and try it on your own, either during this video or after. So let's start with a nice, interesting function. Let's say f of x is equal to the absolute value of x. So that's pretty straightforward. Now let's try to create a scaled version of f of x. So we could say g of x is equal to, well, I'll start with just absolute value of x. So it's the same as f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So that's pretty straightforward. Now let's try to create a scaled version of f of x. So we could say g of x is equal to, well, I'll start with just absolute value of x. So it's the same as f of x. So it just traced g of x right on top of f. But now let's multiply it by some constant. Let's multiply it by two. So notice the difference between g of x and f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it's the same as f of x. So it just traced g of x right on top of f. But now let's multiply it by some constant. Let's multiply it by two. So notice the difference between g of x and f of x. And you can see that g of x is just two times f of x. In fact, we can write it this way. We can write g of x is equal to two times f of x."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So notice the difference between g of x and f of x. And you can see that g of x is just two times f of x. In fact, we can write it this way. We can write g of x is equal to two times f of x. We get to the exact same place. But you can see that as our x increases, g of x increases twice as fast, at least for positive x's, on the right-hand side. And actually, as x decreases, g of x also increases twice as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We can write g of x is equal to two times f of x. We get to the exact same place. But you can see that as our x increases, g of x increases twice as fast, at least for positive x's, on the right-hand side. And actually, as x decreases, g of x also increases twice as fast. So is that just a coincidence that we have a two here and it increased twice as fast? Well, let's put a three here. Well, now it looks like it's increasing three times as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And actually, as x decreases, g of x also increases twice as fast. So is that just a coincidence that we have a two here and it increased twice as fast? Well, let's put a three here. Well, now it looks like it's increasing three times as fast. And it does that in both directions. Now what if we were to put a 0.5 here? 0.5."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, now it looks like it's increasing three times as fast. And it does that in both directions. Now what if we were to put a 0.5 here? 0.5. Well, now it looks like it's increasing half as fast. And that makes sense because we are just multiplying, we are scaling how much our f of x is. So before, when x equals one, we got to one."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "0.5. Well, now it looks like it's increasing half as fast. And that makes sense because we are just multiplying, we are scaling how much our f of x is. So before, when x equals one, we got to one. But now when x equals one, we only get to 1 1\u20442. Before, when x equals five, we got to five. Now when we get to x equals five, we only get to 2.5."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So before, when x equals one, we got to one. But now when x equals one, we only get to 1 1\u20442. Before, when x equals five, we got to five. Now when we get to x equals five, we only get to 2.5. So we're increasing half as fast, or we have half the slope. Now an interesting question to think about is what would happen if, instead of it just being an absolute value of x, let's say we were to have a non-zero y-intercept. So let's say, I don't know, plus six."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now when we get to x equals five, we only get to 2.5. So we're increasing half as fast, or we have half the slope. Now an interesting question to think about is what would happen if, instead of it just being an absolute value of x, let's say we were to have a non-zero y-intercept. So let's say, I don't know, plus six. So notice, then when we change this constant out front, it not only changes the slope, but it changes the y-intercept because we're multiplying this entire expression by 0.5. So if you multiply it by one, we're back to where we got before. And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's say, I don't know, plus six. So notice, then when we change this constant out front, it not only changes the slope, but it changes the y-intercept because we're multiplying this entire expression by 0.5. So if you multiply it by one, we're back to where we got before. And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two. And we see that. It not only doubles the slope, but it also increases the y-intercept. If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And now if we multiply it by two, this should increase the y-intercept because remember, we're multiplying both of these terms by two. And we see that. It not only doubles the slope, but it also increases the y-intercept. If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before. And we can see this more generally if we just put a general constant here and we can add a slider. And actually, let me make the constant go from zero to 10 with a step of, I don't know, 0.05. That's just how much does it increase every time you change the slider."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "If we go to 0.5, not only did it decrease the slope by a factor of 1\u20442, or I guess you could say multiply the slope by 1\u20442, but it also made our y-intercept be half of what it was before. And we can see this more generally if we just put a general constant here and we can add a slider. And actually, let me make the constant go from zero to 10 with a step of, I don't know, 0.05. That's just how much does it increase every time you change the slider. And notice, when we increase our constant, not only are we getting narrower because the magnitude of the slope is being scaled, but our y-intercept increases. And then as k decreases, our y-intercept is being scaled down and our slope is being scaled down. Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's just how much does it increase every time you change the slider. And notice, when we increase our constant, not only are we getting narrower because the magnitude of the slope is being scaled, but our y-intercept increases. And then as k decreases, our y-intercept is being scaled down and our slope is being scaled down. Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x? So instead of k times f of x, what if we did f of k times x? Another way to think about it is g of x is now equal to the absolute value of kx plus six. What do you think is going to happen?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now that's one way that we could go about scaling, but what if instead of multiplying our entire function by some constant, we instead just replace the x with a constant times x? So instead of k times f of x, what if we did f of k times x? Another way to think about it is g of x is now equal to the absolute value of kx plus six. What do you think is going to happen? Pause this video and think about it. Well, now when we increase k, notice it has no impact on our y-intercept because it's not scaling the y-intercept, but it does have an impact on slope. When k goes from one to two, once again, we are now increasing twice as fast."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What do you think is going to happen? Pause this video and think about it. Well, now when we increase k, notice it has no impact on our y-intercept because it's not scaling the y-intercept, but it does have an impact on slope. When k goes from one to two, once again, we are now increasing twice as fast. And then when k goes from one to 1 1\u20442, we're now increasing half as fast. Now this is with an absolute value function. What if we did it with a different type of function?"}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When k goes from one to two, once again, we are now increasing twice as fast. And then when k goes from one to 1 1\u20442, we're now increasing half as fast. Now this is with an absolute value function. What if we did it with a different type of function? Let's say we did it with a quadratic. So two minus x squared, and let me scroll down a little bit. And so you can see when k equals one, these are the same."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What if we did it with a different type of function? Let's say we did it with a quadratic. So two minus x squared, and let me scroll down a little bit. And so you can see when k equals one, these are the same. And now if we increase our k, let's say we increase our k to two, notice our parabola is, in this case, decreasing as we get further and further from zero at a faster and faster rate. That's because what you would have seen at x equals two, you're now seeing at x equals one because you are multiplying two times that. And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so you can see when k equals one, these are the same. And now if we increase our k, let's say we increase our k to two, notice our parabola is, in this case, decreasing as we get further and further from zero at a faster and faster rate. That's because what you would have seen at x equals two, you're now seeing at x equals one because you are multiplying two times that. And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate. It's a changing rate, but it's a lower changing rate. I guess you could put it that way. And we could also try just to see what happens with our parabola here."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so, and then if we go between zero and one, notice on either side of zero, our parabola is decreasing at a lower rate. It's a changing rate, but it's a lower changing rate. I guess you could put it that way. And we could also try just to see what happens with our parabola here. If instead of doing kx, we once again put the k out front, what is that going to do? And notice, that is changing not only how fast the curve changes at different points, but it's now also changing the y-intercept because we are now scaling that y-intercept. So I'll leave you there."}, {"video_title": "Scaling functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we could also try just to see what happens with our parabola here. If instead of doing kx, we once again put the k out front, what is that going to do? And notice, that is changing not only how fast the curve changes at different points, but it's now also changing the y-intercept because we are now scaling that y-intercept. So I'll leave you there. This is just the beginning of thinking about scaling. I really want you to build an intuitive sense of what is going on here and really think about mathematically why it makes sense. And go on to Desmos and play around with it yourself and also try other types of functions and see what happens."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The midline is a line, a horizontal line, where half of the function is above it and half of the function is below it. Then I want you to think about the amplitude. How far does this function vary from that midline? Either how far above does it go or how far does it go below? And it should be the same amount because the midline should be between the highest and the lowest points. And then finally, think about what the period of this function is. How much do you have to have a change in x to get to the same point in the cycle of this periodic function?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Either how far above does it go or how far does it go below? And it should be the same amount because the midline should be between the highest and the lowest points. And then finally, think about what the period of this function is. How much do you have to have a change in x to get to the same point in the cycle of this periodic function? So I encourage you to pause the video now and think about those questions. So let's tackle the midline first. So one way to think about it is, well, how high does this function go?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "How much do you have to have a change in x to get to the same point in the cycle of this periodic function? So I encourage you to pause the video now and think about those questions. So let's tackle the midline first. So one way to think about it is, well, how high does this function go? Well, the highest y value for this function we see is four. It keeps hitting four on a fairly regular basis. And we'll talk about how regular that is when we talk about the period."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So one way to think about it is, well, how high does this function go? Well, the highest y value for this function we see is four. It keeps hitting four on a fairly regular basis. And we'll talk about how regular that is when we talk about the period. And what's the lowest value that this function gets to? Well, it gets to y equals negative two. So what's halfway between four and negative two?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we'll talk about how regular that is when we talk about the period. And what's the lowest value that this function gets to? Well, it gets to y equals negative two. So what's halfway between four and negative two? Well, you could eyeball it or you could count or you could literally just take the average between four and negative two. So four, so we could, the midline is going to be the horizontal line y is equal to four plus negative two over two, just literally the mean, the arithmetic mean between four and negative two, the average of four and negative two, which is just going to be equal to one. So the line y equals one is the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's halfway between four and negative two? Well, you could eyeball it or you could count or you could literally just take the average between four and negative two. So four, so we could, the midline is going to be the horizontal line y is equal to four plus negative two over two, just literally the mean, the arithmetic mean between four and negative two, the average of four and negative two, which is just going to be equal to one. So the line y equals one is the midline. So that's the midline right over here. And you see that, you see that it's kind of cutting the function and it's where you have half of the function is above it and half of the function is below it. So that's the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the line y equals one is the midline. So that's the midline right over here. And you see that, you see that it's kind of cutting the function and it's where you have half of the function is above it and half of the function is below it. So that's the midline. Midline. Now let's think about the amplitude. Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's the midline. Midline. Now let's think about the amplitude. Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline. And the midline's in the middle, so it's going to be the same amount whether you go above or below. So if you, one way to say it is, well, at this maximum point right over here, how far above the midline is this? How far above the midline is this?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is how much this function varies from the midline, either above the midline or below the midline. And the midline's in the middle, so it's going to be the same amount whether you go above or below. So if you, one way to say it is, well, at this maximum point right over here, how far above the midline is this? How far above the midline is this? Well, to get from one to four, you have to go, you're three above the midline, or another way of thinking about it, this maximum point is y equals four minus y equals one. Well, you had a, your y can go as much as three above the midline. Or you could say your y value could be as much as three below the midline."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "How far above the midline is this? Well, to get from one to four, you have to go, you're three above the midline, or another way of thinking about it, this maximum point is y equals four minus y equals one. Well, you had a, your y can go as much as three above the midline. Or you could say your y value could be as much as three below the midline. That's this point right over here. One minus three is negative one. So your amplitude right over here is equal to three."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or you could say your y value could be as much as three below the midline. That's this point right over here. One minus three is negative one. So your amplitude right over here is equal to three. You can vary as much as three, either above the midline or below the midline. Finally, the period. And when I think about the period, I try to look for a relatively convenient spot on the curve."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So your amplitude right over here is equal to three. You can vary as much as three, either above the midline or below the midline. Finally, the period. And when I think about the period, I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's at a nice, when x is at negative two, y is at one, it's at a nice integer value. And so what I want to do is keep traveling along this curve until I get to the same y value, but not just the same y value, but the y value, I get the same y value and I'm also traveling in the same direction. So for example, let's travel along this curve."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And when I think about the period, I try to look for a relatively convenient spot on the curve. And I'm calling this a convenient spot because it's at a nice, when x is at negative two, y is at one, it's at a nice integer value. And so what I want to do is keep traveling along this curve until I get to the same y value, but not just the same y value, but the y value, I get the same y value and I'm also traveling in the same direction. So for example, let's travel along this curve. So essentially, our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here?"}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So for example, let's travel along this curve. So essentially, our x is increasing. Our x keeps increasing. Now you might say, hey, have I completed a cycle here? Because once again, y is equal to one. You haven't completed a cycle here because notice over here, our y is increasing as x increases, while here our y is decreasing as x increases. Our slope is positive here, our slope is negative here."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now you might say, hey, have I completed a cycle here? Because once again, y is equal to one. You haven't completed a cycle here because notice over here, our y is increasing as x increases, while here our y is decreasing as x increases. Our slope is positive here, our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals one, or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our slope is positive here, our slope is negative here. So this isn't the same point on the cycle. We need to get to the point where y once again equals one, or we could say, especially in this case, we're at the midline again, but our slope is increasing. So let's just keep going. Let's just keep going. So that gets us to right over there. So notice, now we have completed one cycle."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's just keep going. Let's just keep going. So that gets us to right over there. So notice, now we have completed one cycle. So the change in x needed to complete one cycle, that is your period. So to go from negative two to zero, your period is two. So your period here is two, and you could do it again."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So notice, now we have completed one cycle. So the change in x needed to complete one cycle, that is your period. So to go from negative two to zero, your period is two. So your period here is two, and you could do it again. So we're at that point. Let's see, we wanna get back to a point where we're at the midline, and I just happened to start right over here at the midline. I could've started really at any point."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So your period here is two, and you could do it again. So we're at that point. Let's see, we wanna get back to a point where we're at the midline, and I just happened to start right over here at the midline. I could've started really at any point. You wanna get to the same point, but also where the slope is the same. We're at the same point in the cycle once again. So I could go, so if I travel one, I'm at the midline again, but I'm now going down."}, {"video_title": "Midline, amplitude and period of a function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I could've started really at any point. You wanna get to the same point, but also where the slope is the same. We're at the same point in the cycle once again. So I could go, so if I travel one, I'm at the midline again, but I'm now going down. So I have to go further. Now I am back at that same point in the cycle. I'm at y equals one, and the slope is positive."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The longest day of the year in Juneau, Alaska is June 21st. It's 1,096.5 minutes long. Half a year later, when the days are at their shortest, the days are about 382.5 minutes long. If it's not a leap year, the year is 365 days long. And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If it's not a leap year, the year is 365 days long. And June 21st is the 172nd day of the year. Write a trigonometric function that models the length, L, of the t-th day of the year. So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be L as a function of t, assuming it's not a leap year. So I encourage you to pause this video and try to do this on your own before I try to work through it. So let me give a go at it. So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So instead of first starting at L of t, I'm going to actually start with L as a function of u, where u is another variable. I'll just use this kind of an intermediary variable that'll help set it up in a little bit of a simpler way. Where u is days after June 21st. So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's just think about this a little bit. June 21st, if we're thinking about in terms of u, u is going to be equal to 0, because it's 0 days after June 21st. But if we're thinking about in terms of t, June 21st is the 172nd day of the year. So 172. So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So 172. So what's the relationship between u and t? Well, it's shifted by 172 days. u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "u is going to be equal to t minus 172. Notice when t is 172, u is equal to 0. So let's figure out L of u first, and then later we can just substitute u with t minus 172. So first of all, what's happening when u is equal to 0? Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So first of all, what's happening when u is equal to 0? Let me write all this down. So what is happening when u is equal to 0? Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, u equals 0 is June 21st, and that's the maximum point. So what trig function hits its maximum point when the input into the trig function is 0? Well, sine of 0 is 0, while cosine of 0 is 1. Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Cosine hits its maximum point. So it seems a little bit easier to model this with a cosine. So it's going to be some amplitude times cosine of, and let's say I'll write some coefficient c right over here. Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Actually, let me just use a b, since I already used an a. Some coefficient over here times our u plus some constant. That'll shift the entire function up or down. So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is the form that our function of u is going to take. And now we just have to figure out what each of these parts are. So first let's think about the amplitude and what the midline is going to be. The midline is essentially how much we're shifting the function up. So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The midline is essentially how much we're shifting the function up. So let's get our calculator out. So the midline is going to be halfway between these two numbers, so we could say 1,096.5 plus 382.5 divided by 2 gets us to 739.5. So that's what c is equal to. c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's what c is equal to. c is equal to 739.5. Now the amplitude is how much do we vary from that midline? So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we could take 1,096 minus this, or we could take this minus 382.5. So let's do that. So let's take 1,096.5 minus what we just got, 739.5. And we get 357. So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we get 357. So this is how much we vary from that midline. So a is equal to 357. So this right over here is equal to 357. So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this right over here is equal to 357. So what's b going to be equal to? And for that, I always think about, well, what's the behavior of the function? What's the period of the function going to be? Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's the period of the function going to be? Let me make a little table here. So when u, let's put some different inputs from u. When u is 0, we're 0 days after June 21. We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens?"}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When u is 0, we're 0 days after June 21. We're at our maximum point. And we already said that what we want the cosine function to evaluate to at that point is essentially we want it to evaluate as 357 times cosine of 0 plus 739.5. Now what happens? What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what happens? What's a full period? Well, a full period is a year. At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "At a year, we get to the same point in the year, which is, I guess, a little bit of common sense. So you go all the way to 365. When u is 365, we should have completed a period. We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We should be back to that maximum point. So this should essentially be 357 times cosine of 2 pi. If we were just thinking in terms of a traditional trig function, if we just had a theta in here, you complete a period every 2 pi. So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this should be equivalent to what I'm writing out right over here, plus 739.5. So one way to think about it is b times 365 should be equal to 2 pi. Notice, this is going to be b times 365. So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's write that down. b times 365, that's the input into the cosine function, needs to be equal to 2 pi. Or b is equal to 2 pi over 365. b is equal to 2 pi over 365. And we are almost done. We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we are almost done. We figured out what a, b, and c are. Now we just have to substitute u with t minus 172 to get our function of t. So let's just do that. So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5. And we are done. So it seems like a very complicated expression, but if you just break it down and kind of think about it, make the point that we're talking about, the extreme point, either the minimum or the maximum, make that something, make that when the input into our function is 0 or 2 pi."}, {"video_title": "Day length in Alaska Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we get, we deserve a little bit of a drum roll now, L of t is equal to a, which is 357, times cosine of b, 2 pi over 365, times not u, but now we're going to write it in terms of t. We want to think about day of the year, not days after June 21st. So times t minus 172, and then finally plus our midline, plus 739.5. And we are done. So it seems like a very complicated expression, but if you just break it down and kind of think about it, make the point that we're talking about, the extreme point, either the minimum or the maximum, make that something, make that when the input into our function is 0 or 2 pi. 0 is actually the easiest one. And then later you can worry about the shift. Hopefully you found that helpful."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We saw that if you took 3x squared minus 4x plus 7, and you divided by x minus 1, you got 3x minus 1 with a remainder of 6. And when we do polynomial long division, how do we know when we got to our remainder? Well, when we get to an expression that has a lower degree than the thing that is the divisor, the thing that we're dividing into the other thing. And so in this example, we could have rewritten what we just did right over here as our f of x. Let me just write it right over here. So we could have said 3x squared minus 4x plus 7 is equal to x minus 1 times the quotient right over here, or you could say the quotient times x minus 1. So it's going to be equal to this business."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And so in this example, we could have rewritten what we just did right over here as our f of x. Let me just write it right over here. So we could have said 3x squared minus 4x plus 7 is equal to x minus 1 times the quotient right over here, or you could say the quotient times x minus 1. So it's going to be equal to this business. It's going to be equal to 3x minus 1 times the divisor. Times x minus 1. And then you're still going to, and this isn't going to, when you multiply these two things, you're not going to get exactly this."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it's going to be equal to this business. It's going to be equal to 3x minus 1 times the divisor. Times x minus 1. And then you're still going to, and this isn't going to, when you multiply these two things, you're not going to get exactly this. You still have to add the remainder. So plus the remainder. And actually, let me write the actual remainder down."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then you're still going to, and this isn't going to, when you multiply these two things, you're not going to get exactly this. You still have to add the remainder. So plus the remainder. And actually, let me write the actual remainder down. So plus 6. And the analogy here is exactly the analogy to when you did traditional division. If I were to say, actually let me just show you the analogy."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And actually, let me write the actual remainder down. So plus 6. And the analogy here is exactly the analogy to when you did traditional division. If I were to say, actually let me just show you the analogy. If I were to say 25 divided by 4, you would say, okay, well 4 goes into 25 6 times. 6 times 4 is 24. You would subtract, and then you would get a remainder 1."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If I were to say, actually let me just show you the analogy. If I were to say 25 divided by 4, you would say, okay, well 4 goes into 25 6 times. 6 times 4 is 24. You would subtract, and then you would get a remainder 1. Or another way of saying this is you could say that 25, 25 is equal to 6 times 4, is equal to 6 times 4, 6 times 4 plus 1. Plus 1. And so we just did the exact same thing here, but we just did it with expressions."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You would subtract, and then you would get a remainder 1. Or another way of saying this is you could say that 25, 25 is equal to 6 times 4, is equal to 6 times 4, 6 times 4 plus 1. Plus 1. And so we just did the exact same thing here, but we just did it with expressions. And so, once again, I haven't started the proof yet. I just wanted to make you feel comfortable with what I just wrote right over here. If I divided this expression into this polynomial, and I were to get this quotient, that's the same thing as saying this polynomial could be equal to 3x minus 1 times x minus 1 plus 6."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And so we just did the exact same thing here, but we just did it with expressions. And so, once again, I haven't started the proof yet. I just wanted to make you feel comfortable with what I just wrote right over here. If I divided this expression into this polynomial, and I were to get this quotient, that's the same thing as saying this polynomial could be equal to 3x minus 1 times x minus 1 plus 6. Now this is true in general. Let's abstract a little bit. This is our f of x."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If I divided this expression into this polynomial, and I were to get this quotient, that's the same thing as saying this polynomial could be equal to 3x minus 1 times x minus 1 plus 6. Now this is true in general. Let's abstract a little bit. This is our f of x. So that is f of x. So f of x is going to be equal to whatever the quotient is. And let me call that q of x."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This is our f of x. So that is f of x. So f of x is going to be equal to whatever the quotient is. And let me call that q of x. So, this is a different color. So I'm going to call that q of x. q of x. This right over here is q of x."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And let me call that q of x. So, this is a different color. So I'm going to call that q of x. q of x. This right over here is q of x. So f of x is going to be equal to the quotient, q of x, times, this is our x minus a. In this case, a is 1, but I'm just trying to generalize it a little bit. So x minus a, and then plus the remainder."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This right over here is q of x. So f of x is going to be equal to the quotient, q of x, times, this is our x minus a. In this case, a is 1, but I'm just trying to generalize it a little bit. So x minus a, and then plus the remainder. Plus the remainder. And we know that the remainder is going to be a constant, because the remainder is going to have a lower degree than x minus a. And x minus a is first degree."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So x minus a, and then plus the remainder. Plus the remainder. And we know that the remainder is going to be a constant, because the remainder is going to have a lower degree than x minus a. And x minus a is first degree. So in order to have a lower degree, this has to be zero's degree. This has to be a constant. So this is true in general."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And x minus a is first degree. So in order to have a lower degree, this has to be zero's degree. This has to be a constant. So this is true in general. This is true for any polynomial, f of x, divided by any x minus a. So this is just true. So this is true for any, f of x and x minus a."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is true in general. This is true for any polynomial, f of x, divided by any x minus a. So this is just true. So this is true for any, f of x and x minus a. Now, what is going to happen if we evaluate f of a? Well, if f of x is indeed, can be written like this, well we could write f of, let me do a in a new color so it sticks out. We could write f of a would be equal to, would be equal to q of a, so q of a, q of a times, I think you might see where this is going, times a minus a, times a minus a, minus a plus r. Well, what's that going to be equal to?"}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is true for any, f of x and x minus a. Now, what is going to happen if we evaluate f of a? Well, if f of x is indeed, can be written like this, well we could write f of, let me do a in a new color so it sticks out. We could write f of a would be equal to, would be equal to q of a, so q of a, q of a times, I think you might see where this is going, times a minus a, times a minus a, minus a plus r. Well, what's that going to be equal to? What's all this business going to be equal to? Well, a minus a is zero. And q of a, I don't care what q of a is, if you're going to multiply it by zero, all of this is going to be zero."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We could write f of a would be equal to, would be equal to q of a, so q of a, q of a times, I think you might see where this is going, times a minus a, times a minus a, minus a plus r. Well, what's that going to be equal to? What's all this business going to be equal to? Well, a minus a is zero. And q of a, I don't care what q of a is, if you're going to multiply it by zero, all of this is going to be zero. So f of a, f of a is going to be equal to, is going to be equal to r. And so you're done. This is the proof of the polynomial remainder theorem. Any function, if when you divide it by x minus a, you get the quotient q of x and the remainder r, it can then be written in this way."}, {"video_title": "Polynomial remainder theorem proof Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And q of a, I don't care what q of a is, if you're going to multiply it by zero, all of this is going to be zero. So f of a, f of a is going to be equal to, is going to be equal to r. And so you're done. This is the proof of the polynomial remainder theorem. Any function, if when you divide it by x minus a, you get the quotient q of x and the remainder r, it can then be written in this way. And if it's written in this way, and you evaluate it at f of a, and you put the a over here, you're going to see that f of a is going to be whatever that remainder was. And that is the polynomial remainder theorem. And we're done."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we're told that the negative of the cube root of y is equal to 4 times the cube root of y plus 5. So in all of these, it's helpful to just be able to isolate the cube root, isolate the radical in the equation, and then solve from there. So let's see if we can isolate the radical. So the simplest thing to do, if we want all of the radical onto the left-hand side of the equation, we can subtract 4 times the cube root of y from both sides of this equation. So let's subtract 4 times the cube root of y from both sides of this equation. 4 times the cube root of y from both sides. And so your left-hand side, you already have negative 1 times the cube root of y."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the simplest thing to do, if we want all of the radical onto the left-hand side of the equation, we can subtract 4 times the cube root of y from both sides of this equation. So let's subtract 4 times the cube root of y from both sides of this equation. 4 times the cube root of y from both sides. And so your left-hand side, you already have negative 1 times the cube root of y. Then you're going to subtract 4 more of the cube root of y. So you're going to have negative 5 times the cube root of y. That's your left-hand side."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And so your left-hand side, you already have negative 1 times the cube root of y. Then you're going to subtract 4 more of the cube root of y. So you're going to have negative 5 times the cube root of y. That's your left-hand side. Now the right-hand side, these two guys canceled out. That was the whole point behind subtracting this value. So that cancels out."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "That's your left-hand side. Now the right-hand side, these two guys canceled out. That was the whole point behind subtracting this value. So that cancels out. And you're just left with a 5 there. You're just left with this 5 right over there. Now we've almost isolated this cube root of y."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So that cancels out. And you're just left with a 5 there. You're just left with this 5 right over there. Now we've almost isolated this cube root of y. We just have to divide both sides of the equation by negative 5. So we just divide both sides of this equation by negative 5. And these cancel out."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now we've almost isolated this cube root of y. We just have to divide both sides of the equation by negative 5. So we just divide both sides of this equation by negative 5. And these cancel out. That was the whole point. And we are left with the cube root of y is equal to 5 divided by negative 5 is negative 1. Now the cube root of y is equal to negative 1."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And these cancel out. That was the whole point. And we are left with the cube root of y is equal to 5 divided by negative 5 is negative 1. Now the cube root of y is equal to negative 1. Well, the easiest way to solve this is let's take both sides of this equation to the third power. And another way, this statement right here is the exact same statement. This is the exact same statement as saying y to the 1 third is equal to negative 1."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now the cube root of y is equal to negative 1. Well, the easiest way to solve this is let's take both sides of this equation to the third power. And another way, this statement right here is the exact same statement. This is the exact same statement as saying y to the 1 third is equal to negative 1. These are just two different ways of writing the same thing. This is equivalent to taking the 1 third power. So if we take both sides of this equation to the third power, that's like taking both sides of this equation to the third power."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "This is the exact same statement as saying y to the 1 third is equal to negative 1. These are just two different ways of writing the same thing. This is equivalent to taking the 1 third power. So if we take both sides of this equation to the third power, that's like taking both sides of this equation to the third power. And you can see here, y to the 1 third to the third, and then to the third, that's like saying y to the 1 third times 3 power, or y to the first power. That's the whole point of it. If you take the cube root of y to the third power, that's just going to be y."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So if we take both sides of this equation to the third power, that's like taking both sides of this equation to the third power. And you can see here, y to the 1 third to the third, and then to the third, that's like saying y to the 1 third times 3 power, or y to the first power. That's the whole point of it. If you take the cube root of y to the third power, that's just going to be y. So the left-hand side becomes y. And then the right-hand side, what's negative 1 to the third power? Negative 1 times negative 1 is 1, times negative 1 again is negative 1."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "If you take the cube root of y to the third power, that's just going to be y. So the left-hand side becomes y. And then the right-hand side, what's negative 1 to the third power? Negative 1 times negative 1 is 1, times negative 1 again is negative 1. So we get y is equal to negative 1 as our solution. Now let's make sure that it actually works. Let's go back to our original equation."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Negative 1 times negative 1 is 1, times negative 1 again is negative 1. So we get y is equal to negative 1 as our solution. Now let's make sure that it actually works. Let's go back to our original equation. And I'll put negative 1 in for our y's. We had the negative of the cube root of this time negative 1 has to be equal to 4 times the cube root of negative 1 plus 5. Let's verify that this is the case."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Let's go back to our original equation. And I'll put negative 1 in for our y's. We had the negative of the cube root of this time negative 1 has to be equal to 4 times the cube root of negative 1 plus 5. Let's verify that this is the case. The cube root of negative 1 is negative 1. Negative 1 to the third power is negative 1. So this is equal to the negative of negative 1 has to be equal to 4 times the cube root of negative 1 is negative 1 plus 5."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Let's verify that this is the case. The cube root of negative 1 is negative 1. Negative 1 to the third power is negative 1. So this is equal to the negative of negative 1 has to be equal to 4 times the cube root of negative 1 is negative 1 plus 5. The negative of negative 1 is just positive 1. So 1 needs to be equal to 4 times negative 1, negative 4 plus 5. This is true."}, {"video_title": "Solving radical equations 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this is equal to the negative of negative 1 has to be equal to 4 times the cube root of negative 1 is negative 1 plus 5. The negative of negative 1 is just positive 1. So 1 needs to be equal to 4 times negative 1, negative 4 plus 5. This is true. Negative 4 plus 5 is 1. So this works out. This is our solution."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So we just have a bunch of real parts and imaginary parts that we can then add up together. So we have 2 minus 3i, and then we're subtracting this entire quantity. And to get rid of the parentheses, we can just distribute the negative sign, or another way to think about it, we can say that this is negative 1 times all of this. So we can just distribute the negative sign, and negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6, and then negative 1 times negative 18i, well, that's just going to be positive. Positive 18i."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So we can just distribute the negative sign, and negative 1 times 6 is negative 6. Let me do these in magenta. So this is negative 6, and then negative 1 times negative 18i, well, that's just going to be positive. Positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2, and then we have a minus 6, so we have 2 minus 6, and we want to add the imaginary parts."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "Positive 18i. Negative times a negative is a positive. And now we want to add the real parts, and we want to add the imaginary parts. So here's a real part here, 2, and then we have a minus 6, so we have 2 minus 6, and we want to add the imaginary parts. We have a negative, let me do that in a different color, we have a negative 3i right over here, so negative 3i, or minus 3i right over there, and then we have a plus 18i, or positive 18i. So then we have a positive 18i. If you add the real parts, 2 minus 6 is negative 4, and you add the imaginary parts, if I have negative 3 of something, and to that I add 18 of something, well, that's just going to leave me with 15 of that something."}, {"video_title": "Subtracting complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So here's a real part here, 2, and then we have a minus 6, so we have 2 minus 6, and we want to add the imaginary parts. We have a negative, let me do that in a different color, we have a negative 3i right over here, so negative 3i, or minus 3i right over there, and then we have a plus 18i, or positive 18i. So then we have a positive 18i. If you add the real parts, 2 minus 6 is negative 4, and you add the imaginary parts, if I have negative 3 of something, and to that I add 18 of something, well, that's just going to leave me with 15 of that something. Or another way you could think about it, if I have 18 of something, and I subtract 3 of that something, I'll have 15 of that something. And in this case, the something is i, is the imaginary unit. So this is going to be plus 15i, and we are done."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And let's actually just do it. I'll show the initial expression and the form that I want it in, or that we want it in, and then we can talk a little bit about why would we ever actually want to do that. So let's say my expression is 1 30 seconds times 2 to the t power. So this is fairly straightforward exponential expression. But let's say we want it in the form a times b, and this is where it's going to get hairy, a times b to the t over 10th power minus 1. And so you're probably immediately saying, well, why would I ever want to take something nice and simple like this and turn it into this beastly thing right over here? And the answer is when you get into higher math and you start doing your physics and your chemistry, you're going to see maybe you got a result like this, but then you look in your textbook or your professor, it has a result like this."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So this is fairly straightforward exponential expression. But let's say we want it in the form a times b, and this is where it's going to get hairy, a times b to the t over 10th power minus 1. And so you're probably immediately saying, well, why would I ever want to take something nice and simple like this and turn it into this beastly thing right over here? And the answer is when you get into higher math and you start doing your physics and your chemistry, you're going to see maybe you got a result like this, but then you look in your textbook or your professor, it has a result like this. You say, well, how do I transfer from this to this? Or actually, sometimes when you transfer to a form like this, obviously I've just arbitrarily written this form here, but sometimes when you write it in another form that might even be a little hairier, it can give you an intuition on the underlying processes that that expression is trying to describe. So if you can take that on a leap of faith, let's actually try to do it."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And the answer is when you get into higher math and you start doing your physics and your chemistry, you're going to see maybe you got a result like this, but then you look in your textbook or your professor, it has a result like this. You say, well, how do I transfer from this to this? Or actually, sometimes when you transfer to a form like this, obviously I've just arbitrarily written this form here, but sometimes when you write it in another form that might even be a little hairier, it can give you an intuition on the underlying processes that that expression is trying to describe. So if you can take that on a leap of faith, let's actually try to do it. At minimum, it's going to make you a lot better at exponent properties. So see if you can rewrite this in this form. So I'm assuming you took a go at it."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So if you can take that on a leap of faith, let's actually try to do it. At minimum, it's going to make you a lot better at exponent properties. So see if you can rewrite this in this form. So I'm assuming you took a go at it. So let's try to do it together. So the first thing that I might want to do is, well, let's see if we can. Let's see."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So I'm assuming you took a go at it. So let's try to do it together. So the first thing that I might want to do is, well, let's see if we can. Let's see. What would I want to do? The first thing I want to do is take this t and get it into a t over 10. So to do that, we essentially just need to multiply by 10 and divide by 10."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "Let's see. What would I want to do? The first thing I want to do is take this t and get it into a t over 10. So to do that, we essentially just need to multiply by 10 and divide by 10. So let's multiply by 10 and then also divide by 10. Then we haven't changed the value up here. So we can rewrite this as 1 32nd times 2 to the, so let me circle t, let me do this in a different color, t over 10 times 10."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So to do that, we essentially just need to multiply by 10 and divide by 10. So let's multiply by 10 and then also divide by 10. Then we haven't changed the value up here. So we can rewrite this as 1 32nd times 2 to the, so let me circle t, let me do this in a different color, t over 10 times 10. All right. So we got a t over 10 over here. But then I have this times 10."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So we can rewrite this as 1 32nd times 2 to the, so let me circle t, let me do this in a different color, t over 10 times 10. All right. So we got a t over 10 over here. But then I have this times 10. So how do I deal with this? Well, one thing that I could do, let me actually just write this the other way around. Let me write it as 10 times t over 10."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "But then I have this times 10. So how do I deal with this? Well, one thing that I could do, let me actually just write this the other way around. Let me write it as 10 times t over 10. So hopefully what I just did here isn't a huge stretch here. I just literally multiplied and divided by 10. So I have this t over 10."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "Let me write it as 10 times t over 10. So hopefully what I just did here isn't a huge stretch here. I just literally multiplied and divided by 10. So I have this t over 10. But when I write it this way, an exponent property might jump out at you. If I have a to the b and then I raise that to the c, that's going to be a to the bc. Or another way around, a to the bc is going to be a to the b to the c. And so this piece right over here, I can rewrite it as 2 to the 10th and then raise that to the t over 10 power."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So I have this t over 10. But when I write it this way, an exponent property might jump out at you. If I have a to the b and then I raise that to the c, that's going to be a to the bc. Or another way around, a to the bc is going to be a to the b to the c. And so this piece right over here, I can rewrite it as 2 to the 10th and then raise that to the t over 10 power. Once again, 2 to the 10th and then raise that to the t over 10, that's going to be the same thing as 2 to the 10 times t over 10. And of course, we still have the 1 over 32 over here. And I'm tempted to write that as 2 to the negative 5th power, but I won't do that just yet."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "Or another way around, a to the bc is going to be a to the b to the c. And so this piece right over here, I can rewrite it as 2 to the 10th and then raise that to the t over 10 power. Once again, 2 to the 10th and then raise that to the t over 10, that's going to be the same thing as 2 to the 10 times t over 10. And of course, we still have the 1 over 32 over here. And I'm tempted to write that as 2 to the negative 5th power, but I won't do that just yet. So let's see. What's 2 to the 10th power? Actually, let's just keep it as 2 to the 10th power just for simplicity right now."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And I'm tempted to write that as 2 to the negative 5th power, but I won't do that just yet. So let's see. What's 2 to the 10th power? Actually, let's just keep it as 2 to the 10th power just for simplicity right now. Later, you might know that that's going to be 1,024. But let's see what else we can do. So we know this is going to be some number."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "Actually, let's just keep it as 2 to the 10th power just for simplicity right now. Later, you might know that that's going to be 1,024. But let's see what else we can do. So we know this is going to be some number. Actually, let me just write it out as 1,024. So we have 1 over 32 times 1,024 to the t over 10 power. So it seems like we're getting close."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So we know this is going to be some number. Actually, let me just write it out as 1,024. So we have 1 over 32 times 1,024 to the t over 10 power. So it seems like we're getting close. We did, if there was no minus 1 here, we're essentially done. But now there's this minus 1. So how do we deal with that?"}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So it seems like we're getting close. We did, if there was no minus 1 here, we're essentially done. But now there's this minus 1. So how do we deal with that? Well, we can do a similar type of strategy. We can subtract 1, and then we could add 1. Then we're not actually changing the value."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So how do we deal with that? Well, we can do a similar type of strategy. We can subtract 1, and then we could add 1. Then we're not actually changing the value. Just as we multiplied by 10 and divided by 10, we're not changing the value up here. If you subtract 1 and add 1 to the exponent, you're not changing its value. And so what is this going to be?"}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "Then we're not actually changing the value. Just as we multiplied by 10 and divided by 10, we're not changing the value up here. If you subtract 1 and add 1 to the exponent, you're not changing its value. And so what is this going to be? We want to leave this minus 1 here, but we want to get rid of this plus 1 somehow. And here we just have to remind ourselves that if we have a to the b times a to the c, that's going to be equal to a to the b plus c. If you have the same base raised to different exponents and you multiply them, you could just add the exponents. And so you could also go the other way around."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And so what is this going to be? We want to leave this minus 1 here, but we want to get rid of this plus 1 somehow. And here we just have to remind ourselves that if we have a to the b times a to the c, that's going to be equal to a to the b plus c. If you have the same base raised to different exponents and you multiply them, you could just add the exponents. And so you could also go the other way around. If you have a to the b plus c, you could break it up into a to the b times a to the c. So this business right over here, this is 1024 to the t over 10 minus 1 plus 1. So we can break this up as 1024 to the t over 10 minus 1. That's this part here."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And so you could also go the other way around. If you have a to the b plus c, you could break it up into a to the b times a to the c. So this business right over here, this is 1024 to the t over 10 minus 1 plus 1. So we can break this up as 1024 to the t over 10 minus 1. That's this part here. And then times 1024 to the 1. Let me make this in a different color. So let's see, green."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "That's this part here. And then times 1024 to the 1. Let me make this in a different color. So let's see, green. So this right over here. So times 1024 to the 1 power. That's this 1 power right over here."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So let's see, green. So this right over here. So times 1024 to the 1 power. That's this 1 power right over here. And of course, we still have the 1 over 32. All right, so now we're really close. We have the 1024 to the t over 10 minus 1."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "That's this 1 power right over here. And of course, we still have the 1 over 32. All right, so now we're really close. We have the 1024 to the t over 10 minus 1. We have t over 10 minus 1. And now we just have to simplify. We can rewrite."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "We have the 1024 to the t over 10 minus 1. We have t over 10 minus 1. And now we just have to simplify. We can rewrite. This is going to be equal to, we could just bring the 1024. 1024 to the first power, that's just 1024. So that's going to be 1024 over, I could put all these commas here if I like, 1024 over this 32."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "We can rewrite. This is going to be equal to, we could just bring the 1024. 1024 to the first power, that's just 1024. So that's going to be 1024 over, I could put all these commas here if I like, 1024 over this 32. Let me do it in that magenta color. Over this 32 times, home stretch, times 1024 to the t over 10 minus 1 power. And now we could just simplify this."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So that's going to be 1024 over, I could put all these commas here if I like, 1024 over this 32. Let me do it in that magenta color. Over this 32 times, home stretch, times 1024 to the t over 10 minus 1 power. And now we could just simplify this. You might recognize 1024, we already saw that that was the same thing as 2 to the 10th power. 32 is the same thing as 2 to the 5th power. So 2 to the 10th divided by 2 to the 5th, actually this is another exponent property at play here, although you could just divide the numbers."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "And now we could just simplify this. You might recognize 1024, we already saw that that was the same thing as 2 to the 10th power. 32 is the same thing as 2 to the 5th power. So 2 to the 10th divided by 2 to the 5th, actually this is another exponent property at play here, although you could just divide the numbers. If you have a to the b over a to the c, this is going to be equal to a to the b minus c. So this is going to be 2 to the 10 minus 5, or this whole thing right over here. This thing, see I wanted to do that in a different color. This thing is just going to be 2 to the 5th power, or 32."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "So 2 to the 10th divided by 2 to the 5th, actually this is another exponent property at play here, although you could just divide the numbers. If you have a to the b over a to the c, this is going to be equal to a to the b minus c. So this is going to be 2 to the 10 minus 5, or this whole thing right over here. This thing, see I wanted to do that in a different color. This thing is just going to be 2 to the 5th power, or 32. So this is going to be 32 times 1024. We're in the home stretch here. 1024 to the t over 10 minus 1."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "This thing is just going to be 2 to the 5th power, or 32. So this is going to be 32 times 1024. We're in the home stretch here. 1024 to the t over 10 minus 1. So once again, normally in our lives we like to make things simpler, and I'm a big advocate of that. It's a good life philosophy. But this is a case where we really did make it more complicated."}, {"video_title": "Rewriting an exponential expression in a hairier way Algebra II Khan Academy.mp3", "Sentence": "1024 to the t over 10 minus 1. So once again, normally in our lives we like to make things simpler, and I'm a big advocate of that. It's a good life philosophy. But this is a case where we really did make it more complicated. We started with 132 times 2 to the t. Actually, there's other ways we could have written that. And we turned it into this thing with this somewhat hairier exponent, but it's a useful skill to have because you might get a result like what we originally started with, and then someone else might get a result like this. And it's very important to realize, hey, you actually got the same result."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me do my best attempt at graphing that. And to start off, I'm going to graph with the simplest function, or the simplest version of this, or the root of this, which is just cosine of x. So let me just graph, and eventually you can kind of, let me just, let me graph cosine of x. So this is my y-axis. And I want to have some space here so I can eventually graph this entire thing. So let me, so let's say that this is negative 1, this is negative 2, this is positive 1, this is positive 2, and let's say that this right over here is, this right over here is 2 pi. 2 pi, and then of course that could be pi right over there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So this is my y-axis. And I want to have some space here so I can eventually graph this entire thing. So let me, so let's say that this is negative 1, this is negative 2, this is positive 1, this is positive 2, and let's say that this right over here is, this right over here is 2 pi. 2 pi, and then of course that could be pi right over there. Now, the first thing I'm going to do, let me copy this, because I could use it later. I could use it later to graph the whole thing. So, so let's start off."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "2 pi, and then of course that could be pi right over there. Now, the first thing I'm going to do, let me copy this, because I could use it later. I could use it later to graph the whole thing. So, so let's start off. So I'm just going to graph y is equal to cosine of x. So when x is equal to zero, and I'm just going to do it between the interval zero and 2 pi. Obviously it's a periodic function, it'll keep going in the negative and the positive directions."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So, so let's start off. So I'm just going to graph y is equal to cosine of x. So when x is equal to zero, and I'm just going to do it between the interval zero and 2 pi. Obviously it's a periodic function, it'll keep going in the negative and the positive directions. So what happens when x is equal to zero? What is cosine of x? Well, cosine of zero is 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Obviously it's a periodic function, it'll keep going in the negative and the positive directions. So what happens when x is equal to zero? What is cosine of x? Well, cosine of zero is 1. What about when, what about when x is equal to pi? What is cosine of pi? Well, cosine of pi is negative 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine of zero is 1. What about when, what about when x is equal to pi? What is cosine of pi? Well, cosine of pi is negative 1. Cosine of pi is negative 1. And then what's cosine of 2 pi? Well, that's 1 again."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine of pi is negative 1. Cosine of pi is negative 1. And then what's cosine of 2 pi? Well, that's 1 again. We get back, we've completed a period, or we've completed an entire cycle, and 2 pi is the period of cosine of x. So this is one cycle right over here. I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, that's 1 again. We get back, we've completed a period, or we've completed an entire cycle, and 2 pi is the period of cosine of x. So this is one cycle right over here. I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi. Now, what I want to think about is, what happens to this graph? Instead of graphing, instead of graphing y equals cosine of x, and let me draw some graph paper again. Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "I could keep going if I wanted to, but the whole point, I just wanted to graph this one cycle between zero and 2 pi. Now, what I want to think about is, what happens to this graph? Instead of graphing, instead of graphing y equals cosine of x, and let me draw some graph paper again. Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x. So the only difference between that and that is now I'm multiplying the x by 1 3rd. What's going to happen to the graph over here? How is this going to change instead of being an x if it's a 1 3rd x?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Instead of drawing y is equal to cosine of x, I'm going to draw y is equal to cosine of 1 3rd x. So the only difference between that and that is now I'm multiplying the x by 1 3rd. What's going to happen to the graph over here? How is this going to change instead of being an x if it's a 1 3rd x? What's going to happen over here? And now I'm going to do it over the entire interval between zero and 6 pi. So let me just make sure I have enough space."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "How is this going to change instead of being an x if it's a 1 3rd x? What's going to happen over here? And now I'm going to do it over the entire interval between zero and 6 pi. So let me just make sure I have enough space. 3 pi, 4 pi, 5 pi, and 6 pi. What's going to happen to this graph? Well, there's a couple of ways to think about it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me just make sure I have enough space. 3 pi, 4 pi, 5 pi, and 6 pi. What's going to happen to this graph? Well, there's a couple of ways to think about it. The easiest might just be to say, well, to complete an entire cycle, we're going to go 1 3rd as fast, or we're going to go 3 times slower. Or if you just want to think about the period here, what's the period of cosine of 1 3rd x? Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about it. The easiest might just be to say, well, to complete an entire cycle, we're going to go 1 3rd as fast, or we're going to go 3 times slower. Or if you just want to think about the period here, what's the period of cosine of 1 3rd x? Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here. So it's the absolute value of 1 3rd, which is just 1 3rd. So the period is 2 pi over 1 3rd, which is the same thing as 2 pi times 3, which is 6 pi, which gels with the intuition. It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the period is going to be 2 pi divided by the absolute value of this coefficient right over here. So it's the absolute value of 1 3rd, which is just 1 3rd. So the period is 2 pi over 1 3rd, which is the same thing as 2 pi times 3, which is 6 pi, which gels with the intuition. It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi. Because whatever we take x, we're taking 1 3rd of it. So to get to 2 pi, you can't just have x equals 2 pi. x now has to equal 6 pi to get 2 pi inputted into the cosine function."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It's going to take 3 times as much time to get whatever we input into the cosine function to get back to 2 pi. Because whatever we take x, we're taking 1 3rd of it. So to get to 2 pi, you can't just have x equals 2 pi. x now has to equal 6 pi to get 2 pi inputted into the cosine function. So the period is now 6 pi. At x is equal to 0, 1 3rd times 0 is 0, and the cosine of 0 is 1. When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "x now has to equal 6 pi to get 2 pi inputted into the cosine function. So the period is now 6 pi. At x is equal to 0, 1 3rd times 0 is 0, and the cosine of 0 is 1. When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi. Cosine of 2 pi is equal to 1. And if you want to go in between, over here, to go in between, we tried pi. But over here, we could try 3 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "When x is equal to 6 pi, you have 6 pi divided by 3 is 2 pi. Cosine of 2 pi is equal to 1. And if you want to go in between, over here, to go in between, we tried pi. But over here, we could try 3 pi. When x is 3 pi, you have cosine of 1 3rd of 3 pi. That's cosine of pi. Cosine of pi is negative 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "But over here, we could try 3 pi. When x is 3 pi, you have cosine of 1 3rd of 3 pi. That's cosine of pi. Cosine of pi is negative 1. So when x is equal to 3 pi, we have cosine of 1 3rd times 3 pi is negative 1. So it's going to look something like this. It's going to look something like this."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Cosine of pi is negative 1. So when x is equal to 3 pi, we have cosine of 1 3rd times 3 pi is negative 1. So it's going to look something like this. It's going to look something like this. Drawing my best attempt to draw it. So it's going to look something like this. So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It's going to look something like this. Drawing my best attempt to draw it. So it's going to look something like this. So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3. Or you could see that the period is 3 times longer. The period here was 2 pi. Period here was 2 pi."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So you see, to go from y equals cosine of x to y equals cosine of 1 3rd x, it essentially stretched out this function by a factor of 3. Or you could see that the period is 3 times longer. The period here was 2 pi. Period here was 2 pi. All right. Well, there's only one more transformation we need in order to get to the function that they're asking us about. We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Period here was 2 pi. All right. Well, there's only one more transformation we need in order to get to the function that they're asking us about. We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x. So let's try to draw that. So let me put my axis here again. And let me label it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "We just have to, instead of having a cosine of 1 3rd x, we just have to negative 2.5 cosine of 1 3rd x. So let's try to draw that. So let me put my axis here again. And let me label it. So that's cosine 2 pi, 3 pi, 4 pi, 5 pi, and 6 pi. And our goal now is to draw the graph of y is equal to. And we're just doing it over this between 0 and 6 pi here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And let me label it. So that's cosine 2 pi, 3 pi, 4 pi, 5 pi, and 6 pi. And our goal now is to draw the graph of y is equal to. And we're just doing it over this between 0 and 6 pi here. We only did it between 0 and 2 pi here. Obviously, they're all periodic. They all keep going on and on."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And we're just doing it over this between 0 and 6 pi here. We only did it between 0 and 2 pi here. Obviously, they're all periodic. They all keep going on and on. But now we want to graph y is equal to negative 2.5 times cosine of 1 3rd x. So given this change, we're now multiplying by negative 2.5. What is going to be?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "They all keep going on and on. But now we want to graph y is equal to negative 2.5 times cosine of 1 3rd x. So given this change, we're now multiplying by negative 2.5. What is going to be? Well, actually, let's think about a few things. What was the amplitude in the first two graphs right over here? So what was the amplitude in these first two graphs?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "What is going to be? Well, actually, let's think about a few things. What was the amplitude in the first two graphs right over here? So what was the amplitude in these first two graphs? Well, there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these cases, the minimum is negative 1, maximum is 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So what was the amplitude in these first two graphs? Well, there's two ways to think about it. You could say the amplitude is half the difference between the minimum and the maximum points. In either of these cases, the minimum is negative 1, maximum is 1. The difference is 2. Half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "In either of these cases, the minimum is negative 1, maximum is 1. The difference is 2. Half of that is 1. Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is once again 1. What's going to be the amplitude for this thing right over here? Well, the amplitude is going to be the absolute value of what's multiplying the cosine function."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Or you could just say it's the absolute value of the coefficient here, which is implicitly a 1. And the absolute value of 1 is once again 1. What's going to be the amplitude for this thing right over here? Well, the amplitude is going to be the absolute value of what's multiplying the cosine function. So the amplitude in this case, I'll do it in green, is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well, let's think about it."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is going to be the absolute value of what's multiplying the cosine function. So the amplitude in this case, I'll do it in green, is going to be equal to the absolute value of negative 2.5, which is equal to 2.5. So given that, how is multiplying by negative 2.5 going to transform this graph right over here? Well, let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point, it would go up by a factor of 2 and 1 half. But it's a negative 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, let's think about it. If it was multiplying by just a positive 2.5, you would stretch it out. At each point, it would go up by a factor of 2 and 1 half. But it's a negative 2.5. So at each point, you're going to stretch it out, and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got to 1 in this case."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "But it's a negative 2.5. So at each point, you're going to stretch it out, and then you're going to flip it over the x-axis. So let's do that. So when x was 0, you got to 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So when x was 0, you got to 1 in this case. But now we're going to multiply that by negative 2.5, which means you're going to get to negative 2.5. So let me draw negative 2.5 right over there. So that's negative 2.5. That would be negative. Let me make it clear. This would be negative 3 right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So that's negative 2.5. That would be negative. Let me make it clear. This would be negative 3 right over here. This would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "This would be negative 3 right over here. This would be positive 3. So that number right over there is negative 2.5. And let me draw a dotted line there. It could serve to be useful. Now, when cosine of 1 third x is 0, it doesn't matter what you multiply it by. You're still going to get 0 right over here."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And let me draw a dotted line there. It could serve to be useful. Now, when cosine of 1 third x is 0, it doesn't matter what you multiply it by. You're still going to get 0 right over here. Now, when cosine of 1 third x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well, cosine of 1 third x we see is negative 1. Negative 1 times negative 2.5 is positive 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "You're still going to get 0 right over here. Now, when cosine of 1 third x was negative 1, which was the case when x is equal to 3 pi, what's going to happen over here? Well, cosine of 1 third x we see is negative 1. Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right. Let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "Negative 1 times negative 2.5 is positive 2.5. So we're going to get to positive 2.5, which is right. Let me draw a dotted line over here. We're going to get to positive 2.5, which is right over there. And then when cosine of 1 third x is equal to 0, doesn't matter what we multiply it by. We get to 0. And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5?"}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "We're going to get to positive 2.5, which is right over there. And then when cosine of 1 third x is equal to 0, doesn't matter what we multiply it by. We get to 0. And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5? Well, it's going to be negative 2.5. So we're going to get back over here. So we're ready to draw our graph."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And then finally, when cosine of 1 third x, when x is at 6 pi, cosine of 1 third x is equal to 1, what's that going to be when you multiply it by negative 2.5? Well, it's going to be negative 2.5. So we're going to get back over here. So we're ready to draw our graph. It looks something. Actually, let me do that magenta color, since that's the color I wrote this in. It will look like this."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So we're ready to draw our graph. It looks something. Actually, let me do that magenta color, since that's the color I wrote this in. It will look like this. I can draw it as a solid line. So it will look like that. So you saw what happened."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "It will look like this. I can draw it as a solid line. So it will look like that. So you saw what happened. By putting this 1 third here, it stretched out the graph. It increased the period by a factor of 3. And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So you saw what happened. By putting this 1 third here, it stretched out the graph. It increased the period by a factor of 3. And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit. But now it's a negative. So not only do you increase the amplitude, but you flip it over. So it is indeed the case that the amplitude here is 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "And then multiplying it by negative 2.5, if you just multiplied by 2.5, you would just multiply that out a little bit. But now it's a negative. So not only do you increase the amplitude, but you flip it over. So it is indeed the case that the amplitude here is 2.5. We vary 2.5 from our middle position. Or you could say that the difference between the minimum and the maximum is 5. So half of that is 2.5."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So it is indeed the case that the amplitude here is 2.5. We vary 2.5 from our middle position. Or you could say that the difference between the minimum and the maximum is 5. So half of that is 2.5. But it isn't just multiplying this graph by 2.5. If you multiply this graph by 2.5, you'd get something that looked something like that. But because we had a negative, we had to flip it over the x-axis."}, {"video_title": "Example Amplitude and period cosine transformations Trigonometry Khan Academy.mp3", "Sentence": "So half of that is 2.5. But it isn't just multiplying this graph by 2.5. If you multiply this graph by 2.5, you'd get something that looked something like that. But because we had a negative, we had to flip it over the x-axis. And we got this here. So this amplitude is 2.5. But it's a flipped over version of this graph."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "She walked a total of 27 kilometers. What is the distance Sloan walked in the first day of the trip? And it says to round our final answer to the nearest kilometer. So like always, have a go at this and see if you can figure out how much she walked on the first day. All right, well let's just call the amount that she walked on the first day A. And then using A, let's see if we can set up an expression for how much she walked in total. And then that should be equal to 27, and then hopefully we're going to be able to solve for A."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "So like always, have a go at this and see if you can figure out how much she walked on the first day. All right, well let's just call the amount that she walked on the first day A. And then using A, let's see if we can set up an expression for how much she walked in total. And then that should be equal to 27, and then hopefully we're going to be able to solve for A. So on the first day, she walks A kilometers. Now how about the second day? Well, they tell us that each day, she walked 20% more than the distance she walked the day before."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "And then that should be equal to 27, and then hopefully we're going to be able to solve for A. So on the first day, she walks A kilometers. Now how about the second day? Well, they tell us that each day, she walked 20% more than the distance she walked the day before. So on the next day, she's going to walk 20% more than A kilometers, so that's 1.2 times A. And what about the day after that, her third day? Well, that's just going to be 1.2 times this the second day, and so that's going to be 1.2 times 1.2, or we could say 1.2 squared times A."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "Well, they tell us that each day, she walked 20% more than the distance she walked the day before. So on the next day, she's going to walk 20% more than A kilometers, so that's 1.2 times A. And what about the day after that, her third day? Well, that's just going to be 1.2 times this the second day, and so that's going to be 1.2 times 1.2, or we could say 1.2 squared times A. And then how much on the fourth day? And that's, she went on a four-day hiking trip, so that's the last day. Well, that's going to be 1.2 times the third day, so that's going to be 1.2 to the third power times A."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "Well, that's just going to be 1.2 times this the second day, and so that's going to be 1.2 times 1.2, or we could say 1.2 squared times A. And then how much on the fourth day? And that's, she went on a four-day hiking trip, so that's the last day. Well, that's going to be 1.2 times the third day, so that's going to be 1.2 to the third power times A. So this is an expression in A on how much she walked over the four days, and we know that she walked a total of 27 kilometers. So this is going to be equal to 27 kilometers. Now you could solve for A over here."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "Well, that's going to be 1.2 times the third day, so that's going to be 1.2 to the third power times A. So this is an expression in A on how much she walked over the four days, and we know that she walked a total of 27 kilometers. So this is going to be equal to 27 kilometers. Now you could solve for A over here. You could factor out the A, and you could say A times one plus 1.2 plus 1.2 squared plus 1.2 to the third power is equal to 27. And then you could say that A is equal to 27 over one plus 1.2 plus 1.2 squared plus 1.2 to the third power, and we would need a calculator to evaluate this. But I'm going to do a different technique, a technique that would work even if you had 20 terms here."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "Now you could solve for A over here. You could factor out the A, and you could say A times one plus 1.2 plus 1.2 squared plus 1.2 to the third power is equal to 27. And then you could say that A is equal to 27 over one plus 1.2 plus 1.2 squared plus 1.2 to the third power, and we would need a calculator to evaluate this. But I'm going to do a different technique, a technique that would work even if you had 20 terms here. You, in theory, could also do this with 20 terms, but it gets a lot more complicated, or if you had 200 terms. So the other way to approach this is use the formula for a finite geometric series. What does it evaluate to?"}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "But I'm going to do a different technique, a technique that would work even if you had 20 terms here. You, in theory, could also do this with 20 terms, but it gets a lot more complicated, or if you had 200 terms. So the other way to approach this is use the formula for a finite geometric series. What does it evaluate to? And just as a reminder, the sum of first n terms, it's going to be the first term, which we could call A, minus the first term times our common ratio, in this case, our common ratio is 1.2 because every successive term is 1.2 times the first. So our first term times our common ratio to the nth power, all of that over one minus the common ratio. In other videos, we explain where this comes from."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "What does it evaluate to? And just as a reminder, the sum of first n terms, it's going to be the first term, which we could call A, minus the first term times our common ratio, in this case, our common ratio is 1.2 because every successive term is 1.2 times the first. So our first term times our common ratio to the nth power, all of that over one minus the common ratio. In other videos, we explain where this comes from. We prove this, but here we can just apply it. We already know what our A is. I used that as our variable."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "In other videos, we explain where this comes from. We prove this, but here we can just apply it. We already know what our A is. I used that as our variable. Our common ratio in this situation is going to be equal to 1.2, and our n is going to be equal to four. Another way I like to think about it is it's our first term, which we see right over there, minus the term that we did not get to. If we were to have a fifth term, it would have been that fifth term that we're subtracting because we aren't getting to a fourth power here."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "I used that as our variable. Our common ratio in this situation is going to be equal to 1.2, and our n is going to be equal to four. Another way I like to think about it is it's our first term, which we see right over there, minus the term that we did not get to. If we were to have a fifth term, it would have been that fifth term that we're subtracting because we aren't getting to a fourth power here. The fifth term would have been to the fourth power. All of that over one minus the common ratio. And so this left-hand side of our equation, we could rewrite as our first term minus our first term times our common ratio, 1.2, to the fourth power, all of that over one minus our common ratio."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "If we were to have a fifth term, it would have been that fifth term that we're subtracting because we aren't getting to a fourth power here. The fifth term would have been to the fourth power. All of that over one minus the common ratio. And so this left-hand side of our equation, we could rewrite as our first term minus our first term times our common ratio, 1.2, to the fourth power, all of that over one minus our common ratio. And then that could be equal to 27. Let me scroll down a little bit so we have some more space to then solve this. And so let's see, I can simplify this a little bit."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "And so this left-hand side of our equation, we could rewrite as our first term minus our first term times our common ratio, 1.2, to the fourth power, all of that over one minus our common ratio. And then that could be equal to 27. Let me scroll down a little bit so we have some more space to then solve this. And so let's see, I can simplify this a little bit. We could, this is going to be equal to negative 0.2. Our numerator, we can factor out an a, and so this is going to be equal to a times one minus 1.2 to the fourth power. And let's see, we can multiply both the numerator and the denominator by a negative one, and so this would get us to a times, a times, and I'll put the a out of the fraction, a times, so I'll just swap the order here and get rid of this negative."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "And so let's see, I can simplify this a little bit. We could, this is going to be equal to negative 0.2. Our numerator, we can factor out an a, and so this is going to be equal to a times one minus 1.2 to the fourth power. And let's see, we can multiply both the numerator and the denominator by a negative one, and so this would get us to a times, a times, and I'll put the a out of the fraction, a times, so I'll just swap the order here and get rid of this negative. 1.2 to the fourth power minus one over 0.2 is equal to 27. Again, all I did is I took the a out of the fraction, so it's out here, and I multiplied the numerator and the denominator by a negative. The numerator multiplied by a negative would swap these two and then multiplying negative 0.2 times a negative is just positive 0.2."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "And let's see, we can multiply both the numerator and the denominator by a negative one, and so this would get us to a times, a times, and I'll put the a out of the fraction, a times, so I'll just swap the order here and get rid of this negative. 1.2 to the fourth power minus one over 0.2 is equal to 27. Again, all I did is I took the a out of the fraction, so it's out here, and I multiplied the numerator and the denominator by a negative. The numerator multiplied by a negative would swap these two and then multiplying negative 0.2 times a negative is just positive 0.2. And so now I can just multiply both sides times the reciprocal of this. So I'll do it here. So 0.2 over 1.2 to the fourth minus one, and then here 0.2 over 1.2 to the fourth minus one."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "The numerator multiplied by a negative would swap these two and then multiplying negative 0.2 times a negative is just positive 0.2. And so now I can just multiply both sides times the reciprocal of this. So I'll do it here. So 0.2 over 1.2 to the fourth minus one, and then here 0.2 over 1.2 to the fourth minus one. That cancels with that, that cancels with that. That's exactly why I did that. And we're left with a is equal to, it is equal to, I'll just write it in yellow, 27 times 0.2, all of that over 1.2 to the fourth minus one, and this expression should give us the exact same value as that expression we just saw, but this is useful even if we had 100 terms, we could use this."}, {"video_title": "Geometric series word problems hike Algebra 2 Khan Academy.mp3", "Sentence": "So 0.2 over 1.2 to the fourth minus one, and then here 0.2 over 1.2 to the fourth minus one. That cancels with that, that cancels with that. That's exactly why I did that. And we're left with a is equal to, it is equal to, I'll just write it in yellow, 27 times 0.2, all of that over 1.2 to the fourth minus one, and this expression should give us the exact same value as that expression we just saw, but this is useful even if we had 100 terms, we could use this. And so I'll get the calculator out. This will give us, so I will, actually I'll evaluate this denominator first. So I'll have 1.2 to the fourth power, which is equal to minus one is equal to, now it's in the denominator, so I could just take the reciprocal of it, and then multiply that times 27 times 0.2 is equal to 5.029."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We're told use the interactive graph below to sketch a graph of y is equal to negative two times three to the x plus five. And so this is clearly an exponential function right over here. Let's think about the behavior as x is, when x is very negative or when x is very positive. When x is very negative, three to a very negative number, like you said, let's say you had three to the negative third power, that would be 1 27th, or three to the negative fourth power, that'd be 1 81st. So this is going to get smaller and smaller and smaller. It's going to approach zero as x becomes more negative. And since this is approaching zero, this whole thing right over here is going to approach zero."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "When x is very negative, three to a very negative number, like you said, let's say you had three to the negative third power, that would be 1 27th, or three to the negative fourth power, that'd be 1 81st. So this is going to get smaller and smaller and smaller. It's going to approach zero as x becomes more negative. And since this is approaching zero, this whole thing right over here is going to approach zero. And so this whole expression is, if this first part's approaching zero, then this whole expression is going to approach five. So we're gonna have a horizontal asymptote that we're gonna approach as we go to the left. As we go to, as x gets more and more negative, we're going to approach positive five."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And since this is approaching zero, this whole thing right over here is going to approach zero. And so this whole expression is, if this first part's approaching zero, then this whole expression is going to approach five. So we're gonna have a horizontal asymptote that we're gonna approach as we go to the left. As we go to, as x gets more and more negative, we're going to approach positive five. And then as x gets larger and larger and larger, three to the x is growing exponentially. But then we're multiplying it times negative two, so it's going to become more and more and more negative, and then we add a five. And so what we have here, well this doesn't, this looks like a line."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "As we go to, as x gets more and more negative, we're going to approach positive five. And then as x gets larger and larger and larger, three to the x is growing exponentially. But then we're multiplying it times negative two, so it's going to become more and more and more negative, and then we add a five. And so what we have here, well this doesn't, this looks like a line. We wanna graph an exponential. So let's go pick the exponential in terms of x. There you have it."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so what we have here, well this doesn't, this looks like a line. We wanna graph an exponential. So let's go pick the exponential in terms of x. There you have it. And so we can move three things. We can move this point. It doesn't even just have to be the y-intercept, although that's a convenient thing to figure out."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "There you have it. And so we can move three things. We can move this point. It doesn't even just have to be the y-intercept, although that's a convenient thing to figure out. We can move this point here, and we can move the asymptote. And maybe the asymptote's the first interesting thing. We said as x becomes more and more and more and more negative y is going to approach five."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It doesn't even just have to be the y-intercept, although that's a convenient thing to figure out. We can move this point here, and we can move the asymptote. And maybe the asymptote's the first interesting thing. We said as x becomes more and more and more and more negative y is going to approach five. So let me put this up here. So that's our asymptote. It doesn't look like it quite yet, but when we try out some values for x and the corresponding y's, and we move these points accordingly, hopefully our exponential is going to look right."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We said as x becomes more and more and more and more negative y is going to approach five. So let me put this up here. So that's our asymptote. It doesn't look like it quite yet, but when we try out some values for x and the corresponding y's, and we move these points accordingly, hopefully our exponential is going to look right. So let's think about, let's pick some convenient x's. So let's think about when x is equal to zero. If x is equal to zero, three to the zeroth power is one."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It doesn't look like it quite yet, but when we try out some values for x and the corresponding y's, and we move these points accordingly, hopefully our exponential is going to look right. So let's think about, let's pick some convenient x's. So let's think about when x is equal to zero. If x is equal to zero, three to the zeroth power is one. Negative two times one is negative two, plus three is three. So when x is equal to zero, y is three. Now let's think about when x is equal to one, what, and I'm just picking that because it's easy to compute."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "If x is equal to zero, three to the zeroth power is one. Negative two times one is negative two, plus three is three. So when x is equal to zero, y is three. Now let's think about when x is equal to one, what, and I'm just picking that because it's easy to compute. Three to the first power is three, times negative two is negative six, plus five is negative one. So when x is one, y is negative one. And so let's see, does this, is this consistent with what we just described?"}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now let's think about when x is equal to one, what, and I'm just picking that because it's easy to compute. Three to the first power is three, times negative two is negative six, plus five is negative one. So when x is one, y is negative one. And so let's see, does this, is this consistent with what we just described? When x is very negative, we should be approaching, we should be approaching positive five, and that looks like the case as we move to the left. We're getting closer and closer and closer to five. In fact, it looks like they overlap, but it's really we're just getting closer and closer and closer, because this term, this term right over here is getting smaller and smaller and smaller as x becomes more and more and more negative."}, {"video_title": "Graphing exponential functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so let's see, does this, is this consistent with what we just described? When x is very negative, we should be approaching, we should be approaching positive five, and that looks like the case as we move to the left. We're getting closer and closer and closer to five. In fact, it looks like they overlap, but it's really we're just getting closer and closer and closer, because this term, this term right over here is getting smaller and smaller and smaller as x becomes more and more and more negative. But then as x becomes more and more positive, this term becomes really negative, because we're multiplying it times a negative two, and we see that it becomes really negative. So I feel pretty good about what we've just graphed. We've graphed the horizontal asymptote, it makes sense, and we've picked two points that sit on this, on the graph of this exponential."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "Now, factoring is something that we've been doing for many years now. You can go all the way back to when you were thinking about how would I factor the number 12? Well, I could write the number 12 as three times four. I could also write it as two times six. These are all legitimate factors. Or I could try to do a prime factorization of 12 where I'm trying to write it as the product of, you could view it as its most basic constituents, which would be the prime numbers. And so we've done stuff like, well, 12 can be expressed as two times six."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "I could also write it as two times six. These are all legitimate factors. Or I could try to do a prime factorization of 12 where I'm trying to write it as the product of, you could view it as its most basic constituents, which would be the prime numbers. And so we've done stuff like, well, 12 can be expressed as two times six. Two is prime, but then six can be expressed as two times three. And so 12 could be expressed as two times two times three, which we see right over here. This is all review."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "And so we've done stuff like, well, 12 can be expressed as two times six. Two is prime, but then six can be expressed as two times three. And so 12 could be expressed as two times two times three, which we see right over here. This is all review. And this would be a prime factorization. And we saw an analog when we first learned it in Algebra 1. In Algebra 1, we learn things like, and sometimes this might be in a Math 1 class or even in a pre-algebra class, you'll learn things like, hey, how do I factor x squared plus six x?"}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "This is all review. And this would be a prime factorization. And we saw an analog when we first learned it in Algebra 1. In Algebra 1, we learn things like, and sometimes this might be in a Math 1 class or even in a pre-algebra class, you'll learn things like, hey, how do I factor x squared plus six x? And you might recognize that, hey, x squared could be rewritten as x times x. And six x, that really just means six times x. And so both of them have x as a factor."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "In Algebra 1, we learn things like, and sometimes this might be in a Math 1 class or even in a pre-algebra class, you'll learn things like, hey, how do I factor x squared plus six x? And you might recognize that, hey, x squared could be rewritten as x times x. And six x, that really just means six times x. And so both of them have x as a factor. And so we might wanna factor that out. And so we could rewrite this entire expression as x times x plus six. What we just did is we factored out these x's that I am circling in blue."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "And so both of them have x as a factor. And so we might wanna factor that out. And so we could rewrite this entire expression as x times x plus six. What we just did is we factored out these x's that I am circling in blue. So in general, this idea of factoring, if you're thinking about numbers, you're writing one number as the product of other numbers. If you're thinking about expressions, you're writing an expression as the product of other expressions. Well, now as we go a little bit more advanced into algebra, we're gonna start thinking about doing this with higher-order expressions."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "What we just did is we factored out these x's that I am circling in blue. So in general, this idea of factoring, if you're thinking about numbers, you're writing one number as the product of other numbers. If you're thinking about expressions, you're writing an expression as the product of other expressions. Well, now as we go a little bit more advanced into algebra, we're gonna start thinking about doing this with higher-order expressions. So we've done it with just an x or just an x squared, but now we're gonna start thinking about, well, what happens if we have something to the third power, fourth power, sixth power, 10th power, 100th power? But it's really the same ideas. And we could start with monomials, which is a fancy word for just a single term."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, now as we go a little bit more advanced into algebra, we're gonna start thinking about doing this with higher-order expressions. So we've done it with just an x or just an x squared, but now we're gonna start thinking about, well, what happens if we have something to the third power, fourth power, sixth power, 10th power, 100th power? But it's really the same ideas. And we could start with monomials, which is a fancy word for just a single term. So let's say I had something like six x to the seventh. What are the different ways that I could factor this? Pause this video and think about it."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "And we could start with monomials, which is a fancy word for just a single term. So let's say I had something like six x to the seventh. What are the different ways that I could factor this? Pause this video and think about it. Can I express this as the product of two other things? Well, I could rewrite this as this is being equal to two x to the third times what? Well, let's see."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and think about it. Can I express this as the product of two other things? Well, I could rewrite this as this is being equal to two x to the third times what? Well, let's see. What do I have to multiply two by to get to six? I have to multiply it by three. And what do I have to multiply x to the third by to get to x to the seventh?"}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's see. What do I have to multiply two by to get to six? I have to multiply it by three. And what do I have to multiply x to the third by to get to x to the seventh? I could multiply it times three x to the fourth. Notice two times three is six. X to the third times x to the fourth is x to the seventh."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "And what do I have to multiply x to the third by to get to x to the seventh? I could multiply it times three x to the fourth. Notice two times three is six. X to the third times x to the fourth is x to the seventh. We add exponents when we're multiplying things with the same base. But this isn't the only way to factor it. Just as we saw that three times four wasn't the only way to factor 12, you could also express this as maybe being equal to x to the sixth times what?"}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "X to the third times x to the fourth is x to the seventh. We add exponents when we're multiplying things with the same base. But this isn't the only way to factor it. Just as we saw that three times four wasn't the only way to factor 12, you could also express this as maybe being equal to x to the sixth times what? Well, we would still have to multiply by six then. And then we'd have to multiply by another x. So we could write this as x to the sixth times six x."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "Just as we saw that three times four wasn't the only way to factor 12, you could also express this as maybe being equal to x to the sixth times what? Well, we would still have to multiply by six then. And then we'd have to multiply by another x. So we could write this as x to the sixth times six x. So there's oftentimes multiple ways to factor a higher degree monomial like this. And there is also an analog to doing something like a prime factorization. When you're trying to really decompose, rewrite this expression as a product of its simplest parts."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "So we could write this as x to the sixth times six x. So there's oftentimes multiple ways to factor a higher degree monomial like this. And there is also an analog to doing something like a prime factorization. When you're trying to really decompose, rewrite this expression as a product of its simplest parts. How would you do that for six x to the seventh? Well, you could rewrite that. You could say six x to the seventh."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "When you're trying to really decompose, rewrite this expression as a product of its simplest parts. How would you do that for six x to the seventh? Well, you could rewrite that. You could say six x to the seventh. Well, that's equal to, we could think about the six first. We know that the prime factorization of six is two times three, two times three. And then x to the seventh is just seven x's multiplied by each other."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "You could say six x to the seventh. Well, that's equal to, we could think about the six first. We know that the prime factorization of six is two times three, two times three. And then x to the seventh is just seven x's multiplied by each other. So times x times x times x times x times x. How many is that? That's five, six, and seven."}, {"video_title": "Introduction to factoring higher degree monomials Algebra 2 Khan Academy.mp3", "Sentence": "And then x to the seventh is just seven x's multiplied by each other. So times x times x times x times x times x. How many is that? That's five, six, and seven. And so some of what we were doing, when I said two x to the third, what we really thought about is, okay, I had a two, and then I had x times x times x. And then what do I have to multiply that? Well, I have to multiply that by three x to the fourth."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "we're asked to subtract negative two x squared plus four x minus one from six x squared plus three x minus nine. And like always, I encourage you to pause the video and see if you can give it a go. All right, now let's work through this together. So I could rewrite this as six x squared plus three x minus nine minus, minus this expression right over here. So I'll put that in parentheses, minus negative two x squared, negative two x squared, plus four x minus one. Now what can we do from here? Well, we can distribute this negative sign."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I could rewrite this as six x squared plus three x minus nine minus, minus this expression right over here. So I'll put that in parentheses, minus negative two x squared, negative two x squared, plus four x minus one. Now what can we do from here? Well, we can distribute this negative sign. We can distribute this negative sign. And then if we did that, we would get the six x squared plus three x minus nine won't change, so we still have that. Six x squared plus three x minus nine."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we can distribute this negative sign. We can distribute this negative sign. And then if we did that, we would get the six x squared plus three x minus nine won't change, so we still have that. Six x squared plus three x minus nine. But then if I distribute the negative sign, the negative of negative two x squared is positive two x squared. So that's going to be positive two, good, give it a little more space, positive two x squared. And then subtract, and then the negative of positive four x is, I'm going to subtract four x now."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Six x squared plus three x minus nine. But then if I distribute the negative sign, the negative of negative two x squared is positive two x squared. So that's going to be positive two, good, give it a little more space, positive two x squared. And then subtract, and then the negative of positive four x is, I'm going to subtract four x now. And then the negative of negative one, or the opposite of negative one, is gonna be positive one. So I've just distributed the negative sign. And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then subtract, and then the negative of positive four x is, I'm going to subtract four x now. And then the negative of negative one, or the opposite of negative one, is gonna be positive one. So I've just distributed the negative sign. And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say. So I have an x squared term. Here it's six x squared. Here I have a two x squared term."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "And now I can add terms that have the same degree on our x, the same degree terms, I guess you could say. So I have an x squared term. Here it's six x squared. Here I have a two x squared term. So I can add those two together. Six x squared plus two x squared. If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have?"}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Here I have a two x squared term. So I can add those two together. Six x squared plus two x squared. If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have? I'm now going to have eight x squareds. Eight x squareds. Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "If I have six x squareds, and then I have another two x squareds, how many x squares am I now going to have? I'm now going to have eight x squareds. Eight x squareds. Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared. Then we can add the x terms. You could view these as the first degree terms. Three x, we have three x, and then we have minus four x."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Or six x squared plus two x squared, we add the coefficients, the six and the two, to get eight, eight x squared. Then we can add the x terms. You could view these as the first degree terms. Three x, we have three x, and then we have minus four x. So three x minus four x, if I have three of something and I take away four of them, I'm now going to have negative one of that thing. So, or you could say that the coefficients, three minus four would be negative one. So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "Three x, we have three x, and then we have minus four x. So three x minus four x, if I have three of something and I take away four of them, I'm now going to have negative one of that thing. So, or you could say that the coefficients, three minus four would be negative one. So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x. That's the same thing as negative one x. And then finally, I can deal with our constant terms. I have, I'm subtracting a nine, and then I'm adding a one."}, {"video_title": "Polynomial subtraction Mathematics II High School Math Khan Academy.mp3", "Sentence": "So I now have negative one x. I could write it as negative one x, but I might as well just write it as negative x. That's the same thing as negative one x. And then finally, I can deal with our constant terms. I have, I'm subtracting a nine, and then I'm adding a one. So you could say, well, what's negative nine plus one? Well, that's going to be negative eight. That's going to be negative eight."}, {"video_title": "Multiplying monomials by polynomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have our rectangle here, and it's broken up into these three smaller rectangles. And we see for all of these rectangles, the height here is four units. And then the widths are expressed in terms, or at least the first two are expressed in terms of x. And then this last one has a width of two. So what's the area of the entire rectangle? I encourage you to pause the video and think about it. What's the area of this blue, this blue that looks like a square, but let's just call it a rectangle, which all squares are rectangles, so that's safe."}, {"video_title": "Multiplying monomials by polynomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then this last one has a width of two. So what's the area of the entire rectangle? I encourage you to pause the video and think about it. What's the area of this blue, this blue that looks like a square, but let's just call it a rectangle, which all squares are rectangles, so that's safe. Well, it's gonna be the height times the width. So the area here is going to be the height, which is four, times the width, which is x squared. And then to that, we wanna add the area of this, I guess we could say this salmon-colored rectangle."}, {"video_title": "Multiplying monomials by polynomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "What's the area of this blue, this blue that looks like a square, but let's just call it a rectangle, which all squares are rectangles, so that's safe. Well, it's gonna be the height times the width. So the area here is going to be the height, which is four, times the width, which is x squared. And then to that, we wanna add the area of this, I guess we could say this salmon-colored rectangle. And well, that's gonna be the height, four, times the width, three x. So we could say four times three x, we could write it like that. But what is four times three x?"}, {"video_title": "Multiplying monomials by polynomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then to that, we wanna add the area of this, I guess we could say this salmon-colored rectangle. And well, that's gonna be the height, four, times the width, three x. So we could say four times three x, we could write it like that. But what is four times three x? Well, it's going to be 12 x. I have three x four times, I have 12 x's. So that's going to be 12 x. 12 x is the area of this salmon-colored rectangle."}, {"video_title": "Multiplying monomials by polynomials area model Mathematics II High School Math Khan Academy.mp3", "Sentence": "But what is four times three x? Well, it's going to be 12 x. I have three x four times, I have 12 x's. So that's going to be 12 x. 12 x is the area of this salmon-colored rectangle. And then finally, the area of this green rectangle, we actually can figure out it exactly. We don't even have to express it in terms of a variable. Its height is four, its width is two, so the area's gonna be four times two, or eight."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Each taco costs C dollars to make and is sold for P dollars. Write an equation that relates T, C, and P. So pause this video and see if you can do that. All right, now let's work this together. So let's just remind ourselves what's going on here. So we have the number of tacos sold per day is T. So T equals number of tacos sold per day. They tell us the net profit of $300. So we could say 300 is equal to net profit."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So let's just remind ourselves what's going on here. So we have the number of tacos sold per day is T. So T equals number of tacos sold per day. They tell us the net profit of $300. So we could say 300 is equal to net profit. They tell us that each taco costs C dollars to make. So C is equal to cost per taco. And then P is what each taco is sold for."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So we could say 300 is equal to net profit. They tell us that each taco costs C dollars to make. So C is equal to cost per taco. And then P is what each taco is sold for. Right over there is sold for P dollars. So P is equal to the price per taco. So how do we figure out what the net profit is going to be?"}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And then P is what each taco is sold for. Right over there is sold for P dollars. So P is equal to the price per taco. So how do we figure out what the net profit is going to be? Well, we could write it this way. Net profit is going to be equal to the total amount of money that you bring in minus the total amount of money that you have to spend. So what is going to be the total amount that you bring in?"}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So how do we figure out what the net profit is going to be? Well, we could write it this way. Net profit is going to be equal to the total amount of money that you bring in minus the total amount of money that you have to spend. So what is going to be the total amount that you bring in? Well, it's going to be the amount you sell times, well, it's going to be the number of tacos you sell times the price per taco. So T times P, this is how much money you're bringing in, but of course, you have to also subtract out your cost. Well, what's going to be your cost?"}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So what is going to be the total amount that you bring in? Well, it's going to be the amount you sell times, well, it's going to be the number of tacos you sell times the price per taco. So T times P, this is how much money you're bringing in, but of course, you have to also subtract out your cost. Well, what's going to be your cost? Well, it's going to be the number of tacos times the cost per taco. That's how much we're going to spend. So the number of tacos times the cost per taco."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Well, what's going to be your cost? Well, it's going to be the number of tacos times the cost per taco. That's how much we're going to spend. So the number of tacos times the cost per taco. Now here, I've written everything as variables. Well, I wrote net profit out, but they told us that net profit is equal to 300. So I could write it like this."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So the number of tacos times the cost per taco. Now here, I've written everything as variables. Well, I wrote net profit out, but they told us that net profit is equal to 300. So I could write it like this. 300 is equal to T times P. I'll do it in those same colors. T, P minus T, C. Minus T, C. Now it's completely possible that they didn't give us net profit as $300 per day. Instead, they said that is one of the variables and they gave one of the other variables."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So I could write it like this. 300 is equal to T times P. I'll do it in those same colors. T, P minus T, C. Minus T, C. Now it's completely possible that they didn't give us net profit as $300 per day. Instead, they said that is one of the variables and they gave one of the other variables. And that would have been okay. We could have used the same logic. Whatever they didn't give us, we could have set up as a variable and whatever they did give us, maybe they gave us the price per taco, we could have put that as a given number."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Instead, they said that is one of the variables and they gave one of the other variables. And that would have been okay. We could have used the same logic. Whatever they didn't give us, we could have set up as a variable and whatever they did give us, maybe they gave us the price per taco, we could have put that as a given number. Now I know what some of y'all are thinking. You didn't get exactly this. You might've thought about it the other way."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Whatever they didn't give us, we could have set up as a variable and whatever they did give us, maybe they gave us the price per taco, we could have put that as a given number. Now I know what some of y'all are thinking. You didn't get exactly this. You might've thought about it the other way. You might've thought about it as the profit is going to be equal to the number of tacos times the profit per taco. And what's going to be the profit per taco? Well, that's going to be how much each taco is sold for minus how much each taco costs."}, {"video_title": "Modeling with multiple variables Taco stand Modeling Algebra 2 Khan Academy.mp3", "Sentence": "You might've thought about it the other way. You might've thought about it as the profit is going to be equal to the number of tacos times the profit per taco. And what's going to be the profit per taco? Well, that's going to be how much each taco is sold for minus how much each taco costs. And this also would have been completely credible. And if you look at these two, they're actually algebraically equivalent. If you factor out T here, you get this expression on the right."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "We measure that the initial mass of a sample of carbon-14 is 741 grams. Write a function that models the mass of the carbon-14 sample remaining t years since the initial measurement. All right, so like always, pause the video and see if you can come up with this function m that is going to be a function of t, the years since the initial measurement. All right, let's work through it together. What I like to do is I always like to start up with a little bit of a table to get a sense of things. So let's think about t, how much time, how many years have passed since the initial measurement, and what the amount of mass we're going to have. Well, we know that the initial mass of a sample of carbon-14 is 741 grams."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, let's work through it together. What I like to do is I always like to start up with a little bit of a table to get a sense of things. So let's think about t, how much time, how many years have passed since the initial measurement, and what the amount of mass we're going to have. Well, we know that the initial mass of a sample of carbon-14 is 741 grams. So at t equals zero, our mass is 741. Now, what's another interesting t that we could think about? Well, we know every 5,730 years, we lose exactly half of our mass of carbon-14, every 5,730 years."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we know that the initial mass of a sample of carbon-14 is 741 grams. So at t equals zero, our mass is 741. Now, what's another interesting t that we could think about? Well, we know every 5,730 years, we lose exactly half of our mass of carbon-14, every 5,730 years. So let's think about what happens when t is 5,730. Well, we're going to lose half of our mass. So we're going to multiply this times 1 half."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we know every 5,730 years, we lose exactly half of our mass of carbon-14, every 5,730 years. So let's think about what happens when t is 5,730. Well, we're going to lose half of our mass. So we're going to multiply this times 1 half. So this is going to be 741 times 1 half. I'm not even going to calculate what that is right now. And then let's say we have another 5,730 years take place."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're going to multiply this times 1 half. So this is going to be 741 times 1 half. I'm not even going to calculate what that is right now. And then let's say we have another 5,730 years take place. So that's going to be, and I'm just going to write two times 5,730, I could calculate it, what's going to be 10,000, 11,460, or something like that. All right, but let's just go with two times 5,000, is it 10,000, yeah, 10,000 plus 1,400, so 11,400 plus 60, yeah, so 11,460. But let's just leave it like this."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then let's say we have another 5,730 years take place. So that's going to be, and I'm just going to write two times 5,730, I could calculate it, what's going to be 10,000, 11,460, or something like that. All right, but let's just go with two times 5,000, is it 10,000, yeah, 10,000 plus 1,400, so 11,400 plus 60, yeah, so 11,460. But let's just leave it like this. Well, then it's going to be this times 1 half. So it's going to be 741 times 1 half times 1 half. So we're going to multiply by 1 half again."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "But let's just leave it like this. Well, then it's going to be this times 1 half. So it's going to be 741 times 1 half times 1 half. So we're going to multiply by 1 half again. And so this is the same thing as 741 times 1 half squared. And then let's just think about, if we wait another 5,730 years, so three times 5,730, well, then it's going to be 1 half times this. So it's going to be 741."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "So we're going to multiply by 1 half again. And so this is the same thing as 741 times 1 half squared. And then let's just think about, if we wait another 5,730 years, so three times 5,730, well, then it's going to be 1 half times this. So it's going to be 741. This times 1 half is going to be 1 half to the third power. So you might notice a little bit of a pattern here. However many half-lives we have, we're going to multiply, we're going to raise 1 half to that power and then multiply it times our initial mass."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's going to be 741. This times 1 half is going to be 1 half to the third power. So you might notice a little bit of a pattern here. However many half-lives we have, we're going to multiply, we're going to raise 1 half to that power and then multiply it times our initial mass. This is 1 half-life has gone by, 2 half-lives, we have an exponent of 2, 3 half-lives, we multiply by 3, sorry, we multiply by 1 half three times. So what's going to be a general way to express m of t? Well, m of t is going to be our initial value, 741, times, and you might already be identifying this as an exponential function, we're going to multiply times this number, which we could call our common ratio, as many half-lives have passed by."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "However many half-lives we have, we're going to multiply, we're going to raise 1 half to that power and then multiply it times our initial mass. This is 1 half-life has gone by, 2 half-lives, we have an exponent of 2, 3 half-lives, we multiply by 3, sorry, we multiply by 1 half three times. So what's going to be a general way to express m of t? Well, m of t is going to be our initial value, 741, times, and you might already be identifying this as an exponential function, we're going to multiply times this number, which we could call our common ratio, as many half-lives have passed by. So how do we know how many half-lives have passed by? Well, we could take t and we can divide it by the half-life. And try to test this out."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, m of t is going to be our initial value, 741, times, and you might already be identifying this as an exponential function, we're going to multiply times this number, which we could call our common ratio, as many half-lives have passed by. So how do we know how many half-lives have passed by? Well, we could take t and we can divide it by the half-life. And try to test this out. When t equals zero, it's going to be 1 half to the zero power, which is just 1, and we're just going to have 741. When t is equal to 5,730, this exponent is going to be 1, which we want it to be. We're just going to multiply our initial value by 1 half once."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "And try to test this out. When t equals zero, it's going to be 1 half to the zero power, which is just 1, and we're just going to have 741. When t is equal to 5,730, this exponent is going to be 1, which we want it to be. We're just going to multiply our initial value by 1 half once. When this exponent is 2 times 5,730, well, then the exponent is going to be 2, and we're going to multiply by 1 half twice. It's going to be 1 half to the second power. And it's going to work for everything in between."}, {"video_title": "Constructing exponential models half life Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're just going to multiply our initial value by 1 half once. When this exponent is 2 times 5,730, well, then the exponent is going to be 2, and we're going to multiply by 1 half twice. It's going to be 1 half to the second power. And it's going to work for everything in between. When we are a fraction of a half-life along, we're going to get a non-integer exponent, and that too will work out. And so this is our function. We are done."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "For example, we know if we took the number four and raised it to the third power, this is equivalent to taking three fours and multiplying them. Or you could also view it as starting with a one and then multiplying the one by four or multiplying that by four three times. But either way, this is going to result in four times four is 16, times four is 64. We also know a little bit about negative exponents. So for example, if I were to take four to the negative three power, we know this negative tells us to take the reciprocal, one over four to the third. And we already know four to the third is 64, so this is going to be 1 64th. Now let's think about fractional exponents."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We also know a little bit about negative exponents. So for example, if I were to take four to the negative three power, we know this negative tells us to take the reciprocal, one over four to the third. And we already know four to the third is 64, so this is going to be 1 64th. Now let's think about fractional exponents. So we're gonna think about what is four to the 1 1\u20442 power. I encourage you to pause the video and at least take a guess about what you think this is. So the mathematical convention here, the mathematical definition that most people use or that frankly all people use here, is that four to the 1 1\u20442 power is the exact same thing as the square root of four."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now let's think about fractional exponents. So we're gonna think about what is four to the 1 1\u20442 power. I encourage you to pause the video and at least take a guess about what you think this is. So the mathematical convention here, the mathematical definition that most people use or that frankly all people use here, is that four to the 1 1\u20442 power is the exact same thing as the square root of four. And we'll talk in the future about why this is, and the reason why this is defined this way is it has all sorts of neat and elegant properties when you start manipulating the actual exponents. But what is the square root of four, especially the principal root, mean? Well, that means, well, what is a number that if I were to multiply it by itself, or if I were to have two of those numbers and I were to multiply them times each other, that same number, I'm gonna get four."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the mathematical convention here, the mathematical definition that most people use or that frankly all people use here, is that four to the 1 1\u20442 power is the exact same thing as the square root of four. And we'll talk in the future about why this is, and the reason why this is defined this way is it has all sorts of neat and elegant properties when you start manipulating the actual exponents. But what is the square root of four, especially the principal root, mean? Well, that means, well, what is a number that if I were to multiply it by itself, or if I were to have two of those numbers and I were to multiply them times each other, that same number, I'm gonna get four. Well, what times itself is equal to four? Well, that's of course equal to two. And just to get a sense of why this starts to work out well, remember, we could have also written that four is equal to two squared."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well, that means, well, what is a number that if I were to multiply it by itself, or if I were to have two of those numbers and I were to multiply them times each other, that same number, I'm gonna get four. Well, what times itself is equal to four? Well, that's of course equal to two. And just to get a sense of why this starts to work out well, remember, we could have also written that four is equal to two squared. So you're starting to see something interesting. Four to the 1\u20442 is equal to two. Two squared is equal to four."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And just to get a sense of why this starts to work out well, remember, we could have also written that four is equal to two squared. So you're starting to see something interesting. Four to the 1\u20442 is equal to two. Two squared is equal to four. So let's get a couple more examples of this just so you make sure you get what's going on. I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think nine to the 1\u20442 power is going to be?"}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Two squared is equal to four. So let's get a couple more examples of this just so you make sure you get what's going on. I encourage you to pause it as much as necessary and try to figure it out yourself. So based on what I just told you, what do you think nine to the 1\u20442 power is going to be? Well, that's just the square root of nine, the principal root of nine. That's just going to be equal to three. And likewise, we could have also said that three squared is, or let me write it this way, that nine is equal to three squared."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So based on what I just told you, what do you think nine to the 1\u20442 power is going to be? Well, that's just the square root of nine, the principal root of nine. That's just going to be equal to three. And likewise, we could have also said that three squared is, or let me write it this way, that nine is equal to three squared. These are both true statements. Let's do one more like this. What is 25 to the 1\u20442 going to be?"}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And likewise, we could have also said that three squared is, or let me write it this way, that nine is equal to three squared. These are both true statements. Let's do one more like this. What is 25 to the 1\u20442 going to be? Well, this is just going to be five. Five times five is 25, or you could say that 25 is equal to five squared. Now, let's think about what happens when we take something to the 1\u20443 power."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "What is 25 to the 1\u20442 going to be? Well, this is just going to be five. Five times five is 25, or you could say that 25 is equal to five squared. Now, let's think about what happens when we take something to the 1\u20443 power. So let's imagine taking eight to the 1\u20443 power. So the definition here is that taking something to the 1\u20443 power is the same thing as taking the cube root of, is the same thing as taking the cube root of that number. And the cube root is saying, well, what number, if I had three of that number and I multiply them, that I'm going to get eight."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now, let's think about what happens when we take something to the 1\u20443 power. So let's imagine taking eight to the 1\u20443 power. So the definition here is that taking something to the 1\u20443 power is the same thing as taking the cube root of, is the same thing as taking the cube root of that number. And the cube root is saying, well, what number, if I had three of that number and I multiply them, that I'm going to get eight. So something times something times something is eight. Well, we already know that eight is equal to 2\u2043. So the cube root of eight, or eight to the 1\u20443, is just going to be equal to two."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And the cube root is saying, well, what number, if I had three of that number and I multiply them, that I'm going to get eight. So something times something times something is eight. Well, we already know that eight is equal to 2\u2043. So the cube root of eight, or eight to the 1\u20443, is just going to be equal to two. This says, hey, give me the number that if I take, if I say that number times that number times that number, I'm gonna get eight. Well, that number is two, because two to the 3\u20443 is eight. Do a few more examples of that."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So the cube root of eight, or eight to the 1\u20443, is just going to be equal to two. This says, hey, give me the number that if I take, if I say that number times that number times that number, I'm gonna get eight. Well, that number is two, because two to the 3\u20443 is eight. Do a few more examples of that. What is 64 to the 1\u20443 power? Well, we already know that four times four times four is 64. So this is going to be four."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Do a few more examples of that. What is 64 to the 1\u20443 power? Well, we already know that four times four times four is 64. So this is going to be four. And we already wrote over here that 64 is the same thing as four to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So this is going to be four. And we already wrote over here that 64 is the same thing as four to the third. I think you're starting to see a little bit of a pattern here, a little bit of symmetry here. And we can extend this idea to arbitrary rational exponents. So what's happening, what happens if I were to raise, let's say I had, let me think of a good number here. So let's say I have 32. I have the number 32, and I raise it to the 1\u2075 power."}, {"video_title": "Basic fractional exponents Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And we can extend this idea to arbitrary rational exponents. So what's happening, what happens if I were to raise, let's say I had, let me think of a good number here. So let's say I have 32. I have the number 32, and I raise it to the 1\u2075 power. So this says, hey, give me the number that if I were to multiply that number, or I were to repeatedly multiply that number five times, what is that, I would get 32. Well, 32 is the same thing as two times two times two times two times two. So two is that number that if I were to multiply it five times, then I'm going to get 32."}, {"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": "Modeling with combined functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Ify is building a tree tower, which is a tower built on top of a tree. The tree is currently five meters tall, and Ify has found, I don't know ify or ify, and Ify has found it is growing by 0.1 meters a month, or a tenth of a meter a month. The tower is currently two meters tall, so this tower that sits on top of the tree is two meters tall currently, and Ify or ify adds to it at about 0.2 meters a month, so I guess he or she, I don't know, ify, well, let's just say he, is continuously building this tower on this continuously growing tree, which is fascinating. All right, the function a of m returns the tree's height in meters m months from now. Fascinating. The function b of m returns the tower's height in meters m months from now. So this is the tree's height, a of m is the tree's height, b of m is the tower's height."}, {"video_title": "Modeling with combined functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "All right, the function a of m returns the tree's height in meters m months from now. Fascinating. The function b of m returns the tower's height in meters m months from now. So this is the tree's height, a of m is the tree's height, b of m is the tower's height. Find the formula of the two functions. So a of m, so they tell us the tree is currently five meters tall, so it's gonna be five meters tall right at the start, and then every month it is growing by 0.1 meters, so it's going to be five plus 0.1 times m, and this m here, this is not meters, this is actually the month. Remember, m is the number of months."}, {"video_title": "Modeling with combined functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is the tree's height, a of m is the tree's height, b of m is the tower's height. Find the formula of the two functions. So a of m, so they tell us the tree is currently five meters tall, so it's gonna be five meters tall right at the start, and then every month it is growing by 0.1 meters, so it's going to be five plus 0.1 times m, and this m here, this is not meters, this is actually the month. Remember, m is the number of months. So after zero months, which is right now, where this is just going to be five, after one month it's going to be 5.1, after two months it's going to be 5.2, which is exactly what we want. All right, now let's think about the tower. So the formula for b of m, so the tower is currently two meters tall, so it's currently two meters tall, and it grows at 2.10 of a meter per month, so 2.10 times the number of months, and once again, this m right over here is not meters."}, {"video_title": "Modeling with combined functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Remember, m is the number of months. So after zero months, which is right now, where this is just going to be five, after one month it's going to be 5.1, after two months it's going to be 5.2, which is exactly what we want. All right, now let's think about the tower. So the formula for b of m, so the tower is currently two meters tall, so it's currently two meters tall, and it grows at 2.10 of a meter per month, so 2.10 times the number of months, and once again, this m right over here is not meters. I'm not writing the units here. We're just assuming that whatever this returns is in meters. This m right over here is the number of months that pass by, the number of months from now."}, {"video_title": "Modeling with combined functions Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So the formula for b of m, so the tower is currently two meters tall, so it's currently two meters tall, and it grows at 2.10 of a meter per month, so 2.10 times the number of months, and once again, this m right over here is not meters. I'm not writing the units here. We're just assuming that whatever this returns is in meters. This m right over here is the number of months that pass by, the number of months from now. All right, the function c of m returns the vertical distance between the ground and the top end of the tower. Makes sense, that would be from the bottom of the tree to the top of the tower. Write the formula c of m in terms of a of m and b of m. Well, the total height is going to be the height of the tree, which is a of m, plus the height of the tower, plus b of m. That's what c of m is going to be, and then they say write the formula of c of m in terms of m. Well, we just need to add these two functions."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Let's say that we have a polynomial P of X, and we can factor it, and we can put it in the form X minus one times X plus two times X minus three times X plus four. And what we are concerned with are the zeros of this polynomial. And you might say, what is a zero of a polynomial? Well, those are the X values that are going to make the polynomial equal to zero. So another way to think about it is, for what X values is P of X going to be equal to zero? Or another way you can think about it is, for what X values is this expression going to be equal to zero? So for what X values is X minus one times X plus two times X minus three times X plus four going to be equal to zero?"}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, those are the X values that are going to make the polynomial equal to zero. So another way to think about it is, for what X values is P of X going to be equal to zero? Or another way you can think about it is, for what X values is this expression going to be equal to zero? So for what X values is X minus one times X plus two times X minus three times X plus four going to be equal to zero? I encourage you to pause this video, think about that a little bit before we work through it together. Well, the key realization here is, if you have the product of a bunch of expressions, if any one of them is equal to zero, it doesn't matter what the others are, because zero times anything else is going to be equal to zero. So the fancy term for that is the zero product property."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So for what X values is X minus one times X plus two times X minus three times X plus four going to be equal to zero? I encourage you to pause this video, think about that a little bit before we work through it together. Well, the key realization here is, if you have the product of a bunch of expressions, if any one of them is equal to zero, it doesn't matter what the others are, because zero times anything else is going to be equal to zero. So the fancy term for that is the zero product property. But all it says is, hey, if you can find an X value that makes any one of these expressions equal to zero, well, that's going to make the entire expression going to be, it's going to make the entire expression equal to zero. So the zeros of this polynomial are going to be the X values that could make X minus one equal to zero. So X minus one equals zero, well, we know what X value would make that happen if X is equal to one."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So the fancy term for that is the zero product property. But all it says is, hey, if you can find an X value that makes any one of these expressions equal to zero, well, that's going to make the entire expression going to be, it's going to make the entire expression equal to zero. So the zeros of this polynomial are going to be the X values that could make X minus one equal to zero. So X minus one equals zero, well, we know what X value would make that happen if X is equal to one. If you add one to both sides here, X equals one. So X equals one is a zero of this polynomial. Another way to say that is P of one, when X equals one, that whole polynomial is going to be equal to zero."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So X minus one equals zero, well, we know what X value would make that happen if X is equal to one. If you add one to both sides here, X equals one. So X equals one is a zero of this polynomial. Another way to say that is P of one, when X equals one, that whole polynomial is going to be equal to zero. How do I know that? Well, if I put a one in right over here, this expression right over here, X minus one, that is going to be equal to zero. So you're gonna have zero times a bunch of other stuff, which is going to be equal to zero."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Another way to say that is P of one, when X equals one, that whole polynomial is going to be equal to zero. How do I know that? Well, if I put a one in right over here, this expression right over here, X minus one, that is going to be equal to zero. So you're gonna have zero times a bunch of other stuff, which is going to be equal to zero. And so by the same idea, we can figure out what the other zeros are. What would make this part equal to zero? What X value would make X plus two equal to zero?"}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So you're gonna have zero times a bunch of other stuff, which is going to be equal to zero. And so by the same idea, we can figure out what the other zeros are. What would make this part equal to zero? What X value would make X plus two equal to zero? Well, X equals negative two. X equals negative two would make X plus two equal to zero. So X equals negative two is another zero of this polynomial."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "What X value would make X plus two equal to zero? Well, X equals negative two. X equals negative two would make X plus two equal to zero. So X equals negative two is another zero of this polynomial. And we could keep going. What would make X minus three equal to zero? Well, if X is equal to three, that would make X minus three equal to zero, and that would then make the entire expression equal to zero."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So X equals negative two is another zero of this polynomial. And we could keep going. What would make X minus three equal to zero? Well, if X is equal to three, that would make X minus three equal to zero, and that would then make the entire expression equal to zero. And then last but not least, what would make X plus four equal to zero? Well, if X is equal to negative four. And just like that, we have found four zeros for this polynomial."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, if X is equal to three, that would make X minus three equal to zero, and that would then make the entire expression equal to zero. And then last but not least, what would make X plus four equal to zero? Well, if X is equal to negative four. And just like that, we have found four zeros for this polynomial. When X equals one, the polynomial is equal to zero. When X equals negative two, the polynomial is equal to zero. When X equals three, the polynomial is equal to zero."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And just like that, we have found four zeros for this polynomial. When X equals one, the polynomial is equal to zero. When X equals negative two, the polynomial is equal to zero. When X equals three, the polynomial is equal to zero. And when X equals negative four, the polynomial is equal to zero. And one of the interesting things about the zeros of a polynomial, you could actually use that to start to sketch out what the graph might look like. So for example, we know that this polynomial is going to take on the value zero at these zeros."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "When X equals three, the polynomial is equal to zero. And when X equals negative four, the polynomial is equal to zero. And one of the interesting things about the zeros of a polynomial, you could actually use that to start to sketch out what the graph might look like. So for example, we know that this polynomial is going to take on the value zero at these zeros. So let me just draw a rough sketch right over here. So if this is my X axis, that's my Y axis. So that's my Y axis."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So for example, we know that this polynomial is going to take on the value zero at these zeros. So let me just draw a rough sketch right over here. So if this is my X axis, that's my Y axis. So that's my Y axis. And so let's see, at X equals one, so let me just do it this way. So we have one, two, three, and four. And then you have negative one, negative two, negative three, and then last but not least, negative four."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So that's my Y axis. And so let's see, at X equals one, so let me just do it this way. So we have one, two, three, and four. And then you have negative one, negative two, negative three, and then last but not least, negative four. We know that this polynomial, P of X, is going to be equal to zero at X equals one. So it's going to intersect the X axis right there. It's going to be equal to zero at X equals negative two, so right over there."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then you have negative one, negative two, negative three, and then last but not least, negative four. We know that this polynomial, P of X, is going to be equal to zero at X equals one. So it's going to intersect the X axis right there. It's going to be equal to zero at X equals negative two, so right over there. At X equals three, right over there. And X equals negative four. Now we don't know exactly what the graph looks like just based on this."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "It's going to be equal to zero at X equals negative two, so right over there. At X equals three, right over there. And X equals negative four. Now we don't know exactly what the graph looks like just based on this. We could try out some values on either side to figure out, hey, is it above the X axis or below the X axis for X values less than negative four? We can try things out like that, but we know it intersects the X axis at these points. So it might look something like this."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Now we don't know exactly what the graph looks like just based on this. We could try out some values on either side to figure out, hey, is it above the X axis or below the X axis for X values less than negative four? We can try things out like that, but we know it intersects the X axis at these points. So it might look something like this. This is a very rough sketch. It might look something like this. We don't know without doing a little bit more work."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So it might look something like this. This is a very rough sketch. It might look something like this. We don't know without doing a little bit more work. But ahead of time, I took a look at what this looks like. I went on to Desmos and I graphed it. And you can see it looks exactly as what we would expect."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We don't know without doing a little bit more work. But ahead of time, I took a look at what this looks like. I went on to Desmos and I graphed it. And you can see it looks exactly as what we would expect. The graph of this polynomial intersects the X axis at X equals negative four. Actually, let me color code it. X equals negative four, and that is that zero right over there."}, {"video_title": "Zeros of polynomials introduction Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And you can see it looks exactly as what we would expect. The graph of this polynomial intersects the X axis at X equals negative four. Actually, let me color code it. X equals negative four, and that is that zero right over there. X equals negative two, that's this zero right there. X equals one, right over there. And then X equals three, right over there."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "Their responses are shown below. So Theodore factored 24x to the fifth as being equal to 8x third times 3x squared. And Claire factored 24x to the fifth as being equal to 4x times 6x to the fourth. And then they ask us which of the students factored 24x to the fifth correctly? So I encourage you, pause the video and see if you can figure this out. Which of them factored it correctly? All right, now let's first look at Theodore."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "And then they ask us which of the students factored 24x to the fifth correctly? So I encourage you, pause the video and see if you can figure this out. Which of them factored it correctly? All right, now let's first look at Theodore. So he factored it into these two monomials, 8x to the third and 3x squared. Well, let's just see, if we were to multiply these two things, do we get 24x to the fifth? So if you multiply 8 times 3, you do indeed get 24."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, now let's first look at Theodore. So he factored it into these two monomials, 8x to the third and 3x squared. Well, let's just see, if we were to multiply these two things, do we get 24x to the fifth? So if you multiply 8 times 3, you do indeed get 24. And then all you have to do is multiply the x terms, or the powers of x. You have x to the third times x squared. That indeed does equal x to the fifth."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "So if you multiply 8 times 3, you do indeed get 24. And then all you have to do is multiply the x terms, or the powers of x. You have x to the third times x squared. That indeed does equal x to the fifth. So Theodore did factor it correctly. This is one factorization, I guess you could say, of 24x to the fifth. Now let's look at Claire."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "That indeed does equal x to the fifth. So Theodore did factor it correctly. This is one factorization, I guess you could say, of 24x to the fifth. Now let's look at Claire. So Claire, if we were to take just the coefficients, 4 times 6 is indeed equal to 24. And then if we were to look at the powers of x, we have x to the first power here times x to the fourth power, which is going to be x to the fifth power. So Claire also factored it correctly."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now let's look at Claire. So Claire, if we were to take just the coefficients, 4 times 6 is indeed equal to 24. And then if we were to look at the powers of x, we have x to the first power here times x to the fourth power, which is going to be x to the fifth power. So Claire also factored it correctly. And this just goes to show you that there's more than one possible factorization of a monomial like 24x to the fifth. I could come up with another one. I could write something like 24x to the fifth."}, {"video_title": "Monomial factorization Mathematics II High School Math Khan Academy.mp3", "Sentence": "So Claire also factored it correctly. And this just goes to show you that there's more than one possible factorization of a monomial like 24x to the fifth. I could come up with another one. I could write something like 24x to the fifth. I could say that that is 12x, I could write that as 12x to the third times, let's see, what would have to be left? 12 times 2 is 24, so 2x squared. That's another possible factorization."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Right over here, I've drawn a unit circle, and when we say unit circle, we're talking about a circle with radius one. So for example, this point right over here is the point one comma zero, x is equal to one, y is zero. This point is the point zero comma one. This is the point negative one comma zero, and this is the point zero comma negative one. That the radius over here, the distance from the center of the circle, which is at the origin, to any point in the circle, or any point on the circle, I should say, this radius is equal to one. So the unit circle definition of trig functions leverages this unit circle. That's why it's called the unit circle definition."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "This is the point negative one comma zero, and this is the point zero comma negative one. That the radius over here, the distance from the center of the circle, which is at the origin, to any point in the circle, or any point on the circle, I should say, this radius is equal to one. So the unit circle definition of trig functions leverages this unit circle. That's why it's called the unit circle definition. And we saw that if we define an angle as the kind of bottom side of the angle being along the positive x-axis, and then the other side of that angle, thinking about where it intersects the unit circle, so let's say that this is the angle theta, we define sine of theta and cosine of theta, or cosine theta and sine of theta, as the x and y coordinates of this point at which this side of the angle, the side that is not the positive x-axis, where that intersects the unit circle. So for example, this point right over here, if we would call this the x-coordinate of this point, so this value right over here, we would call that cosine of theta, we would call that cosine of theta, and the y-coordinate of that point, which is this point right over here, we would call that sine of theta. And in previous videos on the unit circle, we talked about why this is really just a natural extension of the Sohcahtoa definition."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "That's why it's called the unit circle definition. And we saw that if we define an angle as the kind of bottom side of the angle being along the positive x-axis, and then the other side of that angle, thinking about where it intersects the unit circle, so let's say that this is the angle theta, we define sine of theta and cosine of theta, or cosine theta and sine of theta, as the x and y coordinates of this point at which this side of the angle, the side that is not the positive x-axis, where that intersects the unit circle. So for example, this point right over here, if we would call this the x-coordinate of this point, so this value right over here, we would call that cosine of theta, we would call that cosine of theta, and the y-coordinate of that point, which is this point right over here, we would call that sine of theta. And in previous videos on the unit circle, we talked about why this is really just a natural extension of the Sohcahtoa definition. What's useful is it starts to work for negative angles, it works for angles, it even works for 90-degree angles, it works for angles more than 90 degrees, it works for angles less than 90 degrees. So it's really, really, really useful. But what I wanna do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "And in previous videos on the unit circle, we talked about why this is really just a natural extension of the Sohcahtoa definition. What's useful is it starts to work for negative angles, it works for angles, it even works for 90-degree angles, it works for angles more than 90 degrees, it works for angles less than 90 degrees. So it's really, really, really useful. But what I wanna do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity. It almost falls out of the fact that this point right over here is on a circle, a circle of radius one. So what is the equation of a circle with radius one centered at the origin? Well, the equation of that is x squared, x squared, we have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "But what I wanna do is leverage what we already know about the unit circle definition of trig functions to help prove the Pythagorean identity. It almost falls out of the fact that this point right over here is on a circle, a circle of radius one. So what is the equation of a circle with radius one centered at the origin? Well, the equation of that is x squared, x squared, we have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem. The equation of a circle, of a unit circle centered at the origin is x squared plus y squared, plus y squared is equal to, is equal to one, is equal to the radius squared. This distance right over here is equal to one. Well, we've already said that we're defining cosine of theta as the x coordinate of this point, and we're defining sine of theta as the y coordinate of this point, and this point is sitting on the circle."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, the equation of that is x squared, x squared, we have other videos where we really prove this using the distance formula, which is really just an application of the Pythagorean theorem. The equation of a circle, of a unit circle centered at the origin is x squared plus y squared, plus y squared is equal to, is equal to one, is equal to the radius squared. This distance right over here is equal to one. Well, we've already said that we're defining cosine of theta as the x coordinate of this point, and we're defining sine of theta as the y coordinate of this point, and this point is sitting on the circle. It has to satisfy this relationship right over here. So that means, well, if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value, and it has to satisfy this relationship, that means that cosine of theta squared plus sine squared of theta, plus sine squared of theta needs to be equal to, needs to be equal to one, or sine squared theta plus cosine squared of theta needs to be equal to one, and that's just from the point. This is the x, cosine theta is the x coordinate, sine theta is the y coordinate."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, we've already said that we're defining cosine of theta as the x coordinate of this point, and we're defining sine of theta as the y coordinate of this point, and this point is sitting on the circle. It has to satisfy this relationship right over here. So that means, well, if we're defining cosine of theta to be the x, to be this x value, sine of theta to be the y value, and it has to satisfy this relationship, that means that cosine of theta squared plus sine squared of theta, plus sine squared of theta needs to be equal to, needs to be equal to one, or sine squared theta plus cosine squared of theta needs to be equal to one, and that's just from the point. This is the x, cosine theta is the x coordinate, sine theta is the y coordinate. They have to satisfy this relationship, which defines a circle, so cosine squared theta plus sine squared theta is one. Now, this is called, as we've seen in other videos, this is called the Pythagorean identity, and you say, why is that useful? Well, using this, if you know sine of theta, you can figure out what cosine of theta is going to be, or vice versa, and if you know one of cosine of theta, and then you can, say you know cosine theta, then you use this to figure out sine of theta, then you can figure out tangent of theta, because tangent of theta is just sine over cosine, but if you're a little bit confused as to why this is called the Pythagorean identity, well, it really just falls out of where even the equation of a circle even came from."}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "This is the x, cosine theta is the x coordinate, sine theta is the y coordinate. They have to satisfy this relationship, which defines a circle, so cosine squared theta plus sine squared theta is one. Now, this is called, as we've seen in other videos, this is called the Pythagorean identity, and you say, why is that useful? Well, using this, if you know sine of theta, you can figure out what cosine of theta is going to be, or vice versa, and if you know one of cosine of theta, and then you can, say you know cosine theta, then you use this to figure out sine of theta, then you can figure out tangent of theta, because tangent of theta is just sine over cosine, but if you're a little bit confused as to why this is called the Pythagorean identity, well, it really just falls out of where even the equation of a circle even came from. If we look at this point right over here, if we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well, to think about that, we can construct a right triangle, so this distance right over here, this distance right over here, and so that we can deal with any quadrant. I'll make it the absolute value of the cosine theta is this distance right over here, and this distance right over here is the absolute value of the sine of theta, the absolute value of the sine of theta, and I obviously don't have to take the absolute value for this first quadrant here, but if I went into the other quadrants and I were to set up a similar right triangle, then the absolute value is at play, and so what do we know from the Pythagorean theorem?"}, {"video_title": "Pythagorean trig identity from unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, using this, if you know sine of theta, you can figure out what cosine of theta is going to be, or vice versa, and if you know one of cosine of theta, and then you can, say you know cosine theta, then you use this to figure out sine of theta, then you can figure out tangent of theta, because tangent of theta is just sine over cosine, but if you're a little bit confused as to why this is called the Pythagorean identity, well, it really just falls out of where even the equation of a circle even came from. If we look at this point right over here, if we look at this point right over here, which we're saying is the x coordinate is cosine theta and the y coordinate is sine of theta, what is the distance between that point and the origin? Well, to think about that, we can construct a right triangle, so this distance right over here, this distance right over here, and so that we can deal with any quadrant. I'll make it the absolute value of the cosine theta is this distance right over here, and this distance right over here is the absolute value of the sine of theta, the absolute value of the sine of theta, and I obviously don't have to take the absolute value for this first quadrant here, but if I went into the other quadrants and I were to set up a similar right triangle, then the absolute value is at play, and so what do we know from the Pythagorean theorem? This is a right triangle here. The hypotenuse has length one, so we know that this expression squared, the absolute value of cosine of theta squared plus this expression squared, which is this length, plus the absolute value of the sine of theta squared needs to be equal to the length of the hypotenuse squared, which is the same thing, which is going to be equal to one squared, or we could say, this is the same thing. If you're gonna square something, the sine, if it's negative, it's gonna be negative times a negative, so it's just gonna be positive, so this is going to be the same thing as saying that the cosine squared theta plus sine squared theta, plus sine squared theta is equal to one."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Derek sent a chain letter to his friends, asking them to forward the letter to more friends. The group of people who received the email gains nine-tenths of its size every three weeks, and can be modeled by a function P, which depends on the amount of time, T, in weeks. Derek initially sent the chain letter to 40 friends. Write a function that models the group of people who received the email T weeks since Derek initially sent the chain letter. So pause the video if you want to have a go at this. All right, now, the way I like to think about these, let's just create a table with values for T and our function P, which is a function of T, for some values that we can just pull out of the description here. So when T is zero, when it's been zero weeks since Derek initially sent the chain letter, how many people have gotten it?"}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Write a function that models the group of people who received the email T weeks since Derek initially sent the chain letter. So pause the video if you want to have a go at this. All right, now, the way I like to think about these, let's just create a table with values for T and our function P, which is a function of T, for some values that we can just pull out of the description here. So when T is zero, when it's been zero weeks since Derek initially sent the chain letter, how many people have gotten it? What do they tell us? He initially, Derek initially sent the chain letter to 40 friends. So T equals zero, P of T, or P of zero, is 40."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So when T is zero, when it's been zero weeks since Derek initially sent the chain letter, how many people have gotten it? What do they tell us? He initially, Derek initially sent the chain letter to 40 friends. So T equals zero, P of T, or P of zero, is 40. Now, what's an interesting time period here? It says that the email, the number of people who've received the email gains nine-tenths, or increases by nine-tenths every three weeks, every three weeks. So after three weeks, so three weeks have gone by, so I'm just adding three to T, what is P of T going to be?"}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So T equals zero, P of T, or P of zero, is 40. Now, what's an interesting time period here? It says that the email, the number of people who've received the email gains nine-tenths, or increases by nine-tenths every three weeks, every three weeks. So after three weeks, so three weeks have gone by, so I'm just adding three to T, what is P of T going to be? Well, they tell us it's going to gain nine-tenths of its size. So it's going to be 40 plus nine-tenths times 40, which is going to be equal to what? Well, that's equal to 40, if we factor out a 40, 40 times one plus nine-tenths, or you could say this is equal to 40 times, whoops, 40 times 1.9."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So after three weeks, so three weeks have gone by, so I'm just adding three to T, what is P of T going to be? Well, they tell us it's going to gain nine-tenths of its size. So it's going to be 40 plus nine-tenths times 40, which is going to be equal to what? Well, that's equal to 40, if we factor out a 40, 40 times one plus nine-tenths, or you could say this is equal to 40 times, whoops, 40 times 1.9. Or another way of thinking about it, after three weeks, we've grown 90%. That's another way of saying that the number of people who receive the email gains nine-tenths of its size. You could say the group of people who receive the email grows 90% every three weeks."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, that's equal to 40, if we factor out a 40, 40 times one plus nine-tenths, or you could say this is equal to 40 times, whoops, 40 times 1.9. Or another way of thinking about it, after three weeks, we've grown 90%. That's another way of saying that the number of people who receive the email gains nine-tenths of its size. You could say the group of people who receive the email grows 90% every three weeks. And so if we go another three weeks, so plus another three weeks, I could say, well, let me just write this as six weeks. Well, how many people will have received the email? Well, it's going to be this number, and it's going to be grown another 90%."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "You could say the group of people who receive the email grows 90% every three weeks. And so if we go another three weeks, so plus another three weeks, I could say, well, let me just write this as six weeks. Well, how many people will have received the email? Well, it's going to be this number, and it's going to be grown another 90%. So we're going to multiply it times 1.9 again. So it's going to be 40 times 1.9 times 1.9. We're going to grow by another nine-tenths."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, it's going to be this number, and it's going to be grown another 90%. So we're going to multiply it times 1.9 again. So it's going to be 40 times 1.9 times 1.9. We're going to grow by another nine-tenths. Growing by nine-tenths is the same thing as multiplying by one and nine-tenths. The one is what you already are, and then you're growing by another nine-tenths. So this is the same thing as 40 times 1.9 squared."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "We're going to grow by another nine-tenths. Growing by nine-tenths is the same thing as multiplying by one and nine-tenths. The one is what you already are, and then you're growing by another nine-tenths. So this is the same thing as 40 times 1.9 squared. You go another three weeks, nine weeks, we're going to grow another 90%. So you're going to multiply by, you're going to take this number and multiply by 1.9 again, which is going to be 1.9 to the third power. And so what's going on over here?"}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is the same thing as 40 times 1.9 squared. You go another three weeks, nine weeks, we're going to grow another 90%. So you're going to multiply by, you're going to take this number and multiply by 1.9 again, which is going to be 1.9 to the third power. And so what's going on over here? Well, we can see it's an exponential function. We have our initial value, and every three weeks, we're multiplying by 1.9. So 1.9 would be our common ratio."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "And so what's going on over here? Well, we can see it's an exponential function. We have our initial value, and every three weeks, we're multiplying by 1.9. So 1.9 would be our common ratio. So we could say that P of t is equal to our initial value, 40, times our common ratio, 1.9, and we multiply by 1.9 every three weeks. So we could just say how many three-week periods have passed by? Well, we will take t and divide it by three."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So 1.9 would be our common ratio. So we could say that P of t is equal to our initial value, 40, times our common ratio, 1.9, and we multiply by 1.9 every three weeks. So we could just say how many three-week periods have passed by? Well, we will take t and divide it by three. t divided by three is the number of three-week periods that have gone by. And there you have it. And notice, t equals zero, 1.9 to the zeroth power is one."}, {"video_title": "Constructing exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, we will take t and divide it by three. t divided by three is the number of three-week periods that have gone by. And there you have it. And notice, t equals zero, 1.9 to the zeroth power is one. So 40 times one. t equals three, that's going to be 1.9 to the first power, three over three. And so we're going to grow by 90%, and so on and so forth."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "For what, for what c is x minus five a factor, factor of, of p of x? And I encourage you to pause the video and try to work through it. And that's p is actually a lowercase p. P of x. All right, so I'm assuming you have attempted this. And we just have to realize, okay, if x minus five is a factor of p of x, that means you could write x, you could write p of x, let me do this in that green color, that means you could write p of x as being equal to x minus five, x minus five times some other business, times some other polynomial. So times, times some other polynomial. I don't know, let's just call it g of x maybe."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "All right, so I'm assuming you have attempted this. And we just have to realize, okay, if x minus five is a factor of p of x, that means you could write x, you could write p of x, let me do this in that green color, that means you could write p of x as being equal to x minus five, x minus five times some other business, times some other polynomial. So times, times some other polynomial. I don't know, let's just call it g of x maybe. G of x. And this would be, well it would be a second degree polynomial. And so what would p of five have to be?"}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "I don't know, let's just call it g of x maybe. G of x. And this would be, well it would be a second degree polynomial. And so what would p of five have to be? Well p of five would have to be zero. Five would have to be a root of this polynomial. You see it right over here."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "And so what would p of five have to be? Well p of five would have to be zero. Five would have to be a root of this polynomial. You see it right over here. If you replace, if you replace this with a five, then this is going to be a five, and this is going to be a five. And it doesn't matter what g of five is, this is going to be zero. So if x minus five, x minus five is, is a factor, is a factor, if and only if, if and only if, p of five is equal to zero."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "You see it right over here. If you replace, if you replace this with a five, then this is going to be a five, and this is going to be a five. And it doesn't matter what g of five is, this is going to be zero. So if x minus five, x minus five is, is a factor, is a factor, if and only if, if and only if, p of five is equal to zero. P of five is equal to zero. You could say that five is a root of the polynomial p. P of five is equal to zero. So let's just set p of five equal to zero, and then try to solve for c. Alright, so we're going to get five to the third power."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "So if x minus five, x minus five is, is a factor, is a factor, if and only if, if and only if, p of five is equal to zero. P of five is equal to zero. You could say that five is a root of the polynomial p. P of five is equal to zero. So let's just set p of five equal to zero, and then try to solve for c. Alright, so we're going to get five to the third power. So this right over here, five to the third. So p of five, p, let me just write it down. P of five is equal to, five to the third is 125, plus two times five squared."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "So let's just set p of five equal to zero, and then try to solve for c. Alright, so we're going to get five to the third power. So this right over here, five to the third. So p of five, p, let me just write it down. P of five is equal to, five to the third is 125, plus two times five squared. So it's two times 25 plus 50, plus c times five, plus five c, plus 10. That needs to be equal to zero. And let's see, if we add, if we were to add 125 to 50 to 10, we are going to get 175, 185."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "P of five is equal to, five to the third is 125, plus two times five squared. So it's two times 25 plus 50, plus c times five, plus five c, plus 10. That needs to be equal to zero. And let's see, if we add, if we were to add 125 to 50 to 10, we are going to get 175, 185. So we get 185 plus five c, plus five c is equal to zero. Subtract 185 from both sides, you get five c is equal to negative 185, negative 185, or c, c is equal to negative 185 over five, which is going to be, let's see, it's negative 185 over five, which is equal to, let's see, five goes into 180, let's see, it goes into, let's see, five, let's see, it's gonna be 30, five times 30 is 150. Five times 30 is 150, and then we'll have another 35 to go."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "And let's see, if we add, if we were to add 125 to 50 to 10, we are going to get 175, 185. So we get 185 plus five c, plus five c is equal to zero. Subtract 185 from both sides, you get five c is equal to negative 185, negative 185, or c, c is equal to negative 185 over five, which is going to be, let's see, it's negative 185 over five, which is equal to, let's see, five goes into 180, let's see, it goes into, let's see, five, let's see, it's gonna be 30, five times 30 is 150. Five times 30 is 150, and then we'll have another 35 to go. So it's gonna be 37, negative 37. Is that right? Five times 30 is 150."}, {"video_title": "Constructing a polynomial that has a certain factor Algebra II Khan Academy.mp3", "Sentence": "Five times 30 is 150, and then we'll have another 35 to go. So it's gonna be 37, negative 37. Is that right? Five times 30 is 150. Five times seven is 35, yep. Negative 37, and we're done. If this was x to the third plus two x squared, minus 37x plus 10, then x minus five would be a factor of this polynomial."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And they want us to put it in here. This is actually a screenshot from the exercise on Khan Academy, but I'm just going to write on it. If you're doing it on Khan Academy, you would type it in. But pause this video and see if you can answer this first part. All right, so one intersection point is clearly identifiable from the graph. I see two intersection points, I see that one, and I see that one there. This second one seems clearly identifiable because when I look at the grid, it looks clearly to be at a value of x equals two and y equals one, it seems to be the point two comma one."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "But pause this video and see if you can answer this first part. All right, so one intersection point is clearly identifiable from the graph. I see two intersection points, I see that one, and I see that one there. This second one seems clearly identifiable because when I look at the grid, it looks clearly to be at a value of x equals two and y equals one, it seems to be the point two comma one. So it's two comma one there. And what's interesting about these intersection points is, because it's a point that sits on the graph of both of these curves, that means that it satisfies both of these equations, that it's a solution to both of these equations. So the other one is, find the other intersection point, your answer must be exact."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "This second one seems clearly identifiable because when I look at the grid, it looks clearly to be at a value of x equals two and y equals one, it seems to be the point two comma one. So it's two comma one there. And what's interesting about these intersection points is, because it's a point that sits on the graph of both of these curves, that means that it satisfies both of these equations, that it's a solution to both of these equations. So the other one is, find the other intersection point, your answer must be exact. So they want us to figure out this intersection point right over here. Well, to do that, we're going to have to try to solve this system of equations. And this is interesting because this is a system of equations where one of the equations is not linear, where it is a quadratic."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So the other one is, find the other intersection point, your answer must be exact. So they want us to figure out this intersection point right over here. Well, to do that, we're going to have to try to solve this system of equations. And this is interesting because this is a system of equations where one of the equations is not linear, where it is a quadratic. So let's see how we can go about doing that. So let me write down the equations. I have y is equal to three x squared minus six x plus one."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And this is interesting because this is a system of equations where one of the equations is not linear, where it is a quadratic. So let's see how we can go about doing that. So let me write down the equations. I have y is equal to three x squared minus six x plus one. And our next equation right over here, y minus x plus one is equal to zero. Well, one way to tackle, and this is one way to tackle any system of equations, is through substitution. So if I can rewrite this linear equation as in terms of y, if I can solve for y, then I can substitute what y equals back into my first equation, into my quadratic one, and then hopefully I can solve for x."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "I have y is equal to three x squared minus six x plus one. And our next equation right over here, y minus x plus one is equal to zero. Well, one way to tackle, and this is one way to tackle any system of equations, is through substitution. So if I can rewrite this linear equation as in terms of y, if I can solve for y, then I can substitute what y equals back into my first equation, into my quadratic one, and then hopefully I can solve for x. So let's solve for y here. And actually, let me color code it because this one is in red, and this one is the line in that blue color. So let's just solve for y."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So if I can rewrite this linear equation as in terms of y, if I can solve for y, then I can substitute what y equals back into my first equation, into my quadratic one, and then hopefully I can solve for x. So let's solve for y here. And actually, let me color code it because this one is in red, and this one is the line in that blue color. So let's just solve for y. The easiest way to solve for y is to add x to both sides and subtract one from both sides. That was hard to see. So and subtract one from both sides."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's just solve for y. The easiest way to solve for y is to add x to both sides and subtract one from both sides. That was hard to see. So and subtract one from both sides. And so we are going to get y, and then all the rest of the stuff cancels out, is equal to x minus one. And so now we can substitute x minus one back in for y. And so we get x minus one is equal to three x squared minus six x plus one."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So and subtract one from both sides. And so we are going to get y, and then all the rest of the stuff cancels out, is equal to x minus one. And so now we can substitute x minus one back in for y. And so we get x minus one is equal to three x squared minus six x plus one. Now we wanna get a zero on one side of this equation, so let's subtract x. I'll do this in a neutral color now. Let's subtract x from both sides, and let's add one to both sides. And then what do we get?"}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And so we get x minus one is equal to three x squared minus six x plus one. Now we wanna get a zero on one side of this equation, so let's subtract x. I'll do this in a neutral color now. Let's subtract x from both sides, and let's add one to both sides. And then what do we get? On the left-hand side, we just get zero. And on the right-hand side, we get three x squared minus seven x plus two. So this is equal to zero."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And then what do we get? On the left-hand side, we just get zero. And on the right-hand side, we get three x squared minus seven x plus two. So this is equal to zero. Now we could try to factor this. Let's see, is there an obvious way to factor it? Can I think of two numbers, a times b, that's equal to the product of three and two?"}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So this is equal to zero. Now we could try to factor this. Let's see, is there an obvious way to factor it? Can I think of two numbers, a times b, that's equal to the product of three and two? Three times two. And if this looks unfamiliar, you can review factoring by grouping. And can I think of the same two a plus b where it's going to be equal to negative seven?"}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "Can I think of two numbers, a times b, that's equal to the product of three and two? Three times two. And if this looks unfamiliar, you can review factoring by grouping. And can I think of the same two a plus b where it's going to be equal to negative seven? And actually, negative six and negative one work. So what I can do is I can rewrite this whole thing as zero is equal to three x squared. And then instead of negative seven x, I can write negative six x and minus x."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And can I think of the same two a plus b where it's going to be equal to negative seven? And actually, negative six and negative one work. So what I can do is I can rewrite this whole thing as zero is equal to three x squared. And then instead of negative seven x, I can write negative six x and minus x. And then I have my plus two. I'm just factoring by grouping. For those of you who are not familiar with this technique, you could also use a quadratic formula."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And then instead of negative seven x, I can write negative six x and minus x. And then I have my plus two. I'm just factoring by grouping. For those of you who are not familiar with this technique, you could also use a quadratic formula. So then zero is equal to, so if I group these first two, I can factor out a three x. So I'm going to get three x times x minus two. And then these second two, I can factor out, in these second two, I can factor out a negative one."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "For those of you who are not familiar with this technique, you could also use a quadratic formula. So then zero is equal to, so if I group these first two, I can factor out a three x. So I'm going to get three x times x minus two. And then these second two, I can factor out, in these second two, I can factor out a negative one. So I have negative one times x minus two. And then I can factor out a negative two. I'll scroll down a little bit so I have some space."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "And then these second two, I can factor out, in these second two, I can factor out a negative one. So I have negative one times x minus two. And then I can factor out a negative two. I'll scroll down a little bit so I have some space. So I have zero is equal to, if I factor out an x minus two, I'm going to get x minus two times three x minus one. And so a solution would be a situation where either of these is equal to zero, or I'll scroll down a little bit more. So x minus two could be equal to zero, or three x minus one is equal to zero."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "I'll scroll down a little bit so I have some space. So I have zero is equal to, if I factor out an x minus two, I'm going to get x minus two times three x minus one. And so a solution would be a situation where either of these is equal to zero, or I'll scroll down a little bit more. So x minus two could be equal to zero, or three x minus one is equal to zero. The point at which x minus two equals zero is when x is equal to two. And for three x minus one equals zero, add one to both sides, you get three x is equal to one, or x is equal to 1 3rd. So we figured out the, we already saw the solution where x is equals two."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So x minus two could be equal to zero, or three x minus one is equal to zero. The point at which x minus two equals zero is when x is equal to two. And for three x minus one equals zero, add one to both sides, you get three x is equal to one, or x is equal to 1 3rd. So we figured out the, we already saw the solution where x is equals two. That's this point right over here. We already typed that in. But now we figured out the x value of the other solution."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So we figured out the, we already saw the solution where x is equals two. That's this point right over here. We already typed that in. But now we figured out the x value of the other solution. So this is x is equal to 1 3rd right over here. So our x value is 1 3rd. But we still have to figure out the y value."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "But now we figured out the x value of the other solution. So this is x is equal to 1 3rd right over here. So our x value is 1 3rd. But we still have to figure out the y value. Well, the y value is going to be the corresponding y we get for that x in either equation. And I like to focus on the simpler of the two equations. So we could figure out what is x when, or what is y when x is equal to 1 3rd using this equation."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "But we still have to figure out the y value. Well, the y value is going to be the corresponding y we get for that x in either equation. And I like to focus on the simpler of the two equations. So we could figure out what is x when, or what is y when x is equal to 1 3rd using this equation. We could have used the original one, but this is even simpler. It's already solved for y. So y is equal to 1 3rd minus one."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So we could figure out what is x when, or what is y when x is equal to 1 3rd using this equation. We could have used the original one, but this is even simpler. It's already solved for y. So y is equal to 1 3rd minus one. I'm just substituting that 1 3rd back into this. And so you get y is equal to negative 2 3rds. And it looks like that as well."}, {"video_title": "Quadratic systems a line and a parabola Equations Algebra 2 Khan Academy.mp3", "Sentence": "So y is equal to 1 3rd minus one. I'm just substituting that 1 3rd back into this. And so you get y is equal to negative 2 3rds. And it looks like that as well. Y is equal to negative 2 3rds right over there. So this is the point 1 3rd comma negative 2 3rds. And we're done."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Or a lot of times, if you're reading a math or a science book, they're gonna do some proof or something like that and they're gonna say, oh, well, you know, it's obviously from this step to this step and you're gonna try to follow it. And you're just like, well, does that make sense? So it's a really useful muscle to be able to see, like, does, do these steps or how, whoever manipulated the polynomial, does it make sense to you? And especially if it's you, it's super useful to be able to find if there are errors and to correct them. It'll just give you a better critical eye for this type of thing. So let's just start with this one. We have four x minus three times x minus two squared."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And especially if it's you, it's super useful to be able to find if there are errors and to correct them. It'll just give you a better critical eye for this type of thing. So let's just start with this one. We have four x minus three times x minus two squared. And it looks like this person, over five steps, tries to expand it out. And so what I encourage you to do, pause the video right now and see if they did it correctly. And if they didn't do it correctly, try to identify on what step they messed up."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We have four x minus three times x minus two squared. And it looks like this person, over five steps, tries to expand it out. And so what I encourage you to do, pause the video right now and see if they did it correctly. And if they didn't do it correctly, try to identify on what step they messed up. All right, so assuming you've had a go at it, let's do this together. So as we go from the first expression to the second, to step one, what do they do? Well, they just expanded out x minus two squared."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And if they didn't do it correctly, try to identify on what step they messed up. All right, so assuming you've had a go at it, let's do this together. So as we go from the first expression to the second, to step one, what do they do? Well, they just expanded out x minus two squared. X minus two squared is just x minus two times x minus two. They haven't done anything to the four x minus three yet. And what are they doing in step, so that seems correct."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, they just expanded out x minus two squared. X minus two squared is just x minus two times x minus two. They haven't done anything to the four x minus three yet. And what are they doing in step, so that seems correct. So in step two, it looks like they're just trying to multiply x minus two times x minus two. So you have x times x, which would be x squared. You have x times negative two, which would be negative two x."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And what are they doing in step, so that seems correct. So in step two, it looks like they're just trying to multiply x minus two times x minus two. So you have x times x, which would be x squared. You have x times negative two, which would be negative two x. You have negative two times x, which would be negative two x. And then you have negative two times negative two, which would be positive four. So it looks like they multiplied this out correctly."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "You have x times negative two, which would be negative two x. You have negative two times x, which would be negative two x. And then you have negative two times negative two, which would be positive four. So it looks like they multiplied this out correctly. So step two, we're still doing good. All right, now what do they do in step three? And this whole time, four x minus three, they haven't really touched it yet."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it looks like they multiplied this out correctly. So step two, we're still doing good. All right, now what do they do in step three? And this whole time, four x minus three, they haven't really touched it yet. So they're just trying to simplify it. And all they did is they added these two middle terms. Minus negative two x minus two x is gonna be negative four x."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And this whole time, four x minus three, they haven't really touched it yet. So they're just trying to simplify it. And all they did is they added these two middle terms. Minus negative two x minus two x is gonna be negative four x. So this still looks correct. The x squared didn't change, the plus four didn't change, they just added these middle two terms. Now as we go to this next step four, well now they're trying to multiply these two expressions."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Minus negative two x minus two x is gonna be negative four x. So this still looks correct. The x squared didn't change, the plus four didn't change, they just added these middle two terms. Now as we go to this next step four, well now they're trying to multiply these two expressions. So they're doing some algebraic multiplication. So let's see if we can figure this out. So we have four x times, let me do this in a new color."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now as we go to this next step four, well now they're trying to multiply these two expressions. So they're doing some algebraic multiplication. So let's see if we can figure this out. So we have four x times, let me do this in a new color. I'm getting bored of that magenta. All right, so we have four x times x squared, which is indeed four x to the third power. Then you have four x times negative four x, which is gonna be negative 16 x squared."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So we have four x times, let me do this in a new color. I'm getting bored of that magenta. All right, so we have four x times x squared, which is indeed four x to the third power. Then you have four x times negative four x, which is gonna be negative 16 x squared. So they did that right. Then you have four x times four, which is going to be 16 x, and they wrote that right over there. Then you're going to have negative three times x squared, which is negative three x squared."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Then you have four x times negative four x, which is gonna be negative 16 x squared. So they did that right. Then you have four x times four, which is going to be 16 x, and they wrote that right over there. Then you're going to have negative three times x squared, which is negative three x squared. We see that right over there. Then you're going to have negative three times negative four x, which is going to be positive 12 x, and they wrote negative 12 x. So they forgot, they saw a negative, negative, but they still put a negative there."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Then you're going to have negative three times x squared, which is negative three x squared. We see that right over there. Then you're going to have negative three times negative four x, which is going to be positive 12 x, and they wrote negative 12 x. So they forgot, they saw a negative, negative, but they still put a negative there. Negative three times negative four x, is negative times negative, is gonna be a positive, positive 12 x. So they made an error here, and then they said negative three times positive four, which is indeed negative 12. So this part is right."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So they forgot, they saw a negative, negative, but they still put a negative there. Negative three times negative four x, is negative times negative, is gonna be a positive, positive 12 x. So they made an error here, and then they said negative three times positive four, which is indeed negative 12. So this part is right. So the error, this thing should read positive, positive 12 x. So the error they made is in step four. Step four is the error, and then that ended up giving them the wrong answer here, because they did a minus 12 x instead of, if this was a negative 12 x, then negative 12 x plus 16 x got you this four x, but we know it's supposed to be plus 12 x."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this part is right. So the error, this thing should read positive, positive 12 x. So the error they made is in step four. Step four is the error, and then that ended up giving them the wrong answer here, because they did a minus 12 x instead of, if this was a negative 12 x, then negative 12 x plus 16 x got you this four x, but we know it's supposed to be plus 12 x. So it really should be, it really should be 28. This should be 20, this should be 28 x right over here, but they really messed up. If you take that error, they did the step right, but step four is where they actually made the error."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Step four is the error, and then that ended up giving them the wrong answer here, because they did a minus 12 x instead of, if this was a negative 12 x, then negative 12 x plus 16 x got you this four x, but we know it's supposed to be plus 12 x. So it really should be, it really should be 28. This should be 20, this should be 28 x right over here, but they really messed up. If you take that error, they did the step right, but step four is where they actually made the error. So let's keep going. Let's give ourselves a little bit more practice at looking at ways to manipulate polynomials and see if they're valid. So here, this comes from an exercise on Khan Academy."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "If you take that error, they did the step right, but step four is where they actually made the error. So let's keep going. Let's give ourselves a little bit more practice at looking at ways to manipulate polynomials and see if they're valid. So here, this comes from an exercise on Khan Academy. Let's see which of these are valid identities, which of these are valid statements. So this first one, two x plus y times four x minus two y is all of this business right over here. Let's just multiply it out."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So here, this comes from an exercise on Khan Academy. Let's see which of these are valid identities, which of these are valid statements. So this first one, two x plus y times four x minus two y is all of this business right over here. Let's just multiply it out. Two x times four x is going to be eight x squared. Two x times negative two y is going to be negative, negative four x y, and then, let me switch colors, y times four x is going to be plus four x y, and then y times negative two y is going to be, is going to be minus two y squared. And so let's see, these two, did I do that right?"}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's just multiply it out. Two x times four x is going to be eight x squared. Two x times negative two y is going to be negative, negative four x y, and then, let me switch colors, y times four x is going to be plus four x y, and then y times negative two y is going to be, is going to be minus two y squared. And so let's see, these two, did I do that right? Let's see, two x, two x times negative two y is negative four x y, and then I had four x times y is positive four x y, so these two are going to cancel out. So this is already, this is looking shady. So all we're left with is eight x squared minus two y squared."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And so let's see, these two, did I do that right? Let's see, two x, two x times negative two y is negative four x y, and then I had four x times y is positive four x y, so these two are going to cancel out. So this is already, this is looking shady. So all we're left with is eight x squared minus two y squared. If we factor two out, if we factor two out, it's going to be two times four x squared minus y squared. So this is, this is not a true, this is not a true statement right over here. Now let's try this one."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So all we're left with is eight x squared minus two y squared. If we factor two out, if we factor two out, it's going to be two times four x squared minus y squared. So this is, this is not a true, this is not a true statement right over here. Now let's try this one. n plus two squared, n plus two squared minus n squared is equal to this. Well, what's n plus two squared? That's going to be n squared plus four n, it's going to be two n plus two n, it's going to be plus four n plus four, and then we're going to subtract out an n squared, these cancel, so you're going to have four n plus four, which is equal to four times n plus one."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now let's try this one. n plus two squared, n plus two squared minus n squared is equal to this. Well, what's n plus two squared? That's going to be n squared plus four n, it's going to be two n plus two n, it's going to be plus four n plus four, and then we're going to subtract out an n squared, these cancel, so you're going to have four n plus four, which is equal to four times n plus one. So this one right over here works out. This is a true, I guess, in the language of this question, it is a valid identity, or you could say it's a true statement, this equation is true. And then we have this last one right over here."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "That's going to be n squared plus four n, it's going to be two n plus two n, it's going to be plus four n plus four, and then we're going to subtract out an n squared, these cancel, so you're going to have four n plus four, which is equal to four times n plus one. So this one right over here works out. This is a true, I guess, in the language of this question, it is a valid identity, or you could say it's a true statement, this equation is true. And then we have this last one right over here. And once again, let's see if we can multiply it out. Let me go down here into the black space. So if I have a times two a, that's going to be two a squared and then a times one is going to be plus a, and then if I have b times two a, it's going to be plus two a b, and then finally if I have b times one, it's going to be plus b."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then we have this last one right over here. And once again, let's see if we can multiply it out. Let me go down here into the black space. So if I have a times two a, that's going to be two a squared and then a times one is going to be plus a, and then if I have b times two a, it's going to be plus two a b, and then finally if I have b times one, it's going to be plus b. And then out here, we are subtracting a b. So over here we're going to subtract a b, these characters are going to cancel out, and then we're left with two a squared plus a plus two a b. And it looks like they factor out an a, so let's see if we can factor out an a ourselves."}, {"video_title": "Analyzing polynomial manipulations Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So if I have a times two a, that's going to be two a squared and then a times one is going to be plus a, and then if I have b times two a, it's going to be plus two a b, and then finally if I have b times one, it's going to be plus b. And then out here, we are subtracting a b. So over here we're going to subtract a b, these characters are going to cancel out, and then we're left with two a squared plus a plus two a b. And it looks like they factor out an a, so let's see if we can factor out an a ourselves. So if we factor out an a, we're going to be left with, this first term is going to be two a, this right over here, if we factor out an a, is going to be plus one, and this if we factor out an a is going to be two b, is going to be two b, and that's exactly what they wrote over here, they just wrote it in a different order. A times two a, two a, plus two b, plus two b, plus one, plus one. So this is legit."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Let's get some practice solving some exponential equations and we have one right over here. We have 26 to the nine x plus five power equals one. So pause the video and see if you can tell me what x is going to be. Well, the key here is to realize that 26 to the zeroth power, to the zeroth power, is equal to one. Anything to the zeroth power is going to be equal to one. Zero to zeroth power, we can discuss at some other time, but anything other than zero to the zeroth power is going to be one. So we just have to say, well, nine x plus five needs to be equal to zero."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Well, the key here is to realize that 26 to the zeroth power, to the zeroth power, is equal to one. Anything to the zeroth power is going to be equal to one. Zero to zeroth power, we can discuss at some other time, but anything other than zero to the zeroth power is going to be one. So we just have to say, well, nine x plus five needs to be equal to zero. Nine x plus five needs to be equal to zero. And this is pretty straightforward to solve. Subtract five from both sides and we get nine x is equal to negative five."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So we just have to say, well, nine x plus five needs to be equal to zero. Nine x plus five needs to be equal to zero. And this is pretty straightforward to solve. Subtract five from both sides and we get nine x is equal to negative five. Divide both sides by nine and we are left with x is equal to negative five. Let's do another one of these and let's make it a little bit more interesting. Let's say we have the exponential equation two to the three x plus five power is equal to 64 to the x minus seventh power."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Subtract five from both sides and we get nine x is equal to negative five. Divide both sides by nine and we are left with x is equal to negative five. Let's do another one of these and let's make it a little bit more interesting. Let's say we have the exponential equation two to the three x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video and see if you can tell me what x is going to be or what x needs to be to satisfy this exponential equation. All right, so you might at first say, oh, maybe three x plus five needs to be equal to x minus seven but that wouldn't work because these are two different bases. You have two to the three x plus five power then you have 64 to the x minus seven."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Let's say we have the exponential equation two to the three x plus five power is equal to 64 to the x minus seventh power. Once again, pause the video and see if you can tell me what x is going to be or what x needs to be to satisfy this exponential equation. All right, so you might at first say, oh, maybe three x plus five needs to be equal to x minus seven but that wouldn't work because these are two different bases. You have two to the three x plus five power then you have 64 to the x minus seven. So the key here is to express both of these with the same base. And lucky for us, 64 is a power of two. Two to the, let's see, two to the third is eight."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "You have two to the three x plus five power then you have 64 to the x minus seven. So the key here is to express both of these with the same base. And lucky for us, 64 is a power of two. Two to the, let's see, two to the third is eight. So it's gonna be two to the third times two to the third. Eight times eight is 64. So it's two to the sixth is equal to 64."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Two to the, let's see, two to the third is eight. So it's gonna be two to the third times two to the third. Eight times eight is 64. So it's two to the sixth is equal to 64. And you can verify that. Take six twos and multiply them together. You're going to get 64."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So it's two to the sixth is equal to 64. And you can verify that. Take six twos and multiply them together. You're going to get 64. This is just a little bit easier for me. Eight times eight, and this is the same thing as two to the sixth power is 64. And I knew it was two to the sixth power because I just added the exponents because I had the same base."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "You're going to get 64. This is just a little bit easier for me. Eight times eight, and this is the same thing as two to the sixth power is 64. And I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64. Let me rewrite the whole thing. So this is two to the three x plus five power is equal to, instead of writing 64, I am going to write two to the sixth power."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "And I knew it was two to the sixth power because I just added the exponents because I had the same base. All right, so I can rewrite 64. Let me rewrite the whole thing. So this is two to the three x plus five power is equal to, instead of writing 64, I am going to write two to the sixth power. Two to the sixth power. And then that to the x minus seventh power. X minus seventh power."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So this is two to the three x plus five power is equal to, instead of writing 64, I am going to write two to the sixth power. Two to the sixth power. And then that to the x minus seventh power. X minus seventh power. And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers. A to the bc power. So this equation I can rewrite as two to the three x plus five is equal to two to the, and I just multiply six times x minus seven."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "X minus seventh power. And to simplify this a little bit, we just have to remind ourselves that if I raise something to one power and then I raise that to another power, this is the same thing as raising my base to the product of these powers. A to the bc power. So this equation I can rewrite as two to the three x plus five is equal to two to the, and I just multiply six times x minus seven. So it's going to be six x, six x minus, six times seven is 42. I'll just write the whole thing in yellow. So six x minus 42."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So this equation I can rewrite as two to the three x plus five is equal to two to the, and I just multiply six times x minus seven. So it's going to be six x, six x minus, six times seven is 42. I'll just write the whole thing in yellow. So six x minus 42. I just multiplied the six times the entire expression, x minus seven. And so now it's interesting. I have two to the three x plus five power has to be equal to two to the six x minus 42 power."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So six x minus 42. I just multiplied the six times the entire expression, x minus seven. And so now it's interesting. I have two to the three x plus five power has to be equal to two to the six x minus 42 power. So these need to be the same exponent. So three x plus five needs to be equal to six x minus 42. So there we go."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "I have two to the three x plus five power has to be equal to two to the six x minus 42 power. So these need to be the same exponent. So three x plus five needs to be equal to six x minus 42. So there we go. It sets up a nice little linear equation for us. Three x plus five is equal to six x minus 42. Let's see."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "So there we go. It sets up a nice little linear equation for us. Three x plus five is equal to six x minus 42. Let's see. We could get all of our, since, I'll put all my x's on the right-hand side since I have more x's on the right already. So let me subtract three x from both sides. And let me, I want to get rid of this 42 here."}, {"video_title": "Solving exponential equations using exponent properties High School Math Khan Academy.mp3", "Sentence": "Let's see. We could get all of our, since, I'll put all my x's on the right-hand side since I have more x's on the right already. So let me subtract three x from both sides. And let me, I want to get rid of this 42 here. So let's add 42 to both sides. And we are going to be left with five plus 42 is 47, is equal to, 47 is equal to three x. Now we just divide both sides by three, and we are left with x is equal to 47 over three."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We're told this table defines function f. All right, for every x, they give us the corresponding f of x. According to the table, is f even, odd, or neither? So pause this video and see if you can figure that out on your own. All right, now let's work on this together. So let's just remind ourselves the definition of even and odd. One definition that we can think of is that f of x, if f of x is equal to f of negative x, then we are dealing with an even function. And if f of x is equal to the negative of f of negative x, or another way of saying that, if f of negative x, if f of negative x, instead of it being equal to f of x, it's equal to negative f of x, these last two are equivalent, then in these situations, we are dealing with an odd function."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work on this together. So let's just remind ourselves the definition of even and odd. One definition that we can think of is that f of x, if f of x is equal to f of negative x, then we are dealing with an even function. And if f of x is equal to the negative of f of negative x, or another way of saying that, if f of negative x, if f of negative x, instead of it being equal to f of x, it's equal to negative f of x, these last two are equivalent, then in these situations, we are dealing with an odd function. And if neither of these are true, then we're dealing with neither. So what about what's going on over here? So let's see."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And if f of x is equal to the negative of f of negative x, or another way of saying that, if f of negative x, if f of negative x, instead of it being equal to f of x, it's equal to negative f of x, these last two are equivalent, then in these situations, we are dealing with an odd function. And if neither of these are true, then we're dealing with neither. So what about what's going on over here? So let's see. F of negative seven is equal to negative one. What about f of the negative of negative seven? Well, that would be f of seven."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's see. F of negative seven is equal to negative one. What about f of the negative of negative seven? Well, that would be f of seven. And we see f of seven here is also equal to negative one. So at least in that case and that case, if we think of x as seven, f of x is equal to f of negative x. So it works for that."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, that would be f of seven. And we see f of seven here is also equal to negative one. So at least in that case and that case, if we think of x as seven, f of x is equal to f of negative x. So it works for that. It also works for negative three and three. F of three is equal to f of negative three. They're both equal to two."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it works for that. It also works for negative three and three. F of three is equal to f of negative three. They're both equal to two. And you can see, and you can kind of visualize in your head that we have this symmetry around the y-axis. And so this looks like an even function. So I will circle that in."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "They're both equal to two. And you can see, and you can kind of visualize in your head that we have this symmetry around the y-axis. And so this looks like an even function. So I will circle that in. Let's do another example. So here, once again, the table defines function f. It's a different function f. Is this function even, odd, or neither? So pause this video and try to think about it."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So I will circle that in. Let's do another example. So here, once again, the table defines function f. It's a different function f. Is this function even, odd, or neither? So pause this video and try to think about it. All right, so let's just try a few examples. So here we have f of five is equal to two. F of five is equal to two."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to think about it. All right, so let's just try a few examples. So here we have f of five is equal to two. F of five is equal to two. What is f of negative five? F of negative five, not only is it not equal to two, it would have to be equal to two if this was an even function. And it would be equal to negative two if this was an odd function, but it's neither."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "F of five is equal to two. What is f of negative five? F of negative five, not only is it not equal to two, it would have to be equal to two if this was an even function. And it would be equal to negative two if this was an odd function, but it's neither. So we very clearly see, just looking at that data point, that this can neither be even nor odd. So I would say neither or neither right over here. Let's do one more example."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And it would be equal to negative two if this was an odd function, but it's neither. So we very clearly see, just looking at that data point, that this can neither be even nor odd. So I would say neither or neither right over here. Let's do one more example. Once again, the table defines function f. According to the table, is it even, odd, or neither? Pause the video again and try to answer it. All right, so actually, let's just start over here."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do one more example. Once again, the table defines function f. According to the table, is it even, odd, or neither? Pause the video again and try to answer it. All right, so actually, let's just start over here. So we have f of four is equal to negative eight. What is f of negative four? And the whole idea here is I wanna say, okay, if f of x is equal to something, what is f of negative x?"}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, so actually, let's just start over here. So we have f of four is equal to negative eight. What is f of negative four? And the whole idea here is I wanna say, okay, if f of x is equal to something, what is f of negative x? Well, they luckily give us f of negative four. It is equal to eight. So it looks like it's not equal to f of x."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And the whole idea here is I wanna say, okay, if f of x is equal to something, what is f of negative x? Well, they luckily give us f of negative four. It is equal to eight. So it looks like it's not equal to f of x. It's equal to the negative of f of x. This is equal to the negative of f of four. So on that data point alone, at least that data point satisfies it being odd."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So it looks like it's not equal to f of x. It's equal to the negative of f of x. This is equal to the negative of f of four. So on that data point alone, at least that data point satisfies it being odd. It's equal to the negative of f of x. But now let's try the other points just to make sure. So f of one is equal to five."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So on that data point alone, at least that data point satisfies it being odd. It's equal to the negative of f of x. But now let's try the other points just to make sure. So f of one is equal to five. What is f of negative one? Well, it is equal to negative five. Once again, f of negative x is equal to the negative of f of x."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So f of one is equal to five. What is f of negative one? Well, it is equal to negative five. Once again, f of negative x is equal to the negative of f of x. So that checks out. And then f of zero, well, f of zero is, of course, equal to zero. But, of course, if you say, what is the negative of f of, if you say, what, f of negative of zero, well, that's still f of zero."}, {"video_title": "Even and odd functions Tables Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Once again, f of negative x is equal to the negative of f of x. So that checks out. And then f of zero, well, f of zero is, of course, equal to zero. But, of course, if you say, what is the negative of f of, if you say, what, f of negative of zero, well, that's still f of zero. And then if you were to take the negative of zero, that's still zero. So you could view this, this is consistent still with being odd. This you could view as the negative of f of negative zero, which, of course, is still going to be zero."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "So the first question I'll ask you, if you do one revolution, if you have an angle that went all the way around once, how many radians is that? Well, we know that that is 2 pi radians. Now, that exact same angle, if we were to measure it in degrees, how many degrees is that? Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, you've heard of people doing a 360, doing one full revolution. That is equal to 360 degrees. Now, can we simplify this? It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "It's important to write this little superscript circle. That's literally the units under question. Sometimes it doesn't look like a unit, but it is a unit. You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "You could literally write degrees instead of that little symbol. And the units right here, of course, are the word radians. Now, can we simplify this a little bit? Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, sure. Both 2 pi and 360 are divisible by 2, so let's divide things by 2. And if we do that, what do we get for what pi radians are equal to? Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, on the left-hand side here, we're just left with pi radians. And on the right-hand side here, 360 divided by 2 is 180, and we have still the units, which are degrees. So we get pi radians are equal to 180 degrees, which actually answered the first part of our question. We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We wanted to convert pi radians. Well, we just figured out pi radians are equal to 100, 180 degrees. Pi radians are equal to 180 degrees. And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "And if you want to think about it, we know pi radians are halfway around a circle like that, and that's the same thing as 180 degrees. So now let's think about the second part of it. We want to convert negative pi over 3 radians. Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees?"}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Let me do this in a new color. Negative pi over 3. So negative pi over 3 radians. How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "How can we convert that to degrees? What do we get based on this information right over here? Well, to figure this out, we need to know how many degrees there are per radian. If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "If we need to multiply this times degrees, and I'm going to write the word out because if I just wrote a little circle here, it would be hard to visualize that as a unit. Degrees per radian. So how many degrees are there per radian? Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "Well, we know that for every 180 degrees, we have pi radians. Or you could say that there are 180 over pi degrees per radian. And this is going to work out. We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "We have however many radians we have times the number of degrees per radian. So of course, the units are going to work out. Radians cancel out. The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out."}, {"video_title": "Example Converting radians to degrees Trigonometry Khan Academy.mp3", "Sentence": "The pi also cancels out. So you're left with negative 180 divided by 3, leaving us with negative 60. And we don't want to forget the units. We could write them out. That's the only units that's left. Degrees, which we could write out. We could write out the word degrees or we could just put that little symbol there."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So there's two changes here. Instead of two to the x, we have two to the negative x. And then we're not leaving that alone. We then subtract five. So let's take them step by step. So let's first think about what y equals two to the negative x will look like. Well, any input we now put into it, x, we're now going to take the negative of it."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "We then subtract five. So let's take them step by step. So let's first think about what y equals two to the negative x will look like. Well, any input we now put into it, x, we're now going to take the negative of it. So if I input a two, it's like taking the opposite of the two and then inputting that into two to the x. And so what we're essentially going to do is flip this graph over the y-axis. So here we have the point two comma four."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, any input we now put into it, x, we're now going to take the negative of it. So if I input a two, it's like taking the opposite of the two and then inputting that into two to the x. And so what we're essentially going to do is flip this graph over the y-axis. So here we have the point two comma four. Over here we're going to have the point negative two comma four. When x is zero, they're going to give us the same value. So they're both going to have the same y-intercept."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So here we have the point two comma four. Over here we're going to have the point negative two comma four. When x is zero, they're going to give us the same value. So they're both going to have the same y-intercept. And so our graph is going to look like, our graph is going to look something like this. There are going to be mirror images flipped around the y-axis. And so it's going to look like that."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So they're both going to have the same y-intercept. And so our graph is going to look like, our graph is going to look something like this. There are going to be mirror images flipped around the y-axis. And so it's going to look like that. That is the graph of y is equal to two to the negative x. And then we have to worry about the subtracting five from it. Well, you're subtracting five from your final y value, or you're subtracting five to get your y value now, where your y value is going to be five lower is I guess the best way to say it."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so it's going to look like that. That is the graph of y is equal to two to the negative x. And then we have to worry about the subtracting five from it. Well, you're subtracting five from your final y value, or you're subtracting five to get your y value now, where your y value is going to be five lower is I guess the best way to say it. So this is going to shift the graph down by five. So instead of having the y-intercept there, it's going to be five lower. One, two, one, each hash mark is two."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, you're subtracting five from your final y value, or you're subtracting five to get your y value now, where your y value is going to be five lower is I guess the best way to say it. So this is going to shift the graph down by five. So instead of having the y-intercept there, it's going to be five lower. One, two, one, each hash mark is two. So this is one, two, three, four, five. It's going to be right over there. So shift down by five."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "One, two, one, each hash mark is two. So this is one, two, three, four, five. It's going to be right over there. So shift down by five. Two, four, five. It's going to look like that. And then the asymptote, instead of the asymptote going towards y equals zero, the asymptote is going to be at y is equal to negative five."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So shift down by five. Two, four, five. It's going to look like that. And then the asymptote, instead of the asymptote going towards y equals zero, the asymptote is going to be at y is equal to negative five. So the asymptote is going to be y equals negative five. So it should look something like, it should look something like, something like what I'm drawing right now. So something like, something like that."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then the asymptote, instead of the asymptote going towards y equals zero, the asymptote is going to be at y is equal to negative five. So the asymptote is going to be y equals negative five. So it should look something like, it should look something like, something like what I'm drawing right now. So something like, something like that. So now we can look at which choices. So this should be the graph of y equals two to the negative x minus five. So let's see which of these choices depict that."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So something like, something like that. So now we can look at which choices. So this should be the graph of y equals two to the negative x minus five. So let's see which of these choices depict that. So this first choice actually seems to be spot on. It's exactly what we drew. But we could look at the other ones just in case."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's see which of these choices depict that. So this first choice actually seems to be spot on. It's exactly what we drew. But we could look at the other ones just in case. Well, this looks like, what did they do over here? It looks like they, instead of flipping over the y-axis, they flipped over the x-axis and then they shifted down. So that's not right."}, {"video_title": "Transforming exponential graphs Mathematics III High School Math Khan Academy.mp3", "Sentence": "But we could look at the other ones just in case. Well, this looks like, what did they do over here? It looks like they, instead of flipping over the y-axis, they flipped over the x-axis and then they shifted down. So that's not right. Here it looks like they got what we got, but then they flipped it over the x-axis. And this looks like they flipped it over the y-axis, but then they shifted, instead of shifting down by five, it looks like they shifted to the left by five. So we should feel pretty good, especially because we essentially drew this before even looking at the choice."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "The thickness of wood scraped off in millimeters as a function of the speed of the sandpaper in meters per second, T of v, so this is the thickness scraped off, so that's how much, so this is the thickness, or how much is scraped off, and it is a function of speed. Actually, since they're using v, and also they're getting negative values, so we care about the direction, I'll even call that, it's actually the velocity. So this is how much is scraped off as a function of velocity, it's shown below. So if the velocity is greater than zero, that means that the sandpaper is moving to the right, that makes sense, that's the standard convention. And if the velocity is less than zero, it means the sandpaper is moving to the left. Fair enough. The function is even."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So if the velocity is greater than zero, that means that the sandpaper is moving to the right, that makes sense, that's the standard convention. And if the velocity is less than zero, it means the sandpaper is moving to the left. Fair enough. The function is even. What is the significance of the evenness of this function? Well, the fact that it's even means that T of v is equal to T of negative v. So that tells us that if our velocity is 8 meters per second to the left, we're going to get as much scraped off as if we go 8 millimeters per second to the right, and we see that right over here. So that is equal to that."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "The function is even. What is the significance of the evenness of this function? Well, the fact that it's even means that T of v is equal to T of negative v. So that tells us that if our velocity is 8 meters per second to the left, we're going to get as much scraped off as if we go 8 millimeters per second to the right, and we see that right over here. So that is equal to that. If we go at 6 meters per second to the left, we're going to get just as much scraped off as we go 6 meters per second to the right. So these two are going to be the same. So what it's really telling us, and we could say do it for 4 meters per second and negative 4, is it doesn't matter if we go to the left or the right."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So that is equal to that. If we go at 6 meters per second to the left, we're going to get just as much scraped off as we go 6 meters per second to the right. So these two are going to be the same. So what it's really telling us, and we could say do it for 4 meters per second and negative 4, is it doesn't matter if we go to the left or the right. What really matters is the magnitude of the velocity or the absolute value of it, but it doesn't matter if we're going to the left or the right. Whether we're going to the left or the right, for a given magnitude of velocity, we are going to get the same amount scraped off. Now let's see which of these choices are consistent with what I just said."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So what it's really telling us, and we could say do it for 4 meters per second and negative 4, is it doesn't matter if we go to the left or the right. What really matters is the magnitude of the velocity or the absolute value of it, but it doesn't matter if we're going to the left or the right. Whether we're going to the left or the right, for a given magnitude of velocity, we are going to get the same amount scraped off. Now let's see which of these choices are consistent with what I just said. Moving the sandpaper faster scrapes off more wood. Well, that's true. We see as the speed increases or the magnitude of the speed increases, we scrape off more wood."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Now let's see which of these choices are consistent with what I just said. Moving the sandpaper faster scrapes off more wood. Well, that's true. We see as the speed increases or the magnitude of the speed increases, we scrape off more wood. As the magnitude of the speed, this negative 8, you might say, hey, that's lower than negative 2, but the magnitude is larger. We're going 8 meters per second to the left and we're scraping off more. So this is a true statement, but it's not the significance of the evenness of the function."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "We see as the speed increases or the magnitude of the speed increases, we scrape off more wood. As the magnitude of the speed, this negative 8, you might say, hey, that's lower than negative 2, but the magnitude is larger. We're going 8 meters per second to the left and we're scraping off more. So this is a true statement, but it's not the significance of the evenness of the function. This could have been true even if this was a 7, but then this function would no longer be even. The piece of wood is 6 millimeters thick. Well, we actually don't get any of that from the function."}, {"video_title": "Interpreting features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So this is a true statement, but it's not the significance of the evenness of the function. This could have been true even if this was a 7, but then this function would no longer be even. The piece of wood is 6 millimeters thick. Well, we actually don't get any of that from the function. Moving the sandpaper to the right has the same effect as moving it to the left. Well, that seems pretty close to what I had said earlier, that for a given speed to the right or to the left, we get the same amount that is taken off of the piece of sandpaper or the piece of wood. So this looks like our answer."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So what we have here are two different polynomials, P1 and P2, and they have been expressed in factored form, and you can also see their graphs. This is the graph of y is equal to P1 of x in blue, and the graph of y is equal to P2x in white. What we're going to do in this video is continue our study of zeros, but we're gonna look at a special case when something interesting happens with the zeros. So let's just first look at P1's zeros. So I'll set up a little table here because it'll be useful. So the first column, let's just make it the zeros, the x values at which our polynomial is equal to zero, and that's pretty easy to figure out from factored form. When x is equal to one, the whole thing's going to be equal to zero because zero times anything is zero."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let's just first look at P1's zeros. So I'll set up a little table here because it'll be useful. So the first column, let's just make it the zeros, the x values at which our polynomial is equal to zero, and that's pretty easy to figure out from factored form. When x is equal to one, the whole thing's going to be equal to zero because zero times anything is zero. When x is equal to two, by the same argument, and when x is equal to three. And we can see it here on the graph. When x equals one, the graph of y is equal to P1 intersects the x-axis."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to one, the whole thing's going to be equal to zero because zero times anything is zero. When x is equal to two, by the same argument, and when x is equal to three. And we can see it here on the graph. When x equals one, the graph of y is equal to P1 intersects the x-axis. It does it again at the next zero, x equals two, and at the next zero, x equals three. We can also see the property that between consecutive zeros, our function, our polynomial, maintains the same sign. So between these first two, or actually before this first zero, it's negative, then between these first two, it's positive, then the next two, it's negative, and then after that, it is positive."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "When x equals one, the graph of y is equal to P1 intersects the x-axis. It does it again at the next zero, x equals two, and at the next zero, x equals three. We can also see the property that between consecutive zeros, our function, our polynomial, maintains the same sign. So between these first two, or actually before this first zero, it's negative, then between these first two, it's positive, then the next two, it's negative, and then after that, it is positive. Now what about P2? Well, P2 is interesting because if you were to multiply this out, it would have the same degree as P1. In either case, you would have an x to the third term."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So between these first two, or actually before this first zero, it's negative, then between these first two, it's positive, then the next two, it's negative, and then after that, it is positive. Now what about P2? Well, P2 is interesting because if you were to multiply this out, it would have the same degree as P1. In either case, you would have an x to the third term. You would have a third degree polynomial. But how many zeros, how many distinct, unique zeros does P2 have? Pause this video and think about that."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "In either case, you would have an x to the third term. You would have a third degree polynomial. But how many zeros, how many distinct, unique zeros does P2 have? Pause this video and think about that. Well, let's just list them out. So our zeros. Well, once again, if x equals one, this whole expression's going to be equal to zero, so we have a zero at x equals one, and we can see that our white graph also intersects the x-axis at x equals one."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and think about that. Well, let's just list them out. So our zeros. Well, once again, if x equals one, this whole expression's going to be equal to zero, so we have a zero at x equals one, and we can see that our white graph also intersects the x-axis at x equals one. And then if x is equal to three, this whole thing's going to be equal to zero, and we can see that it intersects the x-axis at x equals three. And then notice, this next part of the expression would say, oh, well, we have a zero at x equals three, but we already said that. So we actually have two zeros for a third degree polynomial."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, once again, if x equals one, this whole expression's going to be equal to zero, so we have a zero at x equals one, and we can see that our white graph also intersects the x-axis at x equals one. And then if x is equal to three, this whole thing's going to be equal to zero, and we can see that it intersects the x-axis at x equals three. And then notice, this next part of the expression would say, oh, well, we have a zero at x equals three, but we already said that. So we actually have two zeros for a third degree polynomial. So something very interesting is happening. In some ways, you could say that, hey, it's trying to reinforce that we have a zero at x minus three and this notion of having multiple parts of our factored form that would all point to the same zero, that is the idea of multiplicity. So let me write this word down."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So we actually have two zeros for a third degree polynomial. So something very interesting is happening. In some ways, you could say that, hey, it's trying to reinforce that we have a zero at x minus three and this notion of having multiple parts of our factored form that would all point to the same zero, that is the idea of multiplicity. So let me write this word down. So multiplicity. Multiplicity. I'll write it out there, and I will write it over here."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let me write this word down. So multiplicity. Multiplicity. I'll write it out there, and I will write it over here. Multiplicity. And so for each of these zeros, we have a multiplicity of one. They're only deduced one time when you look at it at factored form."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "I'll write it out there, and I will write it over here. Multiplicity. And so for each of these zeros, we have a multiplicity of one. They're only deduced one time when you look at it at factored form. Only one of the factors points to each of those zeros. So they all have a multiplicity of one. For P2, the first zero has a multiple of one."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "They're only deduced one time when you look at it at factored form. Only one of the factors points to each of those zeros. So they all have a multiplicity of one. For P2, the first zero has a multiple of one. Only one of the expressions points to a zero of one or would become zero if x would be equal to one. But notice, out of our factors, when we have it in factored form, out of our factored expressions, or our expression factors, I should say, two of them become zero when x is equal to three. This one and this one are going to become zero, and so here we have a multiplicity of two."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "For P2, the first zero has a multiple of one. Only one of the expressions points to a zero of one or would become zero if x would be equal to one. But notice, out of our factors, when we have it in factored form, out of our factored expressions, or our expression factors, I should say, two of them become zero when x is equal to three. This one and this one are going to become zero, and so here we have a multiplicity of two. And I encourage you to pause this video again and look at the behavior of graphs and see if you can see a difference between the behavior of the graph when we have a multiplicity of one versus when we have a multiplicity of two. All right, now let's look through it together. We can look at P1 where all of the zeros have a multiplicity of one, and you can see every time we have a zero, we are crossing the x-axis."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "This one and this one are going to become zero, and so here we have a multiplicity of two. And I encourage you to pause this video again and look at the behavior of graphs and see if you can see a difference between the behavior of the graph when we have a multiplicity of one versus when we have a multiplicity of two. All right, now let's look through it together. We can look at P1 where all of the zeros have a multiplicity of one, and you can see every time we have a zero, we are crossing the x-axis. Not only are we intersecting it, but we are crossing it. We're crossing the x-axis there. We're crossing it again, and we're crossing it again."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We can look at P1 where all of the zeros have a multiplicity of one, and you can see every time we have a zero, we are crossing the x-axis. Not only are we intersecting it, but we are crossing it. We're crossing the x-axis there. We're crossing it again, and we're crossing it again. So at all of these, we have a sign change around that zero. But what happens here? Well, on the first zero that has a multiplicity of one that only makes one of the factors equal zero, we have a sign change just like we saw with P1."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We're crossing it again, and we're crossing it again. So at all of these, we have a sign change around that zero. But what happens here? Well, on the first zero that has a multiplicity of one that only makes one of the factors equal zero, we have a sign change just like we saw with P1. But what happens at x equals three where we have a multiplicity of two? Well, there we intersect the x-axis still. P of three is zero, but notice we don't have a sign change."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, on the first zero that has a multiplicity of one that only makes one of the factors equal zero, we have a sign change just like we saw with P1. But what happens at x equals three where we have a multiplicity of two? Well, there we intersect the x-axis still. P of three is zero, but notice we don't have a sign change. We were positive before, and we are positive after. We touch the x-axis right there, but then we go back up. And the general idea, and I encourage you to just test this out and think about why this is true, is that if you have an odd multiplicity, and let me write this down."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "P of three is zero, but notice we don't have a sign change. We were positive before, and we are positive after. We touch the x-axis right there, but then we go back up. And the general idea, and I encourage you to just test this out and think about why this is true, is that if you have an odd multiplicity, and let me write this down. If the multiplicity is odd, so if it's one, three, five, seven, et cetera, then you're going to have a sign change, sign change, while if it is even, as the case of two or four or six, you're going to have no sign change, no sign, no sign change. One way to think about it, in an example where you have a multiplicity of two, so let's just use this zero here where x is equal to three. When x is less than three, both of these are going to be negative, and a negative times a negative is a positive."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And the general idea, and I encourage you to just test this out and think about why this is true, is that if you have an odd multiplicity, and let me write this down. If the multiplicity is odd, so if it's one, three, five, seven, et cetera, then you're going to have a sign change, sign change, while if it is even, as the case of two or four or six, you're going to have no sign change, no sign, no sign change. One way to think about it, in an example where you have a multiplicity of two, so let's just use this zero here where x is equal to three. When x is less than three, both of these are going to be negative, and a negative times a negative is a positive. And when x is greater than three, both of them are going to be positive. And so in either case, you have a positive. So notice, you saw no sign change."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "When x is less than three, both of these are going to be negative, and a negative times a negative is a positive. And when x is greater than three, both of them are going to be positive. And so in either case, you have a positive. So notice, you saw no sign change. Another thing to appreciate is thinking about the number of zeros relative to the degree of the polynomial. And what you see is is that the number of zeros, number of zeros, is at most equal to the degree of the polynomial. So it is going to be less than or equal to the degree of the polynomial."}, {"video_title": "Multiplicity of zeros of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So notice, you saw no sign change. Another thing to appreciate is thinking about the number of zeros relative to the degree of the polynomial. And what you see is is that the number of zeros, number of zeros, is at most equal to the degree of the polynomial. So it is going to be less than or equal to the degree of the polynomial. And why is that the case? Well, you might not, all your zeros might have a multiplicity of one, in which case, the number of zeros is going to be equal to the degree of the polynomial. But if you have a zero that has a higher than one multiplicity, well, then you're going to have fewer distinct zeros."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "The graph here shows this is y is equal to f of x, the solid blue line, and this is y is equal to g of x, as a dashed red line. And they ask us, what is g of x in terms of f of x? And like always, pause the video and see if you can give a go at it, and then we're gonna do it together. Alright, so when you immediately look like it, it looks like g of x is kind of a thinned up version of f of x. It seems like if you were to compress it towards the center, that's what g of x looks like. But let's put a little bit more meat on that bone. And see if we can identify corresponding points."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Alright, so when you immediately look like it, it looks like g of x is kind of a thinned up version of f of x. It seems like if you were to compress it towards the center, that's what g of x looks like. But let's put a little bit more meat on that bone. And see if we can identify corresponding points. So for example, if we were to look at f of negative six, that's, so f of negative six, that seems like it corresponds, or it gives us the same value as f of negative six. And we want to find the corresponding point, so we've hit this minimum point, we're coming back up. Hit the minimum point, we're coming back up."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And see if we can identify corresponding points. So for example, if we were to look at f of negative six, that's, so f of negative six, that seems like it corresponds, or it gives us the same value as f of negative six. And we want to find the corresponding point, so we've hit this minimum point, we're coming back up. Hit the minimum point, we're coming back up. It seems like the corresponding point right over there is g of negative three. So let's write that down. Let's see, it looks like f of negative six, and it is equal to g of negative three."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Hit the minimum point, we're coming back up. It seems like the corresponding point right over there is g of negative three. So let's write that down. Let's see, it looks like f of negative six, and it is equal to g of negative three. These are corresponding points. If you apply the transportation at the point f equals, at the point negative six comma f of negative six, you get to the point negative three, g of negative three right over there. Let's do a couple of more."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's see, it looks like f of negative six, and it is equal to g of negative three. These are corresponding points. If you apply the transportation at the point f equals, at the point negative six comma f of negative six, you get to the point negative three, g of negative three right over there. Let's do a couple of more. If you look at f is, so f of two looks like it corresponds to g of one. So one, f of two corresponds to g of one. So let's write that down."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's do a couple of more. If you look at f is, so f of two looks like it corresponds to g of one. So one, f of two corresponds to g of one. So let's write that down. F of two looks like it corresponds to g of one. And once again, I'm looking at where the function's at the same value, and also optically, I'm just looking at, well, it looks like it's the same part of the function. If we assume g of x is a squeezed version of f of x."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's write that down. F of two looks like it corresponds to g of one. And once again, I'm looking at where the function's at the same value, and also optically, I'm just looking at, well, it looks like it's the same part of the function. If we assume g of x is a squeezed version of f of x. And so in general, it looks like, it looks like for a given x, so we could say f of x is going to be equal to g of, well, whatever you have in here, it seems like we have half the value over here. So g of x over two. Or if you wanted to think of it the other way, if you wanted to think of it the other way, if you want to say g of x is going to be f of, well, whatever we have here, it's f of twice that."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "If we assume g of x is a squeezed version of f of x. And so in general, it looks like, it looks like for a given x, so we could say f of x is going to be equal to g of, well, whatever you have in here, it seems like we have half the value over here. So g of x over two. Or if you wanted to think of it the other way, if you wanted to think of it the other way, if you want to say g of x is going to be f of, well, whatever we have here, it's f of twice that. So f of two x. And we see that that is one of the choices, that g of x is equal to f of two x. Whatever the x that you input into g of x, you get that same value out of the function when you input two times that into f of x."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Or if you wanted to think of it the other way, if you wanted to think of it the other way, if you want to say g of x is going to be f of, well, whatever we have here, it's f of twice that. So f of two x. And we see that that is one of the choices, that g of x is equal to f of two x. Whatever the x that you input into g of x, you get that same value out of the function when you input two times that into f of x. And these seem to validate that. It looks optically like that, that we shrunk it down. One way to think about it, when you take, when you multiply the input into a function by a number larger than one, it's going to compress, it's going to make things happen faster."}, {"video_title": "Compressing functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Whatever the x that you input into g of x, you get that same value out of the function when you input two times that into f of x. And these seem to validate that. It looks optically like that, that we shrunk it down. One way to think about it, when you take, when you multiply the input into a function by a number larger than one, it's going to compress, it's going to make things happen faster. The input to the function is going to increase or become negative faster, so it thins it up. And if that doesn't make intuitive sense, you can also just try some of these values. And I encourage you to try more."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this first one says this is the graph of function f, fair enough. Function g is defined as g of x is equal to f of negative x, also fair enough. What is the graph of g? And on Khan Academy, it's multiple choice, but I thought for the sake of this video, it'd be fun to think about what g would look like without having any choices, just sketching it out. So pause this video and try to think about it, at least in your head. All right, now let's work through this together. So we've already gone over that g of x is equal to f of negative x."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And on Khan Academy, it's multiple choice, but I thought for the sake of this video, it'd be fun to think about what g would look like without having any choices, just sketching it out. So pause this video and try to think about it, at least in your head. All right, now let's work through this together. So we've already gone over that g of x is equal to f of negative x. So whatever the value of f is at a certain value, we would expect g to take on that value at the negative of that. So for example, we can see that f of four is equal to two, so we would expect g of negative four to be equal to two. Because once again, g of negative four, we could write it over here, g of negative four is going to be equal to f of the negative of negative four, which is equal to f of four."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we've already gone over that g of x is equal to f of negative x. So whatever the value of f is at a certain value, we would expect g to take on that value at the negative of that. So for example, we can see that f of four is equal to two, so we would expect g of negative four to be equal to two. Because once again, g of negative four, we could write it over here, g of negative four is going to be equal to f of the negative of negative four, which is equal to f of four. And so we could keep going with that. What would g of negative two be? Well, that would be the same thing as f of two, which is zero, so it would be right over there."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Because once again, g of negative four, we could write it over here, g of negative four is going to be equal to f of the negative of negative four, which is equal to f of four. And so we could keep going with that. What would g of negative two be? Well, that would be the same thing as f of two, which is zero, so it would be right over there. What would g of zero be? Well, that would be the same thing as f of zero, because the negative of zero is zero, and f of zero is right over there. It looks like negative two."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, that would be the same thing as f of two, which is zero, so it would be right over there. What would g of zero be? Well, that would be the same thing as f of zero, because the negative of zero is zero, and f of zero is right over there. It looks like negative two. And so you can already see where this is going. And we've already talked about it in previous videos, that if you replace your x with a negative x, you're essentially reflecting over the y-axis. So g is going to look something like this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It looks like negative two. And so you can already see where this is going. And we've already talked about it in previous videos, that if you replace your x with a negative x, you're essentially reflecting over the y-axis. So g is going to look something like this. It is going to look something like this. Once again, g of negative six would be the same thing as f of six. And so that would be the graph of g. And if you're doing this on Khan Academy, you'd pick the choice that looks like this, that would give a reflection over the y-axis."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g is going to look something like this. It is going to look something like this. Once again, g of negative six would be the same thing as f of six. And so that would be the graph of g. And if you're doing this on Khan Academy, you'd pick the choice that looks like this, that would give a reflection over the y-axis. Let's do another example. So here, once again, this is the graph of the function f. And then they say, what is the graph of g? And so pause this video, and at least try to sketch it on your mind what g should look like."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so that would be the graph of g. And if you're doing this on Khan Academy, you'd pick the choice that looks like this, that would give a reflection over the y-axis. Let's do another example. So here, once again, this is the graph of the function f. And then they say, what is the graph of g? And so pause this video, and at least try to sketch it on your mind what g should look like. All right, so in this situation, they didn't replace the x with negative x in f of x. Instead, g of x is equal to the negative of all of f of x. In fact, we could rewrite g of x like this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so pause this video, and at least try to sketch it on your mind what g should look like. All right, so in this situation, they didn't replace the x with negative x in f of x. Instead, g of x is equal to the negative of all of f of x. In fact, we could rewrite g of x like this. We could say that g of x is equal to, notice, all of this right over here, that was our definition of f of x. So g of x is equal to the negative of f of x. So instead of it being f of negative x, it's equal to the negative of f of x."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "In fact, we could rewrite g of x like this. We could say that g of x is equal to, notice, all of this right over here, that was our definition of f of x. So g of x is equal to the negative of f of x. So instead of it being f of negative x, it's equal to the negative of f of x. So one way to think about it is we can see that f of zero is two, but g of zero is going to be the negative of that. So it's going to be equal to negative two. And so you could keep going with that."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So instead of it being f of negative x, it's equal to the negative of f of x. So one way to think about it is we can see that f of zero is two, but g of zero is going to be the negative of that. So it's going to be equal to negative two. And so you could keep going with that. You could see that whatever f of a certain value is, g of that value would be the negative of that. So it would be down here. And so g of x would be a reflection of f of x about the x-axis."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so you could keep going with that. You could see that whatever f of a certain value is, g of that value would be the negative of that. So it would be down here. And so g of x would be a reflection of f of x about the x-axis. So g of x is going to look something, something like, something like that, a reflection about the x-axis. And so once again, I'm kind of getting you to pick the choice that would actually look like that. Let's do another example."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so g of x would be a reflection of f of x about the x-axis. So g of x is going to look something, something like, something like that, a reflection about the x-axis. And so once again, I'm kind of getting you to pick the choice that would actually look like that. Let's do another example. This is strangely fun. All right. So here we're told functions f, so that's in solid in this blue color, and g dashed, so that's right over there, are graphed."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do another example. This is strangely fun. All right. So here we're told functions f, so that's in solid in this blue color, and g dashed, so that's right over there, are graphed. What is the equation of g in terms of f? So pause this video and try to think about it. So the key is to realize how do we transform f of x, actually they labeled it over here, this is f of x right over here, in order to get g. So f of negative x would be a reflection of f above, about the y-axis."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So here we're told functions f, so that's in solid in this blue color, and g dashed, so that's right over there, are graphed. What is the equation of g in terms of f? So pause this video and try to think about it. So the key is to realize how do we transform f of x, actually they labeled it over here, this is f of x right over here, in order to get g. So f of negative x would be a reflection of f above, about the y-axis. And so it would intersect there. It would have this straight portion like this. And I'm just experimenting right now."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So the key is to realize how do we transform f of x, actually they labeled it over here, this is f of x right over here, in order to get g. So f of negative x would be a reflection of f above, about the y-axis. And so it would intersect there. It would have this straight portion like this. And I'm just experimenting right now. It'd have the straight portion like this. And then it would go up, and let's see, f of negative x, so when you input six into it, that would be f of negative six, which is six, so it would go up there. So f of negative x would look something, something like this."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And I'm just experimenting right now. It'd have the straight portion like this. And then it would go up, and let's see, f of negative x, so when you input six into it, that would be f of negative six, which is six, so it would go up there. So f of negative x would look something, something like this. Something like that. So the purple is f of negative x. Now that doesn't quite get us to g, but it gets us a little bit closer, because it looks like if I were to take the reflection of f of negative x, f of negative x about the x-axis, it looks like I'm going to get to g. And so how do you reflect something about the x-axis?"}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So f of negative x would look something, something like this. Something like that. So the purple is f of negative x. Now that doesn't quite get us to g, but it gets us a little bit closer, because it looks like if I were to take the reflection of f of negative x, f of negative x about the x-axis, it looks like I'm going to get to g. And so how do you reflect something about the x-axis? Well we saw it in the example just now. You multiply the entire function by a negative. So we could say that g is equal to the negative of f of negative x."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now that doesn't quite get us to g, but it gets us a little bit closer, because it looks like if I were to take the reflection of f of negative x, f of negative x about the x-axis, it looks like I'm going to get to g. And so how do you reflect something about the x-axis? Well we saw it in the example just now. You multiply the entire function by a negative. So we could say that g is equal to the negative of f of negative x. It's equal to the negative of this. So we're doing both reflections. We're flipping over the y-axis, and we're flipping over the x-axis to get to g. Let's do one more example."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So we could say that g is equal to the negative of f of negative x. It's equal to the negative of this. So we're doing both reflections. We're flipping over the y-axis, and we're flipping over the x-axis to get to g. Let's do one more example. So once again, they've graphed f, they've graphed g, and they've said f is defined as this right over here. What is the equation of g? So they're not just asking it in terms of f. They just wanna know what is the equation of g. Pause this video and try to think about it."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We're flipping over the y-axis, and we're flipping over the x-axis to get to g. Let's do one more example. So once again, they've graphed f, they've graphed g, and they've said f is defined as this right over here. What is the equation of g? So they're not just asking it in terms of f. They just wanna know what is the equation of g. Pause this video and try to think about it. Well, you can see pretty clearly that this is a reflection across the y-axis. And a reflection across the y-axis, you can see pretty clearly that g of x is equal to f of negative x. F of negative x. How do we know that?"}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So they're not just asking it in terms of f. They just wanna know what is the equation of g. Pause this video and try to think about it. Well, you can see pretty clearly that this is a reflection across the y-axis. And a reflection across the y-axis, you can see pretty clearly that g of x is equal to f of negative x. F of negative x. How do we know that? Well, for whenever we take f of x, and we get that value, g at the negative of that value takes on the same function value, I guess I could say. Or another way to think about it is we could just pick this point, negative eight. F of negative eight is equal to a little over four."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "How do we know that? Well, for whenever we take f of x, and we get that value, g at the negative of that value takes on the same function value, I guess I could say. Or another way to think about it is we could just pick this point, negative eight. F of negative eight is equal to a little over four. But g of eight is equal to a little over four, is equal to that same value. And so what is the equation of g? Well, we just have to rewrite this so that we can write it out as an equation."}, {"video_title": "Reflecting functions examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "F of negative eight is equal to a little over four. But g of eight is equal to a little over four, is equal to that same value. And so what is the equation of g? Well, we just have to rewrite this so that we can write it out as an equation. And so we could write out g of x is equal to, if I were to replace all of the x's here with a negative x, what would I get? I would get four times the square root of two minus, instead of an x, I will have a negative x. And then the minus eight is outside of the radical."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let's say that we have the equation six plus three w is equal to the square root of two w plus 12 plus two w. See if you could pause the video and solve for w. And it might have more than one solution, so keep that in mind. All right, now let's work through this together. So the first thing I'd like to do whenever I see one of these radical equations is just isolate the radical on one side of the equation. So let's subtract two w from both sides. I want to get rid of that two w from the right-hand side. I just want the radical sign. And if I subtract two w from both sides, what am I left with?"}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract two w from both sides. I want to get rid of that two w from the right-hand side. I just want the radical sign. And if I subtract two w from both sides, what am I left with? Well, on the left-hand side, I am left with six plus three w minus two w. Well, three of something, take away two of them, you're gonna be left with w. Six plus w is equal to, these cancel out, we're left with the square root of two w plus 12. Now to get rid of the radical, we're going to square both sides. And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And if I subtract two w from both sides, what am I left with? Well, on the left-hand side, I am left with six plus three w minus two w. Well, three of something, take away two of them, you're gonna be left with w. Six plus w is equal to, these cancel out, we're left with the square root of two w plus 12. Now to get rid of the radical, we're going to square both sides. And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution. What do I mean by that? Well, we're gonna get the same result whether we square this or whether we square that, because when you square a negative, it becomes a positive. But those are fundamentally two different equations."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we've seen before that this process right over here, it's a little bit tricky, because when you're squaring a radical in a radical equation like this, and then you solve, you might find an extraneous solution. What do I mean by that? Well, we're gonna get the same result whether we square this or whether we square that, because when you square a negative, it becomes a positive. But those are fundamentally two different equations. We only want the solutions that satisfy the one that doesn't have the negative there. So that's why we're gonna test our solutions to make sure that they're valid for our original equation. So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "But those are fundamentally two different equations. We only want the solutions that satisfy the one that doesn't have the negative there. So that's why we're gonna test our solutions to make sure that they're valid for our original equation. So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12. Now we can subtract two w and 12 from both sides, so let's do that. So then we can get into kind of a standard quadratic form. So let's subtract two w from both sides, and let's subtract 12 from both sides."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So if we square both sides, on the left-hand side, we're going to have, well, it's gonna be w squared plus two times their product, so two times six times w, so that's 12w plus six squared, 36, is equal to, now if you take the square root and square it, you're gonna be left with two w plus 12. Now we can subtract two w and 12 from both sides, so let's do that. So then we can get into kind of a standard quadratic form. So let's subtract two w from both sides, and let's subtract 12 from both sides. So subtract 12 from the right, subtract 12 here. And once again, I just wanna get rid of this on the right-hand side. And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract two w from both sides, and let's subtract 12 from both sides. So subtract 12 from the right, subtract 12 here. And once again, I just wanna get rid of this on the right-hand side. And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero. And let's see, to solve this, let's use this factor bar, there are two numbers that add up to 10, and whose product is 24. Well, what jumps out at me is six and four. So we can rewrite this as w plus four times w plus six is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I am going to be left with, I am going to be left with, on the left-hand side, it's gonna be w squared, see, 12w minus two w is plus 10w, and then 36 minus 12 is plus 24, is equal to zero. And let's see, to solve this, let's use this factor bar, there are two numbers that add up to 10, and whose product is 24. Well, what jumps out at me is six and four. So we can rewrite this as w plus four times w plus six is equal to zero. And so, if I have the product of two things equaling zero, well, to solve this, either one or both of them could be equal to zero. Zero times anything is going to be zero. So w plus four is equal to zero, or w plus six is equal to zero."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we can rewrite this as w plus four times w plus six is equal to zero. And so, if I have the product of two things equaling zero, well, to solve this, either one or both of them could be equal to zero. Zero times anything is going to be zero. So w plus four is equal to zero, or w plus six is equal to zero. And over here, if you subtract four from both sides, you get w is equal to negative four, or subtract six from both sides here, w is equal to negative six. Now let's verify that these actually are solutions to our original equation. Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So w plus four is equal to zero, or w plus six is equal to zero. And over here, if you subtract four from both sides, you get w is equal to negative four, or subtract six from both sides here, w is equal to negative six. Now let's verify that these actually are solutions to our original equation. Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four. So this would be, this is negative 12 here, this is negative eight here, this is negative eight here. So you have six plus negative 12, which is negative six, is equal to the square root of negative eight plus 12 is four plus negative eight, so that would be negative six is equal to two plus negative eight, which is absolutely true. So this is definitely a solution."}, {"video_title": "Solving square-root equations two solutions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Remember, our original equation was six, I'll rewrite it here, our original equation was six plus three w is equal to the square root of two w plus 12 plus two w. So let's see, if w is equal to negative four, if w is equal to negative four right over, let me do this in a different, is equal to negative four, so that's going to be six plus three times negative four is equal to the square root of two times negative four plus 12 plus two times negative four. So this would be, this is negative 12 here, this is negative eight here, this is negative eight here. So you have six plus negative 12, which is negative six, is equal to the square root of negative eight plus 12 is four plus negative eight, so that would be negative six is equal to two plus negative eight, which is absolutely true. So this is definitely a solution. And let's try w is equal to negative six. So w is equal to negative six, so we're going to get, if we look up here, we're going to have six plus three times negative six is equal to the square root of two times negative six plus 12 plus two w. So this is going to be negative 18, this is going to be negative 12, this is negative 12, negative 12 plus 12 is zero, square root of zero, this is all zero, and then two times, and actually let me, I shouldn't have written a w there, I should have written a two times negative six. So back to what I was doing, this right over here is negative 18, this is two times negative six plus 12, this is all zero, square root of zero is zero, and then this is negative 12."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "you are by now probably used to the idea of measuring angles in degrees. We use it in everyday language. We've done some examples on this playlist where if you had an angle like that, you might call that a 30-degree angle. If you have an angle like this, you could call that a 90-degree angle, and we'd often use this symbol just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you have an angle like this, you could call that a 90-degree angle, and we'd often use this symbol just like that. If you were to go 180 degrees, you'd essentially form a straight line. Let me make these proper angles. If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If you go 360 degrees, you've essentially done one full rotation. If you watch figure skating on the Olympics and someone does a rotation, they'll say, oh, they did a 360, or especially in some skateboarding competitions and things like that. But the one thing to realize, and it might not be obvious right from the get-go, is this whole notion of degrees, this is a human-constructed system. This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is not the only way that you can measure angles. If you think about it, you say, well, why do we call a full rotation 360 degrees? There are some possible theories, and I encourage you to think about them. Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Why does 360 degrees show up in our culture as a full rotation? Well, there's a couple of theories there. One is ancient calendars, and even our calendar is close to this, but ancient calendars were based on 360 days in a year. Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Some ancient astronomers observed that things seemed to move 1 360th of the sky per day. Another theory is the ancient Babylonians liked equilateral triangles a lot, and they had a base 60 number system, so they had 60 symbols. We only have 10. We have a base 10. They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We have a base 10. They had 60. In our system, we like to divide things into 10. They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "They probably like to divide things into 60. If you had a circle and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections because you have a base 60 number system, then you might end up with 360 degrees. What I want to think about in this video is an alternate way of measuring angles. That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That alternate way, even though it might not seem as intuitive to you from the get-go, in some ways is much more mathematically pure than degrees. It's not based on these cultural artifacts of base 60 number systems or astronomical patterns. To some degree, an alien on another planet would not use degrees, especially if the degrees are motivated by these astronomical phenomena, but they might use what we're going to define as a radian. There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "There's a certain degree of purity here. Radians. Let's just cut to the chase and define what a radian is. Let me draw a circle here. My best attempt at drawing a circle. Not bad. Let me draw the center of the circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw a circle here. My best attempt at drawing a circle. Not bad. Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw the center of the circle. Now let me draw this radius. You might already notice the word radius is very close to the word radians, and that's not a coincidence. Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's say that this circle has a radius of length r. Let's construct an angle. I'll call that angle theta. Let's construct an angle theta. Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's call this angle right over here theta. Let's just say, for the sake of argument, that this angle is just the exact right measure. If you look at the arc that subtends this angle, that seems like a very fancy word. Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta. Let me write that down."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw the angle. If you look at the arc that subtends the angle, that's a fancy word, but that's really just talking about the arc along the circle that intersects the two sides of the angles. This arc right over here subtends the angle theta. Let me write that down. Subtends this arc. Subtends angle theta. Let's say theta is the exact right size."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me write that down. Subtends this arc. Subtends angle theta. Let's say theta is the exact right size. This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's say theta is the exact right size. This arc is also the same length as the radius of the circle. This arc is also of length r. Given that, if you were defining a new type of angle measurement, and you wanted to call it a radian, which is very close to a radius, how many radians would you define this angle to be? The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The most obvious one, if you view a radian as another way of saying radiuses or radii, you say, look, this is subtended by an arc of one radius, so why don't we call this right over here one radian? Which is exactly how a radian is defined. When you have a circle and you have an angle of one radian, the arc that subtends it is exactly one radius long, which you can imagine might be a little bit useful as we start to interpret more and more types of circles. When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When you give a degree, you really have to do a little bit of math and think about the circumference and all of that to think about how many radiuses are subtending that angle. Here, the angle in radians tells you exactly how many arc lengths that is subtending the angle. Let's do a couple of thought experiments here. Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here. That's the center. We'll start right over there."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Given that, what would be the angle in radians if we were to go... Let me draw another circle here. Let me draw another circle here. That's the center. We'll start right over there. What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We'll start right over there. What would happen if I had an angle? What angle, if I wanted to measure in radians, what angle would this be in radians? You can almost think of it as radiuses. What would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You can almost think of it as radiuses. What would that angle be? Going one full revolution in degrees, that would be 360 degrees. Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Based on this definition, what would this be in radians? Let's think about the arc that subtends this angle. The arc that subtends this angle is the entire circumference of this circle. It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's the entire circumference of this circle. What's the circumference of a circle in terms of radiuses? If this has length r, if the radius is length r, what's the circumference of the circle in terms of r? We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We know that. That's going to be 2 pi r. Going back to this angle, the length of the arc that subtends this angle is how many radiuses this is? What's 2 pi radiuses this is? It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's 2 pi times r. This angle right over here, I'll call this a different angle, x. x in this case is going to be 2 pi radians. It is subtended by an arc length of 2 pi radiuses. If the radius was one unit, then this would be 2 pi times 1, 2 pi radiuses. Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this. A full revolution in degrees is 360 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Given that, let's start to think about how we can convert between radians and degrees and vice versa. If I were to have, and we can just follow up over here, if we do one full revolution, that is 2 pi radians, how many degrees is this going to be equal to? We already know this. A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "A full revolution in degrees is 360 degrees. I could either write out the word degrees or I can use this little degree notation there. Actually, let me write out the word degrees. It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees. 180 degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It might make things a little bit clearer that we're using units in both cases. If we wanted to simplify this a little bit, we could divide both sides by 2, in which case we would get on the left-hand side, we would get pi radians would be equal to how many degrees? It would be equal to 180 degrees. 180 degrees. I could write it that way or I could write it that way. You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "180 degrees. I could write it that way or I could write it that way. You see over here, this is 180 degrees. You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses. We call this pi radians."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You also see if you were to draw a circle around here, we've gone halfway around the circle. The arc length or the arc that subtends the angle is half the circumference. Half the circumference are pi radiuses. We call this pi radians. Pi radians is 180 degrees. From this, we can come up with conversions. One radian would be how many degrees?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We call this pi radians. Pi radians is 180 degrees. From this, we can come up with conversions. One radian would be how many degrees? To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1. I'll just write it singular now."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "One radian would be how many degrees? To do that, we would just have to divide both sides by pi. On the left-hand side, you'd be left with 1. I'll just write it singular now. One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you. This isn't some voodoo."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll just write it singular now. One radian is equal to, I'm just dividing both sides. Let me make it clear what I'm doing here just to show you. This isn't some voodoo. I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This isn't some voodoo. I'm just dividing both sides by pi here. On the left-hand side, you're left with 1. On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have one degree, how many radians is that?"}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "On the right-hand side, you're left with 180 over pi degrees. One radian is equal to 180 over pi degrees, which is starting to make it an interesting way to convert them. Let's think about it the other way. If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees. Now we want to think about one degree."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "If I were to have one degree, how many radians is that? Let's start off with, let me rewrite this thing over here. We said pi radians is equal to 180 degrees. Now we want to think about one degree. Let's solve for one degree. One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now we want to think about one degree. Let's solve for one degree. One degree, we can divide both sides by 180. We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree. This might seem confusing and daunting, and it was for me the first time I was exposed to it, especially because we're not exposed to this in our everyday life. What we're going to see over the next few examples is that as long as we keep in mind this whole idea that 2 pi radians is equal to 360 degrees or that pi radians is equal to 180 degrees, which is the two things that I do keep in my mind, we can always re-derive these two things."}, {"video_title": "Introduction to radians Unit circle definition of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We are left with pi over 180 radians is equal to one degree. Pi over 180 radians is equal to one degree. This might seem confusing and daunting, and it was for me the first time I was exposed to it, especially because we're not exposed to this in our everyday life. What we're going to see over the next few examples is that as long as we keep in mind this whole idea that 2 pi radians is equal to 360 degrees or that pi radians is equal to 180 degrees, which is the two things that I do keep in my mind, we can always re-derive these two things. You might say, hey, how do I remember if it's pi over 180 or 180 over pi to convert the two things? Well, just remember, which is hopefully intuitive, that 2 pi radians is equal to 360 degrees. We'll work through a bunch of examples in the next video to just make sure that we're used to converting one way or the other."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Write a trigonometric function that models the temperature, capital T, in Johannesburg, lowercase t, hours after midnight. So let's see if we can start to think about what a graph might look like of all of this. So this, let's say this is our temperature axis in Celsius degrees, so that is temperature, temperature, and I'm actually gonna first, I'm gonna do two different functions. So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that's my temperature axis, and then this right over here is my time in hours. So that's lowercase t, time in hours, and let's think about the range of temperatures. So the daily low temperature's around three degrees Celsius, so let's actually, and the high is 18, so let's make this right over here 18, this right over here is three, and we can also think about the midpoint between 18 and three that we hit at both 10 a.m. and 10 p.m. 18 plus three is 21, divided by two is 10.5. So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this. The daily high is around 18 degrees Celsius."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the midpoint, or we could say the midline of our trigonometric function is going to be 10.5 degrees Celsius, so let's, let me draw the midline. So we're gonna essentially oscillate around this right over here. We're gonna oscillate around this. The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that. So we're gonna oscillate around this midline."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The daily high is around 18 degrees Celsius. The daily high is around 18 degrees Celsius, and the daily low is around three degrees. The three degrees Celsius, just like that. So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we're gonna oscillate around this midline. We're gonna hit the lows and the highs. Now to simplify things, because we hit this 10.5 degrees at 10 a.m. and 10 p.m., to simplify this, I'm not going to tackle their question that they want immediately, the hour in terms of t hours after midnight. I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'm gonna define a new function, f of t, f of lowercase t, which is equal to the temperature, temperature in Johannesburg, where we're assuming everything is in Johannesburg, temperature t hours after, I'm gonna say t hours after 10 a.m. The reason why I'm picking 10 a.m. is because we know that the temperature is right at the midline at 10 a.m., t hours after 10 a.m. Because if I want to graph f of t at t equals zero, that means we're at 10 a.m., that means that we're halfway between, they tell us, we're halfway between the daily low and the daily high. Now what is the period of this trigonometric function going to be? Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, after 24 hours, we're back to, we're going to be back to 10 a.m. So our period is going to be 24 hours. So let me put 24 hours there, and then this is, halfway is 12 hours. So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what happens after 12 hours? After 12 hours, we're back at 10 p.m., where we're back at the midway between our lows and our highs. And then, after 24 hours, we're back at 10 a.m. again. So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So those are going to be points on f of t. And now let's think about what'll happen as we go beyond, as we start at 10 a.m. and go forward. So as we go, start at 10 a.m. and go forward, they tell us that the hottest part, the hottest part, the highest temperatures are in the afternoon. The afternoon is going to be around here. So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we should be going up in temperature, and the highest point is actually going to be halfway between these two. So it's going to be six hours after 10 a.m., which is 4 p.m. So that's going to be the high at 4 p.m. So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let me draw a curve, draw our curve like this. So it'll look like this. And then our low, so now we're at 10 p.m. And then you go six hours after 10 p.m., you're now at 4 a.m., which is gonna be the low. This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is 18 hours after 10 a.m. After 10 a.m., you're gonna be at your low temperature, roughly right over there. And your curve will look something like this. So what would be, before we even try to model T of t, what would be an expression, and obviously we keep going, we keep going like that, and we could even go hours before 10 a.m. This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is obviously, this keeps on cycling on and on and on forever. Now, what would be an expression for F of t? And I encourage you once again to pause the video and try to think about that. Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, one thing that you could say, well, you say this could be a sine or a cosine function. Actually, you could model it with either of them, but it's always easiest to do the simplest one. Which function is essentially at its midline, at its midline, when the argument to the function is zero? Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, the sine of zero is zero, and if we didn't shift this function up or down, the midline of just a sine function, without it being shifted, is zero. So sine of zero is zero, and then sine begins to increase and oscillate like this. So it feels like sine is a good candidate to model it with. Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary? What's our maximum variance from our midline?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Once again, you could model it with either, but I have a feeling this is gonna be a little bit simpler. Now, let's think about the amplitude. Well, how much do we vary? What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline. So our amplitude is 7.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What's our maximum variance from our midline? So here we are 7.5 above our midline. Here we are 7.5 below our midline. So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So our amplitude is 7.5. And actually, let me just do that in a different color, just so you see where things are coming from. So this is 7.5, this is 7.5. So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it. Our period is 24, 24 hours."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So our amplitude looks like it's 7.5. And now, what is our period? Well, we've already talked about it. Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our period is 24, 24 hours. This distance right over here is 24 hours, which makes complete sense. After 24 hours, you're at the same point in the day. So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we would divide two pi by the period, divided by 24 times t. And if you forget, hey, divide two pi by the period here, you could just remind yourself that what t value will make us go from, so when t is equal to zero, the whole argument to the sine function is going to be zero. That's when we're over here. And then when t is equal to 24, the whole argument's going to be two pi. So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we would have made one rotation around the unit circle if we think about the input into the sine function. Now, we're almost done. If I were to just graph this, this would have a midline around zero, but we see that we've shifted everything up by 10.5. So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we have to shift everything up by 10.5. Now, this is, we've just successfully modeled it, and we could simplify a little bit. We could write this as pi over 12 instead of two pi over 24. But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be? We're gonna have to shift this a little bit."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But this right over here models the temperature in Johannesburg T hours after 10 a.m. After 10 a.m. That's not what they wanted. They want us to model, they want us to model the temperature T hours after midnight. So what would T of T be? We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're gonna have to shift this a little bit. So let's just think about it a little. Let me just write it out. So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T of T, so T of T, this is now we're modeling T hours after midnight. So we're gonna have the same amplitude. We're just gonna have the same variance from the midline. So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's gonna be 7.5 times sine of, actually, we do the same color so you see what I'm changing and not changing, times the sine of, instead of two pi over 24, I'll just write pi over 12. Instead of writing T, I'm gonna shift T either to the right or the left. And actually, you could shift in either direction because this is a periodic function. We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We're gonna have to think about how much we're shifting it. So T is gonna be plus or minus something right over here. I'm gonna shift it plus 10.5, plus 10.5. Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now this is always a little bit, at least in my brain, I have to think about this in a lot of different ways so that I make sure that I'm shifting it in the right direction. So here at 10 a.m., we were at this point. When T is equal to zero, this is zero hours after 10 a.m. But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But in this function, when is 10 a.m.? Well, in this function, 10 a.m., let me write it this way. 10 a.m. is 10 hours after midnight. So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T, capital T of 10, this is 10 hours after midnight, should be equal to, should be equal to f of zero because here, the argument is hours after 10 a.m. So this is 10 a.m., this right over here represents temperature at 10 a.m. And this over here, because capital, this capital T function, this is hours after midnight. This is also temperature at 10 a.m. So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that?"}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we want T of 10 to be the same thing as f of zero. Or a same, another way of thinking about it, when f of zero, this whole argument is zero. So we want this whole argument to be zero when T is equal to 10. So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So how would we do that? Well, if this is T minus 10, notice, T of 10, you put a 10 here, this whole thing becomes zero. This whole thing becomes zero, and you're left with 10.5. And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And over here, f of zero, well, the same thing. This whole thing becomes zero, and all you're left with over here is 10.5. So T of 10 should be f of zero. So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent. But let's actually graph this."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if we wanted to graph it, we've already answered their question. If we put a 10 here, the argument to the sign becomes zero. These two things are going to be equivalent. But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But let's actually graph this. So T of 10, so if we're graphing capital T, T of 10, so this is six, 12, let's see. So this is, let me be, so 10 is going to be someplace around here. So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense."}, {"video_title": "Modeling temperature through the day Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So T of 10 is going to be the same thing as f of zero. So it's going to be like that. And then it's just going to, and then we've essentially just shifted everything to the right, everything to the right by 10. And that makes sense. Because zero after, whatever hours you are after 10 a.m., it's going to be 10 more hours to get to that same point after midnight. So your curve is going to look, so this is going to be shifted by 10, this is going to be shifted by 10, and your curve's going to look something like, let me see, this is going to be shifted by 10, so you're going to get, this is going to be at 16 hours, so let's see, it's going to look something like that. And of course, it'll keep oscillating like that."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "In your mathematical careers, you might encounter people who say it is wrong to say that i is equal to the principal square root of negative 1. And if you ask them why is this wrong, they'll show up with this kind of line of logic that actually seems pretty reasonable. They will tell you that, okay, well let's just start with negative 1. We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root of negative 1. And they'd be right."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "We know from definition that negative 1 is equal to i times i. Everything seems pretty straightforward right now. And then they'll say, well look, if you take this, if you assume this part right here, then we can replace each of these i's with the square root of negative 1. And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look, just from straight up properties of the principal square root function, they'll tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so, if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And they'd be right. So then this would be the same thing as the square root of negative 1 times the square root of negative 1. And then they would tell you that, hey, look, just from straight up properties of the principal square root function, they'll tell you that the square root of a times b is the same thing as the principal square root of a times the principal square root of b. And so, if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b. Based on this property of the radical, of the principal root, root, they'll say that this over here is the same thing as the square root of negative one times negative one. Of negative one times negative one. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And so, if you have the principal square root of a times the principal square root of b, that's the same thing as the square root of a times b. Based on this property of the radical, of the principal root, root, they'll say that this over here is the same thing as the square root of negative one times negative one. Of negative one times negative one. If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I had the principal root of the products, over here I have this on the right. And then from that, we all know that negative one times negative one is one, so this should be equal to the principal square root of one."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "If I have the principal root of the product of two things, that's the same thing as the product of each of their principal roots. I'm doing this in the other order here. Here I had the principal root of the products, over here I have this on the right. And then from that, we all know that negative one times negative one is one, so this should be equal to the principal square root of one. And then the principal square root of one, remember this radical means principal square root, positive square root, that is just going to be positive one. And they'll say this is wrong. Clearly, clearly negative one and positive one are not the same thing."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "And then from that, we all know that negative one times negative one is one, so this should be equal to the principal square root of one. And then the principal square root of one, remember this radical means principal square root, positive square root, that is just going to be positive one. And they'll say this is wrong. Clearly, clearly negative one and positive one are not the same thing. And they'll argue, therefore you can't make this substitution that I, that we did in this step. And what you should then point out is that this was not the incorrect step. That it is true, negative one is not equal to one, but the faulty line of reasoning here was in using this property, was in using this property when both A and B are negative."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "Clearly, clearly negative one and positive one are not the same thing. And they'll argue, therefore you can't make this substitution that I, that we did in this step. And what you should then point out is that this was not the incorrect step. That it is true, negative one is not equal to one, but the faulty line of reasoning here was in using this property, was in using this property when both A and B are negative. If both A and B are negative, this will never be true. So A, A and B both cannot, cannot be, both cannot be negative. In fact, normally when this property is given, sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time, but they'll usually give a little bit of a constraint there."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "That it is true, negative one is not equal to one, but the faulty line of reasoning here was in using this property, was in using this property when both A and B are negative. If both A and B are negative, this will never be true. So A, A and B both cannot, cannot be, both cannot be negative. In fact, normally when this property is given, sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time, but they'll usually give a little bit of a constraint there. They'll usually say for, for A, A and B greater than or equal to zero. So that's where they list this property. This is true for A and B greater than or equal to zero."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "In fact, normally when this property is given, sometimes it's given a little bit in the footnotes or you might not even notice it because it's not relevant when you're learning it the first time, but they'll usually give a little bit of a constraint there. They'll usually say for, for A, A and B greater than or equal to zero. So that's where they list this property. This is true for A and B greater than or equal to zero. And in particular, it's false if both A and B are both, if they are both negative. Now, I've said that, I've just spent the last three minutes saying that people who tell you that this is wrong are wrong. But with that said, I do, I will say that you have to be a little bit careful about it."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "This is true for A and B greater than or equal to zero. And in particular, it's false if both A and B are both, if they are both negative. Now, I've said that, I've just spent the last three minutes saying that people who tell you that this is wrong are wrong. But with that said, I do, I will say that you have to be a little bit careful about it. You have to be a little bit careful about it. When we take traditional principal square roots, so when you take this principal square root of four, we know that this is positive two. That four actually has two square roots."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "But with that said, I do, I will say that you have to be a little bit careful about it. You have to be a little bit careful about it. When we take traditional principal square roots, so when you take this principal square root of four, we know that this is positive two. That four actually has two square roots. There's negative two is also a square root of, also is a square root, square root of four. If you have negative two times negative two, it's also equal to four. This radical symbol here means principal square root."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "That four actually has two square roots. There's negative two is also a square root of, also is a square root, square root of four. If you have negative two times negative two, it's also equal to four. This radical symbol here means principal square root. Or when we're just dealing with real numbers, non-imaginary, non-complex numbers, you can really view it as the positive square root. This has two square roots, positive and negative two. If you have this radical symbol right here, principal square roots, it means the positive square root of two."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "This radical symbol here means principal square root. Or when we're just dealing with real numbers, non-imaginary, non-complex numbers, you can really view it as the positive square root. This has two square roots, positive and negative two. If you have this radical symbol right here, principal square roots, it means the positive square root of two. So when you, when you start thinking about taking square roots of negative numbers, or even in the future we'll do imaginary numbers and complex numbers and all the rest, you have to expand the definition of what this radical means. So when you are taking the square root of, when you are taking the square root really of any, of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "If you have this radical symbol right here, principal square roots, it means the positive square root of two. So when you, when you start thinking about taking square roots of negative numbers, or even in the future we'll do imaginary numbers and complex numbers and all the rest, you have to expand the definition of what this radical means. So when you are taking the square root of, when you are taking the square root really of any, of any negative number, you're really saying that this is no longer the traditional principal square root function. You're now talking that this is the principal complex square root function. Or this is now defined for complex inputs or the domain. It can also generate imaginary or complex outputs. Or I guess you could call that, you could call that the range."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "You're now talking that this is the principal complex square root function. Or this is now defined for complex inputs or the domain. It can also generate imaginary or complex outputs. Or I guess you could call that, you could call that the range. And if you assume that, then really straight from, straight from this, you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only, and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true, so we can apply this, we can apply, we can apply this, we can apply when, when x, when x is greater than or equal to zero."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "Or I guess you could call that, you could call that the range. And if you assume that, then really straight from, straight from this, you get that negative, the square root of negative x is going to be equal to i times the square root of x. And this is only, and I'm going to make this clear because I just told you that this will be false if both a and b are negative. So this is only true, so we can apply this, we can apply, we can apply this, we can apply when, when x, when x is greater than or equal to zero. So if x is greater than or equal to zero, then negative x is clearly a negative number, or I guess it could also be zero. It's a negative number, and then we can apply this right over here. If x was less than zero, then we would be doing all of this nonsense up here, and then we would start to get nonsensical answers."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "So this is only true, so we can apply this, we can apply, we can apply this, we can apply when, when x, when x is greater than or equal to zero. So if x is greater than or equal to zero, then negative x is clearly a negative number, or I guess it could also be zero. It's a negative number, and then we can apply this right over here. If x was less than zero, then we would be doing all of this nonsense up here, and then we would start to get nonsensical answers. And if you look at it this way, and you say, hey, look, I can be the square root of negative one if we're taking the, if it's the principal branch of the complex square root function, then you could rewrite this right over here as the square root of negative one times, times the square root, times the square root of x. And so really, the real fault in this logic, why, what, when people say, hey, one negative one can't be equal to one, the real fault is using this property, is using this property when both a and b, where both of these are negative numbers, that will come up with something that is unambiguously false. If you expand your definition of the complex, or expand your definition of the principal root to include negative numbers in the domain, and including, and to include imaginary numbers, then you can do this."}, {"video_title": "i as the principal root of -1 (a little technical) Precalculus Khan Academy.mp3", "Sentence": "If x was less than zero, then we would be doing all of this nonsense up here, and then we would start to get nonsensical answers. And if you look at it this way, and you say, hey, look, I can be the square root of negative one if we're taking the, if it's the principal branch of the complex square root function, then you could rewrite this right over here as the square root of negative one times, times the square root, times the square root of x. And so really, the real fault in this logic, why, what, when people say, hey, one negative one can't be equal to one, the real fault is using this property, is using this property when both a and b, where both of these are negative numbers, that will come up with something that is unambiguously false. If you expand your definition of the complex, or expand your definition of the principal root to include negative numbers in the domain, and including, and to include imaginary numbers, then you can do this. You can say the square root of negative x is neg, is the square root of negative one times the, or you say the, the principal square root of negative x, I should be particular with my words, is the same thing as the principal square root of negative one times the principal square root of x when x is greater than or equal to zero. And I, I, I, I don't want to confuse you. If x is greater than or equal to zero, this is clearly, this negative x, that is clearly a negative, or I guess you say a non-positive number."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "All right, now let's work through this together. So the first thing to just remind ourselves is what is an extraneous solution? Well, that's a solution that we get, or we think we get, but it's really just a byproduct of how we solved it, but it isn't going to be an actual solution of our original equation. Now, how do these extraneous solutions pop up? Well, it pops up when you take the square of both sides. So for this equation right over here, to get rid of the radical, I'd wanna square both sides of it. If I square both sides, the left-hand side will become three x plus 25, and the right-hand side, if I square this, is going to be what?"}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "Now, how do these extraneous solutions pop up? Well, it pops up when you take the square of both sides. So for this equation right over here, to get rid of the radical, I'd wanna square both sides of it. If I square both sides, the left-hand side will become three x plus 25, and the right-hand side, if I square this, is going to be what? It's going to be d squared plus four, four dx plus x squared. So that's just squaring both sides of this. But notice, there's actually a different equation than this one, that if you squared both sides, you would also get this."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "If I square both sides, the left-hand side will become three x plus 25, and the right-hand side, if I square this, is going to be what? It's going to be d squared plus four, four dx plus x squared. So that's just squaring both sides of this. But notice, there's actually a different equation than this one, that if you squared both sides, you would also get this. What is that different equation? Well, the different equation is if you took the negative of one of these sides. So for example, if you started the original equation, the negative square root of three x plus 25, plus 25, is equal to d plus two x."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "But notice, there's actually a different equation than this one, that if you squared both sides, you would also get this. What is that different equation? Well, the different equation is if you took the negative of one of these sides. So for example, if you started the original equation, the negative square root of three x plus 25, plus 25, is equal to d plus two x. You square both sides of this, you still get this purple equation. Because you square a negative, you get a positive. So both of them, when you square both sides, get us over here."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "So for example, if you started the original equation, the negative square root of three x plus 25, plus 25, is equal to d plus two x. You square both sides of this, you still get this purple equation. Because you square a negative, you get a positive. So both of them, when you square both sides, get us over here. And so when you solve this purple equation, this is a quadratic right over here, you just rearrange it a little bit, you get into standard quadratic form, you'll get two solutions. And it turns out, one of the solutions is going to be for this yellow equation, and one of the solutions is going to be for the purple equation. And the solution that is for the purple equation is going to be an extraneous solution for the yellow equation."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "So both of them, when you square both sides, get us over here. And so when you solve this purple equation, this is a quadratic right over here, you just rearrange it a little bit, you get into standard quadratic form, you'll get two solutions. And it turns out, one of the solutions is going to be for this yellow equation, and one of the solutions is going to be for the purple equation. And the solution that is for the purple equation is going to be an extraneous solution for the yellow equation. It's actually not going to be a solution for the yellow equation. So when they say which value for d makes x equals negative three an extraneous solution for this yellow equation, that's the same thing as saying what value of d makes x equals negative three a solution for this? So a solution, a solution for this."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "And the solution that is for the purple equation is going to be an extraneous solution for the yellow equation. It's actually not going to be a solution for the yellow equation. So when they say which value for d makes x equals negative three an extraneous solution for this yellow equation, that's the same thing as saying what value of d makes x equals negative three a solution for this? So a solution, a solution for this. If it's a solution for this, it's going to be an extraneous solution for that, because these are two different equations. We have, this is, we're taking the negative of just one side of this equation to get this one. If you took the negative of both sides of this, then that becomes the same thing, because you can multiply both sides of an equation times a negative value."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "So a solution, a solution for this. If it's a solution for this, it's going to be an extraneous solution for that, because these are two different equations. We have, this is, we're taking the negative of just one side of this equation to get this one. If you took the negative of both sides of this, then that becomes the same thing, because you can multiply both sides of an equation times a negative value. So a solution for this, which is equivalently a solution, which is equivalent to a solution if I, instead of putting the negative on the left-hand side, if I multiplied the right-hand side by the negative. But anyway, let's think about which value for d makes x equals negative three a solution for this. Well, let's substitute x equals negative three here, and then we just have to solve for d. If x equals negative three, this is going to be negative the square root of three times negative three is negative nine plus 25 is equal to d, two times negative three is negative six."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "If you took the negative of both sides of this, then that becomes the same thing, because you can multiply both sides of an equation times a negative value. So a solution for this, which is equivalently a solution, which is equivalent to a solution if I, instead of putting the negative on the left-hand side, if I multiplied the right-hand side by the negative. But anyway, let's think about which value for d makes x equals negative three a solution for this. Well, let's substitute x equals negative three here, and then we just have to solve for d. If x equals negative three, this is going to be negative the square root of three times negative three is negative nine plus 25 is equal to d, two times negative three is negative six. So d minus six. And so now we can square both sides of, we can square, actually, let's do it this way. We can, I don't want to square both sides, because we lose some information."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "Well, let's substitute x equals negative three here, and then we just have to solve for d. If x equals negative three, this is going to be negative the square root of three times negative three is negative nine plus 25 is equal to d, two times negative three is negative six. So d minus six. And so now we can square both sides of, we can square, actually, let's do it this way. We can, I don't want to square both sides, because we lose some information. There's going to be the negative square root of negative nine plus 25 is 16, is equal to d minus six. So this is going to be equal to the, this is going to be equal to negative four. Principal root of 16 is four, we have the negative out front, is equal to d minus six."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "We can, I don't want to square both sides, because we lose some information. There's going to be the negative square root of negative nine plus 25 is 16, is equal to d minus six. So this is going to be equal to the, this is going to be equal to negative four. Principal root of 16 is four, we have the negative out front, is equal to d minus six. And then add six to both sides, you get two is equal to, two is equal to d. So if d is equal to two here, if d is equal to two, then a solution to this purple equation is going to be x equals negative three. And so that would be an extraneous solution, because if x equals negative three satisfies this over here, it's definitely going to satisfy this over here, but it's not going to satisfy this up here. And you can verify this, if this is equal to two, try out x equals negative three."}, {"video_title": "Extraneous solutions of radical equations (example 2) High School Math Khan Academy.mp3", "Sentence": "Principal root of 16 is four, we have the negative out front, is equal to d minus six. And then add six to both sides, you get two is equal to, two is equal to d. So if d is equal to two here, if d is equal to two, then a solution to this purple equation is going to be x equals negative three. And so that would be an extraneous solution, because if x equals negative three satisfies this over here, it's definitely going to satisfy this over here, but it's not going to satisfy this up here. And you can verify this, if this is equal to two, try out x equals negative three. You're going to get, on the left-hand side, you're going to get 16. And on the right-hand side, you're going to get two minus six, which is equal to negative four. Two minus six, which is negative four, so this does not work out."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "In previous videos, we've already looked at using area models to think about multiplying expressions, like multiplying x plus seven times x plus three. In those videos, we saw that we could think about it as finding the area of a rectangle where we could break up the length of the rectangle as part of the length has length x, and then the rest of it has length seven. So this would be seven here. And then the total length of this side would be x plus seven and then the total length of this side would be x plus, and then you have three right over here. And what area models did is they helped us visualize why we multiply the different terms or how we multiply the different terms. Because if we're looking for the entire area, the entire area is going to be x plus seven, x plus seven, times x plus three, times x plus three. And then of course, we can break that down into the subrectangles."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And then the total length of this side would be x plus seven and then the total length of this side would be x plus, and then you have three right over here. And what area models did is they helped us visualize why we multiply the different terms or how we multiply the different terms. Because if we're looking for the entire area, the entire area is going to be x plus seven, x plus seven, times x plus three, times x plus three. And then of course, we can break that down into the subrectangles. This rectangle, and this is actually going to be a square, would have an area of x squared. This one over here will have an area of seven x, seven times x. This one over here will have an area of three x."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And then of course, we can break that down into the subrectangles. This rectangle, and this is actually going to be a square, would have an area of x squared. This one over here will have an area of seven x, seven times x. This one over here will have an area of three x. And then this one over here will have an area of three times seven, or 21. And so we can figure out that the ultimate product here is going to be x squared plus seven x plus three x plus 21. That's going to be the area of the entire rectangle."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "This one over here will have an area of three x. And then this one over here will have an area of three times seven, or 21. And so we can figure out that the ultimate product here is going to be x squared plus seven x plus three x plus 21. That's going to be the area of the entire rectangle. Of course, we could add the seven x to the three x to get to 10 x. But some of you might be wondering, well, this is all nice when I have plus seven and plus three. I can think about positive lengths."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "That's going to be the area of the entire rectangle. Of course, we could add the seven x to the three x to get to 10 x. But some of you might be wondering, well, this is all nice when I have plus seven and plus three. I can think about positive lengths. I can think about positive areas. But what if it wasn't that way? What if we were dealing with negatives instead?"}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "I can think about positive lengths. I can think about positive areas. But what if it wasn't that way? What if we were dealing with negatives instead? For example, if we now try to do the same thing, we could say, all right, this top length right over here would be x minus seven. So let's just keep going with it. And let's call this length negative seven up here."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "What if we were dealing with negatives instead? For example, if we now try to do the same thing, we could say, all right, this top length right over here would be x minus seven. So let's just keep going with it. And let's call this length negative seven up here. So it has a negative seven length, and we're not necessarily used to thinking about lengths as negative. Let's just go with it. And then the height right over here, it would be x minus three."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And let's call this length negative seven up here. So it has a negative seven length, and we're not necessarily used to thinking about lengths as negative. Let's just go with it. And then the height right over here, it would be x minus three. So we could write an x there for that part of the height. And for this part of the height, we could put a negative three. So let's see, if we kept going with what we did last time, the area here would be x squared."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And then the height right over here, it would be x minus three. So we could write an x there for that part of the height. And for this part of the height, we could put a negative three. So let's see, if we kept going with what we did last time, the area here would be x squared. The area here would be negative seven times x. So that would be negative seven x. This green area would be negative three x."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "So let's see, if we kept going with what we did last time, the area here would be x squared. The area here would be negative seven times x. So that would be negative seven x. This green area would be negative three x. And then this orange area would be negative three times negative seven, which is positive 21. And then we would say that the entire product is x squared minus seven x minus three x plus 21. And we could, of course, add these two together to get negative 10x."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "This green area would be negative three x. And then this orange area would be negative three times negative seven, which is positive 21. And then we would say that the entire product is x squared minus seven x minus three x plus 21. And we could, of course, add these two together to get negative 10x. But does this make sense? Well, one way to think about it is that a negative area is an area that you would take away from the total area. So if x happens to be a positive number here, then this pink area would be negative."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And we could, of course, add these two together to get negative 10x. But does this make sense? Well, one way to think about it is that a negative area is an area that you would take away from the total area. So if x happens to be a positive number here, then this pink area would be negative. And so we would take it away from the whole. And that's exactly what is happening in this expression. And it's worth mentioning that even before when this wasn't a negative seven, when it was a positive seven, and this was a positive seven x, it's completely possible that x is negative, in which case you would have had a negative area anyway."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "So if x happens to be a positive number here, then this pink area would be negative. And so we would take it away from the whole. And that's exactly what is happening in this expression. And it's worth mentioning that even before when this wasn't a negative seven, when it was a positive seven, and this was a positive seven x, it's completely possible that x is negative, in which case you would have had a negative area anyway. But to appreciate that this will all work out even with negative numbers, I'll give an example if x were equal to 10. That will help us make sense of things. So if x were equal to 10, we would get an area model that looks like this."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "And it's worth mentioning that even before when this wasn't a negative seven, when it was a positive seven, and this was a positive seven x, it's completely possible that x is negative, in which case you would have had a negative area anyway. But to appreciate that this will all work out even with negative numbers, I'll give an example if x were equal to 10. That will help us make sense of things. So if x were equal to 10, we would get an area model that looks like this. We're having 10 minus seven, so I'll put minus seven right over here, times 10 minus three. Now you can figure out in your heads what's that going to be. 10 minus seven is three."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "So if x were equal to 10, we would get an area model that looks like this. We're having 10 minus seven, so I'll put minus seven right over here, times 10 minus three. Now you can figure out in your heads what's that going to be. 10 minus seven is three. 10 minus three is seven. So this should all add up to positive 21. Let's make sure that's actually occurring."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "10 minus seven is three. 10 minus three is seven. So this should all add up to positive 21. Let's make sure that's actually occurring. So this blue area is going to be 10 times 10, which is 100. This pink area now is 10 times negative seven, so it's negative 70. So we're gonna take it away from the total area."}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "Let's make sure that's actually occurring. So this blue area is going to be 10 times 10, which is 100. This pink area now is 10 times negative seven, so it's negative 70. So we're gonna take it away from the total area. This green area is negative three times 10, so that's negative 30. And then negative three times negative seven, this orange area is positive 21. Does that all work out?"}, {"video_title": "Area model for multiplying polynomials with negative terms.mp3", "Sentence": "So we're gonna take it away from the total area. This green area is negative three times 10, so that's negative 30. And then negative three times negative seven, this orange area is positive 21. Does that all work out? So let's see, if we take this positive area, 100 minus 70 minus 30, and then add 21, 100 minus 70 is going to be 30. Minus 30 again is zero, and then you just have 21 left over, which is exactly what you would expect. You could actually move this pink area over and subtract it from this blue area, and then you could take this green area, and then you could turn it vertical, and then that would subtract out the rest of the blue area, and then all you would have left is this orange area."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "In a previous video, when we were looking at a very simple case of compounding interest, we got the expression 1 plus 1 over n to the nth power. And the way we got this, we saw an example where a loan shark is charging 100% interest, and that's where this 1 is. And then if they only compound once in the year, so it's 100% over the year, then n is 1. So you get 1 plus 100% over 1 to the first power. You're going to have to pay back twice the original amount of money. If n is 2, 1 plus 1 half is 1 and a half to the second power gets you 2.25. If you compound half the interest, so 100% over 2, but you compound it twice, and then we kept going and going and going, we saw interesting things happen."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So you get 1 plus 100% over 1 to the first power. You're going to have to pay back twice the original amount of money. If n is 2, 1 plus 1 half is 1 and a half to the second power gets you 2.25. If you compound half the interest, so 100% over 2, but you compound it twice, and then we kept going and going and going, we saw interesting things happen. And I want to review that right over here using this calculator. I want to see what happens as we get larger and larger and larger n's. In that last video, we went as high as n is equal to 365, and it seemed to be approaching a magical number."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "If you compound half the interest, so 100% over 2, but you compound it twice, and then we kept going and going and going, we saw interesting things happen. And I want to review that right over here using this calculator. I want to see what happens as we get larger and larger and larger n's. In that last video, we went as high as n is equal to 365, and it seemed to be approaching a magical number. But now let's go even further. So let's type in, and let's throw some really large numbers here. 1 plus 1 over, let's do a million."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "In that last video, we went as high as n is equal to 365, and it seemed to be approaching a magical number. But now let's go even further. So let's type in, and let's throw some really large numbers here. 1 plus 1 over, let's do a million. 1, 2, 3, 1, 2, 3, so that's a million to the millionth power. 1, 2, 3, 1, 2, 3. Did I get the right numbers here?"}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "1 plus 1 over, let's do a million. 1, 2, 3, 1, 2, 3, so that's a million to the millionth power. 1, 2, 3, 1, 2, 3. Did I get the right numbers here? Yeah, that looks right. And before I even press enter, which is exciting, let's just think about what's going on here. This part that we have here as n gets larger and larger, it's getting closer and closer to 1, but never quite exactly 1."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Did I get the right numbers here? Yeah, that looks right. And before I even press enter, which is exciting, let's just think about what's going on here. This part that we have here as n gets larger and larger, it's getting closer and closer to 1, but never quite exactly 1. This is 1 and 1 millionth, so it's very close to 1 but not exactly 1. And then we're going to raise that thing to the millionth power. And normally when you raise something to the millionth power, that's just going to be unbounded, just become some huge number."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "This part that we have here as n gets larger and larger, it's getting closer and closer to 1, but never quite exactly 1. This is 1 and 1 millionth, so it's very close to 1 but not exactly 1. And then we're going to raise that thing to the millionth power. And normally when you raise something to the millionth power, that's just going to be unbounded, just become some huge number. But there's a clue that 1 to the millionth power would just be 1. If we're getting really close to 1, well, maybe this won't just be some unbounded number. And when we calculate it, we see that that's the case."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And normally when you raise something to the millionth power, that's just going to be unbounded, just become some huge number. But there's a clue that 1 to the millionth power would just be 1. If we're getting really close to 1, well, maybe this won't just be some unbounded number. And when we calculate it, we see that that's the case. It's 2.71828 and it just keeps going. Now let's go even higher. Let's do 1 plus 1 over, and actually I can now use scientific notation."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And when we calculate it, we see that that's the case. It's 2.71828 and it just keeps going. Now let's go even higher. Let's do 1 plus 1 over, and actually I can now use scientific notation. Let's say 1 times 10 to the seventh power. So that's literally, this right over here is 10 million, to the ten millionth power. So what do we get here?"}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Let's do 1 plus 1 over, and actually I can now use scientific notation. Let's say 1 times 10 to the seventh power. So that's literally, this right over here is 10 million, to the ten millionth power. So what do we get here? So now we went 2.718281692. Let's go even larger. So let's get our last entry here."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So what do we get here? So now we went 2.718281692. Let's go even larger. So let's get our last entry here. So let's go instead of the seventh power, let's go to the eighth power. So now we're 1 plus 1 over 100 million to the hundred millionth power. I don't even know if this calculator can handle this."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So let's get our last entry here. So let's go instead of the seventh power, let's go to the eighth power. So now we're 1 plus 1 over 100 million to the hundred millionth power. I don't even know if this calculator can handle this. And we get 2.71828181487. And you see that we are quickly approaching, or maybe not so quickly, we have to raise this to a very large power, to the number e in our calculator. You see we've already gotten 1, 2, 3, 4, 5, 6, 7 digits to the right of the decimal point by taking it to the hundred millionth power."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "I don't even know if this calculator can handle this. And we get 2.71828181487. And you see that we are quickly approaching, or maybe not so quickly, we have to raise this to a very large power, to the number e in our calculator. You see we've already gotten 1, 2, 3, 4, 5, 6, 7 digits to the right of the decimal point by taking it to the hundred millionth power. So we are approaching this number. So one way to talk about it is we could say that the limit as n approaches infinity, as n becomes larger and larger, it's not becoming unbounded, it's not going to infinity, it seems to be approaching this number. And we will call this number, we will call this magical and mystical number e. We'll call this number e. And we see from our calculator that this number, and these are almost as famous digits as the digits for pi, we're getting 2.7182818, and it just keeps going and going and going, never, never repeating."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "You see we've already gotten 1, 2, 3, 4, 5, 6, 7 digits to the right of the decimal point by taking it to the hundred millionth power. So we are approaching this number. So one way to talk about it is we could say that the limit as n approaches infinity, as n becomes larger and larger, it's not becoming unbounded, it's not going to infinity, it seems to be approaching this number. And we will call this number, we will call this magical and mystical number e. We'll call this number e. And we see from our calculator that this number, and these are almost as famous digits as the digits for pi, we're getting 2.7182818, and it just keeps going and going and going, never, never repeating. So it's an infinite string of digits, never, never repeating. Just like pi. Pi, you remember, is the ratio of the circumference to the diameter of the circle."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And we will call this number, we will call this magical and mystical number e. We'll call this number e. And we see from our calculator that this number, and these are almost as famous digits as the digits for pi, we're getting 2.7182818, and it just keeps going and going and going, never, never repeating. So it's an infinite string of digits, never, never repeating. Just like pi. Pi, you remember, is the ratio of the circumference to the diameter of the circle. e is another one of these crazy numbers that shows up in the universe. And in other videos on Khan Academy, we go into depth why this is so magical and mystical. Already this is kind of cool, that I can take an infinite, if I just add one over a number to one and take it to that number, and I make that number larger and larger and larger, it's approaching this number."}, {"video_title": "e as a limit Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Pi, you remember, is the ratio of the circumference to the diameter of the circle. e is another one of these crazy numbers that shows up in the universe. And in other videos on Khan Academy, we go into depth why this is so magical and mystical. Already this is kind of cool, that I can take an infinite, if I just add one over a number to one and take it to that number, and I make that number larger and larger and larger, it's approaching this number. But what's even crazier about it is we'll see that this number, which you can view one way of it, it's coming out of this compound interest, that number, pi, the imaginary unit, which is defined as that imaginary unit squared is a negative one, that they all fit together in this magical, mystical way, and we'll see that again in future videos. But just for the sake of e, what you can imagine, what's happening here is going to our previous example of borrowing a dollar and trying to charge 100% over a year, that when our n was one, that means you're just charging over one period. When n is two, you're charging over two periods, and then compounding, or you're compounding over two periods."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "What I want to do in this video is get some practice or become familiar with what different angle measures in radians actually represent. And to get our familiarity, we're going to start with a ray that starts at the origin and moves along, and not moves, and points along the positive x-axis. So we're going to start with this magenta ray. And we're going to rotate it around the origin counterclockwise by different angle measures and think about what quadrant do we fall into if we start with this and we were to rotate by counterclockwise by 3 pi over 5 radians. And then if we start with this and we were to rotate counterclockwise by 2 pi over 7 radians. Or if we were to start with this and then rotate counterclockwise by 3 radians. So I encourage you to pause the video and think about starting with this, if we were to rotate counterclockwise by each of these, what quadrant are we going to end up in?"}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "And we're going to rotate it around the origin counterclockwise by different angle measures and think about what quadrant do we fall into if we start with this and we were to rotate by counterclockwise by 3 pi over 5 radians. And then if we start with this and we were to rotate counterclockwise by 2 pi over 7 radians. Or if we were to start with this and then rotate counterclockwise by 3 radians. So I encourage you to pause the video and think about starting with this, if we were to rotate counterclockwise by each of these, what quadrant are we going to end up in? So assume you have had, you paused the video and you tried it out on your own. So let's try this first one, 3 pi over 5. So 3 pi over 5, so we're going to start rotating."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and think about starting with this, if we were to rotate counterclockwise by each of these, what quadrant are we going to end up in? So assume you have had, you paused the video and you tried it out on your own. So let's try this first one, 3 pi over 5. So 3 pi over 5, so we're going to start rotating. So if we go straight up, if we rotate it essentially, if we want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us 2 pi over 2. So that would have been a counterclockwise rotation of pi over 2 radians. Now is 3 pi over 5 greater or less than that?"}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So 3 pi over 5, so we're going to start rotating. So if we go straight up, if we rotate it essentially, if we want to think in degrees, if you rotate it counterclockwise 90 degrees, that is going to get us 2 pi over 2. So that would have been a counterclockwise rotation of pi over 2 radians. Now is 3 pi over 5 greater or less than that? Well 3 pi over 5, 3 pi over 5 is greater than, or I guess another way to say it is, 3 pi over 6 is less than 3 pi over 5. You make the denominator smaller, making the fraction larger. So 3 pi over 6 is the same thing as pi over 2."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Now is 3 pi over 5 greater or less than that? Well 3 pi over 5, 3 pi over 5 is greater than, or I guess another way to say it is, 3 pi over 6 is less than 3 pi over 5. You make the denominator smaller, making the fraction larger. So 3 pi over 6 is the same thing as pi over 2. So let me write it this way. So 3 pi over 2 is less than 3 pi over 5. So it's definitely past this, so we're going to go past this."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So 3 pi over 6 is the same thing as pi over 2. So let me write it this way. So 3 pi over 2 is less than 3 pi over 5. So it's definitely past this, so we're going to go past this. Now does that get us all the way over here? If we were to go, essentially be pointed in the opposite direction, instead of being pointed to the right, making a full, I guess you could say, 180 degree counterclockwise rotation, that would be pi radians. That would be pi radians."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So it's definitely past this, so we're going to go past this. Now does that get us all the way over here? If we were to go, essentially be pointed in the opposite direction, instead of being pointed to the right, making a full, I guess you could say, 180 degree counterclockwise rotation, that would be pi radians. That would be pi radians. But this thing is less than pi. Pi would be 5 pi over 5. So this is less than pi radians."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "That would be pi radians. But this thing is less than pi. Pi would be 5 pi over 5. So this is less than pi radians. So we are going to sit, we are going to sit someplace, someplace, and I'm just estimating it, we are going to sit someplace like that. And so we are going to sit in the second quadrant. Now let's think about 2 pi over 7."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So this is less than pi radians. So we are going to sit, we are going to sit someplace, someplace, and I'm just estimating it, we are going to sit someplace like that. And so we are going to sit in the second quadrant. Now let's think about 2 pi over 7. So 2 pi over 7, do we even get past pi over 2? Well, pi over 2 here would be 3.5 pi over 7. So we don't even get to pi over 2."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about 2 pi over 7. So 2 pi over 7, do we even get past pi over 2? Well, pi over 2 here would be 3.5 pi over 7. So we don't even get to pi over 2. We're going to end up, we're going to end up someplace, someplace over here. This thing is greater than zero, so we're going to definitely start moving counterclockwise. But we're not even going to get to, this thing is less than pi over 2."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So we don't even get to pi over 2. We're going to end up, we're going to end up someplace, someplace over here. This thing is greater than zero, so we're going to definitely start moving counterclockwise. But we're not even going to get to, this thing is less than pi over 2. So this is going to throw us in the first quadrant. Now what about 3 radians? So one way to think about it is 3 is a little bit less than pi."}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "But we're not even going to get to, this thing is less than pi over 2. So this is going to throw us in the first quadrant. Now what about 3 radians? So one way to think about it is 3 is a little bit less than pi. Right? 3 is less than pi, but it's greater than pi over 2. How do we know that?"}, {"video_title": "Rotation by radians and quadrants Trigonometry Khan Academy.mp3", "Sentence": "So one way to think about it is 3 is a little bit less than pi. Right? 3 is less than pi, but it's greater than pi over 2. How do we know that? Well, pi is, pi is approximately 3.14159, and it just keeps going on and on and on forever. So 3 is definitely closer to that than it is to half of that. So it's going to be between pi over 2 and pi."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have a nice little equation here that has some rational expressions in it. And like always, pause the video and see if you can figure out which x's satisfy this equation. All right, let's work through it together. Now, when I see things in the denominator like this, my instinct is to try to not have denominators like this. And so what we could do is, to get rid of this x minus one in the denominator on the left-hand side, we can multiply both sides of the equation times x minus one. x minus one. So we're gonna multiply both sides by x minus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now, when I see things in the denominator like this, my instinct is to try to not have denominators like this. And so what we could do is, to get rid of this x minus one in the denominator on the left-hand side, we can multiply both sides of the equation times x minus one. x minus one. So we're gonna multiply both sides by x minus one. And once again, the whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then, to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one. So, x plus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we're gonna multiply both sides by x minus one. And once again, the whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then, to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one. So, x plus one. Multiply both sides times x plus one. And so, what is that going to give us? Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where, for the x's where that's defined, for x not being equal to one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So, x plus one. Multiply both sides times x plus one. And so, what is that going to give us? Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where, for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down. So we have x plus, I think I'm gonna need some space, so let me make sure I don't write too big."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where, for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down. So we have x plus, I think I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to, now, if we multiply both of these times three over x plus one, the x plus one's going to cancel with the x plus one and we're gonna be left with three times x minus one. So that is going to be three x minus three, three x minus three. And then, minus, minus one times both of these."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have x plus, I think I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to, now, if we multiply both of these times three over x plus one, the x plus one's going to cancel with the x plus one and we're gonna be left with three times x minus one. So that is going to be three x minus three, three x minus three. And then, minus, minus one times both of these. So one times x minus one times x plus one. So minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied it times each of these terms."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then, minus, minus one times both of these. So one times x minus one times x plus one. So minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied it times each of these terms. When I multiplied it times this first term, the x plus one and the x plus one cancelled, so I just had to multiply three times x minus one. And then for the second term, I just multiplied it times both of these. And now you might recognize this, if you have something x plus one times x minus one, that's going to be x squared minus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "All I did is I multiplied, took the x minus one times x plus one, multiplied it times each of these terms. When I multiplied it times this first term, the x plus one and the x plus one cancelled, so I just had to multiply three times x minus one. And then for the second term, I just multiplied it times both of these. And now you might recognize this, if you have something x plus one times x minus one, that's going to be x squared minus one. So I could rewrite all of this right over here as being equal to, as being equal to x squared minus one. And once again, that's because this is the same thing as x squared minus one. And since I'm subtracting an x squared minus one, actually let me just, I don't want to do too much on one step, so let's go to the next step."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And now you might recognize this, if you have something x plus one times x minus one, that's going to be x squared minus one. So I could rewrite all of this right over here as being equal to, as being equal to x squared minus one. And once again, that's because this is the same thing as x squared minus one. And since I'm subtracting an x squared minus one, actually let me just, I don't want to do too much on one step, so let's go to the next step. So I could multiply this out. So I could multiply x times negative two x, which would give us negative two x squared. X times four, which is going to give us plus four x."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And since I'm subtracting an x squared minus one, actually let me just, I don't want to do too much on one step, so let's go to the next step. So I could multiply this out. So I could multiply x times negative two x, which would give us negative two x squared. X times four, which is going to give us plus four x. And I could multiply one times negative two x, so I'm gonna subtract two x. And then one times four, which is going to be plus four. And then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "X times four, which is going to give us plus four x. And I could multiply one times negative two x, so I'm gonna subtract two x. And then one times four, which is going to be plus four. And then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one. And over here we can simplify it a little bit. This is going to be, that is four x minus two x is going to be, let me make sure I didn't write it over, yep, four x minus two x, so that would be two x. And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to, subtracting a two."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then that is going to be equal to, that is going to be equal to, we have three x minus three, and then we can distribute this negative sign, so we could say minus x squared plus one. And over here we can simplify it a little bit. This is going to be, that is four x minus two x is going to be, let me make sure I didn't write it over, yep, four x minus two x, so that would be two x. And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to, subtracting a two. So we can rewrite everything as, we'll do it in a neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x, plus three x minus two. Now we can try to get all of this business onto the right-hand side. So let's subtract it from both sides."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to, subtracting a two. So we can rewrite everything as, we'll do it in a neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x, plus three x minus two. Now we can try to get all of this business onto the right-hand side. So let's subtract it from both sides. So we'll add x squared to both sides, add x squared, that gets rid of this white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with, let's see, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's subtract it from both sides. So we'll add x squared to both sides, add x squared, that gets rid of this white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with, let's see, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to cancel with that, that, that is equal to zero. I don't like having this negative on the x squared, so let's multiply both sides times negative one. And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get, I'm going to get positive x squared plus x minus six is equal to zero."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to cancel with that, that, that is equal to zero. I don't like having this negative on the x squared, so let's multiply both sides times negative one. And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get, I'm going to get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this, and actually let me just do it right over here so that we can see the original problem. So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so if I do that, so if I just take the negative of both sides, so if I just multiply that times negative one, same thing as taking the negative of both sides, I'm going to get, I'm going to get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this, and actually let me just do it right over here so that we can see the original problem. So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term. Well, positive three and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that right? Yeah, three times negative two is negative six."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So if I were to factor this, what two numbers, their product is negative six, they're going to have different signs since their product is negative, and they add up to one, the coefficient on the first degree term. Well, positive three and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that right? Yeah, three times negative two is negative six. Three x minus two x is positive x. All right, so I just factored, I just wrote this in this quadratic and factored form. And so the way that you get this equaling zero is if either one of those equals zero."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Yeah, three times negative two is negative six. Three x minus two x is positive x. All right, so I just factored, I just wrote this in this quadratic and factored form. And so the way that you get this equaling zero is if either one of those equals zero. x plus three equals zero, or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides, you get, that's going to happen if x is equal to negative three or over here if you add two to both sides, x is equal to two. So either one of these will satisfy, but we want to be careful."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so the way that you get this equaling zero is if either one of those equals zero. x plus three equals zero, or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides, you get, that's going to happen if x is equal to negative three or over here if you add two to both sides, x is equal to two. So either one of these will satisfy, but we want to be careful. We want to make sure that our original equation isn't going to be undefined for either one of these. And negative three does not make either of the denominators equal to zero, so that's cool. And positive two does not make either of the denominators equal to zero, so it looks like we're in good shape."}, {"video_title": "Equations with rational expressions (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So either one of these will satisfy, but we want to be careful. We want to make sure that our original equation isn't going to be undefined for either one of these. And negative three does not make either of the denominators equal to zero, so that's cool. And positive two does not make either of the denominators equal to zero, so it looks like we're in good shape. There's two solutions to that equation. If one of them made any of the denominators equal to zero, then they would have been extraneous solutions. They would have been solutions for some of our intermediate steps, but not for the actual original equations with the expressions as they were written."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Here is his work. Is Jaden's work correct? If not, what is the first step where Jaden made a mistake? So pause this video and review Jaden's work and see if it's correct, or if it's not correct, tell me where it's not correct. All right, now let's work this together. So let's see, just to remind ourselves what Jaden's trying to do, he's trying to decide whether f of x is even, odd, or neither, and f of x is expressed, or it's defined, as x minus the cube root of x. So let's see, the first thing that Jaden did is he's trying to figure out what is f of negative x?"}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and review Jaden's work and see if it's correct, or if it's not correct, tell me where it's not correct. All right, now let's work this together. So let's see, just to remind ourselves what Jaden's trying to do, he's trying to decide whether f of x is even, odd, or neither, and f of x is expressed, or it's defined, as x minus the cube root of x. So let's see, the first thing that Jaden did is he's trying to figure out what is f of negative x? Because remember, if f of negative x is equal to f of x, we are even, and if f of negative x is equal to negative f of x, then we are odd. So it makes sense for him to find the expression for f of negative x. So he tries to evaluate f of negative x, and when he does that, everywhere where he sees an x in f of x, he replaces it with a negative x, so that seems good."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's see, the first thing that Jaden did is he's trying to figure out what is f of negative x? Because remember, if f of negative x is equal to f of x, we are even, and if f of negative x is equal to negative f of x, then we are odd. So it makes sense for him to find the expression for f of negative x. So he tries to evaluate f of negative x, and when he does that, everywhere where he sees an x in f of x, he replaces it with a negative x, so that seems good. And then let's see, this becomes a negative x, that makes sense, minus, and then a negative x under the radical, and this is a cube root right over here, that's the same thing as negative one times x. The cube root of negative one is negative one, so he takes that negative out of the radical, out of the cube root, so this makes sense. And so then he has a negative x, and you subtract a negative, you get a positive, so then that makes sense."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So he tries to evaluate f of negative x, and when he does that, everywhere where he sees an x in f of x, he replaces it with a negative x, so that seems good. And then let's see, this becomes a negative x, that makes sense, minus, and then a negative x under the radical, and this is a cube root right over here, that's the same thing as negative one times x. The cube root of negative one is negative one, so he takes that negative out of the radical, out of the cube root, so this makes sense. And so then he has a negative x, and you subtract a negative, you get a positive, so then that makes sense. And then the next thing he says, or he's trying to do, is check if f of negative x is equal to f of x, or f of negative x. He's gonna check whether this is equal to one of them. And so here, Jaden says negative x plus the cube root of x, so that's what f of negative x, what he evaluated it to be, isn't the same as f of x, and let's see, is that the case?"}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so then he has a negative x, and you subtract a negative, you get a positive, so then that makes sense. And then the next thing he says, or he's trying to do, is check if f of negative x is equal to f of x, or f of negative x. He's gonna check whether this is equal to one of them. And so here, Jaden says negative x plus the cube root of x, so that's what f of negative x, what he evaluated it to be, isn't the same as f of x, and let's see, is that the case? Is it not the same as f of x? Yep, it's definitely, it's not the same as f of x, or negative f of x, which is equal to negative x minus the cube root of x. Now that seems a little bit fishy."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so here, Jaden says negative x plus the cube root of x, so that's what f of negative x, what he evaluated it to be, isn't the same as f of x, and let's see, is that the case? Is it not the same as f of x? Yep, it's definitely, it's not the same as f of x, or negative f of x, which is equal to negative x minus the cube root of x. Now that seems a little bit fishy. Did he do the right thing right over here? Is negative f of x equal to negative x minus the cube root of x? See, negative of f of x is going to be a negative times this entire expression."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now that seems a little bit fishy. Did he do the right thing right over here? Is negative f of x equal to negative x minus the cube root of x? See, negative of f of x is going to be a negative times this entire expression. It's gonna be a negative out front times x minus the cube root of x, and so this is going to be equal to, you distribute the negative sign, you get negative x plus the cube root of x. So Jaden calculated the wrong negative f of x right over here, so he isn't right that negative x plus the cube root of x, it is actually the same as negative f of x, so he's wrong right over here. So Jaden's mistake is right over here, really it looks like he didn't evaluate negative f of x correctly."}, {"video_title": "Even and odd functions Find the mistake Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "See, negative of f of x is going to be a negative times this entire expression. It's gonna be a negative out front times x minus the cube root of x, and so this is going to be equal to, you distribute the negative sign, you get negative x plus the cube root of x. So Jaden calculated the wrong negative f of x right over here, so he isn't right that negative x plus the cube root of x, it is actually the same as negative f of x, so he's wrong right over here. So Jaden's mistake is right over here, really it looks like he didn't evaluate negative f of x correctly. So Jaden's work, is Jaden's work correct? No, if not, what is the first step where Jaden made a mistake well, it would be step two. What he should have said is, it actually is the same as negative f of x, and so therefore his conclusion should be that f of x is odd."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can reason through that a little bit. All right, now let's work through this together and really all we're going to do is use properties of multiplication and use properties of exponents to essentially rewrite this expression. So we can just view this, if we're just multiplying a bunch of things, it doesn't matter what order we multiply them in. So you could just view this as five times x squared times three times x to the fifth or we could multiply our five and three first. So you could view this as five times three, times three, times x squared, times x squared, times x to the fifth, times x to the fifth. And now what is five times three? I think you know that, that is 15."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "So you could just view this as five times x squared times three times x to the fifth or we could multiply our five and three first. So you could view this as five times three, times three, times x squared, times x squared, times x to the fifth, times x to the fifth. And now what is five times three? I think you know that, that is 15. Now what is x squared times x to the fifth? Now some of you might recognize that exponent properties would come into play here. If I'm multiplying two things like this, we have the same base and different exponents, that this is going to be equal to x to the and we add these two exponents, x to the two plus five power or x to the seventh power."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "I think you know that, that is 15. Now what is x squared times x to the fifth? Now some of you might recognize that exponent properties would come into play here. If I'm multiplying two things like this, we have the same base and different exponents, that this is going to be equal to x to the and we add these two exponents, x to the two plus five power or x to the seventh power. If what I just did seems counterintuitive to you, I'll just remind you. What is x squared? X squared is x times x and what is x to the fifth?"}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "If I'm multiplying two things like this, we have the same base and different exponents, that this is going to be equal to x to the and we add these two exponents, x to the two plus five power or x to the seventh power. If what I just did seems counterintuitive to you, I'll just remind you. What is x squared? X squared is x times x and what is x to the fifth? That is x times x times x times x times x. And if you multiply them all together, what do you get? Well you got seven x's and you're multiplying them all together, that is x to the seventh."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "X squared is x times x and what is x to the fifth? That is x times x times x times x times x. And if you multiply them all together, what do you get? Well you got seven x's and you're multiplying them all together, that is x to the seventh. And so there you have it, five x squared times three x to the fifth is 15 x to the seventh power. So the key is, is look at these coefficients, look at these numbers, the five and the three, multiply those and then for any variable you have, if you have x here, so you have a common base, then you can add those exponents. And what we just did is known as multiplying monomials, which sounds very fancy, but this is a monomial, monomial."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "Well you got seven x's and you're multiplying them all together, that is x to the seventh. And so there you have it, five x squared times three x to the fifth is 15 x to the seventh power. So the key is, is look at these coefficients, look at these numbers, the five and the three, multiply those and then for any variable you have, if you have x here, so you have a common base, then you can add those exponents. And what we just did is known as multiplying monomials, which sounds very fancy, but this is a monomial, monomial. And in the future we'll do multiplying things like polynomials, where we have multiple of these things added together. But that's all it is, multiplying monomials. Let's do one more example and let's use a different variable this time, just to get some variety in there."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "And what we just did is known as multiplying monomials, which sounds very fancy, but this is a monomial, monomial. And in the future we'll do multiplying things like polynomials, where we have multiple of these things added together. But that's all it is, multiplying monomials. Let's do one more example and let's use a different variable this time, just to get some variety in there. Let's say we wanna multiply the monomial, three t to the seventh power, times another monomial, negative four t. Pause this video and see if you can work through that. All right, so I'm gonna do this one a little bit faster. I am going to look at the three and the negative four, and I'm gonna multiply those first, and I'm going to get a negative 12."}, {"video_title": "Multiplying monomials Polynomial arithmetic Algebra 2 Khan Academy.mp3", "Sentence": "Let's do one more example and let's use a different variable this time, just to get some variety in there. Let's say we wanna multiply the monomial, three t to the seventh power, times another monomial, negative four t. Pause this video and see if you can work through that. All right, so I'm gonna do this one a little bit faster. I am going to look at the three and the negative four, and I'm gonna multiply those first, and I'm going to get a negative 12. And then if I were to want to multiply the t to the seventh times t, once again, they're both, the variable t is our base, so that's going to be t to the seventh times t to the first power, that's what t is, that's going to be t to the seven plus one power, or t to the eighth. But there you go, we are done again. We've just multiplied another set of monomials."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Well, a good way to figure out if things are equivalent is to just try to get them all in the same form. So the seventh root of v to the third power, the seventh root of something is the same thing as raising it to the 1 7th power. So this is equivalent to v to the third power raised to the 1 7th power. And if I raise something to an exponent and then raise that to an exponent, well, then that's the same thing as raising it to the product of these two exponents. So this is going to be the same thing as v to the 3 times 1 7th power, which of course is 3 7ths. So we've written it in multiple forms now. Let's see which of these match."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And if I raise something to an exponent and then raise that to an exponent, well, then that's the same thing as raising it to the product of these two exponents. So this is going to be the same thing as v to the 3 times 1 7th power, which of course is 3 7ths. So we've written it in multiple forms now. Let's see which of these match. So v to the third to the 1 7th power, well, that was a form that we have right over here. So that is equivalent. v to the 3 7ths, that's what we have right over here."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's see which of these match. So v to the third to the 1 7th power, well, that was a form that we have right over here. So that is equivalent. v to the 3 7ths, that's what we have right over here. So that one is definitely equivalent. Now let's think about this one. This is the cube root of v to the seventh."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "v to the 3 7ths, that's what we have right over here. So that one is definitely equivalent. Now let's think about this one. This is the cube root of v to the seventh. Is this going to be equivalent? Well, one way to think about it, this is going to be the same thing as v to the 1 3rd power. Actually, no, this wasn't the cube root of v to the 7th."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is the cube root of v to the seventh. Is this going to be equivalent? Well, one way to think about it, this is going to be the same thing as v to the 1 3rd power. Actually, no, this wasn't the cube root of v to the 7th. This was the cube root of v and that to the 7th power. So that's the same thing as v to the 1 3rd power and then that to the 7th power. So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Actually, no, this wasn't the cube root of v to the 7th. This was the cube root of v and that to the 7th power. So that's the same thing as v to the 1 3rd power and then that to the 7th power. So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power. So this is not going to be equivalent for all v's for which this expression is defined. Let's do a few more of these or similar types of problems dealing with roots and fractional exponents. The following equation is true for g greater than or equal to 0 and d is a constant."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So that is the same thing as v to the 7 3rd power, which is clearly different to v to the 3 7ths power. So this is not going to be equivalent for all v's for which this expression is defined. Let's do a few more of these or similar types of problems dealing with roots and fractional exponents. The following equation is true for g greater than or equal to 0 and d is a constant. What is the value of d? Well, if I'm taking the 6th root of something, that's the same thing as raising it to the 1 6th power. So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "The following equation is true for g greater than or equal to 0 and d is a constant. What is the value of d? Well, if I'm taking the 6th root of something, that's the same thing as raising it to the 1 6th power. So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power. And just like we just saw in the last example, that's the same thing as g to the 5 times 1 6th power. This is just our exponent properties. I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So the 6th root of g to the 5th is the same thing as g to the 5th raised to the 1 6th power. And just like we just saw in the last example, that's the same thing as g to the 5 times 1 6th power. This is just our exponent properties. I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents. So that's the same thing as g to the 5 6th power. And so d is 5 6th, 5 over 6. The 6th root of g to the 5th is the same thing as g to the 5 6th power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "I raise something to an exponent and then raise that whole thing to another exponent, I can just multiply the exponents. So that's the same thing as g to the 5 6th power. And so d is 5 6th, 5 over 6. The 6th root of g to the 5th is the same thing as g to the 5 6th power. Let's do one more of these. The following equation is true for x greater than 0 and d is a constant. What is the value of d?"}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "The 6th root of g to the 5th is the same thing as g to the 5 6th power. Let's do one more of these. The following equation is true for x greater than 0 and d is a constant. What is the value of d? All right, this is interesting. And I forgot to tell you in the last one, but pause this video as well and see if you can work it out. Pause for this question as well and see if you can work it out."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "What is the value of d? All right, this is interesting. And I forgot to tell you in the last one, but pause this video as well and see if you can work it out. Pause for this question as well and see if you can work it out. Well, here, let's just start rewriting the root as an exponent. So I can rewrite the whole thing. This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Pause for this question as well and see if you can work it out. Well, here, let's just start rewriting the root as an exponent. So I can rewrite the whole thing. This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power. So that is the same thing as x to the negative 1 7th power. And so that is going to be equal to x to the d. And so d must be equal to negative 1 7th. So the key here is when you're taking the reciprocal of something, that's the same thing as raising it to the negative of that exponent."}, {"video_title": "Rewriting roots as rational exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "This is the same thing as 1 over, instead of writing the 7th root of x, I'll write x to the 1 7th power is equal to x to the d. And if I have 1 over something to a power, that's the same thing as that something raised to the negative of that power. So that is the same thing as x to the negative 1 7th power. And so that is going to be equal to x to the d. And so d must be equal to negative 1 7th. So the key here is when you're taking the reciprocal of something, that's the same thing as raising it to the negative of that exponent. Another way of thinking about it is you could view this as x to the 1 7th to the negative 1 power. And then if you multiply these exponents, you get what we have right over there. But either way, d is equal to negative 1 7th."}, {"video_title": "Finding the greatest common factor of two monomials Algebra I Khan Academy.mp3", "Sentence": "Now we have to be a little bit careful when we talk about greatest in the context of algebraic expressions like this, because it's greatest from the point of view that it includes the most factors of each of these monomials, it's not necessarily the greatest possible number because maybe some of these variables could take on negative values, maybe they're taking on values less than 1, so if you square it, it's actually going to become a smaller number. But I think without getting too much into the weeds there, I think if we just kind of run through the process of it, you'll understand it a little bit better. So to find the greatest common factor, let's just essentially break down each of these numbers into what we could call their prime factorization, but it's kind of a combination of the prime factorization of the numeric parts of the number, plus essentially the factorization of the variable parts. So if we wanted to write 10, we could say, or if we were to write 10cd squared, we can rewrite that as a product of the prime factors of 10, the prime factorization of 10 is just 2 times 5, those are both prime numbers, so 10 can be broken down as 2 times 5, c can only be broken down by c, we don't know anything else that c can be broken into, so c, so 2 times 5 times c, but then the d squared can be rewritten as d times d. This is what I mean by writing this monomial essentially as the product of its constituents, for the numeric part of it, the constituents are the prime factors, and for the rest of it, we're just kind of expanding out the exponents. Now let's do that for 25c to the third d squared. So 25 right here, that's 5 times 5, so this is equal to 5 times 5, and then c to the third, that's times c, times c, times c, and then d squared, times d squared, d squared is times d, times d. So what's their greatest common factor in this context? Well, they both have at least one 5, then they both have at least one c over here, so let's just take up one of the c's right over there, and then they both have two d's."}, {"video_title": "Finding the greatest common factor of two monomials Algebra I Khan Academy.mp3", "Sentence": "So if we wanted to write 10, we could say, or if we were to write 10cd squared, we can rewrite that as a product of the prime factors of 10, the prime factorization of 10 is just 2 times 5, those are both prime numbers, so 10 can be broken down as 2 times 5, c can only be broken down by c, we don't know anything else that c can be broken into, so c, so 2 times 5 times c, but then the d squared can be rewritten as d times d. This is what I mean by writing this monomial essentially as the product of its constituents, for the numeric part of it, the constituents are the prime factors, and for the rest of it, we're just kind of expanding out the exponents. Now let's do that for 25c to the third d squared. So 25 right here, that's 5 times 5, so this is equal to 5 times 5, and then c to the third, that's times c, times c, times c, and then d squared, times d squared, d squared is times d, times d. So what's their greatest common factor in this context? Well, they both have at least one 5, then they both have at least one c over here, so let's just take up one of the c's right over there, and then they both have two d's. So the greatest common factor in this context, the greatest common factor of these two monomials, is going to be the factors that they have in common, so it's going to be equal to this 5 times, we only have one c in common, times, and we have two d's in common, times d, times d, so this is equal to 5cd squared. And so 5d squared, we can kind of view it as the greatest, but I'll put that in quotes, depending on whether c is negative or positive, and d is greater than or less than zero, but this is the greatest common factor of these two monomials. It's divisible into both of them, and it uses the most factors possible."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "That we can actually put input negative numbers in the domain of this function, that we can actually get imaginary or complex results. So we can rewrite negative 52 as negative 1 times 52. So this can be rewritten as the principal square root of negative 1 times 52. And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times the principal square root of negative 1, times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then, if we assume that this is the principal branch of the complex square root function, we can rewrite this. This is going to be equal to the square root of negative 1 times the principal square root of negative 1, times the principal square root of 52. Now, I want to be very, very clear here. You can do what we just did. If we have the principal square root of the product of two things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You can do what we just did. If we have the principal square root of the product of two things, we can rewrite that as the principal square root of each, and then we take the product. But you can only do this if either both of these numbers are positive, or only one of them is negative. You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. You could do this so far."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You cannot do this if both of these were negative. For example, you could not do this. You could not say the principal square root of 52 is equal to negative 1 times negative 52. You could do this so far. I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You could do this so far. I haven't said anything wrong. 52 is definitely negative 1 times negative 52. But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not okay."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But then, since these are both negative, you cannot then say that this is equal to the square root of negative 1 times the square root of negative 52. In fact, I invite you to continue on this train of reasoning. You're going to get a nonsensical answer. This is not okay. You cannot do this right over here. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is not okay. You cannot do this right over here. And the reason why you cannot do this is that this property does not work when both of these numbers are negative. Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, with that said, we can do it if only one of them are negative, or both of them are positive, obviously. Now, the principal square root of negative 1, if we're talking about the principal branch of the complex square root function, is i. So this right over here does simplify to i. And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So let me, so 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then let's think if we can simplify the square root of 52 any. And to do that, we can think about its prime factorization, see if we have any perfect squares sitting in there. So let me, so 52 is 2 times 26, and 26 is 2 times 13. So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to, this is equal to, well we have our i now, the principal square root of negative 1 is i. The other square root of negative 1 is negative i. But the principal square root of negative 1 is i."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we have 2 times 2 there, or 4 there, which is a perfect square. So we can rewrite this as equal to, this is equal to, well we have our i now, the principal square root of negative 1 is i. The other square root of negative 1 is negative i. But the principal square root of negative 1 is i. And then we're going to multiply that times the square root of 4 times 13. And this is equal to, and this is going to be equal to i times the square root of 4, i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But the principal square root of negative 1 is i. And then we're going to multiply that times the square root of 4 times 13. And this is equal to, and this is going to be equal to i times the square root of 4, i times the square root of 4, or the principal square root of 4 times the principal square root of 13. The principal square root of 4 is 2. So this all simplifies, and we can switch the order over here. This is equal to 2 times the square root of 13, 2 times the principal square root of 13 I should say, times i. And I just switched around the order, it makes it a little bit easier to read if I put the i after the numbers over here."}, {"video_title": "Imaginary roots of negative numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "The principal square root of 4 is 2. So this all simplifies, and we can switch the order over here. This is equal to 2 times the square root of 13, 2 times the principal square root of 13 I should say, times i. And I just switched around the order, it makes it a little bit easier to read if I put the i after the numbers over here. But I'm just multiplying i times 2 times the square root of 13. That's the same thing as multiplying 2 times the principal square root of 13 times i. And I think this is about as simplified as we can get here."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can try this before we work through it together. All right, now let's work through this together. Now, the first thing that we might want to do, there's several ways that you could approach this, but the one thing I like to do is get rid of this x here in the denominator, and the easiest way I can think of doing that is by multiplying both sides of this equation by nine minus x. Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Now, when you do that, it's important that you then put the qualifier that the x cannot be equal to the value that would have made this denominator zero, because clearly if somehow you do all this algebraic manipulation and you got x is equal to nine, that still wouldn't be a valid solution, because if you were to substitute nine back into the original equation, you would be dividing by zero in the denominator. So let's just put that right over here. X cannot be equal to nine. And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And so then we can safely move ahead with our algebraic manipulations. So on the left-hand side, as long as x does not equal nine, if we multiply and divide by nine minus x, they cancel out, and we'll just be left with an x plus one. And on the right-hand side, if you multiply 2 3rds times nine minus x, we get 2 3rds times nine is six, and then 2 3rds times negative x is negative 2 3rds x. And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "And once again, let's remind ourselves that x cannot be equal to nine. And then we can get all of our x's on the same side. So let's put that on the left. So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "So let's add 2 3rds x to both sides. So plus 2 3rds, 2 3rds x, plus 2 3rds x. And then what do we have? Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three."}, {"video_title": "Rational equations intro Algebra 2 Khan Academy.mp3", "Sentence": "Well, on the left-hand side, we have one x, which is the same thing as 3 3rds x plus 2 3rds x is going to give us 5 3rds x plus one is equal to six, and then these characters cancel out. And then we can just subtract one from both sides, and we get 5 3rds x, 5 3rds x is equal to five. And then last but not least, we can multiply both sides of this equation times the reciprocal of 5 3rds, which is, of course, 3 5ths, and I'm doing that, so I just have an x isolated on the left-hand side. So times 3 5ths, and we are left with 3 5ths times 5 3rds is, of course, equal to one, so we're left with x is equal to five times 3 5ths is three. And so we're feeling pretty good about x equals three, but we have to make sure that that's consistent with our original expression. Well, if we look up here, if you substitute back x equals three, you don't get a zero in the denominator. X is not equal to nine, x equals three is consistent with that, so we should feel good about our solution."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "She models the relationship between the temperature P of the pizza, this seems like it's going to be interesting, the temperature P of the pizza, in degrees Celsius and time T since she took it out of the freezer in minutes as P is equal to 20 minus 25 times 0.8 to the T power. So that's how she's modeling her temperature of the pizza, P as a function of time. She wants to graph the relationship over the first 25 minutes. So what we're going to do here is not so much focus on the graph itself, although we will look at that. I'm actually just going to use a graphing calculator in order to have access to the graph. But I want to look at the graph in the context of what we are trying to model and carefully think about what should be the labels for the axes, what parts of the graph are interesting. So this right over here is this function graphed on Desmos."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "So what we're going to do here is not so much focus on the graph itself, although we will look at that. I'm actually just going to use a graphing calculator in order to have access to the graph. But I want to look at the graph in the context of what we are trying to model and carefully think about what should be the labels for the axes, what parts of the graph are interesting. So this right over here is this function graphed on Desmos. You can see, I typed it in right over here, P is equal to 20 minus 25 times 0.8 to the T power, exactly what we had down here. Now, remember, this is modeling the temperature of our pizza as a function of time. So to help us remember that, let's put in some labels for our axes."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "So this right over here is this function graphed on Desmos. You can see, I typed it in right over here, P is equal to 20 minus 25 times 0.8 to the T power, exactly what we had down here. Now, remember, this is modeling the temperature of our pizza as a function of time. So to help us remember that, let's put in some labels for our axes. So the graph settings, if I go down here, our x-axis. Now, our x-axis is really the T-axis. That's our independent variable over here."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "So to help us remember that, let's put in some labels for our axes. So the graph settings, if I go down here, our x-axis. Now, our x-axis is really the T-axis. That's our independent variable over here. And what is it measuring? Well, it says it right over here. It's measuring time T in minutes."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "That's our independent variable over here. And what is it measuring? Well, it says it right over here. It's measuring time T in minutes. So we could write it like this, T, which is measured in minutes. And then what about our y-axis? Well, this is really our P-axis, and that's measuring degrees Celsius."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "It's measuring time T in minutes. So we could write it like this, T, which is measured in minutes. And then what about our y-axis? Well, this is really our P-axis, and that's measuring degrees Celsius. So that's our P-axis, and it's measuring degrees Celsius. All right, so let's just look at what our graph looks like so far. So there we have it."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "Well, this is really our P-axis, and that's measuring degrees Celsius. So that's our P-axis, and it's measuring degrees Celsius. All right, so let's just look at what our graph looks like so far. So there we have it. We've put in our axes, and we have already typed this part in, so I can make it so I can focus on the graph itself. Now, are we done? Is this all we need to really think about?"}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "So there we have it. We've put in our axes, and we have already typed this part in, so I can make it so I can focus on the graph itself. Now, are we done? Is this all we need to really think about? Well, the next part to think about is the domain and what part of the y-axis, what part of the range are we really interested in? Well, the first thing to realize is we're modeling something as a function of time. And so we really shouldn't be having negative time here, and we wanna think about the relationship over the first 25 minutes."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "Is this all we need to really think about? Well, the next part to think about is the domain and what part of the y-axis, what part of the range are we really interested in? Well, the first thing to realize is we're modeling something as a function of time. And so we really shouldn't be having negative time here, and we wanna think about the relationship over the first 25 minutes. So let's go back here. And when we look at the range of x-values that we care about, and really that we could think about the part of the domain that we care about, we wanna restrict to x being greater than or equal to zero. And obviously, in this situation, x is really t. And then we can also think about it as less than 25."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "And so we really shouldn't be having negative time here, and we wanna think about the relationship over the first 25 minutes. So let's go back here. And when we look at the range of x-values that we care about, and really that we could think about the part of the domain that we care about, we wanna restrict to x being greater than or equal to zero. And obviously, in this situation, x is really t. And then we can also think about it as less than 25. We don't have to restrict the upper bound, but these are the first 25 minutes that she cares about. So let's do it like that. And now let's look at our graph."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "And obviously, in this situation, x is really t. And then we can also think about it as less than 25. We don't have to restrict the upper bound, but these are the first 25 minutes that she cares about. So let's do it like that. And now let's look at our graph. And the important things to appreciate is that we have the axes, we can see them. And so at time t is equal to zero, we see that we actually have a negative temperature in degrees Celsius, and that makes sense. It came out of the freezer, so it's below freezing."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "And now let's look at our graph. And the important things to appreciate is that we have the axes, we can see them. And so at time t is equal to zero, we see that we actually have a negative temperature in degrees Celsius, and that makes sense. It came out of the freezer, so it's below freezing. And then we see that the pizza's warming up as it gets closer and closer to room temperature, which over here, it looks like it's pretty close to 20 degrees Celsius. And so now it looks like we have been able to graph what Chloe is trying to look at. It looks like we have modeled it well, we have labeled it accordingly, and we have set the ranges of x values and the ranges of y values that we'd wanna look at."}, {"video_title": "Graph labels and scales Modeling Algebra II Khan Academy.mp3", "Sentence": "It came out of the freezer, so it's below freezing. And then we see that the pizza's warming up as it gets closer and closer to room temperature, which over here, it looks like it's pretty close to 20 degrees Celsius. And so now it looks like we have been able to graph what Chloe is trying to look at. It looks like we have modeled it well, we have labeled it accordingly, and we have set the ranges of x values and the ranges of y values that we'd wanna look at. The y values, we just wanna make sure that over the range of x values, and it's really a subset of the domain, not to confuse the term range too much, the subset of the domain of the x values that we care about, that we can see the corresponding y values, and we very clearly can see them. And we're essentially done. We've thought about how to best look at the graph of this model."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. So plus 2x squared minus 7x plus 13x. Now let's just look at the different terms that have different degrees of x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. And then let's look at the x terms here. So we have an 8x right there."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 8x minus 7x. Oh, sorry, you're going to have 14x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. Plus 14x. That's these three terms."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. But it's a constant term. It's not being multiplied by x."}, {"video_title": "Example 2 Adding polynomials Algebra I Khan Academy.mp3", "Sentence": "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. So minus 3. And we've simplified it as far as we can go."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say that the log base x of a is equal to b. Right? That's the same thing as saying, that is the exact same thing as saying that x to the b is equal to a. Fair enough. So what I want to do is experiment. What happens if I multiply this expression by another variable, let's call it c. Right? So I'm going to multiply both sides of this equation times c. I'm going to switch colors just to keep things interesting."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Fair enough. So what I want to do is experiment. What happens if I multiply this expression by another variable, let's call it c. Right? So I'm going to multiply both sides of this equation times c. I'm going to switch colors just to keep things interesting. That's not an x, that's a c. I should probably just do a dot instead. Times c. Right? So I'm going to multiply both sides of this equation times c. So I get c times log base x of a is equal to, multiply both sides of the equation, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm going to multiply both sides of this equation times c. I'm going to switch colors just to keep things interesting. That's not an x, that's a c. I should probably just do a dot instead. Times c. Right? So I'm going to multiply both sides of this equation times c. So I get c times log base x of a is equal to, multiply both sides of the equation, right? Is equal to b times c. Fair enough. I think you realize I have not done anything profound just yet. Let's go back."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm going to multiply both sides of this equation times c. So I get c times log base x of a is equal to, multiply both sides of the equation, right? Is equal to b times c. Fair enough. I think you realize I have not done anything profound just yet. Let's go back. We said that this is the same thing as this. So let's experiment with something. Let's raise this side to the power c. So I'm going to raise this side to the power c. That's a kind of a caret."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's go back. We said that this is the same thing as this. So let's experiment with something. Let's raise this side to the power c. So I'm going to raise this side to the power c. That's a kind of a caret. And when you type exponents, that's what you'd use, a caret, right? So I'm going to raise it to the power c. So then this side is x to the b to the c power is equal to a to the c, right? All I did is I raised both sides of this equation to the c-th power."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's raise this side to the power c. So I'm going to raise this side to the power c. That's a kind of a caret. And when you type exponents, that's what you'd use, a caret, right? So I'm going to raise it to the power c. So then this side is x to the b to the c power is equal to a to the c, right? All I did is I raised both sides of this equation to the c-th power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent? What happens to the exponents? Well, that's just an exponent rule and you just multiply those two exponents."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "All I did is I raised both sides of this equation to the c-th power. And what do we know about when you raise something to an exponent and you raise that whole thing to another exponent? What happens to the exponents? Well, that's just an exponent rule and you just multiply those two exponents. So this just becomes, this just implies that x to the bc is equal to a to the c. What can we do now? Well, I don't know. Let's take the logarithm of both sides."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that's just an exponent rule and you just multiply those two exponents. So this just becomes, this just implies that x to the bc is equal to a to the c. What can we do now? Well, I don't know. Let's take the logarithm of both sides. Or let's just write this. Let's not take the logarithm of both sides. Let's write this as a logarithm expression."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's take the logarithm of both sides. Or let's just write this. Let's not take the logarithm of both sides. Let's write this as a logarithm expression. We know that x to the bc is equal to a to the c. Well, that's the exact same thing as saying that the logarithm base x of a to the c is equal to bc, correct? Because all I did is I rewrote this as a logarithm expression. And I think now you realize that something interesting has happened."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's write this as a logarithm expression. We know that x to the bc is equal to a to the c. Well, that's the exact same thing as saying that the logarithm base x of a to the c is equal to bc, correct? Because all I did is I rewrote this as a logarithm expression. And I think now you realize that something interesting has happened. That bc, this bc, well, of course it's the same thing as this bc, right? So this expression must be equal to this expression. And I think we have another logarithm property."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I think now you realize that something interesting has happened. That bc, this bc, well, of course it's the same thing as this bc, right? So this expression must be equal to this expression. And I think we have another logarithm property. That if I have some kind of coefficient in front of the logarithm, I'm multiplying the logarithm. So if I have c clog base x of a. But that's c times the logarithm base x of a."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I think we have another logarithm property. That if I have some kind of coefficient in front of the logarithm, I'm multiplying the logarithm. So if I have c clog base x of a. But that's c times the logarithm base x of a. That equals the log base x of a to the c. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property. So let's review what we know so far about logarithms."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But that's c times the logarithm base x of a. That equals the log base x of a to the c. So you could take this coefficient and instead make it an exponent on the term inside the logarithm. That is another logarithm property. So let's review what we know so far about logarithms. We know that if I write, let me use a different, let me say, if I write, well, let me just, the letters I've been using, c times logarithm base x of a is equal to logarithm base x of a to the c. We know that. And we know, we just learned, that logarithm base x of a plus logarithm base x of b is equal to the logarithm base x of a times b. Now let me ask you a question."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's review what we know so far about logarithms. We know that if I write, let me use a different, let me say, if I write, well, let me just, the letters I've been using, c times logarithm base x of a is equal to logarithm base x of a to the c. We know that. And we know, we just learned, that logarithm base x of a plus logarithm base x of b is equal to the logarithm base x of a times b. Now let me ask you a question. What happens if instead of a positive sign here, we put a negative sign? Well, you could probably figure it out yourself, but we could do that same exact proof that we did in the beginning, but in this time we'll set it up with a negative. So if I said that, let's just say that log base x of a is equal to l. Let's say that log base x of b is equal to m. And let's say that log base x of a divided by b is equal to n, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now let me ask you a question. What happens if instead of a positive sign here, we put a negative sign? Well, you could probably figure it out yourself, but we could do that same exact proof that we did in the beginning, but in this time we'll set it up with a negative. So if I said that, let's just say that log base x of a is equal to l. Let's say that log base x of b is equal to m. And let's say that log base x of a divided by b is equal to n, right? How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to a. Let me switch colors."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if I said that, let's just say that log base x of a is equal to l. Let's say that log base x of b is equal to m. And let's say that log base x of a divided by b is equal to n, right? How can we write all of these expressions as exponents? Well, this just says that x to the l is equal to a. Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to b. And this is just saying that x to the n is equal to a over b."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me switch colors. That keeps it interesting. This is just saying that x to the m is equal to b. And this is just saying that x to the n is equal to a over b. So what can we do here? Well, what's another way of writing a over b? Well, that's just the same thing as writing x to the l, because that's a, x to the l over x to the m. That's b, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this is just saying that x to the n is equal to a over b. So what can we do here? Well, what's another way of writing a over b? Well, that's just the same thing as writing x to the l, because that's a, x to the l over x to the m. That's b, right? And this we know from our exponent rules, right? This could also be written as x to the l, x to the negative m, right? Or that also equals x to the l minus m. So what do we know?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that's just the same thing as writing x to the l, because that's a, x to the l over x to the m. That's b, right? And this we know from our exponent rules, right? This could also be written as x to the l, x to the negative m, right? Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Right? x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or that also equals x to the l minus m. So what do we know? We know that x to the n is equal to x to the l minus m. Right? x to the n is equal to x to the l minus m. Those equal each other. I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n? I'm going to do it up here, because I think we have stumbled upon another logarithm rule."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "I just made a big equal chain here. So we know that n is equal to l minus m. Well, what does that do for us? Well, what's another way of writing n? I'm going to do it up here, because I think we have stumbled upon another logarithm rule. What's another way of writing n? I did it right here. This is another way of writing n, right?"}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "I'm going to do it up here, because I think we have stumbled upon another logarithm rule. What's another way of writing n? I did it right here. This is another way of writing n, right? So logarithm base x of a over b, this is an x over here, is equal to l. l is this right here. Log base x of a is equal to l. The log base x of a minus m. I wrote m right here. That's log base x of b."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is another way of writing n, right? So logarithm base x of a over b, this is an x over here, is equal to l. l is this right here. Log base x of a is equal to l. The log base x of a minus m. I wrote m right here. That's log base x of b. There you go. I probably didn't have to prove it. You could have probably tried it out with dividing it, putting it in your number, whatever."}, {"video_title": "Proof a log b = log(b^a), log a - log b = log(a b) Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's log base x of b. There you go. I probably didn't have to prove it. You could have probably tried it out with dividing it, putting it in your number, whatever. But you now are hopefully satisfied that we have this new logarithm property right there. I have one more logarithm property to show you, but I don't think I have time to show it in this video. So I will do it in the next video."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And the reason why they say interactive graph, this is a screenshot from the exercise on Khan Academy where you could click and place the zeros. But the key here is let's figure out what X values make P of X equal to zero. Those are the zeros, and then we can plot them. So pause this video and see if you can figure that out. So the key here is to try to factor this expression right over here, this third degree expression, because really we're trying to solve the Xs for which five X to the third plus five X squared minus 30 X is equal to zero. And the way we do that is by factoring this left-hand expression. So the first thing I always look for is a common factor across all of the terms."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure that out. So the key here is to try to factor this expression right over here, this third degree expression, because really we're trying to solve the Xs for which five X to the third plus five X squared minus 30 X is equal to zero. And the way we do that is by factoring this left-hand expression. So the first thing I always look for is a common factor across all of the terms. And it looks like all of the terms are divisible by five X. So let's factor out a five X. So this is going to be five X times, if we take a five X out of five X to the third, we're left with an X squared."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So the first thing I always look for is a common factor across all of the terms. And it looks like all of the terms are divisible by five X. So let's factor out a five X. So this is going to be five X times, if we take a five X out of five X to the third, we're left with an X squared. If we take out a five X out of five X squared, we're left with an X, so plus X. And if we take out a five X of negative 30 X, we're left with a negative six is equal to zero. And now we have five X times this second degree, the second degree expression."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So this is going to be five X times, if we take a five X out of five X to the third, we're left with an X squared. If we take out a five X out of five X squared, we're left with an X, so plus X. And if we take out a five X of negative 30 X, we're left with a negative six is equal to zero. And now we have five X times this second degree, the second degree expression. And to factor that, let's see, what two numbers add up to one? You could view this as a one X here. And their product is equal to negative six."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And now we have five X times this second degree, the second degree expression. And to factor that, let's see, what two numbers add up to one? You could view this as a one X here. And their product is equal to negative six. And let's see, positive three and negative two would do the trick. So I can rewrite this as five X times. So X plus three, X plus three times X minus two."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And their product is equal to negative six. And let's see, positive three and negative two would do the trick. So I can rewrite this as five X times. So X plus three, X plus three times X minus two. And if what I just did looks unfamiliar, I encourage you to review factoring quadratics on Khan Academy. And that is all going to be equal to zero. And so if I try to figure out what X values are going to make this whole expression zero, it could be the X values or the X value that makes five X equal zero."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So X plus three, X plus three times X minus two. And if what I just did looks unfamiliar, I encourage you to review factoring quadratics on Khan Academy. And that is all going to be equal to zero. And so if I try to figure out what X values are going to make this whole expression zero, it could be the X values or the X value that makes five X equal zero. Because if five X is zero, zero times anything else is gonna be zero. So what makes five X equal zero? Well, if we divide both sides by five, you're going to get X is equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And so if I try to figure out what X values are going to make this whole expression zero, it could be the X values or the X value that makes five X equal zero. Because if five X is zero, zero times anything else is gonna be zero. So what makes five X equal zero? Well, if we divide both sides by five, you're going to get X is equal to zero. And it is the case. If X equals zero, this becomes zero. And then doesn't matter what these are, zero times anything is zero."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, if we divide both sides by five, you're going to get X is equal to zero. And it is the case. If X equals zero, this becomes zero. And then doesn't matter what these are, zero times anything is zero. The other possible X value that would make everything zero is the X value that makes X plus three equal to zero. Subtract three from both sides, you get X is equal to negative three. And then the other X value is the X value that makes X minus two equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then doesn't matter what these are, zero times anything is zero. The other possible X value that would make everything zero is the X value that makes X plus three equal to zero. Subtract three from both sides, you get X is equal to negative three. And then the other X value is the X value that makes X minus two equal to zero. Add two to both sides, that's going to be X equals two. So there you have it. We have identified the three X values that make our polynomial equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then the other X value is the X value that makes X minus two equal to zero. Add two to both sides, that's going to be X equals two. So there you have it. We have identified the three X values that make our polynomial equal to zero. And those are going to be the zeros and the X intercepts. So we have one at X equals zero. We have one at X equals negative three."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We have identified the three X values that make our polynomial equal to zero. And those are going to be the zeros and the X intercepts. So we have one at X equals zero. We have one at X equals negative three. We have one at X equals, at X equals two. And the reason why it's, we're done now with this exercise. If you're doing this on Khan Academy, you would have just clicked into these three places."}, {"video_title": "Zeros of polynomials (with factoring) common factor Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We have one at X equals negative three. We have one at X equals, at X equals two. And the reason why it's, we're done now with this exercise. If you're doing this on Khan Academy, you would have just clicked into these three places. But the reason why folks find this to be useful is it helps us start to think about what the graph could be. Because a graph has to intersect the X axis at these points. So the graph might look something like that."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base two of eight. What does this evaluate to? Well, it's asking us, or it will evaluate to the power that I have to, or the exponent that I have to raise our base to, that I have to raise two to, to get to eight. So two to the first power is two, two to the second power is four, two to the third power is eight. So this right over here is going to be equal to three. Fair enough, we did examples like that in the last video."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, it's asking us, or it will evaluate to the power that I have to, or the exponent that I have to raise our base to, that I have to raise two to, to get to eight. So two to the first power is two, two to the second power is four, two to the third power is eight. So this right over here is going to be equal to three. Fair enough, we did examples like that in the last video. Let's do something a little bit more interesting. What is, and I'll color code it, what is log base eight? What is log base eight of two?"}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Fair enough, we did examples like that in the last video. Let's do something a little bit more interesting. What is, and I'll color code it, what is log base eight? What is log base eight of two? Now this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise eight to, to get to two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "What is log base eight of two? Now this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise eight to, to get to two. So let's think about that in another way. So we could say eight, eight to some power, to some power, and that exponent that I'm raising eight to is essentially what this logarithm would evaluate to. Eight to some power is going to be equal to two, is going to be equal to two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise eight to, to get to two. So let's think about that in another way. So we could say eight, eight to some power, to some power, and that exponent that I'm raising eight to is essentially what this logarithm would evaluate to. Eight to some power is going to be equal to two, is going to be equal to two. Well, if two to the third power is eight, eight to the one third power is equal to two. So x is equal to one third. Eight to the one third power is equal to two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Eight to some power is going to be equal to two, is going to be equal to two. Well, if two to the third power is eight, eight to the one third power is equal to two. So x is equal to one third. Eight to the one third power is equal to two. Or you could say the cube root of eight is two. So in this case, x is one third. This logarithm right over here will evaluate to, will evaluate to one third."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Eight to the one third power is equal to two. Or you could say the cube root of eight is two. So in this case, x is one third. This logarithm right over here will evaluate to, will evaluate to one third. Fascinating. Let's mix it up a little bit more. Let's say we had the log base two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "This logarithm right over here will evaluate to, will evaluate to one third. Fascinating. Let's mix it up a little bit more. Let's say we had the log base two. Instead of eight, let's put a one eighth, one eighth right over here. So I'll give you a few seconds to think about that. Well, it's asking us, or this will evaluate to, the exponent that I have to raise two to, to get to one eighth."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's say we had the log base two. Instead of eight, let's put a one eighth, one eighth right over here. So I'll give you a few seconds to think about that. Well, it's asking us, or this will evaluate to, the exponent that I have to raise two to, to get to one eighth. So if we said that this is equal to x, we're essentially saying two to the x power, two to the x power is equal to one eighth, is equal to one eighth. Well, we know two to the third power, let me write this down, we already know that two to the third power is equal to eight. If we want to get to one eighth, which is a reciprocal of eight, we just have to raise two to the negative three power."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, it's asking us, or this will evaluate to, the exponent that I have to raise two to, to get to one eighth. So if we said that this is equal to x, we're essentially saying two to the x power, two to the x power is equal to one eighth, is equal to one eighth. Well, we know two to the third power, let me write this down, we already know that two to the third power is equal to eight. If we want to get to one eighth, which is a reciprocal of eight, we just have to raise two to the negative three power. Two to the negative three power is one over two to the third power, which is the same thing as one over eight. So, if we're asking ourselves, what exponent do we have to raise two to to get to one eighth? Well, we have to raise it to the negative three power."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we want to get to one eighth, which is a reciprocal of eight, we just have to raise two to the negative three power. Two to the negative three power is one over two to the third power, which is the same thing as one over eight. So, if we're asking ourselves, what exponent do we have to raise two to to get to one eighth? Well, we have to raise it to the negative three power. So x is equal to negative three. This logarithm evaluates to negative three. Now let's really, really mix it up."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we have to raise it to the negative three power. So x is equal to negative three. This logarithm evaluates to negative three. Now let's really, really mix it up. What would be the log, what would be the log base eight, base eight of one, one half? What does this evaluate to? Let me clean this up so that we have some space to work with."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now let's really, really mix it up. What would be the log, what would be the log base eight, base eight of one, one half? What does this evaluate to? Let me clean this up so that we have some space to work with. So, as always, we're saying what power do I have to raise eight to to get to one half? So let's think about that a little bit. We already know that eight to the one third power is equal to two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me clean this up so that we have some space to work with. So, as always, we're saying what power do I have to raise eight to to get to one half? So let's think about that a little bit. We already know that eight to the one third power is equal to two. If we want the reciprocal of two right over here, we have to just raise eight to the negative one third. So let me write that down. Eight to the negative one third power is going to be equal to one over eight to the one third power and we already know the cube root of eight or eight to the one third power is equal to two."}, {"video_title": "Fancier logarithm expressions Logarithms Algebra II Khan Academy.mp3", "Sentence": "We already know that eight to the one third power is equal to two. If we want the reciprocal of two right over here, we have to just raise eight to the negative one third. So let me write that down. Eight to the negative one third power is going to be equal to one over eight to the one third power and we already know the cube root of eight or eight to the one third power is equal to two. This is equal to one half. So the log base eight of one half is equal to, well, what power do I have to raise eight to to get to one half is negative one third. Equal to negative one third."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Maybe they're gonna change the title. It seems a little bit too long. But anyway, let's actually just tackle these together. So the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The relationship between the elapsed time t in days since the beginning of spring and the total number of locusts, L of t, so the number of locusts is gonna be a function of the number of days that have elapsed since the beginning of spring, is modeled by the following function. So locusts as a function of time is gonna be 750 times 1.85 to the t-th power."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The relationship between the elapsed time t in days since the beginning of spring and the total number of locusts, L of t, so the number of locusts is gonna be a function of the number of days that have elapsed since the beginning of spring, is modeled by the following function. So locusts as a function of time is gonna be 750 times 1.85 to the t-th power. Complete the following sentence about the daily rate of change of the locust population. Every day the locust population, well, every day, think about what's going to happen. I'll draw a little table just to make it hopefully a little bit clearer."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So locusts as a function of time is gonna be 750 times 1.85 to the t-th power. Complete the following sentence about the daily rate of change of the locust population. Every day the locust population, well, every day, think about what's going to happen. I'll draw a little table just to make it hopefully a little bit clearer. So let me draw a little bit of a table. So we'll put t and L of t. So when t is zero, so when zero days have elapsed, well, this is gonna be 1.85 to the zero-th power. This is gonna be one."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "I'll draw a little table just to make it hopefully a little bit clearer. So let me draw a little bit of a table. So we'll put t and L of t. So when t is zero, so when zero days have elapsed, well, this is gonna be 1.85 to the zero-th power. This is gonna be one. So you're going to have 750 locusts right from the get-go. Then when t equals one, what's going to happen? Well, then this is going to be 750 times 1.85 to the first power."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is gonna be one. So you're going to have 750 locusts right from the get-go. Then when t equals one, what's going to happen? Well, then this is going to be 750 times 1.85 to the first power. So it's gonna be times 1.85. When t is equal to two, what's L of t? It's going to be 750 times 1.85 squared."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, then this is going to be 750 times 1.85 to the first power. So it's gonna be times 1.85. When t is equal to two, what's L of t? It's going to be 750 times 1.85 squared. Well, that's the same thing as 1.85 times 1.85. So notice, and this just comes out of this being an exponential function, every day you have 1.85 times as many as you had the day before. 1.85."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's going to be 750 times 1.85 squared. Well, that's the same thing as 1.85 times 1.85. So notice, and this just comes out of this being an exponential function, every day you have 1.85 times as many as you had the day before. 1.85. We essentially take what we had the day before and we multiply by 1.85. And since 1.85 is larger than one, that's going to grow the number of locusts we have. So this is going to grow."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "1.85. We essentially take what we had the day before and we multiply by 1.85. And since 1.85 is larger than one, that's going to grow the number of locusts we have. So this is going to grow. I'm actually not using, I'm not on the website right now, so that's why normally there would be a dropdown here. So I'm going to grow by a factor of, well, I'm going to grow by a factor of 1.85 every day. Let's do another one of these."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is going to grow. I'm actually not using, I'm not on the website right now, so that's why normally there would be a dropdown here. So I'm going to grow by a factor of, well, I'm going to grow by a factor of 1.85 every day. Let's do another one of these. All right, so this one tells us that Vera is an ecologist who studies the rate of change in the bear population of Siberia over time. The relationship between the elapsed time t in years since Vera began studying the population and the total number of bears, n of t, is modeled by the following function. All right, fair enough."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's do another one of these. All right, so this one tells us that Vera is an ecologist who studies the rate of change in the bear population of Siberia over time. The relationship between the elapsed time t in years since Vera began studying the population and the total number of bears, n of t, is modeled by the following function. All right, fair enough. Got a little exponential thing going on. Complete the following sentence about the yearly rate of change of the bear population. What's interesting about every year that passes, t is in years now, every year that passes is going to be 2 3rds times the year before."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right, fair enough. Got a little exponential thing going on. Complete the following sentence about the yearly rate of change of the bear population. What's interesting about every year that passes, t is in years now, every year that passes is going to be 2 3rds times the year before. I can do that same table that I just did just to make that clear. So let me do that. Whoops."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "What's interesting about every year that passes, t is in years now, every year that passes is going to be 2 3rds times the year before. I can do that same table that I just did just to make that clear. So let me do that. Whoops. Let me make this clear. So table. So this is t and this is n of t. When t is zero, n of t, you're going to have 2,187 bears."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Whoops. Let me make this clear. So table. So this is t and this is n of t. When t is zero, n of t, you're going to have 2,187 bears. So that's the first year that she began studying that population. Zero years since Vera began studying the population. The first year is going to be 2,187 times 2 3rds to the first power."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is t and this is n of t. When t is zero, n of t, you're going to have 2,187 bears. So that's the first year that she began studying that population. Zero years since Vera began studying the population. The first year is going to be 2,187 times 2 3rds to the first power. So times 2 3rds. The second year is going to be 2,187 times 2 3rds to the second power. So that's just 2 3rds times 2 3rds."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "The first year is going to be 2,187 times 2 3rds to the first power. So times 2 3rds. The second year is going to be 2,187 times 2 3rds to the second power. So that's just 2 3rds times 2 3rds. So each successive year, you're going to have 2 3rds of the bear population of the year before. You're multiplying the year before by 2 3rds. So every year the bear population shrinks by a factor of 2 3rds."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So that's just 2 3rds times 2 3rds. So each successive year, you're going to have 2 3rds of the bear population of the year before. You're multiplying the year before by 2 3rds. So every year the bear population shrinks by a factor of 2 3rds. All right, let's do one more of these. So they tell us that Akiba started studying how the number of branches on his tree change over time. All right."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So every year the bear population shrinks by a factor of 2 3rds. All right, let's do one more of these. So they tell us that Akiba started studying how the number of branches on his tree change over time. All right. The relationship between the elapsed time t in years since Akiba started studying his tree and the total number of its branches, n of t, is modeled by the following function. Complete the following sentence about the yearly percent change in the number of branches. Every year, blank percent of branches are added or subtracted from the total number of branches."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right. The relationship between the elapsed time t in years since Akiba started studying his tree and the total number of its branches, n of t, is modeled by the following function. Complete the following sentence about the yearly percent change in the number of branches. Every year, blank percent of branches are added or subtracted from the total number of branches. Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look, each year you're going to have 1.75 times the number of branches you had the year before. And so if you have 1.75 times the number of branches the year before, you have grown by 75%. And I'll make that a little bit clearer."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Every year, blank percent of branches are added or subtracted from the total number of branches. Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look, each year you're going to have 1.75 times the number of branches you had the year before. And so if you have 1.75 times the number of branches the year before, you have grown by 75%. And I'll make that a little bit clearer. So 75% of branches, every year 75% of branches are added to the total number of branches. And I'll just draw that table again like I've done in the last two examples to make that hopefully clear. Okay, so this is t and this is n of t. So t equals zero, you have 42 branches."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I'll make that a little bit clearer. So 75% of branches, every year 75% of branches are added to the total number of branches. And I'll just draw that table again like I've done in the last two examples to make that hopefully clear. Okay, so this is t and this is n of t. So t equals zero, you have 42 branches. t equals one, it's going to be 42 times 1.75. Times 1.75. When t equals two, it's going to be 42 times 1.75 squared."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Okay, so this is t and this is n of t. So t equals zero, you have 42 branches. t equals one, it's going to be 42 times 1.75. Times 1.75. When t equals two, it's going to be 42 times 1.75 squared. 42 times 1.75 times 1.75. So every year you are multiplying times 1.75. So times 1.75, something funky is happening with my pen right over there."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "When t equals two, it's going to be 42 times 1.75 squared. 42 times 1.75 times 1.75. So every year you are multiplying times 1.75. So times 1.75, something funky is happening with my pen right over there. But if you're multiplying by 1.75, if you're growing by a factor of 1.75, this is the same thing as adding 75%. Once again, you are adding 75%. Think about it this way."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So times 1.75, something funky is happening with my pen right over there. But if you're multiplying by 1.75, if you're growing by a factor of 1.75, this is the same thing as adding 75%. Once again, you are adding 75%. Think about it this way. If you just grew by a factor of one, then you're not adding anything. You're staying constant. If you grow by 10%, then you're going to be 1.1 times as large."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Think about it this way. If you just grew by a factor of one, then you're not adding anything. You're staying constant. If you grow by 10%, then you're going to be 1.1 times as large. If you grow by 200%, then you're going to be two times as large. So this right over here, this right over here is, is, is, or if you, let me be careful what I just said. I think I just mistaken."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "If you grow by 10%, then you're going to be 1.1 times as large. If you grow by 200%, then you're going to be two times as large. So this right over here, this right over here is, is, is, or if you, let me be careful what I just said. I think I just mistaken. If you grow by 200%, you're going to be three times as large as you were before. One is constant, and then another 200% would be another twofold. So that would make you three times as large."}, {"video_title": "Interpreting change in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "I think I just mistaken. If you grow by 200%, you're going to be three times as large as you were before. One is constant, and then another 200% would be another twofold. So that would make you three times as large. Don't want to confuse you. My brain recognized that I said something weird right at that end. Alright, hopefully you enjoyed that."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and see if you can have a go at that. All right, now let's work through it together. So if they didn't give us this second piece of information that has a known factor of X plus two, this polynomial would not be so easy to factor. But because we know we have a known factor of X plus two, I could divide that into our expression and figure out what I have left over and then see if I can factor from there. So let's do that. Let's divide X plus two into our polynomial. So it's four X to the third power plus 19 X squared plus 19 X minus six."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "But because we know we have a known factor of X plus two, I could divide that into our expression and figure out what I have left over and then see if I can factor from there. So let's do that. Let's divide X plus two into our polynomial. So it's four X to the third power plus 19 X squared plus 19 X minus six. And we've done this multiple times already. We look at the highest degree terms. X goes into four X to the third, four X squared times, I put that in the X squared or the second degree column."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So it's four X to the third power plus 19 X squared plus 19 X minus six. And we've done this multiple times already. We look at the highest degree terms. X goes into four X to the third, four X squared times, I put that in the X squared or the second degree column. Four X squared times X is four X to the third. Four X squared times two is eight X, eight X squared. And then I wanna subtract these from what I have up here."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "X goes into four X to the third, four X squared times, I put that in the X squared or the second degree column. Four X squared times X is four X to the third. Four X squared times two is eight X, eight X squared. And then I wanna subtract these from what I have up here. So I'll subtract. And then I'm going to be left with 19 X squared minus eight X squared is 11 X squared. And then I will bring down, bring down the 19 X, so plus 19 X."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And then I wanna subtract these from what I have up here. So I'll subtract. And then I'm going to be left with 19 X squared minus eight X squared is 11 X squared. And then I will bring down, bring down the 19 X, so plus 19 X. And so once again, look at X and 11 X squared. X goes into 11 X squared, 11 X times. So that's plus 11 X."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And then I will bring down, bring down the 19 X, so plus 19 X. And so once again, look at X and 11 X squared. X goes into 11 X squared, 11 X times. So that's plus 11 X. 11 X times X is 11 X squared. 11 X times two is 22 X. You just subtract these from what we have in that teal color."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So that's plus 11 X. 11 X times X is 11 X squared. 11 X times two is 22 X. You just subtract these from what we have in that teal color. And we are left with 19 minus 22 of something is negative three of that something. In this case, it's negative three X. And then we bring down that negative six."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "You just subtract these from what we have in that teal color. And we are left with 19 minus 22 of something is negative three of that something. In this case, it's negative three X. And then we bring down that negative six. And then we look at, once again, at the X and the negative three X. X goes into negative three X negative three times. And so negative three times X is negative three X. Negative three times two is negative six."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And then we bring down that negative six. And then we look at, once again, at the X and the negative three X. X goes into negative three X negative three times. And so negative three times X is negative three X. Negative three times two is negative six. And then if we wanna subtract what we have in red from what we have in magenta, so I can just multiply them both by negative. And so everything just cancels out and we have no remainder. And so we can rewrite P of X now."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "Negative three times two is negative six. And then if we wanna subtract what we have in red from what we have in magenta, so I can just multiply them both by negative. And so everything just cancels out and we have no remainder. And so we can rewrite P of X now. We can rewrite P of X as being equal to X plus two times all of this business. Four X squared plus 11 X minus three. Now we're not done yet because we haven't expressed it as a product of linear factors."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And so we can rewrite P of X now. We can rewrite P of X as being equal to X plus two times all of this business. Four X squared plus 11 X minus three. Now we're not done yet because we haven't expressed it as a product of linear factors. This one over here is linear. But four X squared plus 11 X minus three, that's still quadratic. So we have to factor that further."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "Now we're not done yet because we haven't expressed it as a product of linear factors. This one over here is linear. But four X squared plus 11 X minus three, that's still quadratic. So we have to factor that further. And let's see, there's a couple of ways we could approach it. We could use, well, we could try something like the quadratic formula. Or we could factor by grouping."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So we have to factor that further. And let's see, there's a couple of ways we could approach it. We could use, well, we could try something like the quadratic formula. Or we could factor by grouping. And to factor by grouping, and the whole reason why we have to factor by grouping is we have a leading coefficient here that is not one. And so we need to think of two numbers whose product is equal to four times negative three. So we have to think of two numbers."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "Or we could factor by grouping. And to factor by grouping, and the whole reason why we have to factor by grouping is we have a leading coefficient here that is not one. And so we need to think of two numbers whose product is equal to four times negative three. So we have to think of two numbers. Let's just call them A and B. A times B needs to be equal to four times negative three, which is negative 12. And A plus B needs to be equal to 11."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So we have to think of two numbers. Let's just call them A and B. A times B needs to be equal to four times negative three, which is negative 12. And A plus B needs to be equal to 11. And so the best that I can think of, they have to be opposite signs because their product is a negative. So if I had negative, if I had, if I had positive 12 and negative one, that works. If A is equal to negative one and B is equal to positive 12, that works."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And A plus B needs to be equal to 11. And so the best that I can think of, they have to be opposite signs because their product is a negative. So if I had negative, if I had, if I had positive 12 and negative one, that works. If A is equal to negative one and B is equal to positive 12, that works. And so what I wanna do is I wanna take this first degree term right over here, 11x, and I wanna split it into a 12x and a negative one x. So let's do that. So I can, let's just focus on this part, on this part right now, and then I'll put it all back together at the end."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "If A is equal to negative one and B is equal to positive 12, that works. And so what I wanna do is I wanna take this first degree term right over here, 11x, and I wanna split it into a 12x and a negative one x. So let's do that. So I can, let's just focus on this part, on this part right now, and then I'll put it all back together at the end. So I can rewrite all of this business as four x squared. And instead of writing the 11x there, I'm gonna use this blue color, I'm gonna break it up as a 12x, so plus 12x and then minus one x. Notice these two still add up to 11x."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So I can, let's just focus on this part, on this part right now, and then I'll put it all back together at the end. So I can rewrite all of this business as four x squared. And instead of writing the 11x there, I'm gonna use this blue color, I'm gonna break it up as a 12x, so plus 12x and then minus one x. Notice these two still add up to 11x. And then I have my minus three. And then let's see, out of these first two, what can I factor out? Let's see, I can factor out a four x so I can rewrite these first two as, and if this is unfamiliar to you, I encourage you to review factoring by grouping on Khan Academy."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "Notice these two still add up to 11x. And then I have my minus three. And then let's see, out of these first two, what can I factor out? Let's see, I can factor out a four x so I can rewrite these first two as, and if this is unfamiliar to you, I encourage you to review factoring by grouping on Khan Academy. So if we factor out a four x, that's going to be, it's going to be, we're gonna be left with an x here, and then we're gonna be left with a three over here. And then these second two terms, if we factor out a negative one, so I'll write negative one, times we're gonna have an x plus three. And so then we can factor out the x plus three."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "Let's see, I can factor out a four x so I can rewrite these first two as, and if this is unfamiliar to you, I encourage you to review factoring by grouping on Khan Academy. So if we factor out a four x, that's going to be, it's going to be, we're gonna be left with an x here, and then we're gonna be left with a three over here. And then these second two terms, if we factor out a negative one, so I'll write negative one, times we're gonna have an x plus three. And so then we can factor out the x plus three. So let's do that. I'm running out of colors. So I factor out the x plus three, and I am left with x plus three times, times four x. I'm gonna keep these colors the same so you know where I got them from."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And so then we can factor out the x plus three. So let's do that. I'm running out of colors. So I factor out the x plus three, and I am left with x plus three times, times four x. I'm gonna keep these colors the same so you know where I got them from. Four x minus one. This is a very colorful solution that we have over here. And there you have it."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "So I factor out the x plus three, and I am left with x plus three times, times four x. I'm gonna keep these colors the same so you know where I got them from. Four x minus one. This is a very colorful solution that we have over here. And there you have it. I factored the second part into these two factors. And so now I can put it all together. I can rewrite p of x as a product of linear factors."}, {"video_title": "Factoring using polynomial division Algebra 2 Khan Academy.mp3", "Sentence": "And there you have it. I factored the second part into these two factors. And so now I can put it all together. I can rewrite p of x as a product of linear factors. P of x is equal to x plus two times x plus three times four x minus one. Four x minus one. And we are done."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And one way to think about it is, what is x squared plus seven x plus 10 divided by x plus two? What is that going to be? All right, now there's two ways that you could approach this. One way is to try to factor the numerator and see if it has a factor that is common to the denominator. So let's try to do that. So we've done this many, many times. If this looks new to you, I encourage you to review factoring polynomials other places on Khan Academy."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "One way is to try to factor the numerator and see if it has a factor that is common to the denominator. So let's try to do that. So we've done this many, many times. If this looks new to you, I encourage you to review factoring polynomials other places on Khan Academy. But what two numbers add up to seven and when you multiply them, you get 10? Well, that would be two and five. So we could rewrite that numerator as x plus two times x plus five."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "If this looks new to you, I encourage you to review factoring polynomials other places on Khan Academy. But what two numbers add up to seven and when you multiply them, you get 10? Well, that would be two and five. So we could rewrite that numerator as x plus two times x plus five. And then, of course, the denominator, you still have x plus two. And then we clearly see we have a common factor. And so as long as x does not equal negative two, because if x equals negative two, this whole expression is undefined because then you get a zero in the denominator."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "So we could rewrite that numerator as x plus two times x plus five. And then, of course, the denominator, you still have x plus two. And then we clearly see we have a common factor. And so as long as x does not equal negative two, because if x equals negative two, this whole expression is undefined because then you get a zero in the denominator. So as long as x does not equal negative two, well, then we can divide the numerator and the denominator by x plus two. Once again, the reason why I put that constraint is we can't divide the numerator and the denominator by zero. So for any other values of x, this x plus two will be nonzero and we could divide the numerator and the denominator by that and they would cancel out and we would just be left with x plus five."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And so as long as x does not equal negative two, because if x equals negative two, this whole expression is undefined because then you get a zero in the denominator. So as long as x does not equal negative two, well, then we can divide the numerator and the denominator by x plus two. Once again, the reason why I put that constraint is we can't divide the numerator and the denominator by zero. So for any other values of x, this x plus two will be nonzero and we could divide the numerator and the denominator by that and they would cancel out and we would just be left with x plus five. So another way to think about it is this expression, our original expression, could be viewed as x plus five for any x that is not equal to negative two. Now, the other way that we could approach this is through algebraic long division, which is very analogous to the type of long division that you might remember from, I believe it was fourth grade. So what you do is you say, all right, I'm gonna divide x plus two into x squared plus seven x plus 10."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "So for any other values of x, this x plus two will be nonzero and we could divide the numerator and the denominator by that and they would cancel out and we would just be left with x plus five. So another way to think about it is this expression, our original expression, could be viewed as x plus five for any x that is not equal to negative two. Now, the other way that we could approach this is through algebraic long division, which is very analogous to the type of long division that you might remember from, I believe it was fourth grade. So what you do is you say, all right, I'm gonna divide x plus two into x squared plus seven x plus 10. And in this technique, you look at the highest degree terms. So you have an x there and an x squared there. And say, how many times is x going to x squared?"}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "So what you do is you say, all right, I'm gonna divide x plus two into x squared plus seven x plus 10. And in this technique, you look at the highest degree terms. So you have an x there and an x squared there. And say, how many times is x going to x squared? Well, it goes x times. Now, you'd write that in this column because x is just x to the first power. You could view this as the first degree column."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And say, how many times is x going to x squared? Well, it goes x times. Now, you'd write that in this column because x is just x to the first power. You could view this as the first degree column. It's analogous to the place values that we talk about when we first learn numbers or how we regroup or about place value. But here, you could view it as degree places or something like that. And then you take that x and you multiply it times this entire expression."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "You could view this as the first degree column. It's analogous to the place values that we talk about when we first learn numbers or how we regroup or about place value. But here, you could view it as degree places or something like that. And then you take that x and you multiply it times this entire expression. So x times two is two x. Put that in the first degree column. X times x is x squared."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And then you take that x and you multiply it times this entire expression. So x times two is two x. Put that in the first degree column. X times x is x squared. And then what we wanna do is we wanna subtract these things in yellow from what we originally had in blue. So we could do it this way. And then we will be left with seven x minus two x is five x."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "X times x is x squared. And then what we wanna do is we wanna subtract these things in yellow from what we originally had in blue. So we could do it this way. And then we will be left with seven x minus two x is five x. And then x squared minus x squared is just a zero. And then we can bring down this plus 10. And once again, we look at the highest degree term."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And then we will be left with seven x minus two x is five x. And then x squared minus x squared is just a zero. And then we can bring down this plus 10. And once again, we look at the highest degree term. X goes into five x five times. That's a zero degree. It's a constant, so I'll write it in the constant column."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And once again, we look at the highest degree term. X goes into five x five times. That's a zero degree. It's a constant, so I'll write it in the constant column. Five times two is 10. Five times x is five. And then I'll subtract these from what we have up here."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "It's a constant, so I'll write it in the constant column. Five times two is 10. Five times x is five. And then I'll subtract these from what we have up here. And notice, we have no remainder. What's interesting about algebraic long division, we'll probably see it in another video or two, you can actually have a remainder. So those are going to be situations where just the factoring technique alone would not have worked."}, {"video_title": "Dividing polynomials by x (no remainders) Algebra 2 Khan Academy (2).mp3", "Sentence": "And then I'll subtract these from what we have up here. And notice, we have no remainder. What's interesting about algebraic long division, we'll probably see it in another video or two, you can actually have a remainder. So those are going to be situations where just the factoring technique alone would not have worked. In this situation, this model would have been easier. But this is another way to think about it. You say, hey, look, x plus two times x plus five is going to be equal to this."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "Now that we know a little bit about the imaginary unit i, let's see if we can simplify more involved expressions like this one right over here. 2 plus 3i plus 7i squared plus 5i to the third power plus 9i to the fourth power. I encourage you to pause the video right now and try to simplify this on your own. So as you can see here, we have various powers of i. You could view this as i to the first power. We have i squared here. And we already know that i squared is defined to be negative 1."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So as you can see here, we have various powers of i. You could view this as i to the first power. We have i squared here. And we already know that i squared is defined to be negative 1. Then we have i to the third power. i to the third power would just be i times this, or negative i. And we already reviewed this when we first introduced the imaginary unit i, but I'll do it again."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And we already know that i squared is defined to be negative 1. Then we have i to the third power. i to the third power would just be i times this, or negative i. And we already reviewed this when we first introduced the imaginary unit i, but I'll do it again. i to the fourth power is just going to be i times this, which is the same thing as negative 1 times i. That's i to the third power times i again. i times i is negative 1."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And we already reviewed this when we first introduced the imaginary unit i, but I'll do it again. i to the fourth power is just going to be i times this, which is the same thing as negative 1 times i. That's i to the third power times i again. i times i is negative 1. So that's negative 1 times negative 1, which is equal to 1 again. So we can rewrite this whole thing. We could rewrite it as 2 plus 3i."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "i times i is negative 1. So that's negative 1 times negative 1, which is equal to 1 again. So we can rewrite this whole thing. We could rewrite it as 2 plus 3i. 7i squared is going to be the same thing. So i squared is negative 1. So this is the same thing as 7 times negative 1."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "We could rewrite it as 2 plus 3i. 7i squared is going to be the same thing. So i squared is negative 1. So this is the same thing as 7 times negative 1. So that's just going to be minus 7. And then we have 5i to the third power. i to the third power is negative i."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So this is the same thing as 7 times negative 1. So that's just going to be minus 7. And then we have 5i to the third power. i to the third power is negative i. So this could be rewritten as negative i. So this term right over here we could write as minus 5i, or negative 5i, depending on how you want to think about it. And then finally, i to the fourth power."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "i to the third power is negative i. So this could be rewritten as negative i. So this term right over here we could write as minus 5i, or negative 5i, depending on how you want to think about it. And then finally, i to the fourth power. i to the fourth power is just 1. So this is just equal to 1. So this whole term just simplifies to 9."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And then finally, i to the fourth power. i to the fourth power is just 1. So this is just equal to 1. So this whole term just simplifies to 9. So how could we simplify this more? Well, we have several terms that are not imaginary, that they are real numbers. For example, we have this 2 is a real number."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So this whole term just simplifies to 9. So how could we simplify this more? Well, we have several terms that are not imaginary, that they are real numbers. For example, we have this 2 is a real number. Negative 7 is a real number. And 9 is a real number. So we could just add those up."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "For example, we have this 2 is a real number. Negative 7 is a real number. And 9 is a real number. So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So we could just add those up. So 2 plus negative 7 would be negative 5. Negative 5 plus 9 would be 4. So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something, and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So the real numbers add up to 4. And now we have these imaginary numbers. So 3 times i minus 5 times i. So if you have 3 of something, and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients, 3 minus 5 is negative 2. So 3i's minus 5i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further?"}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So if you have 3 of something, and then I were to subtract 5 of that same something from it, now you're going to have negative 2 of that something. Or another way of thinking about it is the coefficients, 3 minus 5 is negative 2. So 3i's minus 5i's, that's going to give you negative 2i. Now you might say, well, can we simplify this any further? Well, no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "Now you might say, well, can we simplify this any further? Well, no, you really can't. This right over here is a real number. 4 is a number that we've been dealing with throughout our mathematical careers. And negative 2i, that's an imaginary number. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "4 is a number that we've been dealing with throughout our mathematical careers. And negative 2i, that's an imaginary number. And so what we really consider this is this 4 minus 2i, we can now consider this entire expression to really be a number. So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. Why is it complex? Well, it has a real part and an imaginary part."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "So this is a number that has a real part and an imaginary part. And numbers like this we call complex numbers. Why is it complex? Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 is 3 plus 0i? And you would be correct."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "Well, it has a real part and an imaginary part. And you might say, well, gee, can't any real number be considered a complex number? For example, if I have the real number 3, can't I just write the real number 3 is 3 plus 0i? And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. And actually, the real numbers are a subset of the complex numbers."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And you would be correct. Any real number is a complex number. You could view this right over here as a complex number. And actually, the real numbers are a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part. 0 is a real number."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And actually, the real numbers are a subset of the complex numbers. Likewise, imaginary numbers are a subset of the complex numbers. For example, you could rewrite i as a real part. 0 is a real number. 0 plus i. So the imaginaries are a subset of complex numbers. Real numbers are a subset of complex numbers."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's greater than negative 3 pi over 2, and it's less than negative pi. And we're also told that sine of theta is equal to 1 half. Just from this information, can we figure out what the tangent of theta is going to be equal to? And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta plus cosine squared theta is equal to 1. So let's do it."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try this on your own. In case you're stumped, I will give you a hint. You should use the Pythagorean identity, the fact that sine squared theta plus cosine squared theta is equal to 1. So let's do it. So we know the Pythagorean identity. Sine squared theta plus cosine squared theta is equal to 1. We know what sine squared theta is."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's do it. So we know the Pythagorean identity. Sine squared theta plus cosine squared theta is equal to 1. We know what sine squared theta is. Sine theta is 1 half, so this could be rewritten as 1 half squared plus cosine squared theta is equal to 1. Or we could write this as 1 fourth plus cosine squared theta is equal to 1. Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "We know what sine squared theta is. Sine theta is 1 half, so this could be rewritten as 1 half squared plus cosine squared theta is equal to 1. Or we could write this as 1 fourth plus cosine squared theta is equal to 1. Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away. That was the whole point. 1 minus 1 fourth is 3 fourths. So what could cosine of theta be?"}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Or we could subtract 1 fourth from both sides, and we get cosine squared theta is equal to, see, you subtract 1 fourth from the left-hand side, and this 1 fourth goes away. That was the whole point. 1 minus 1 fourth is 3 fourths. So what could cosine of theta be? Well, when I square it, I get positive 3 fourths. So it could be the positive or negative square root of 3 fourths. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So what could cosine of theta be? Well, when I square it, I get positive 3 fourths. So it could be the positive or negative square root of 3 fourths. So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2. So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So cosine of theta could be equal to the positive or negative square root of 3 over 4, which is the same thing as the positive or negative square root of 3 over the square root of 4, which is 2. So it's the positive or negative square root of 3 over 2. But how do we know which one of these it actually is? Well, that's where this information becomes useful. That's where this information actually becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta?"}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, that's where this information becomes useful. That's where this information actually becomes useful. Let's draw our unit circle. If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta, you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "If you're saying, well, why am I even worried about cosine of theta? Well, if you know sine of theta, you know cosine of theta. Tangent of theta is just sine of theta over cosine theta. So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it. So unit circle."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So then you will know the tangent of theta. But let's look at the unit circle to figure out which value of cosine we should use. So let me draw it. So unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So unit circle. That's my y-axis. That is my x-axis. And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And I will draw the unit circle in pink. So that's my best attempt at drawing a circle. Please forgive me for its lack of perfect roundness. And it says theta is greater than negative 3 pi over 2. So where's negative 3 pi over 2? So let's see. This is negative pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And it says theta is greater than negative 3 pi over 2. So where's negative 3 pi over 2? So let's see. This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So let's see."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "This is negative pi over 2. So this is one side of the angle. Let me do this in a color. So let's see. This one side of the angle is going to be along the positive x-axis. And we want to figure out what the other side is. So it's going to this right over here."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So let's see. This one side of the angle is going to be along the positive x-axis. And we want to figure out what the other side is. So it's going to this right over here. That's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's going to this right over here. That's negative pi over 2. This is negative pi. So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So it's between negative pi, which is right over here. So let me make that clear. Negative pi is right over here. It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So the angle, our angle theta, is going to put us someplace over here. And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "It's between negative pi and negative 3 pi over 2. Negative 3 pi over 2 is right over here. So the angle, our angle theta, is going to put us someplace over here. And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there. And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And the whole reason I did this, so this whole arc right here, you could think of this as the measure of angle theta right over there. And the whole reason I did that is to think about whether the cosine of theta is going to be positive or negative. We clearly see it's in the second quadrant. The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here, this right over here, actually let me do it in that orange color again, this right over here, that is the cosine of theta. Now is that a positive or negative value? Well, it's clearly a negative value."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "The cosine of theta is the x-coordinate of this point where our angle intersects the unit circle. So this point right over here, this right over here, actually let me do it in that orange color again, this right over here, that is the cosine of theta. Now is that a positive or negative value? Well, it's clearly a negative value. So for the sake of this example, our cosine theta is not the positive one. It is the negative one. So we can write that cosine theta is equal to the negative square root of 3 over 2."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, it's clearly a negative value. So for the sake of this example, our cosine theta is not the positive one. It is the negative one. So we can write that cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta. Well, they told us the sine of theta is 1 half."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "So we can write that cosine theta is equal to the negative square root of 3 over 2. So we figured out cosine theta, but we still have to figure out tangent of theta. And we just have to remind ourselves that the tangent of theta is going to be equal to the sine of theta over the cosine of theta. Well, they told us the sine of theta is 1 half. So it's going to be 1 half over cosine of theta, which is negative square root of 3 over 2. And what does that equal? Well, that's the same thing as 1 half times the reciprocal of this."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, they told us the sine of theta is 1 half. So it's going to be 1 half over cosine of theta, which is negative square root of 3 over 2. And what does that equal? Well, that's the same thing as 1 half times the reciprocal of this. So times negative 2 over the square root of 3. These 2's will cancel out. And we are left with negative 1 over the square root of 3."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "Well, that's the same thing as 1 half times the reciprocal of this. So times negative 2 over the square root of 3. These 2's will cancel out. And we are left with negative 1 over the square root of 3. Now some people don't like a radical in the denominator like this. They don't like an irrational denominator. So we could rationalize the denominator here by multiplying by square root of 3 over square root of 3."}, {"video_title": "Using the Pythagorean trig identity Trig identities and examples Trigonometry Khan Academy.mp3", "Sentence": "And we are left with negative 1 over the square root of 3. Now some people don't like a radical in the denominator like this. They don't like an irrational denominator. So we could rationalize the denominator here by multiplying by square root of 3 over square root of 3. And so that gets us, this will be equal to negative square root of 3 over 3 is the tangent of this angle right over here. And that actually makes sense, because the tangent of the angle is the slope of this line. And we see that it is indeed a negative slope."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "It's the square of x. And 49y squared is also a perfect square. It's the square of 7y. So it looks like we might have a special form here. And to remind ourselves, just think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared, plus a times negative b, which would be negative ab, plus b times a, or a times b again, which would be ab. And then you have b times negative b, so it would be minus b squared."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "So it looks like we might have a special form here. And to remind ourselves, just think about what happens if we take a plus b times a minus b. I'm just doing it in the general case so we can see a pattern here. So over here, this would be a times a, which would be a squared, plus a times negative b, which would be negative ab, plus b times a, or a times b again, which would be ab. And then you have b times negative b, so it would be minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out. And you're left with just a squared minus b squared."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "And then you have b times negative b, so it would be minus b squared. Now these middle two terms cancel out. Negative ab plus ab, they cancel out. And you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x, and b is equal to 7y."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "And you're left with just a squared minus b squared. And that's the exact pattern we have here. We have an a squared minus a b squared. So in this case, a is equal to x, and b is equal to 7y. So we have x squared minus 7y, the whole thing squared. So we can expand this as the difference of squares. Or actually, this thing right over here is the difference of squares."}, {"video_title": "Example 1 Factoring a difference of squares with two variables Algebra II Khan Academy.mp3", "Sentence": "So in this case, a is equal to x, and b is equal to 7y. So we have x squared minus 7y, the whole thing squared. So we can expand this as the difference of squares. Or actually, this thing right over here is the difference of squares. So we can expand this like this. So this will be equal to x plus 7y times x minus 7y. And once again, we're just pattern matching based on this realization right here."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Use the change of base formula to find log base 5 of 100 to the nearest thousandth. So the change of base formula is a useful formula, especially when you're going to use a calculator, because most calculators don't allow you to arbitrarily change the base of your logarithm. They have functions for log base e, which is a natural logarithm, and log base 10. So you generally have to change your base, and that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense or how we can derive it. So the change of base formula just tells us that log... Let me do some colors here. Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So you generally have to change your base, and that's what the change of base formula is. And if we have time, I'll tell you why it makes a lot of sense or how we can derive it. So the change of base formula just tells us that log... Let me do some colors here. Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a. And the reason why this is useful is that we can change our base. Here our base is a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base a of b is the same thing, is the exact same thing as log base x, where x is an arbitrary base, of b over log base... that same base, base x, base x over a. And the reason why this is useful is that we can change our base. Here our base is a, and we can change it to base x. So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if our calculator has a certain base x function, we can convert to that base. It's usually e or base 10. Base 10 is an easy way to go. And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x. And e is obviously the number 2.71 that keeps going on and on and on forever. Now let's apply it to this problem."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And in general, if you just see someone write a logarithm like this, if they just write log of x, they're implying this implies log base 10 of x. If someone writes natural log of x, they are implying log base e of x. And e is obviously the number 2.71 that keeps going on and on and on forever. Now let's apply it to this problem. We have...we need to figure out the logarithm, and I'll use the colors, base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log, I'll make x 10, log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now let's apply it to this problem. We have...we need to figure out the logarithm, and I'll use the colors, base 5 of 100. So this property, this change of base formula, tells us that this is the exact same thing as log, I'll make x 10, log base 10 of 100 divided by log base 10 of 5. And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100, what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And actually, we don't even need a calculator to evaluate this top part. Log base 10 of 100, what power do I have to raise 10 to to get to 100? The second power. So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5. And we can now use our calculator because the log function on a calculator is log base 10. So let's get our calculator out. And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this numerator is just equal to 2, so it simplifies to 2 over log base 10 of 5. And we can now use our calculator because the log function on a calculator is log base 10. So let's get our calculator out. And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10. If they press ln, that means base e. So log, without any other information, is log base 10. So this is log base 10 of 5 is equal to 2 point... and they want it to the nearest thousandth. So 2.861."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we're going to get... we want, let me clear this, 2 divided by... when someone just writes log, they mean base 10. If they press ln, that means base e. So log, without any other information, is log base 10. So this is log base 10 of 5 is equal to 2 point... and they want it to the nearest thousandth. So 2.861. So this is approximately equal to 2.861. And we can verify it, because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 2.861. So this is approximately equal to 2.861. And we can verify it, because in theory, if I raise 5 to this power, I should get 100. And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2. But let's verify it. So if I take 5 to that power, if I take 5 and let me type in... well, let me just type in what we did to the nearest thousandth. 5 to the 2.861, so I'm not putting in all of the digits, what do I get?"}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And it kind of makes sense, because 5 to the second power is 25, 5 to the third power is 125, and this is in between the two, and it's closer to the third power than it is to the second power, and this number is closer to 3 than it is to 2. But let's verify it. So if I take 5 to that power, if I take 5 and let me type in... well, let me just type in what we did to the nearest thousandth. 5 to the 2.861, so I'm not putting in all of the digits, what do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "5 to the 2.861, so I'm not putting in all of the digits, what do I get? I get 99.94. If I put all of these digits in, it should get pretty close to 100. So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100. Now, with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base A, I'll try to be fair to the colors, log base A of B, let's say I set that to be equal to... let's say I set that equal to some number. Let's call that equal to C, or I could call it E for example."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's what makes you feel good, that this is the power that I have to raise 5 to to get to 100. Now, with that out of the way, let's actually think about why this property, why this thing right over here makes sense. So if I write log base A, I'll try to be fair to the colors, log base A of B, let's say I set that to be equal to... let's say I set that equal to some number. Let's call that equal to C, or I could call it E for example. Well, I'll say that's equal to C. That means that A to the Cth power is equal to B. This is an exponential way of writing this truth, this is a logarithmic way of writing this truth. Is equal to B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's call that equal to C, or I could call it E for example. Well, I'll say that's equal to C. That means that A to the Cth power is equal to B. This is an exponential way of writing this truth, this is a logarithmic way of writing this truth. Is equal to B. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this, 10 to the same power will be equal to this because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Is equal to B. Now, we can take the logarithm of any base of both sides of this. Anything you do, if you say 10 to the what power equals this, 10 to the same power will be equal to this because these two things are equal to each other. So let's take the same logarithm of both sides of this, the logarithm with the same base. I'll actually do log base X to prove the general case here. So I'm going to take log base X of both sides of this. So this is log base X of A to the Cth power."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's take the same logarithm of both sides of this, the logarithm with the same base. I'll actually do log base X to prove the general case here. So I'm going to take log base X of both sides of this. So this is log base X of A to the Cth power. I try to be faithful to the colors. Is equal to log base X of B. Is equal to B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base X of A to the Cth power. I try to be faithful to the colors. Is equal to log base X of B. Is equal to B. Let me close it off in orange as well. And we know from our logarithm properties, log of A to the C is the same thing as C times the logarithm of whatever base we are of A. And of course, this is going to be equal to log base X of B."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Is equal to B. Let me close it off in orange as well. And we know from our logarithm properties, log of A to the C is the same thing as C times the logarithm of whatever base we are of A. And of course, this is going to be equal to log base X of B. And if we wanted to solve for C, you just divide both sides by log base X of A. So you get C is equal to, and I'll stick to the color, it says log base X of B, which is this, over log base X of A. And this is what C was."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And of course, this is going to be equal to log base X of B. And if we wanted to solve for C, you just divide both sides by log base X of A. So you get C is equal to, and I'll stick to the color, it says log base X of B, which is this, over log base X of A. And this is what C was. C was log base A of B. Is equal to log base A of B. Let me write it this way."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this is what C was. C was log base A of B. Is equal to log base A of B. Let me write it this way. Let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going. But I want to be fair to the color."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me write it this way. Let me do the original color codes just so it becomes very clear what I'm doing. I think you know where this is going. But I want to be fair to the color. So C is equal to log base X of B over, let me scroll down a little bit, log base X, dividing both sides by that, of A. And we know from here, I can just copy and paste it, this is also equal to C. This is how we defined it. So let me copy it and then let me paste it."}, {"video_title": "Change of base formula Logarithms Algebra II Khan Academy.mp3", "Sentence": "But I want to be fair to the color. So C is equal to log base X of B over, let me scroll down a little bit, log base X, dividing both sides by that, of A. And we know from here, I can just copy and paste it, this is also equal to C. This is how we defined it. So let me copy it and then let me paste it. So this is also equal to C. And we're done. We've proven the change of base formula. Log base A of B is equal to log base X of B divided by log base X of A."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "What I want to do in this video is talk a little bit about polynomial end behavior. And this is really just talking about what happens to a polynomial if as x becomes really large or really, really, really negative. For example, we know we're familiar with quadratic polynomials where y is equal to ax squared plus bx plus c. We know that if a is greater than 0, this is going to be an upward opening parabola of some kind. So it's going to look something like that, the graph of this equation or of this function, you could say. And if a is less than 0, it's going to be a downward opening parabola. It's going to be a downward opening parabola. We spent less time with third degree polynomials, but we've also seen those a little bit."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it's going to look something like that, the graph of this equation or of this function, you could say. And if a is less than 0, it's going to be a downward opening parabola. It's going to be a downward opening parabola. We spent less time with third degree polynomials, but we've also seen those a little bit. So for example, if you have the third degree polynomial, y is equal to ax to the third plus bx squared plus cx plus d. If a is greater than 0, if a is greater than 0, when x is really, really, really negative, this whole thing is going to be really, really, really negative. And then it's going to increase as x becomes less negative, less negative. It's going to do something."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We spent less time with third degree polynomials, but we've also seen those a little bit. So for example, if you have the third degree polynomial, y is equal to ax to the third plus bx squared plus cx plus d. If a is greater than 0, if a is greater than 0, when x is really, really, really negative, this whole thing is going to be really, really, really negative. And then it's going to increase as x becomes less negative, less negative. It's going to do something. It might do a little bit of funky stuff in between. But then as x becomes more and more and more positive, it will become more and more and more positive as well. So it might look something like this, when a is greater than 0."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's going to do something. It might do a little bit of funky stuff in between. But then as x becomes more and more and more positive, it will become more and more and more positive as well. So it might look something like this, when a is greater than 0. But what about when a is less than 0? Well, then, just like here, we would flip it. We would flip it so that, so let me write this."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So it might look something like this, when a is greater than 0. But what about when a is less than 0? Well, then, just like here, we would flip it. We would flip it so that, so let me write this. So if a is less than 0, when x is really negative, you're going to multiply that times a negative a, and you're going to get a positive value. So it's going to look something like this. And then it's going to go like this."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "We would flip it so that, so let me write this. So if a is less than 0, when x is really negative, you're going to multiply that times a negative a, and you're going to get a positive value. So it's going to look something like this. And then it's going to go like this. It might do a little bit of this type of business in between. But then its end behavior, it starts decreasing again. So when we talk about end behavior, we're talking about the idea of what is this function?"}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And then it's going to go like this. It might do a little bit of this type of business in between. But then its end behavior, it starts decreasing again. So when we talk about end behavior, we're talking about the idea of what is this function? What does this polynomial do as x becomes really, really, really, really positive and as x becomes really, really, really, really negative and kind of fully recognizing that some weird things might be happening in the middle? But we just want to think about what happens as extreme values of x. Now, obviously, for a second degree polynomial, nothing really weird happens in the middle."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So when we talk about end behavior, we're talking about the idea of what is this function? What does this polynomial do as x becomes really, really, really, really positive and as x becomes really, really, really, really negative and kind of fully recognizing that some weird things might be happening in the middle? But we just want to think about what happens as extreme values of x. Now, obviously, for a second degree polynomial, nothing really weird happens in the middle. But for a third degree polynomial, we see that some interesting things can start happening in the middle. But the end behavior for a third degree polynomial is that if a is greater than 0, we're starting really small, really low values. And as a becomes positive, we get to really high values."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now, obviously, for a second degree polynomial, nothing really weird happens in the middle. But for a third degree polynomial, we see that some interesting things can start happening in the middle. But the end behavior for a third degree polynomial is that if a is greater than 0, we're starting really small, really low values. And as a becomes positive, we get to really high values. If a is less than 0, we have the opposite. And these are kind of the two prototypes for polynomials. Because from there, we can start thinking about any degree polynomial."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And as a becomes positive, we get to really high values. If a is less than 0, we have the opposite. And these are kind of the two prototypes for polynomials. Because from there, we can start thinking about any degree polynomial. So let's just think about the situation of a fourth degree polynomial. So let's say y is equal to ax to the fourth power plus bx to the third plus cx squared plus dx plus. I don't want to write e because e has other meanings in mathematics."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Because from there, we can start thinking about any degree polynomial. So let's just think about the situation of a fourth degree polynomial. So let's say y is equal to ax to the fourth power plus bx to the third plus cx squared plus dx plus. I don't want to write e because e has other meanings in mathematics. I'll say plus. Well, I'm really running out of letters here. Plus, even f, I don't want to."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I don't want to write e because e has other meanings in mathematics. I'll say plus. Well, I'm really running out of letters here. Plus, even f, I don't want to. I'll just use f, although this isn't the function f. This is just a constant f right over here. So let's just think about what this might look like. Let's think about its end behavior."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Plus, even f, I don't want to. I'll just use f, although this isn't the function f. This is just a constant f right over here. So let's just think about what this might look like. Let's think about its end behavior. And we can think about it relative to a second degree polynomial. So its end behavior, if x is really, really, really, really negative, x to the fourth is still going to be positive. And if a is greater than 0, when x is really, really, really negative, we're going to have really, really positive values, just like a second degree."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Let's think about its end behavior. And we can think about it relative to a second degree polynomial. So its end behavior, if x is really, really, really, really negative, x to the fourth is still going to be positive. And if a is greater than 0, when x is really, really, really negative, we're going to have really, really positive values, just like a second degree. And when x is really positive, same thing. x to the fourth is going to be positive. Times a is still going to be positive."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And if a is greater than 0, when x is really, really, really negative, we're going to have really, really positive values, just like a second degree. And when x is really positive, same thing. x to the fourth is going to be positive. Times a is still going to be positive. So its end behavior is going to look very similar to a second degree polynomial. Now, it might do, in fact, it probably will do some funky stuff in between. It might do something that looks kind of like that in between."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Times a is still going to be positive. So its end behavior is going to look very similar to a second degree polynomial. Now, it might do, in fact, it probably will do some funky stuff in between. It might do something that looks kind of like that in between. But we care about the end behavior. So this is, I guess you could call the stuff that I've dotted line in the middle, this is called the non-end behavior, the middle behavior. This will obviously be different than a second degree polynomial."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It might do something that looks kind of like that in between. But we care about the end behavior. So this is, I guess you could call the stuff that I've dotted line in the middle, this is called the non-end behavior, the middle behavior. This will obviously be different than a second degree polynomial. But what happens at the ends will be the same. And the reason why, when you square something or you raise something to the fourth power, you raise anything to any even power, as long as a is greater than 0, for very large positive values, you're going to get positive values. And for very large negative values, you're going to get very large positive values."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "This will obviously be different than a second degree polynomial. But what happens at the ends will be the same. And the reason why, when you square something or you raise something to the fourth power, you raise anything to any even power, as long as a is greater than 0, for very large positive values, you're going to get positive values. And for very large negative values, you're going to get very large positive values. You take a negative number, raise it to the fourth power or the second power, you're just going to get a positive value. Likewise, if a is less than 0, you're going to have very similar end behavior to this case. For a polynomial where the highest degree term is even, so this is a is less than 0, your end behavior, when a is really, really, really, really negative, this thing's going to be really, really, really positive."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And for very large negative values, you're going to get very large positive values. You take a negative number, raise it to the fourth power or the second power, you're just going to get a positive value. Likewise, if a is less than 0, you're going to have very similar end behavior to this case. For a polynomial where the highest degree term is even, so this is a is less than 0, your end behavior, when a is really, really, really, really negative, this thing's going to be really, really, really positive. We're going to be multiplying it times the negative, so it's going to be really, really, really, really negative. So it'll look like this. And likewise, when x is really, really, really positive, you get the same thing, because you're going to be multiplying a positive times a, which is negative."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "For a polynomial where the highest degree term is even, so this is a is less than 0, your end behavior, when a is really, really, really, really negative, this thing's going to be really, really, really positive. We're going to be multiplying it times the negative, so it's going to be really, really, really, really negative. So it'll look like this. And likewise, when x is really, really, really positive, you get the same thing, because you're going to be multiplying a positive times a, which is negative. And in between, it might do something like that. But its end behavior, you see, is very similar to a second degree polynomial. So if you ignore this, its end behavior is very similar."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And likewise, when x is really, really, really positive, you get the same thing, because you're going to be multiplying a positive times a, which is negative. And in between, it might do something like that. But its end behavior, you see, is very similar to a second degree polynomial. So if you ignore this, its end behavior is very similar. Now, the same is true for a fifth degree, if you were to compare it to a third degree. And the overall idea here is what happens to this value when we get really large x's or really small x's? Are we taking it to an even power, in which case, for either really negative values or really positive values, we're going to get positive values, and then it depends what our coefficient a is."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So if you ignore this, its end behavior is very similar. Now, the same is true for a fifth degree, if you were to compare it to a third degree. And the overall idea here is what happens to this value when we get really large x's or really small x's? Are we taking it to an even power, in which case, for either really negative values or really positive values, we're going to get positive values, and then it depends what our coefficient a is. Or are we taking it to an odd power? So the general idea, and actually, let me just do a fifth degree just to make the clear. So if I had something of the form y is equal to ax to the fifth plus bx to the fourth plus, and it just went all the way."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Are we taking it to an even power, in which case, for either really negative values or really positive values, we're going to get positive values, and then it depends what our coefficient a is. Or are we taking it to an odd power? So the general idea, and actually, let me just do a fifth degree just to make the clear. So if I had something of the form y is equal to ax to the fifth plus bx to the fourth plus, and it just went all the way. I don't even have to write it. This thing, if a is greater than 0, would look something like this. Its end behavior is very similar, or it is similar to a third degree polynomial, where a is greater than 0."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So if I had something of the form y is equal to ax to the fifth plus bx to the fourth plus, and it just went all the way. I don't even have to write it. This thing, if a is greater than 0, would look something like this. Its end behavior is very similar, or it is similar to a third degree polynomial, where a is greater than 0. At the end, it would do this. Now, it might do some craziness like this. I have to get this right."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Its end behavior is very similar, or it is similar to a third degree polynomial, where a is greater than 0. At the end, it would do this. Now, it might do some craziness like this. I have to get this right. So it's 1, 2, 3. It might do some craziness like this in between. But then for really large x's, it will look the same as ax to the third, when a is greater than 0."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "I have to get this right. So it's 1, 2, 3. It might do some craziness like this in between. But then for really large x's, it will look the same as ax to the third, when a is greater than 0. So once again, very, very similar end behavior when a is greater than 0, and very similar end behavior when a is less than 0. It would look like this. At the ends, at a negative value, it will be positive, because this part is going to be really negative, but then it's going to be multiplied by a negative to get a positive."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "But then for really large x's, it will look the same as ax to the third, when a is greater than 0. So once again, very, very similar end behavior when a is greater than 0, and very similar end behavior when a is less than 0. It would look like this. At the ends, at a negative value, it will be positive, because this part is going to be really negative, but then it's going to be multiplied by a negative to get a positive. And for really positive values of x, it will be negative, because once again, this a term is going to be negative. And then what it does in between, where at least for the sake of this video, we're not really thinking about. So the big takeaway here, and this is kind of a little bit of a drum roll here when we're talking about end behavior, if you're looking at an even degree polynomial, it's going to have end behavior like a second degree polynomial."}, {"video_title": "Polynomial end behavior Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "At the ends, at a negative value, it will be positive, because this part is going to be really negative, but then it's going to be multiplied by a negative to get a positive. And for really positive values of x, it will be negative, because once again, this a term is going to be negative. And then what it does in between, where at least for the sake of this video, we're not really thinking about. So the big takeaway here, and this is kind of a little bit of a drum roll here when we're talking about end behavior, if you're looking at an even degree polynomial, it's going to have end behavior like a second degree polynomial. If you ignore what happens in the middle, what happens at really negative values of x and really positive values of x is going to be very similar to a second degree polynomial. And if your degree is odd, you're going to have very similar end behavior to a third degree polynomial. You might do all sorts of craziness in the middle, but for a given a, whether it's greater than 0 or less than 0, you will have end behavior like this or end behavior like that."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "You could argue whether it's going to be more simple or not. And the logarithm property that I'm guessing that we should use for this example right here is the property, if I take log base x of, let me pick some more letters here, of log base x of y to the z power, that this is the same thing as z times log base x of y. So this is a logarithm property. If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base of just the y in this case. So we apply this property over here, and in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "If I'm taking the logarithm of a given base of something to a power, I could take that power out front and multiply that times the log of the base of just the y in this case. So we apply this property over here, and in a second, once I do this problem, we'll talk about why this actually makes a lot of sense and comes straight out of exponent properties. But if we just apply that over here, we get log base 5 of x to the third. Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as, let me do this in a different color, that 3 is the same thing. I'll put it out front, that's the same thing as 3 times the logarithm base 5 of x."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, this is the exponent right over here. That's the same thing as z. So that's going to be the same thing as, let me do this in a different color, that 3 is the same thing. I'll put it out front, that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property. So you could argue that this is a, maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "I'll put it out front, that's the same thing as 3 times the logarithm base 5 of x. And we're done. This is just another way of writing it using this property. So you could argue that this is a, maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that, and I'll just pick some arbitrary letters here, let's say that we know that a to the b power is equal to c. And so if we know that, that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "So you could argue that this is a, maybe this is a simplification because you took the exponent outside of the logarithm and you're multiplying the logarithm by that number now. Now with that out of the way, let's think about why that actually makes sense. So let's say that we know that, and I'll just pick some arbitrary letters here, let's say that we know that a to the b power is equal to c. And so if we know that, that's written as an exponential equation. If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we wanted to write the same truth as a logarithmic equation, we would say logarithm base a of c is equal to b. To what power do I have to raise a to get c? I raise it to the bth power. a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here and raise it to the dth power. So let's take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "a to the b power is equal to c. Fair enough. Now let's take both sides of this equation right over here and raise it to the dth power. So let's take both sides of this equation and raise it to the dth power. Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting the statement. But let me take both sides of this to the dth power. And I should be consistent."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Instead of doing it in place, I'm just going to rewrite it over here. So I wrote the original equation, a to the b is equal to c, which is just rewriting the statement. But let me take both sides of this to the dth power. And I should be consistent. I'll use all capital letters. So this should be a b. Actually, let's say I'm using all lowercase letters."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I should be consistent. I'll use all capital letters. So this should be a b. Actually, let's say I'm using all lowercase letters. So this is a lowercase c. So let me write it this way. a to the, so I'm going to raise this to the dth power. And I'm going to raise this to the dth power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Actually, let's say I'm using all lowercase letters. So this is a lowercase c. So let me write it this way. a to the, so I'm going to raise this to the dth power. And I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now what's interesting over here, is we can now say, what we can do is we can use our, what we know about exponent properties, say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I'm going to raise this to the dth power. Obviously, if these two things are equal to each other, if I raise both sides to the same power, the equality is still going to hold. Now what's interesting over here, is we can now say, what we can do is we can use our, what we know about exponent properties, say, look, if I have a to the b power, and then I raise that to the d power, our exponent properties say that this is the same thing. This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. Actually, let me do that in a different color."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is equal to a to the bd power. This is equal to a to the bd. Let me write it here. Actually, let me do that in a different color. I've already used that green. Let me do this right over here, using what we know about exponent properties. This is the same thing as a to the bd power."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "Actually, let me do that in a different color. I've already used that green. Let me do this right over here, using what we know about exponent properties. This is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we were to write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to to get to c to the dth power?"}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is the same thing as a to the bd power. So we have a to the bd power is equal to c to the dth power. And now this exponential equation, if we were to write it as a logarithmic equation, we would say log base a of c to the dth power is equal to bd. What power do I have to raise a to to get to c to the dth power? To get to this, I have to raise it to the bd power. But what do we know that b is? We already know that b is this thing right over here."}, {"video_title": "Logarithm of a power Logarithms Algebra II Khan Academy.mp3", "Sentence": "What power do I have to raise a to to get to c to the dth power? To get to this, I have to raise it to the bd power. But what do we know that b is? We already know that b is this thing right over here. So if we substitute this in for b, and we can rewrite this as db, we get logarithm base a of c to the dth power is equal to bd, or you could also call that db, if you switch the order. So that's equal to d times b. b is just log base a of c. So there you have it. We just derived this property."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is the graph of g? And if we were doing this on Khan Academy, this is a screenshot from our mobile app, it has multiple choices, but I thought we would just try to sketch it. So pause this video, maybe in your mind, imagine what you think the graph of g is going to look like, or at least how you would tackle it. All right, so g of x is equal to 1 3rd f of x. So for example, we can see here that f of three is equal to negative three. So g of three should be 1 3rd that, so it should be negative one. Likewise, so g of three would be right over there."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, so g of x is equal to 1 3rd f of x. So for example, we can see here that f of three is equal to negative three. So g of three should be 1 3rd that, so it should be negative one. Likewise, so g of three would be right over there. And likewise, g of negative three, what would that be? Well, f of negative three is three. So g of negative three is going to be 1 3rd that, or it's going to be equal to one."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Likewise, so g of three would be right over there. And likewise, g of negative three, what would that be? Well, f of negative three is three. So g of negative three is going to be 1 3rd that, or it's going to be equal to one. F of zero is zero, 1 3rd of that is still zero, so g of zero is still going to be right over there. And we know that's going to happen there and there as well. And so we already have a sense of what this graph is going to look like."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of negative three is going to be 1 3rd that, or it's going to be equal to one. F of zero is zero, 1 3rd of that is still zero, so g of zero is still going to be right over there. And we know that's going to happen there and there as well. And so we already have a sense of what this graph is going to look like. The function g is going to look something like this. I'm just connecting the dots. And they did give us some dots that we can use as reference points."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so we already have a sense of what this graph is going to look like. The function g is going to look something like this. I'm just connecting the dots. And they did give us some dots that we can use as reference points. So the graph of g is going to look something like this. It gets a little bit flattened out or a little bit squooshed or smooshed a little bit to look something like that, and you would pick the choice that looks like that. Let's do another example."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And they did give us some dots that we can use as reference points. So the graph of g is going to look something like this. It gets a little bit flattened out or a little bit squooshed or smooshed a little bit to look something like that, and you would pick the choice that looks like that. Let's do another example. So here, we're told this is the graph of f of x, and it's defined by this expression. What is the graph of g of x, and g of x is this? So pause this video and think about it again."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do another example. So here, we're told this is the graph of f of x, and it's defined by this expression. What is the graph of g of x, and g of x is this? So pause this video and think about it again. All right, now the key realization is is it looks like g of x is, if you were to take all the terms of f of x and multiply it by two, or at least if you were to multiply the absolute value by two, and then if you were to multiply this negative two by two. So it looks like g of x is equal to two times f of x. And we could even set up a little table here."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and think about it again. All right, now the key realization is is it looks like g of x is, if you were to take all the terms of f of x and multiply it by two, or at least if you were to multiply the absolute value by two, and then if you were to multiply this negative two by two. So it looks like g of x is equal to two times f of x. And we could even set up a little table here. This is another way that we can think about it. We can think about x. We can think about f of x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And we could even set up a little table here. This is another way that we can think about it. We can think about x. We can think about f of x. And now we can think about g of x, which should be two times that. So we can see that when x is equal to zero, f of x is equal to one. So g of x should be equal to two, because it's two times f of x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We can think about f of x. And now we can think about g of x, which should be two times that. So we can see that when x is equal to zero, f of x is equal to one. So g of x should be equal to two, because it's two times f of x. So g of x is going to be equal to, or g of zero, I should say, is going to be equal to two. What about when at x equals, well, say when x equals three? When x equals three, f of x is negative two."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of x should be equal to two, because it's two times f of x. So g of x is going to be equal to, or g of zero, I should say, is going to be equal to two. What about when at x equals, well, say when x equals three? When x equals three, f of x is negative two. G of x is going to be two times that, because it's two times f of x. So it's going to be negative four. So g of x, or I should say g of three, is going to be negative four."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x equals three, f of x is negative two. G of x is going to be two times that, because it's two times f of x. So it's going to be negative four. So g of x, or I should say g of three, is going to be negative four. It's going to be right over there. And then maybe let's think about one more point. So f of five is equal to zero."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of x, or I should say g of three, is going to be negative four. It's going to be right over there. And then maybe let's think about one more point. So f of five is equal to zero. G of five is going to be two times that, which is still going to be equal to zero. So it's going to be right over there. And so the graph is going to look something like this."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So f of five is equal to zero. G of five is going to be two times that, which is still going to be equal to zero. So it's going to be right over there. And so the graph is going to look something like this. I'm just really just connecting, I'm just connecting the dots, trying to draw some straight lines. It's going to look something like this. You can see it's kind of stretched in the vertical direction."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so the graph is going to look something like this. I'm just really just connecting, I'm just connecting the dots, trying to draw some straight lines. It's going to look something like this. You can see it's kind of stretched in the vertical direction. So if you were doing this on Khan Academy, it'd be multiple choice. You'd look for the graph that looks like that. Let's do a few more examples."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You can see it's kind of stretched in the vertical direction. So if you were doing this on Khan Academy, it'd be multiple choice. You'd look for the graph that looks like that. Let's do a few more examples. So here we're given a function g is a vertically scaled version of f. So we can see that g is a vertically scaled version of f. The functions are graphed where f is a solid and g is dash. Yeah, we see that. What is the equation of g in terms of f?"}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's do a few more examples. So here we're given a function g is a vertically scaled version of f. So we can see that g is a vertically scaled version of f. The functions are graphed where f is a solid and g is dash. Yeah, we see that. What is the equation of g in terms of f? So pause this video and try to think about it. Well, the way that I would tackle this is once again, let's do it with a table. Let's see the relationship between f and g. So this column is x."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is the equation of g in terms of f? So pause this video and try to think about it. Well, the way that I would tackle this is once again, let's do it with a table. Let's see the relationship between f and g. So this column is x. This column is f of x. And then this column is g of x. Make another column right over here."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's see the relationship between f and g. So this column is x. This column is f of x. And then this column is g of x. Make another column right over here. And so let's see some interesting points. So when, actually, I could pick zero, but zero is maybe less interesting than this point over here. So this is when x is equal to negative three."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Make another column right over here. And so let's see some interesting points. So when, actually, I could pick zero, but zero is maybe less interesting than this point over here. So this is when x is equal to negative three. F of negative three is negative three. What is g of negative three? It looks like it is negative nine."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is when x is equal to negative three. F of negative three is negative three. What is g of negative three? It looks like it is negative nine. When f is, when x is zero, f of zero is negative two. What is g of zero? It is equal to negative six."}, {"video_title": "Scaling functions vertically examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It looks like it is negative nine. When f is, when x is zero, f of zero is negative two. What is g of zero? It is equal to negative six. And so we already see a pattern forming. Whatever f is, g is three times that. Whatever f is, g is three times that."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "We're told a monkey is swinging from a tree. On the first swing, she passes through an arc of 24 meters. With each swing, she passes through an arc one half the length of the previous swing. So what's going on here, let's say this is the top of the rope or the vine that the monkey is swinging from. And so on that first swing, I could draw a little monkey here. So this is my little monkey. So on that first swing, the monkey will go 24 meters."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So what's going on here, let's say this is the top of the rope or the vine that the monkey is swinging from. And so on that first swing, I could draw a little monkey here. So this is my little monkey. So on that first swing, the monkey will go 24 meters. So might do something like this. Then that arc is 24 meters. And then on the second swing, it would be, she'd swing back at an arc half the length of the previous swing."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So on that first swing, the monkey will go 24 meters. So might do something like this. Then that arc is 24 meters. And then on the second swing, it would be, she'd swing back at an arc half the length of the previous swing. So then she would come back, and then it would be half the length, and so maybe swing back over here. And then on the next, so that would be 12. And then on the next swing, she would swing half of that, which would be six meters."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "And then on the second swing, it would be, she'd swing back at an arc half the length of the previous swing. So then she would come back, and then it would be half the length, and so maybe swing back over here. And then on the next, so that would be 12. And then on the next swing, she would swing half of that, which would be six meters. And so she might swing like this. And that makes sense, that's consistent with our experiences swinging from trees, for those of us who've done that. So let's look at the first choice."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "And then on the next swing, she would swing half of that, which would be six meters. And so she might swing like this. And that makes sense, that's consistent with our experiences swinging from trees, for those of us who've done that. So let's look at the first choice. Which expression gives the total length the monkey swings in her first n swings? So pause the video and see if you can do that. And you can express it as, actually express it both two ways."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So let's look at the first choice. Which expression gives the total length the monkey swings in her first n swings? So pause the video and see if you can do that. And you can express it as, actually express it both two ways. Express it as a geometric series, but also express it as the sum of a geometric series if we were to actually evaluate it. So let's do this together. So we already said on the first swing, the monkey goes 24 meters."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "And you can express it as, actually express it both two ways. Express it as a geometric series, but also express it as the sum of a geometric series if we were to actually evaluate it. So let's do this together. So we already said on the first swing, the monkey goes 24 meters. Now on the second swing, and I gave you a hint when I said to express it as a geometric series, she swings half that. Now I could just write a 12 here, but the half is interesting, because that's going to be my common ratio for my geometric series. Every successive swing, the arc length is half the arc length of the last swing."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So we already said on the first swing, the monkey goes 24 meters. Now on the second swing, and I gave you a hint when I said to express it as a geometric series, she swings half that. Now I could just write a 12 here, but the half is interesting, because that's going to be my common ratio for my geometric series. Every successive swing, the arc length is half the arc length of the last swing. So it's going to be 24 times 1 1\u20442. And then on the next swing, it's going to be 24, it's going to be half of this. So it's going to be 24 times 1 1\u20442 times 1 1\u20442."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "Every successive swing, the arc length is half the arc length of the last swing. So it's going to be 24 times 1 1\u20442. And then on the next swing, it's going to be 24, it's going to be half of this. So it's going to be 24 times 1 1\u20442 times 1 1\u20442. So that's 24 times 1 1\u20442 to the second power. And so this would be the first three swings. Notice that the exponent here, we got to the second power."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So it's going to be 24 times 1 1\u20442 times 1 1\u20442. So that's 24 times 1 1\u20442 to the second power. And so this would be the first three swings. Notice that the exponent here, we got to the second power. So the first n swings, we are going to get to 24 times 1 1\u20442 not to the nth power, but to the n minus one power. Notice, after two swings, we only get to 24 times 1 1\u20442 to the first power, after three swings to the second power. So after n swings to the n minus one power."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "Notice that the exponent here, we got to the second power. So the first n swings, we are going to get to 24 times 1 1\u20442 not to the nth power, but to the n minus one power. Notice, after two swings, we only get to 24 times 1 1\u20442 to the first power, after three swings to the second power. So after n swings to the n minus one power. Now, as I said, we don't want to just have this expression. We actually want to know how do we evaluate this? And the way we evaluate this is we look at the formula, which we've explained and we've proven in other videos, the formula for a finite geometric series."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So after n swings to the n minus one power. Now, as I said, we don't want to just have this expression. We actually want to know how do we evaluate this? And the way we evaluate this is we look at the formula, which we've explained and we've proven in other videos, the formula for a finite geometric series. So that tells us, and I'll just write it over here, the sum of first n terms is a, where a is the first term, so that's going to be our 24 in this situation. It's a minus a times our common ratio. I already said that our common ratio is 1 1\u20442 to the nth power."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "And the way we evaluate this is we look at the formula, which we've explained and we've proven in other videos, the formula for a finite geometric series. So that tells us, and I'll just write it over here, the sum of first n terms is a, where a is the first term, so that's going to be our 24 in this situation. It's a minus a times our common ratio. I already said that our common ratio is 1 1\u20442 to the nth power. So one way I like to remember it is, it is our first term minus the first term that we didn't include, or minus what would have been the term right after this. All of that over one minus our common ratio. And there's other ways that you might have seen this written you could factor an a out and you might have seen something like this, a times one minus r to the n, all of that over one minus r. These two are equivalent."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "I already said that our common ratio is 1 1\u20442 to the nth power. So one way I like to remember it is, it is our first term minus the first term that we didn't include, or minus what would have been the term right after this. All of that over one minus our common ratio. And there's other ways that you might have seen this written you could factor an a out and you might have seen something like this, a times one minus r to the n, all of that over one minus r. These two are equivalent. But now, let's use this. So this is going to be equal to, actually I'll use this second form right over here. So our first term a is 24, so we're going to have 24, times one minus our common ratio, which is 1 1\u20442, to the nth power, well we're talking about the first n swing, so I'm just going to leave an n right over there, all of that over one minus our common ratio, one minus 1 1\u20442."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "And there's other ways that you might have seen this written you could factor an a out and you might have seen something like this, a times one minus r to the n, all of that over one minus r. These two are equivalent. But now, let's use this. So this is going to be equal to, actually I'll use this second form right over here. So our first term a is 24, so we're going to have 24, times one minus our common ratio, which is 1 1\u20442, to the nth power, well we're talking about the first n swing, so I'm just going to leave an n right over there, all of that over one minus our common ratio, one minus 1 1\u20442. So we could leave it like that, or we could simplify it a little bit if we like. One minus 1 1\u20442 is equal to 1 1\u20442. 24 divided by 1 1\u20442 is equal to 48, so if you wanted to you could simplify it to 48 times one minus 1 1\u20442 to the nth power."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "So our first term a is 24, so we're going to have 24, times one minus our common ratio, which is 1 1\u20442, to the nth power, well we're talking about the first n swing, so I'm just going to leave an n right over there, all of that over one minus our common ratio, one minus 1 1\u20442. So we could leave it like that, or we could simplify it a little bit if we like. One minus 1 1\u20442 is equal to 1 1\u20442. 24 divided by 1 1\u20442 is equal to 48, so if you wanted to you could simplify it to 48 times one minus 1 1\u20442 to the nth power. So either of these would be legitimate. Now the second part, they say what is the total distance the monkey has traveled when she completes her 25th swing? And they say round your final answer to the nearest meter."}, {"video_title": "Geometric series word problems swing Algebra 2 Khan Academy.mp3", "Sentence": "24 divided by 1 1\u20442 is equal to 48, so if you wanted to you could simplify it to 48 times one minus 1 1\u20442 to the nth power. So either of these would be legitimate. Now the second part, they say what is the total distance the monkey has traveled when she completes her 25th swing? And they say round your final answer to the nearest meter. So pause this video and see if you can work that out. All right, well we can just use this expression here, and we know that we are completing our 25th swing, so n is 25, and so we'll just put a 25 there, so that's going to be 48 times one minus 1 1\u20442 to the 25th power. Now this is going to be a very, very small, very, very small number, so it's actually going to be pretty close to 48 meters, but let's see what this is equal to, and we're going to round to the nearest meter."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "Just as a bit of review, this is going to be equal to x times x, which is x squared, plus x times negative y, which is negative xy, plus y times x, which is plus xy, and then minus y times y, or you could say y times negative y, so it's going to be minus y squared, negative xy, positive xy, so this is just going to simplify to x squared minus y squared. And this is all a review. We covered it when we thought about factoring things that are differences of squares. We thought about this when we were first learning to multiply binomials. And what we're going to do now is essentially just do the same thing, but do it with slightly more complicated expressions. And so another way of expressing what we just did is we could also write something like a plus b times a minus b is going to be equal to what? Well, it's going to be equal to a squared minus b squared."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "We thought about this when we were first learning to multiply binomials. And what we're going to do now is essentially just do the same thing, but do it with slightly more complicated expressions. And so another way of expressing what we just did is we could also write something like a plus b times a minus b is going to be equal to what? Well, it's going to be equal to a squared minus b squared. The only difference between what I did up here and what I did over here is instead of an x, I wrote an a, instead of a y, I wrote a b. So given that, let's see if we can expand and then combine like terms for if I'm multiplying these two expressions. Say I'm multiplying three plus five x to the fourth times three minus five x to the fourth."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "Well, it's going to be equal to a squared minus b squared. The only difference between what I did up here and what I did over here is instead of an x, I wrote an a, instead of a y, I wrote a b. So given that, let's see if we can expand and then combine like terms for if I'm multiplying these two expressions. Say I'm multiplying three plus five x to the fourth times three minus five x to the fourth. Pause this video and see if you can work this out. All right, well, there's two ways to approach it. You could just approach it exactly the way that I approached it up here, but we already know that when we have this pattern where we have something plus something times that same original something minus the other something, well, that's going to be of the form of this thing squared minus this thing squared."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "Say I'm multiplying three plus five x to the fourth times three minus five x to the fourth. Pause this video and see if you can work this out. All right, well, there's two ways to approach it. You could just approach it exactly the way that I approached it up here, but we already know that when we have this pattern where we have something plus something times that same original something minus the other something, well, that's going to be of the form of this thing squared minus this thing squared. And remember, the only reason why I'm applying that is I have a three right over here and here, so the three is playing the role of the a. So let me write that down. That is our a."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "You could just approach it exactly the way that I approached it up here, but we already know that when we have this pattern where we have something plus something times that same original something minus the other something, well, that's going to be of the form of this thing squared minus this thing squared. And remember, the only reason why I'm applying that is I have a three right over here and here, so the three is playing the role of the a. So let me write that down. That is our a. And then the role of the b is being played by five x to the fourth. So that is our b right over there. So this is going to be equal to a squared minus b squared, but our a is three, so it's going to be equal to three squared minus, and then our b is five x to the fourth, minus five x to the fourth squared."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "That is our a. And then the role of the b is being played by five x to the fourth. So that is our b right over there. So this is going to be equal to a squared minus b squared, but our a is three, so it's going to be equal to three squared minus, and then our b is five x to the fourth, minus five x to the fourth squared. Now, what does all of this simplify to? Well, this is going to be equal to three squared is nine, and then minus five x to the fourth squared, let's see, five squared is 25, and then x to the fourth squared, well, that is just going to be x to the fourth times x to the fourth, which is just x to the eighth. Another way to think about it are exponent properties."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "So this is going to be equal to a squared minus b squared, but our a is three, so it's going to be equal to three squared minus, and then our b is five x to the fourth, minus five x to the fourth squared. Now, what does all of this simplify to? Well, this is going to be equal to three squared is nine, and then minus five x to the fourth squared, let's see, five squared is 25, and then x to the fourth squared, well, that is just going to be x to the fourth times x to the fourth, which is just x to the eighth. Another way to think about it are exponent properties. This is the same thing as five squared times x to the fourth squared. If I raise something exponent and then raise that to another exponent, I multiply the exponents. And there you have it."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "Another way to think about it are exponent properties. This is the same thing as five squared times x to the fourth squared. If I raise something exponent and then raise that to another exponent, I multiply the exponents. And there you have it. Let's do another example. Let's say that I were to ask you, what is three y squared plus two y to the fifth times three y squared minus two y to the fifth? Pause this video and see if you can work that out."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "And there you have it. Let's do another example. Let's say that I were to ask you, what is three y squared plus two y to the fifth times three y squared minus two y to the fifth? Pause this video and see if you can work that out. Well, we're going to do it the same way. You can, of course, always just try to expand it out the way we did originally, but we can recognize here that, hey, I have an a plus a b times the a minus a b. So that's going to be equal to our a squared."}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can work that out. Well, we're going to do it the same way. You can, of course, always just try to expand it out the way we did originally, but we can recognize here that, hey, I have an a plus a b times the a minus a b. So that's going to be equal to our a squared. So what's three y squared? Well, that's going to be nine y to the fourth minus our b squared. Well, what's two y to the fifth squared?"}, {"video_title": "Polynomial special products difference of squares Algebra 2 Khan Academy.mp3", "Sentence": "So that's going to be equal to our a squared. So what's three y squared? Well, that's going to be nine y to the fourth minus our b squared. Well, what's two y to the fifth squared? Well, two squared is four, and y to the fifth squared is y to the five times two, y to the 10th power. And there's no further simplification that I could do here. I can't combine any like terms."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And you might see these surface. And they're also kind of fun to do, to realize that you can use the fact that the powers of i cycle through these values. You can use this to really, on the back of an envelope, take arbitrarily high powers of i. So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, is i to the 4th power raised to the 25th power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So let's try, just for fun, let's see what i to the 100th power is. And the realization here is that 100 is a multiple of 4. So you could say that this is the same thing as i to the 4 times 25th power. And this is the same thing, just from our exponent properties, is i to the 4th power raised to the 25th power. If you have something raised to an exponent, and that is raised to an exponent, that's the same thing as multiplying the two exponents. And we know that i to the 4th, that's pretty straightforward. i to the 4th is just 1. i to the 4th is 1, so this is 1."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And this is the same thing, just from our exponent properties, is i to the 4th power raised to the 25th power. If you have something raised to an exponent, and that is raised to an exponent, that's the same thing as multiplying the two exponents. And we know that i to the 4th, that's pretty straightforward. i to the 4th is just 1. i to the 4th is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "i to the 4th is just 1. i to the 4th is 1, so this is 1. So this is equal to 1 to the 25th power, which is just equal to 1. So once again, we use this kind of cycling ability of i when you take its powers to figure out a very high exponent of i. Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4, so you can't just do that that simply. But what you could do is you could write this as a product of two numbers, one that is i to a multiple of 4th power, and then one it isn't."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "Now let's say we try something a little bit stranger. Let's try i to the 501st power. Now in this situation, 501, it's not a multiple of 4, so you can't just do that that simply. But what you could do is you could write this as a product of two numbers, one that is i to a multiple of 4th power, and then one it isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i to the 1st power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "But what you could do is you could write this as a product of two numbers, one that is i to a multiple of 4th power, and then one it isn't. And so you could rewrite this. 500 is a multiple of 4. So you could write this as i to the 500th power times i to the 1st power. You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So you could write this as i to the 500th power times i to the 1st power. You have the same base. When you multiply, you can add exponents. So this would be i to the 501st power. And we know that this is the same thing as i to the 500th power is the same thing as i to the 4th power. 4 times what? 4 times 125 is 500."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So this would be i to the 501st power. And we know that this is the same thing as i to the 500th power is the same thing as i to the 4th power. 4 times what? 4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the 4th to the 125th power. And then that times i to the 1st power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "4 times 125 is 500. So that's this part right over here. i to the 500th is the same thing as i to the 4th to the 125th power. And then that times i to the 1st power. Well, i to the 4th is 1. 1 to the 125th power is just going to be 1. This whole thing is 1."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And then that times i to the 1st power. Well, i to the 4th is 1. 1 to the 125th power is just going to be 1. This whole thing is 1. And so we are just left with i to the 1st. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "This whole thing is 1. And so we are just left with i to the 1st. So this is going to be equal to i. So it seems like a really daunting problem, something that you would have to sit and do all day. But you can use this cycling to realize, look, i to the 500th is just going to be 1. And so i to the 500th 1 is just going to be i times that. So i to any multiple of 4."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So it seems like a really daunting problem, something that you would have to sit and do all day. But you can use this cycling to realize, look, i to the 500th is just going to be 1. And so i to the 500th 1 is just going to be i times that. So i to any multiple of 4. Let me write this generally. So if you have i to any multiple of 4. So this right over here is, well, we'll just restrict k to be non-negative right now."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So i to any multiple of 4. Let me write this generally. So if you have i to any multiple of 4. So this right over here is, well, we'll just restrict k to be non-negative right now. k is greater than or equal to 0. So if we have i to any multiple of 4 right over here, we are going to get 1. Because this is the same thing as i to the 4th power to the kth power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So this right over here is, well, we'll just restrict k to be non-negative right now. k is greater than or equal to 0. So if we have i to any multiple of 4 right over here, we are going to get 1. Because this is the same thing as i to the 4th power to the kth power. And that is the same thing as 1 to the kth power, which is clearly equal to 1. And if we have anything else, if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "Because this is the same thing as i to the 4th power to the kth power. And that is the same thing as 1 to the kth power, which is clearly equal to 1. And if we have anything else, if we have i to the 4k plus 1 power, i to the 4k plus 2 power, we can then just do this technique right over here. So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out, this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So let's try that with a few more problems, just to make it clear that you can do really, really arbitrarily crazy things. So let's take i to the 7,321st power. Now, we just have to figure out, this is going to be some multiple of 4 plus something else. So to do that, well, you could just look at it by sight. That 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "So to do that, well, you could just look at it by sight. That 7,320 is divisible by 4. You can verify that by hand. And then you have that 1 left over. And so this is going to be i to the 7,320 times i to the 1st power. This is a multiple of 4. This right here is a multiple of 4."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And then you have that 1 left over. And so this is going to be i to the 7,320 times i to the 1st power. This is a multiple of 4. This right here is a multiple of 4. And I know that because any 1,000 is a multiple of 4. Any 100 is a multiple of 4. And then 20 is a multiple of 4."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "This right here is a multiple of 4. And I know that because any 1,000 is a multiple of 4. Any 100 is a multiple of 4. And then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the 1st power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And then 20 is a multiple of 4. And so this right over here will simplify to 1. Sorry, that's not i to the i-th power. This is i to the 1st power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1. And we're just going to be left with i to the 1st power, or just i."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "This is i to the 1st power. 7,321 is 7,320 plus 1. And so this part right over here is going to simplify to 1. And we're just going to be left with i to the 1st power, or just i. Let's do another one. i to the 90. Let me try something interesting."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And we're just going to be left with i to the 1st power, or just i. Let's do another one. i to the 90. Let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? It is 96."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "Let me try something interesting. i to the 99th power. So once again, what's the highest multiple of 4 that is less than 99? It is 96. So this is the same thing as i to the 96th power times i to the 3rd power. If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the 4th, and then that to the 16th power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "It is 96. So this is the same thing as i to the 96th power times i to the 3rd power. If you multiply these, same base, add the exponent, you would get i to the 99th power. i to the 96th power, since this is a multiple of 4, this is i to the 4th, and then that to the 16th power. So that's just 1 to the 16th. So this is just 1. And then you're just left with i to the 3rd power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "i to the 96th power, since this is a multiple of 4, this is i to the 4th, and then that to the 16th power. So that's just 1 to the 16th. So this is just 1. And then you're just left with i to the 3rd power. And you could either remember that i to the 3rd power is equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "And then you're just left with i to the 3rd power. And you could either remember that i to the 3rd power is equal to negative i. Or if you forget that, you could just say, look, this is the same thing as i squared times i. This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more, just for the fun of it. Let's take i to the 38th power."}, {"video_title": "Calculating i raised to arbitrary exponents Precalculus Khan Academy.mp3", "Sentence": "This is equal to i squared times i. i squared, by definition, is equal to negative 1. So you have negative 1 times i is equal to negative i. Let me do one more, just for the fun of it. Let's take i to the 38th power. Well, once again, this is equal to i to the 36th times i squared. I'm doing i to the 36th power, since that's the largest multiple of 4 that goes into 38. What's left over is this 2."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "Multiply negative 4x squared by the whole expression, 3x squared plus 25x minus 7. So if you multiply anything times a whole expression, you really just use the distributive property to multiply each term of the expression by the negative 4x squared. 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."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. We're just multiplying negative 4 times x squared times 3 times x squared."}, {"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 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. This is the exact same thing as negative 4 times 3 times x squared times x squared."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. And x squared times x squared, same base, we're taking the product."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. Now let's think about this term over here."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. So let's just multiply the numbers out here."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. So it will be plus negative 100 or we can just say it's minus 100."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. 2 plus 1 is 3."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. We have negative 4x squared."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. We have negative 4."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. I'm just changing the order in which we multiply it."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. There's no simplification to do."}, {"video_title": "Example 3 Multiplying a monomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "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. So we are done."}, {"video_title": "Using multiple logarithm properties to simplify Logarithms Algebra II Khan Academy.mp3", "Sentence": "So the first thing that we realize, and this is the logarithm, one of our logarithm properties, is logarithm for a given base, so let's say that the base is x, of a over b, that is equal to log base x of a minus log base x of b. And here we have 25 to the x over y, so we can simplify, so let me write this down, I'll do this in blue, log base 5 of 25 to the x over y, using this property means that it's the same thing as log base 5 of 25 to the x power minus log base 5 of y. Now, this looks like we can do a little bit of simplifying, and it seems like the relevant logarithm property here is if I have log base x of a to the b power, that's the same thing as b times log base x of a, that this exponent over here can be moved out front, which is what we did right over there. This part right over here can be rewritten as x times the logarithm base 5 of 25, and then of course we have minus log base 5 of y. And this is useful because log base 5 of 25 is actually fairly easy to think about. This part right here is asking us, what power do I have to raise 5 to to get to 25? So we have to raise 5 to the second power to get to 25, so this simplifies to 2."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "What I want to do in this video is prove the change of base formula for logarithms. Change of base formula, which tells us that if I want to figure out the logarithm base a of x, that I can figure this out by taking logarithms with a different base. That this would be equal to the logarithm base b, so some other base, base b of x divided by the logarithm base b of a. This is a really useful result if your calculator only has natural logarithm or log base 10. You can now use this to figure out the logarithm using any base. If you want to figure out the log base 2, let me make it clear. If you want to figure out the logarithm base 3 of 25, you can use your calculator either using log base 10 or log base 2."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is a really useful result if your calculator only has natural logarithm or log base 10. You can now use this to figure out the logarithm using any base. If you want to figure out the log base 2, let me make it clear. If you want to figure out the logarithm base 3 of 25, you can use your calculator either using log base 10 or log base 2. You can say that this is going to be equal to log base 10 of 25, and most calculators have a button for that, divided by log base 10 of 3. This is an application of the change of base formula, but let's actually prove it. Let's say that we want to set logarithm base a of x to be equal to some new variable."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "If you want to figure out the logarithm base 3 of 25, you can use your calculator either using log base 10 or log base 2. You can say that this is going to be equal to log base 10 of 25, and most calculators have a button for that, divided by log base 10 of 3. This is an application of the change of base formula, but let's actually prove it. Let's say that we want to set logarithm base a of x to be equal to some new variable. Let's call that variable equal to y. This right over here, we are just setting that equal to y. This is just another way of saying that a to the y power is equal to x."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's say that we want to set logarithm base a of x to be equal to some new variable. Let's call that variable equal to y. This right over here, we are just setting that equal to y. This is just another way of saying that a to the y power is equal to x. We can rewrite this as a to the y power is equal to x. I'll write the x out here. These two things are equal. This is just another way of restating what we just wrote up here."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is just another way of saying that a to the y power is equal to x. We can rewrite this as a to the y power is equal to x. I'll write the x out here. These two things are equal. This is just another way of restating what we just wrote up here. Let's introduce the logarithm base b. To introduce it, I'm just going to take log base b of both sides of this equation. Let's take logarithm base b of the left-hand side and logarithm base b of the right-hand side."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is just another way of restating what we just wrote up here. Let's introduce the logarithm base b. To introduce it, I'm just going to take log base b of both sides of this equation. Let's take logarithm base b of the left-hand side and logarithm base b of the right-hand side. We know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something. Logarithm base b of a to the y is the same thing as y times the logarithm of base b of a. This is just a traditional logarithm property."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's take logarithm base b of the left-hand side and logarithm base b of the right-hand side. We know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something. Logarithm base b of a to the y is the same thing as y times the logarithm of base b of a. This is just a traditional logarithm property. We prove it elsewhere. We already know it's going to be equal to the right-hand side. It's going to be equal to log base b of x."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is just a traditional logarithm property. We prove it elsewhere. We already know it's going to be equal to the right-hand side. It's going to be equal to log base b of x. Now, let's just solve for y. This is exciting because y was this thing right over here. Now, if we solve for y, we're going to be solving for y in terms of logarithm base b."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "It's going to be equal to log base b of x. Now, let's just solve for y. This is exciting because y was this thing right over here. Now, if we solve for y, we're going to be solving for y in terms of logarithm base b. To solve for y, we just have to divide both sides of this equation by log base b of a. We divide by log base b of a on the left-hand side. We divide by log base b of a on the right-hand side."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now, if we solve for y, we're going to be solving for y in terms of logarithm base b. To solve for y, we just have to divide both sides of this equation by log base b of a. We divide by log base b of a on the left-hand side. We divide by log base b of a on the right-hand side. On the left-hand side, these two characters are going to cancel out. We are left with, and we deserve a drum roll now, that y is equal to log base b of x divided by log base b of a. Let me write it."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "We divide by log base b of a on the right-hand side. On the left-hand side, these two characters are going to cancel out. We are left with, and we deserve a drum roll now, that y is equal to log base b of x divided by log base b of a. Let me write it. Let's copy and paste this so I don't have to keep switching colors. Let me paste this. There you have it."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let me write it. Let's copy and paste this so I don't have to keep switching colors. Let me paste this. There you have it. You have your change of base formula. Remember, y is the same thing as this thing right over here. y is log of a."}, {"video_title": "Change of base formula proof Logarithms Algebra II Khan Academy.mp3", "Sentence": "There you have it. You have your change of base formula. Remember, y is the same thing as this thing right over here. y is log of a. Actually, let me make it clear. y, which is equal to log of a, which is equal to log base a of x, so copy and paste, y, which is equal to this thing, which is how we defined it right over here, y is equal to log base a of x, is we've just shown is also equal to this if we write it in terms of base. If we write it in terms of base b."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Lisa controls the probability in which each candy is picked. She is running out of Honey Bunny, so she wants to program its probabilities so that the probability of getting a different candy twice in a row is greater than 2 1 1 4 times the probability of getting Honey Bunny in one try. So let me read that again. She wants to program its probabilities so that the probability of getting a different candy twice in a row, or really any other candy twice in a row, is greater than 2 1 1 4 times the probability of getting Honey Bunny in one try. Write an inequality that models the situation. Use p to represent the probability of getting Honey Bunny in one try. Solve the inequality and complete the sentence."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "She wants to program its probabilities so that the probability of getting a different candy twice in a row, or really any other candy twice in a row, is greater than 2 1 1 4 times the probability of getting Honey Bunny in one try. Write an inequality that models the situation. Use p to represent the probability of getting Honey Bunny in one try. Solve the inequality and complete the sentence. Remember that the probability must be a number between 0 and 1. So we want to write the inequality that models the problem here. And then we want to complete the sentence."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Solve the inequality and complete the sentence. Remember that the probability must be a number between 0 and 1. So we want to write the inequality that models the problem here. And then we want to complete the sentence. The probability of getting Honey Bunny in one try must be, so they give us a bunch of options, greater than, greater than, or equal to, less than, less than, or equal to. And then we have to put some number here. So to work through this, I've copy and pasted this problem onto my little scratch pad right over here."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And then we want to complete the sentence. The probability of getting Honey Bunny in one try must be, so they give us a bunch of options, greater than, greater than, or equal to, less than, less than, or equal to. And then we have to put some number here. So to work through this, I've copy and pasted this problem onto my little scratch pad right over here. And so let's just think about it a little bit. So they tell us use p to represent the probability of getting Honey Bunny in one try. And they also say she wants to program its probability so that the probability of getting a different candy twice in a row is greater than 2 1 1 4 the probability of getting Honey Bunny in one try."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So to work through this, I've copy and pasted this problem onto my little scratch pad right over here. And so let's just think about it a little bit. So they tell us use p to represent the probability of getting Honey Bunny in one try. And they also say she wants to program its probability so that the probability of getting a different candy twice in a row is greater than 2 1 1 4 the probability of getting Honey Bunny in one try. So if p is the probability of getting Honey Bunny, what's the probability of getting any other candy at once? Well, that's going to be 1 minus p. If you have a probability of p of getting Honey Bunny, then it's 1 minus p of anything but Honey Bunny. Now, what's the probability of getting this twice in a row, of getting anything else twice in a row?"}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And they also say she wants to program its probability so that the probability of getting a different candy twice in a row is greater than 2 1 1 4 the probability of getting Honey Bunny in one try. So if p is the probability of getting Honey Bunny, what's the probability of getting any other candy at once? Well, that's going to be 1 minus p. If you have a probability of p of getting Honey Bunny, then it's 1 minus p of anything but Honey Bunny. Now, what's the probability of getting this twice in a row, of getting anything else twice in a row? Well, you're just going to multiply this probability times itself. It's going to be 1 minus p times 1 minus p. Or we could just write that as 1 minus p squared. So this right over here is the probability of getting a different candy."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, what's the probability of getting this twice in a row, of getting anything else twice in a row? Well, you're just going to multiply this probability times itself. It's going to be 1 minus p times 1 minus p. Or we could just write that as 1 minus p squared. So this right over here is the probability of getting a different candy. Any other candy twice in a row. So prob of any other non-Honey Bunny candy, any other candy twice in a row. Now, they tell us that this probability needs to be greater than 2 1 1 4 times the probability."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this right over here is the probability of getting a different candy. Any other candy twice in a row. So prob of any other non-Honey Bunny candy, any other candy twice in a row. Now, they tell us that this probability needs to be greater than 2 1 1 4 times the probability. 2 1 1 4 is the probability of getting Honey Bunny in one try. So it's greater than 2 1 1 4 times the probability of getting Honey Bunny in one try. Well, that is p. So we have just set up the first part."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now, they tell us that this probability needs to be greater than 2 1 1 4 times the probability. 2 1 1 4 is the probability of getting Honey Bunny in one try. So it's greater than 2 1 1 4 times the probability of getting Honey Bunny in one try. Well, that is p. So we have just set up the first part. We have written an inequality that models the situation. Now let's actually solve this inequality. And so to do that, I will just expand 1 minus p squared out."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, that is p. So we have just set up the first part. We have written an inequality that models the situation. Now let's actually solve this inequality. And so to do that, I will just expand 1 minus p squared out. 1 minus p squared is the same thing as p squared. Well, I'll just multiply it out. So this is going to be 1 squared minus 2p plus p squared."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so to do that, I will just expand 1 minus p squared out. 1 minus p squared is the same thing as p squared. Well, I'll just multiply it out. So this is going to be 1 squared minus 2p plus p squared. And that's going to be greater than 2 1 1 4 p. Now let's see, if we subtract 2 1 1 4 p from both sides, we are going to be left with, and I'm going to reorder this, we're going to get p squared. So you have minus 2p minus 2 1 1 4 p. So that's going to get us minus 4 1 1 4 p. Or let me just write that as 17 over 4p plus 1 is greater than 0. And so let's think about solving this quadratic right over here."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So this is going to be 1 squared minus 2p plus p squared. And that's going to be greater than 2 1 1 4 p. Now let's see, if we subtract 2 1 1 4 p from both sides, we are going to be left with, and I'm going to reorder this, we're going to get p squared. So you have minus 2p minus 2 1 1 4 p. So that's going to get us minus 4 1 1 4 p. Or let me just write that as 17 over 4p plus 1 is greater than 0. And so let's think about solving this quadratic right over here. Under which circumstances is this greater than 0? And to think about it, let's factor it. And actually, before we factor it, let's simplify it a little bit."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And so let's think about solving this quadratic right over here. Under which circumstances is this greater than 0? And to think about it, let's factor it. And actually, before we factor it, let's simplify it a little bit. I don't like having this 17 fourths right over here. So let's multiply both sides times 4. And since 4 is a positive number, it's not going to change the direction of this inequality."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And actually, before we factor it, let's simplify it a little bit. I don't like having this 17 fourths right over here. So let's multiply both sides times 4. And since 4 is a positive number, it's not going to change the direction of this inequality. So we could rewrite this as 4p squared minus 17p plus 4 is greater than 0. And let's see, what are the roots of this? And we could use the quadratic formula if we wanted to do it really quick."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And since 4 is a positive number, it's not going to change the direction of this inequality. So we could rewrite this as 4p squared minus 17p plus 4 is greater than 0. And let's see, what are the roots of this? And we could use the quadratic formula if we wanted to do it really quick. We could probably do it other ways. But negative b, so it's going to be 17 plus or minus the square root of negative 17 squared. So that's 289 minus 4 times a times c. Well, a times c is 16 times 4, so minus 64."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And we could use the quadratic formula if we wanted to do it really quick. We could probably do it other ways. But negative b, so it's going to be 17 plus or minus the square root of negative 17 squared. So that's 289 minus 4 times a times c. Well, a times c is 16 times 4, so minus 64. All of that over 2 times a. All of that over 8. So that's 17 plus or minus, let's see, this is the square root of 225 over 8, which is equal to 17 plus or minus 15 over 8, which is equal to, let's see, 17 minus 15 over 8 is 2 eighths, which is equal to 2 eighths or 1 fourth."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that's 289 minus 4 times a times c. Well, a times c is 16 times 4, so minus 64. All of that over 2 times a. All of that over 8. So that's 17 plus or minus, let's see, this is the square root of 225 over 8, which is equal to 17 plus or minus 15 over 8, which is equal to, let's see, 17 minus 15 over 8 is 2 eighths, which is equal to 2 eighths or 1 fourth. So that's one of them. That's when we take the minus. And if we add 17 plus 15, that gets us to 32 divided by 8 is 4, or 4."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that's 17 plus or minus, let's see, this is the square root of 225 over 8, which is equal to 17 plus or minus 15 over 8, which is equal to, let's see, 17 minus 15 over 8 is 2 eighths, which is equal to 2 eighths or 1 fourth. So that's one of them. That's when we take the minus. And if we add 17 plus 15, that gets us to 32 divided by 8 is 4, or 4. So there's two situations right over here. So let's factor this out. We could write this as p minus 1 fourth times p minus 4 is greater than 0."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And if we add 17 plus 15, that gets us to 32 divided by 8 is 4, or 4. So there's two situations right over here. So let's factor this out. We could write this as p minus 1 fourth times p minus 4 is greater than 0. So under what circumstances is this going to be true? What constraints are this going to be true? Well, if you're taking the product of two terms and they are going to be greater than 0, that means that these two things have to be the same sign."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We could write this as p minus 1 fourth times p minus 4 is greater than 0. So under what circumstances is this going to be true? What constraints are this going to be true? Well, if you're taking the product of two terms and they are going to be greater than 0, that means that these two things have to be the same sign. Or actually, in particular, they both have to be positive or they both have to be negative. So let's look at those two situations. So I'll switch colors here just for fun."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, if you're taking the product of two terms and they are going to be greater than 0, that means that these two things have to be the same sign. Or actually, in particular, they both have to be positive or they both have to be negative. So let's look at those two situations. So I'll switch colors here just for fun. So both positive or both negative. So if they're both positive, that means that p minus 1 fourth has to be greater than 0. And p minus 4 is greater than 0."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So I'll switch colors here just for fun. So both positive or both negative. So if they're both positive, that means that p minus 1 fourth has to be greater than 0. And p minus 4 is greater than 0. Add 1 fourth on both sides right over here, you get p is greater than 1 fourth. And p is greater than 4. So that's the situation where they are both positive."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And p minus 4 is greater than 0. Add 1 fourth on both sides right over here, you get p is greater than 1 fourth. And p is greater than 4. So that's the situation where they are both positive. Now, what about if they're both negative? Well, then you have p minus 1 fourth is less than 0. And p minus 4 is less than 0."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So that's the situation where they are both positive. Now, what about if they're both negative? Well, then you have p minus 1 fourth is less than 0. And p minus 4 is less than 0. Add 1 fourth here. So p needs to be less than 1 fourth. And p needs to be less than 4."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And p minus 4 is less than 0. Add 1 fourth here. So p needs to be less than 1 fourth. And p needs to be less than 4. Now, what does this expression, what does this constraint simplify to? p has to be greater than 1 fourth and p has to be greater than 4. Well, if p is greater than 4, it's definitely going to be greater than 1 fourth."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And p needs to be less than 4. Now, what does this expression, what does this constraint simplify to? p has to be greater than 1 fourth and p has to be greater than 4. Well, if p is greater than 4, it's definitely going to be greater than 1 fourth. So all of this collapses into p needs to be greater than 4. That's the situation where both are positive. p must be greater than 4."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, if p is greater than 4, it's definitely going to be greater than 1 fourth. So all of this collapses into p needs to be greater than 4. That's the situation where both are positive. p must be greater than 4. Now, what about here? Well, if p is less than 1 fourth, it's definitely going to be less than 4. And this is an and right over here."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "p must be greater than 4. Now, what about here? Well, if p is less than 1 fourth, it's definitely going to be less than 4. And this is an and right over here. So this collapses to p is less than 1 fourth. So which one do we go with? p needs to be greater than 4 or p needs to be less than 1 fourth?"}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And this is an and right over here. So this collapses to p is less than 1 fourth. So which one do we go with? p needs to be greater than 4 or p needs to be less than 1 fourth? Well, we need to remind ourselves that we're talking about a probability. To go back to the original problem, we're talking about a probability of someone getting Honey Bunny in one try. A probability must be between 0 and 1."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "p needs to be greater than 4 or p needs to be less than 1 fourth? Well, we need to remind ourselves that we're talking about a probability. To go back to the original problem, we're talking about a probability of someone getting Honey Bunny in one try. A probability must be between 0 and 1. So the probability having to be greater than 4, well, that just doesn't make any sense. That doesn't make any sense in the context of this question. So we have to go with the probability of getting Honey Bunny needs to be less than 1 fourth or less than 0.25, which makes complete sense."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "A probability must be between 0 and 1. So the probability having to be greater than 4, well, that just doesn't make any sense. That doesn't make any sense in the context of this question. So we have to go with the probability of getting Honey Bunny needs to be less than 1 fourth or less than 0.25, which makes complete sense. So let's fill in this information. This was the inequality that modeled the problem. And we got p has to be less than 1 fourth."}, {"video_title": "Quadratic inequality word problem Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we have to go with the probability of getting Honey Bunny needs to be less than 1 fourth or less than 0.25, which makes complete sense. So let's fill in this information. This was the inequality that modeled the problem. And we got p has to be less than 1 fourth. So let's go back to the original problem. The inequality was 1 minus p squared needs to be greater than 2 and 1 fourth. So we could write that multiple times."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Liam opened a savings account and put $6,250 in it. Each year, the account increases by 20%. How many years will it take the account to reach $12,960? Write an equation that models the situation. Use t to represent the number of years since Liam opened the account. So I encourage you to pause this video and actually try to do it on your own first. Try to write this equation that models the situation using the t, variable t in the way they described, and then actually answer the question."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Write an equation that models the situation. Use t to represent the number of years since Liam opened the account. So I encourage you to pause this video and actually try to do it on your own first. Try to write this equation that models the situation using the t, variable t in the way they described, and then actually answer the question. How many years will it take for the account to reach $12,960? Well, let's just think about it. So t is to represent the number of years since Liam opened the account."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Try to write this equation that models the situation using the t, variable t in the way they described, and then actually answer the question. How many years will it take for the account to reach $12,960? Well, let's just think about it. So t is to represent the number of years since Liam opened the account. So let's just say it's been zero years since Liam opened the account. How much is he going to have? Well, he's just going to have $6,250 in it."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So t is to represent the number of years since Liam opened the account. So let's just say it's been zero years since Liam opened the account. How much is he going to have? Well, he's just going to have $6,250 in it. That's how much he starts with. Now let's say it's been one year. One year since he opened the account, how much will he have?"}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Well, he's just going to have $6,250 in it. That's how much he starts with. Now let's say it's been one year. One year since he opened the account, how much will he have? Well, he's going to have $6,250 times, or let's write it this way, plus 20% of $6,250. It grows 20% every year. So this is how much he started the year with, and then he gets another 20% of that $6,250."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "One year since he opened the account, how much will he have? Well, he's going to have $6,250 times, or let's write it this way, plus 20% of $6,250. It grows 20% every year. So this is how much he started the year with, and then he gets another 20% of that $6,250. If we factor out a $6,250, this is equal to $6,250 times 1 plus 20%, or we could write that as 0.2, which is equal to $6,250 times 1.2. Now how much is he going to have at the end of two years? Well, he's going to have the same amount that he had at the end of one year times 1.2, because it grew by 20% again."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So this is how much he started the year with, and then he gets another 20% of that $6,250. If we factor out a $6,250, this is equal to $6,250 times 1 plus 20%, or we could write that as 0.2, which is equal to $6,250 times 1.2. Now how much is he going to have at the end of two years? Well, he's going to have the same amount that he had at the end of one year times 1.2, because it grew by 20% again. So he's going to have the amount that he had at the end of one year times 1.2, which is equal to $6,250 times 1.2, which is equal to $6,250 times 1.2 squared. I think you might see where this is going. I could even write it like this."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Well, he's going to have the same amount that he had at the end of one year times 1.2, because it grew by 20% again. So he's going to have the amount that he had at the end of one year times 1.2, which is equal to $6,250 times 1.2, which is equal to $6,250 times 1.2 squared. I think you might see where this is going. I could even write it like this. Order of operations, you do the exponent first. So what about after three years? So after three years, well, we're just going to compound."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "I could even write it like this. Order of operations, you do the exponent first. So what about after three years? So after three years, well, we're just going to compound. We're going to multiply by 1.2 once again. So then he's going to have 6,250 times 1.2 to the third power. And so after t years, well, we're going to multiply by 1.2 that many times."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So after three years, well, we're just going to compound. We're going to multiply by 1.2 once again. So then he's going to have 6,250 times 1.2 to the third power. And so after t years, well, we're going to multiply by 1.2 that many times. So after t years, in his account, he's going to have 6,250 times 1.2 to the t-th power. Or to the t-th power. I don't want to get confused with a part of your, or I guess the thing that you use to take bites with."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And so after t years, well, we're going to multiply by 1.2 that many times. So after t years, in his account, he's going to have 6,250 times 1.2 to the t-th power. Or to the t-th power. I don't want to get confused with a part of your, or I guess the thing that you use to take bites with. Anyway, so they say write an equation that models the situation. So we want to figure out how many years will it take the account to reach 12,960. So we essentially want to say when is the account going to be 12,960."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "I don't want to get confused with a part of your, or I guess the thing that you use to take bites with. Anyway, so they say write an equation that models the situation. So we want to figure out how many years will it take the account to reach 12,960. So we essentially want to say when is the account going to be 12,960. So we could write 12,960. When is that going to be equal to 6,250 times 1.2 to the t power? So that's the equation right over there that models the situation."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So we essentially want to say when is the account going to be 12,960. So we could write 12,960. When is that going to be equal to 6,250 times 1.2 to the t power? So that's the equation right over there that models the situation. And then we need to think about how we can actually go about solving this thing. Well, a natural thing is to isolate the t variable. Let's divide both sides by 6,250."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So that's the equation right over there that models the situation. And then we need to think about how we can actually go about solving this thing. Well, a natural thing is to isolate the t variable. Let's divide both sides by 6,250. So we could get, and if we flip the two sides, we could get 1.2 to the t power is equal to 12,960 divided by 6,250. And since they're both divisible by 10, why don't we divide them both by 10? So it's 1,296 divided by 625."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Let's divide both sides by 6,250. So we could get, and if we flip the two sides, we could get 1.2 to the t power is equal to 12,960 divided by 6,250. And since they're both divisible by 10, why don't we divide them both by 10? So it's 1,296 divided by 625. And there's several ways that you could solve this problem at this point. One way, if you feel confident that this is going to have an integer answer right over here, you could literally just try to use your calculator and multiply 1.2 enough times to get whatever number this is. And so we could do it that way."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So it's 1,296 divided by 625. And there's several ways that you could solve this problem at this point. One way, if you feel confident that this is going to have an integer answer right over here, you could literally just try to use your calculator and multiply 1.2 enough times to get whatever number this is. And so we could do it that way. And as we'll see, there's a more systematic way of doing it once you learn about logarithms. I'll do that at the end, but I'll do that last just in case you haven't been exposed to logarithms yet. So you could literally say, let me just exit out of everything."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And so we could do it that way. And as we'll see, there's a more systematic way of doing it once you learn about logarithms. I'll do that at the end, but I'll do that last just in case you haven't been exposed to logarithms yet. So you could literally say, let me just exit out of everything. So you could literally say, OK, let's see. 12,960 divided by 6,25 is this value. So let's see how many times we have to multiply by 1.2."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So you could literally say, let me just exit out of everything. So you could literally say, OK, let's see. 12,960 divided by 6,25 is this value. So let's see how many times we have to multiply by 1.2. 1.2 times 1.2 gets us, well, that doesn't get us close enough. So let's try it three times. So let's take that same number."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So let's see how many times we have to multiply by 1.2. 1.2 times 1.2 gets us, well, that doesn't get us close enough. So let's try it three times. So let's take that same number. Let's just take 1.2. Let's raise it. Let's raise 1.2."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So let's take that same number. Let's just take 1.2. Let's raise it. Let's raise 1.2. Let's just do it three times. Times 1.2 times 1.2. That still doesn't get us there."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Let's raise 1.2. Let's just do it three times. Times 1.2 times 1.2. That still doesn't get us there. What if we were to multiply by 1.2 one more time? Well, that actually gets us there. So you see, if you multiply, and we just did this by brute force."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "That still doesn't get us there. What if we were to multiply by 1.2 one more time? Well, that actually gets us there. So you see, if you multiply, and we just did this by brute force. 1.2 to the fourth power will give us this value. So that's one way, kind of a brute force way, of figuring out that t is equal to 4. Another way, and this might be a little bit less intuitive, it might jump out of it."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So you see, if you multiply, and we just did this by brute force. 1.2 to the fourth power will give us this value. So that's one way, kind of a brute force way, of figuring out that t is equal to 4. Another way, and this might be a little bit less intuitive, it might jump out of it. This looks like some type of a power of 5. We know that 5 to the first is 5. 5 squared is 25."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Another way, and this might be a little bit less intuitive, it might jump out of it. This looks like some type of a power of 5. We know that 5 to the first is 5. 5 squared is 25. 5 to the third is 125. 5 to the fourth is 625. And so you might recognize this right over here is 5 to the fourth."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "5 squared is 25. 5 to the third is 125. 5 to the fourth is 625. And so you might recognize this right over here is 5 to the fourth. And it's actually a little bit harder to recognize that this right over here is 6 to the fourth power. And this right over here is 6 fifths. So we could rewrite this as 6 over 5 to the t is equal to 6 to the fourth over 5 to the fourth, which is the same thing as 6 over 5 to the fourth power."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And so you might recognize this right over here is 5 to the fourth. And it's actually a little bit harder to recognize that this right over here is 6 to the fourth power. And this right over here is 6 fifths. So we could rewrite this as 6 over 5 to the t is equal to 6 to the fourth over 5 to the fourth, which is the same thing as 6 over 5 to the fourth power. So here you'd say, well, 6 over 5 to the t needs to be equal to 6 over 5 to the fourth power. t must be equal to 4. Now, this is nice when you can recognize that this is something raised to the fourth power, which isn't easy to do."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So we could rewrite this as 6 over 5 to the t is equal to 6 to the fourth over 5 to the fourth, which is the same thing as 6 over 5 to the fourth power. So here you'd say, well, 6 over 5 to the t needs to be equal to 6 over 5 to the fourth power. t must be equal to 4. Now, this is nice when you can recognize that this is something raised to the fourth power, which isn't easy to do. Or if you know this is an integer, you could just keep multiplying 1.2 if you know it's a low integer. But the systematic way of doing it is to actually use logarithms. And there's many videos on Khan Academy about how to use logarithms."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Now, this is nice when you can recognize that this is something raised to the fourth power, which isn't easy to do. Or if you know this is an integer, you could just keep multiplying 1.2 if you know it's a low integer. But the systematic way of doing it is to actually use logarithms. And there's many videos on Khan Academy about how to use logarithms. But if you're more concerned with, well, gee, if I just want to figure out 1.2 to what power is equal to this thing, what you do, and we prove this in other videos, is if you say, look, let's take the thing that we want 1.2 to some power to be. Let's take the logarithm of that. And actually, you could take a logarithm any base."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And there's many videos on Khan Academy about how to use logarithms. But if you're more concerned with, well, gee, if I just want to figure out 1.2 to what power is equal to this thing, what you do, and we prove this in other videos, is if you say, look, let's take the thing that we want 1.2 to some power to be. Let's take the logarithm of that. And actually, you could take a logarithm any base. Your calculator tends to have a natural log, which is base e and a log base 10. We could just take log base 10. So let's do that."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And actually, you could take a logarithm any base. Your calculator tends to have a natural log, which is base e and a log base 10. We could just take log base 10. So let's do that. So we'll take the logarithm of what we want to get to, 2.0736, and divide that by the thing that we're trying to take the power of to get to this number. So divided by the logarithm of 1.2. And once again, we prove this."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So let's do that. So we'll take the logarithm of what we want to get to, 2.0736, and divide that by the thing that we're trying to take the power of to get to this number. So divided by the logarithm of 1.2. And once again, we prove this. Actually, I wanted to divide. So let me insert a division symbol. So once again, it might look like a little voodoo right here."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "And once again, we prove this. Actually, I wanted to divide. So let me insert a division symbol. So once again, it might look like a little voodoo right here. We've proven it in other videos. But if you wanted to use a calculator to calculate things like this, because sometimes it might be a nice integer number of years. It might be 3 and 1 half years, or it might be 7.1234 years, whatever it might be."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "So once again, it might look like a little voodoo right here. We've proven it in other videos. But if you wanted to use a calculator to calculate things like this, because sometimes it might be a nice integer number of years. It might be 3 and 1 half years, or it might be 7.1234 years, whatever it might be. This will give you a more precise answer. So what do you want to get to? You want to get to 2.0736."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "It might be 3 and 1 half years, or it might be 7.1234 years, whatever it might be. This will give you a more precise answer. So what do you want to get to? You want to get to 2.0736. What are you raising to some power? 1.2. Divide the log of the thing you're trying to get to, divided by the log of what you're taking, what the base that you're trying to raise to a power."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "You want to get to 2.0736. What are you raising to some power? 1.2. Divide the log of the thing you're trying to get to, divided by the log of what you're taking, what the base that you're trying to raise to a power. And you click Enter. And then you get. So this is literally another way of saying that 2.2 to the fourth power is going to be 2.0736."}, {"video_title": "Constructing an exponential equation example Algebra II Khan Academy.mp3", "Sentence": "Divide the log of the thing you're trying to get to, divided by the log of what you're taking, what the base that you're trying to raise to a power. And you click Enter. And then you get. So this is literally another way of saying that 2.2 to the fourth power is going to be 2.0736. So once again, if this looks like voodoo, you don't know what logarithms are. We have videos on Khan Academy on that. But there's multiple ways to tackle it, especially this problem where the answer was a little bit simpler."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "All right, let's work through this together. So the first thing I'd like to do is just see if I can simplify this at all, and maybe by finding some common factors between numerators and denominators, or common factors on either side of the equal sign. So let's factor all of these, all of the numerators and denominators, all the ones on the right-hand side are already done. So this thing up here, I could rewrite this as, let's see, what product is 21? What two numbers when I take their product is 21, positive 21, so they're gonna have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this thing up here, I could rewrite this as, let's see, what product is 21? What two numbers when I take their product is 21, positive 21, so they're gonna have the same sign. And when I add them, I get negative 10. Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both are divisible by three. I could rewrite this as three times x minus four. And these are already factored."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, negative seven and negative three, so this could be rewritten as x minus seven times x minus three. This over here, both are divisible by three. I could rewrite this as three times x minus four. And these are already factored. So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side. And so if I were to multiply both sides by x minus four, so actually let me just, let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And these are already factored. So the one thing that jumps out at me is I have x minus four in the denominator on the left-hand side and on the right-hand side. And so if I were to multiply both sides by x minus four, so actually let me just, let me formally replace this with that. And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form. So let me just scratch that out for now because once I, well, let me multiply by x minus four. So if we multiply both sides by x minus four, and once again, why am I doing this? So I get rid of the x minus fours in the denominator."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And up here, it's not so obvious that it's going to be valuable for me to keep this factored form, so I'm just going to keep it in this yellow form, in the expanded out form. So let me just scratch that out for now because once I, well, let me multiply by x minus four. So if we multiply both sides by x minus four, and once again, why am I doing this? So I get rid of the x minus fours in the denominator. X minus four, and then x minus four. That and that cancels, that and that cancels. And then we're left with, in the numerator, we're left with our x squared minus 10x plus 21, and let's see, divided by three, divided by three is equal to x minus five."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I get rid of the x minus fours in the denominator. X minus four, and then x minus four. That and that cancels, that and that cancels. And then we're left with, in the numerator, we're left with our x squared minus 10x plus 21, and let's see, divided by three, divided by three is equal to x minus five. Let's see, now what we could do, and actually I could have done it in the last step, is I could multiply both sides by three, multiply both sides, do that in another color just so it sticks out a little bit more. So I can multiply both sides by three. So multiply both sides by three."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then we're left with, in the numerator, we're left with our x squared minus 10x plus 21, and let's see, divided by three, divided by three is equal to x minus five. Let's see, now what we could do, and actually I could have done it in the last step, is I could multiply both sides by three, multiply both sides, do that in another color just so it sticks out a little bit more. So I can multiply both sides by three. So multiply both sides by three. On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10x plus 21, and I don't have a denominator anymore. My denominator is one, so I don't need to write it. It's going to be equal to three times, let's distribute the three."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So multiply both sides by three. On the left-hand side, that and that cancels, and I'll just be left with x squared minus 10x plus 21, and I don't have a denominator anymore. My denominator is one, so I don't need to write it. It's going to be equal to three times, let's distribute the three. Three times x is three x. Three times negative five is negative 15. And now I can get this in standard quadratic form by getting all of these terms onto the left-hand side."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's going to be equal to three times, let's distribute the three. Three times x is three x. Three times negative five is negative 15. And now I can get this in standard quadratic form by getting all of these terms onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right, otherwise the equality won't hold. I have to do it from both sides if I want the equality to hold. And I want to get rid of this negative 15, so I can add 15 to both sides."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And now I can get this in standard quadratic form by getting all of these terms onto the left-hand side. The best way to do that, let's subtract three x from the right, but I can't just do it from the right, otherwise the equality won't hold. I have to do it from both sides if I want the equality to hold. And I want to get rid of this negative 15, so I can add 15 to both sides. So let's do that. And what we are left with, scroll down a little bit so I have a little more space. What we are going to be left with is x squared minus 13x and then plus, what is this?"}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I want to get rid of this negative 15, so I can add 15 to both sides. So let's do that. And what we are left with, scroll down a little bit so I have a little more space. What we are going to be left with is x squared minus 13x and then plus, what is this? Plus 36, plus 36. Did I do that right? Yeah, plus 36 is equal to zero."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "What we are going to be left with is x squared minus 13x and then plus, what is this? Plus 36, plus 36. Did I do that right? Yeah, plus 36 is equal to zero. All right, now let's see. We have this quadratic in the standard form. How can we solve this?"}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Yeah, plus 36 is equal to zero. All right, now let's see. We have this quadratic in the standard form. How can we solve this? So the first thing, can we factor this? Product of two numbers, 36. If I add them, I get negative 13."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "How can we solve this? So the first thing, can we factor this? Product of two numbers, 36. If I add them, I get negative 13. They're both going to be negative since they have to have the same sign to get their product to be positive. And let's see, nine and four seem to do the trick, or negative nine and negative four. So x minus four times x minus nine is equal to zero."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "If I add them, I get negative 13. They're both going to be negative since they have to have the same sign to get their product to be positive. And let's see, nine and four seem to do the trick, or negative nine and negative four. So x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. Well, add four to both sides of this. This happens when x is equal to four."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So x minus four times x minus nine is equal to zero. Well, that's going to happen if either x minus four is equal to zero or x minus nine is equal to zero. Well, add four to both sides of this. This happens when x is equal to four. Add nine to both sides of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "This happens when x is equal to four. Add nine to both sides of this. This happens when x is equal to nine. So we could say that the solutions are x equals four or x equals nine. So x is equal to four or x equals nine, but we need to be careful because we have to remember in our original expression, x minus four was a factor of both denominators. And so if we actually tried to test x minus four in the original equation, not one of these intermediary steps, in the original equation, I would end up dividing by zero right over here, and actually I would end up dividing by zero right over there as well. So the original equations, if I tried to substitute four, they don't make sense."}, {"video_title": "Equations with rational expressions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we could say that the solutions are x equals four or x equals nine. So x is equal to four or x equals nine, but we need to be careful because we have to remember in our original expression, x minus four was a factor of both denominators. And so if we actually tried to test x minus four in the original equation, not one of these intermediary steps, in the original equation, I would end up dividing by zero right over here, and actually I would end up dividing by zero right over there as well. So the original equations, if I tried to substitute four, they don't make sense. So this is actually an extraneous solution. It's not going to be a solution to the original equation. The only solution is x is equal to nine."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "What is the graph of g? So pause this video and try to figure that out on your own. All right, now let's work through this. And the way I will think about it, I'll set up a little table here, and I'll have an x column, and then I will have, and then I'll have a, well, actually, just gonna put g of x column. And of course, g of x is equal to f of two x. So when x is, and actually, let me see. When x is equal to, I could pick a point like x equaling zero."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And the way I will think about it, I'll set up a little table here, and I'll have an x column, and then I will have, and then I'll have a, well, actually, just gonna put g of x column. And of course, g of x is equal to f of two x. So when x is, and actually, let me see. When x is equal to, I could pick a point like x equaling zero. So g of zero is going to be f of two times zero. So it's going to be f of two times zero, which is still f of zero, which is going to be equal to a little bit over four, so which is equal to f of zero. And so they're going to both have the same y-intercept."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to, I could pick a point like x equaling zero. So g of zero is going to be f of two times zero. So it's going to be f of two times zero, which is still f of zero, which is going to be equal to a little bit over four, so which is equal to f of zero. And so they're going to both have the same y-intercept. But interesting things are going to happen the further that we get from the y-axis, or as our x increases in either direction away, or as our x gets bigger in either direction, from zero. So let's think about what's going to happen at x equals two. So at x equals two, g of two is going to be equal to f of two times two, two times two, which is equal to f of four."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so they're going to both have the same y-intercept. But interesting things are going to happen the further that we get from the y-axis, or as our x increases in either direction away, or as our x gets bigger in either direction, from zero. So let's think about what's going to happen at x equals two. So at x equals two, g of two is going to be equal to f of two times two, two times two, which is equal to f of four. And we know what f of four is. F of four is equal to zero. So g of two is equal to f of four, which is equal to zero."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So at x equals two, g of two is going to be equal to f of two times two, two times two, which is equal to f of four. And we know what f of four is. F of four is equal to zero. So g of two is equal to f of four, which is equal to zero. So notice, the corresponding point has kind of gotten compressed in, or squeezed in, or squished in in the horizontal direction. And so what you see happening, at least on this side of the graph, is everything's happening a little bit faster. Your, whatever was happening at a certain x is now happening at half of that x."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of two is equal to f of four, which is equal to zero. So notice, the corresponding point has kind of gotten compressed in, or squeezed in, or squished in in the horizontal direction. And so what you see happening, at least on this side of the graph, is everything's happening a little bit faster. Your, whatever was happening at a certain x is now happening at half of that x. So this side of the graph is going to look something, try to draw it a little bit better than that, and it's going to look something like this, like this. Everything's happening twice as fast. And what happens when you go in the negative direction?"}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Your, whatever was happening at a certain x is now happening at half of that x. So this side of the graph is going to look something, try to draw it a little bit better than that, and it's going to look something like this, like this. Everything's happening twice as fast. And what happens when you go in the negative direction? Well, think about what g of negative two is. G of negative two is equal to f of two times negative two, two times negative two, which is equal to f of negative four, which we see is also equal to zero. So g of negative two is zero."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And what happens when you go in the negative direction? Well, think about what g of negative two is. G of negative two is equal to f of two times negative two, two times negative two, which is equal to f of negative four, which we see is also equal to zero. So g of negative two is zero. And you might be thinking, why did you pick two and negative two? Well, the intuition is that things are going to be squeezed in, things are happening twice as fast. So whatever was happening at x equals four is now going to happen at x equals two."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So g of negative two is zero. And you might be thinking, why did you pick two and negative two? Well, the intuition is that things are going to be squeezed in, things are happening twice as fast. So whatever was happening at x equals four is now going to happen at x equals two. Whatever's happening at x equals negative four is now going to happen at x equals negative two. And I saw that we were at very clear points at x equals negative four and x equals four on f, so I just took half of that to pick my x values right over here. And then, so what our graph is going to look like is something like this."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So whatever was happening at x equals four is now going to happen at x equals two. Whatever's happening at x equals negative four is now going to happen at x equals negative two. And I saw that we were at very clear points at x equals negative four and x equals four on f, so I just took half of that to pick my x values right over here. And then, so what our graph is going to look like is something like this. It's going to look something like this. It's gonna look like it's been squished in from the right and the left. Now let's do another example."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then, so what our graph is going to look like is something like this. It's going to look something like this. It's gonna look like it's been squished in from the right and the left. Now let's do another example. So now they've not only given the graph of f, they've given an expression for it. What is the graph of g of x, which is equal to this business? So pause this video and try to figure that out."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now let's do another example. So now they've not only given the graph of f, they've given an expression for it. What is the graph of g of x, which is equal to this business? So pause this video and try to figure that out. All right, the key is to figure out the relationship between f of x and g of x. And what we can see, the main difference is, is instead of an x here, an f of x, we have an x over two. So everywhere there was an x, we have been replaced with an x over two."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to figure that out. All right, the key is to figure out the relationship between f of x and g of x. And what we can see, the main difference is, is instead of an x here, an f of x, we have an x over two. So everywhere there was an x, we have been replaced with an x over two. So another way of thinking about it is g of x is equal to f of, not x, but f of x over two. Or another way of thinking about it, g of x is equal to f of 1 1\u20442 x. And then we can do a similar type of exercise."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So everywhere there was an x, we have been replaced with an x over two. So another way of thinking about it is g of x is equal to f of, not x, but f of x over two. Or another way of thinking about it, g of x is equal to f of 1 1\u20442 x. And then we can do a similar type of exercise. And they've given us some interesting points. The point two, the point, or the point x equals two, the point x equals four, and the point x equals six. So let's think about this."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then we can do a similar type of exercise. And they've given us some interesting points. The point two, the point, or the point x equals two, the point x equals four, and the point x equals six. So let's think about this. Last time, when it was g of x is equal to two x, things were happening twice as fast. Now things are going to happen half as fast. And so what I would do, let me just set up a little table here."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So let's think about this. Last time, when it was g of x is equal to two x, things were happening twice as fast. Now things are going to happen half as fast. And so what I would do, let me just set up a little table here. The interesting x values for me are the ones that if I take half of them, that I'm going to get one of these points. So actually, let me write this. Half 1 1\u20442 x."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so what I would do, let me just set up a little table here. The interesting x values for me are the ones that if I take half of them, that I'm going to get one of these points. So actually, let me write this. Half 1 1\u20442 x. And then I can think about what g of x is equal to f of 1 1\u20442 x is going to be. So I want my 1 1\u20442 x to be, let's see, it could be two, four, and six. Two, four, and six."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Half 1 1\u20442 x. And then I can think about what g of x is equal to f of 1 1\u20442 x is going to be. So I want my 1 1\u20442 x to be, let's see, it could be two, four, and six. Two, four, and six. And why did I pick those again? Well, it's very clear what values f takes on at those points. And so if 1 1\u20442 x is two, then x is equal to four."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Two, four, and six. And why did I pick those again? Well, it's very clear what values f takes on at those points. And so if 1 1\u20442 x is two, then x is equal to four. If 1 1\u20442 x is four, then x is equal to eight. If x is equal to 12, then 1 1\u20442 x is six. And so then we could say, all right, g of four is equal to f of two, which is equal to zero."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so if 1 1\u20442 x is two, then x is equal to four. If 1 1\u20442 x is four, then x is equal to eight. If x is equal to 12, then 1 1\u20442 x is six. And so then we could say, all right, g of four is equal to f of two, which is equal to zero. That's why I picked two, four, and six. It's very easy to evaluate f of two, f of four, and f of six. It gave us those points very clearly."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so then we could say, all right, g of four is equal to f of two, which is equal to zero. That's why I picked two, four, and six. It's very easy to evaluate f of two, f of four, and f of six. It gave us those points very clearly. So g of eight is going to be equal to g, is going to be equal to f of 1\u20442 of eight, or f of four, which is equal to negative four. And then g of 12 is equal to f of six, which is half of 12, which is equal to zero again. So then we could plot these points, and we get a general sense of the shape of the graph."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It gave us those points very clearly. So g of eight is going to be equal to g, is going to be equal to f of 1\u20442 of eight, or f of four, which is equal to negative four. And then g of 12 is equal to f of six, which is half of 12, which is equal to zero again. So then we could plot these points, and we get a general sense of the shape of the graph. So let's see. G of four is equal to zero. G of eight is equal to negative four, all right, over there."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So then we could plot these points, and we get a general sense of the shape of the graph. So let's see. G of four is equal to zero. G of eight is equal to negative four, all right, over there. And then g of 12 is equal to zero again. So everything has been stretched out. So there you go."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "G of eight is equal to negative four, all right, over there. And then g of 12 is equal to zero again. So everything has been stretched out. So there you go. It's been stretched out in at least in the horizontal direction is one way to think about it, in the horizontal direction. And you can see that this point in f corresponds to this point in g. It's gotten twice as far from the origin because everything is growing half as fast. You input an x, you take a half of it, and then you input it into f. And then this point right over here corresponds to this point."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So there you go. It's been stretched out in at least in the horizontal direction is one way to think about it, in the horizontal direction. And you can see that this point in f corresponds to this point in g. It's gotten twice as far from the origin because everything is growing half as fast. You input an x, you take a half of it, and then you input it into f. And then this point right over here corresponds to this point. Instead of happening at four, this vertex point, it's now happening at eight. And last but not least, this point right over here corresponds to this point. Instead of happening at six, it's happening at 12."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You input an x, you take a half of it, and then you input it into f. And then this point right over here corresponds to this point. Instead of happening at four, this vertex point, it's now happening at eight. And last but not least, this point right over here corresponds to this point. Instead of happening at six, it's happening at 12. Everything is getting stretched out. Let's do one more example. F of x is equal to all of this."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Instead of happening at six, it's happening at 12. Everything is getting stretched out. Let's do one more example. F of x is equal to all of this. We have to be careful. We have a cube root over here. And g is a horizontally scaled version of f. The functions are graphed where f is solid and g is dashed."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "F of x is equal to all of this. We have to be careful. We have a cube root over here. And g is a horizontally scaled version of f. The functions are graphed where f is solid and g is dashed. What is the equation of g? So pause this video and see if you can figure that out. All right, let's do this together."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And g is a horizontally scaled version of f. The functions are graphed where f is solid and g is dashed. What is the equation of g? So pause this video and see if you can figure that out. All right, let's do this together. And it looks like they've given us some points that seem to correspond with each other. To go from f to g, it looks like these corresponding points have been squeezed in closer to the origin. And what we can see is that f of negative three, f of negative three seems to be equal to g of negative one."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, let's do this together. And it looks like they've given us some points that seem to correspond with each other. To go from f to g, it looks like these corresponding points have been squeezed in closer to the origin. And what we can see is that f of negative three, f of negative three seems to be equal to g of negative one. And f of six over here, f of six seems to be equal to g of two, g of two. Or another way to think about it, whatever x you input in g, it looks like that's going to be equivalent to three times that x inputted into f. So g of x is equal to f of three x. And so if you wanna know the equation of g, we just evaluate f of three x."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And what we can see is that f of negative three, f of negative three seems to be equal to g of negative one. And f of six over here, f of six seems to be equal to g of two, g of two. Or another way to think about it, whatever x you input in g, it looks like that's going to be equivalent to three times that x inputted into f. So g of x is equal to f of three x. And so if you wanna know the equation of g, we just evaluate f of three x. So f of three x is going to be equal to, and I could just actually put an equal sign like this, f of three x is going to be equal to negative three times the cube root of, instead of an x, I'll put a three x right over there, three x plus two. And then we have plus one. And that's it."}, {"video_title": "Scaling functions horizontally examples Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so if you wanna know the equation of g, we just evaluate f of three x. So f of three x is going to be equal to, and I could just actually put an equal sign like this, f of three x is going to be equal to negative three times the cube root of, instead of an x, I'll put a three x right over there, three x plus two. And then we have plus one. And that's it. That's what g of x is equal to. It's equal to f of three x, which is that. We substituted this x with a three x."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this seems like a very complicated word, but if you break it down, it'll start to make sense, especially when we start to see examples of polynomials. So the first part of this word, let me underline it, we have poly. This comes from Greek for many, and you see poly a lot in the English language, referring to the notion of many of something. So in this case, it's many nomials, and nomial comes from Latin, from the Latin nomen, for name. So you could view this as many names, but in a mathematical context, it's really referring to many terms, and we're gonna talk in a little bit about what a term really is. But to get a tangible sense of what are polynomials and what are not polynomials, let me give you some examples, and then we could write some, maybe more formal rules for them. So an example of a polynomial could be 10x to the seventh power minus nine x squared plus 15x to the third plus nine."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So in this case, it's many nomials, and nomial comes from Latin, from the Latin nomen, for name. So you could view this as many names, but in a mathematical context, it's really referring to many terms, and we're gonna talk in a little bit about what a term really is. But to get a tangible sense of what are polynomials and what are not polynomials, let me give you some examples, and then we could write some, maybe more formal rules for them. So an example of a polynomial could be 10x to the seventh power minus nine x squared plus 15x to the third plus nine. This is a polynomial. Another example of a polynomial, nine a squared minus five. Even if I just have one number, even if I were to just write the number six, that can officially be considered a polynomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So an example of a polynomial could be 10x to the seventh power minus nine x squared plus 15x to the third plus nine. This is a polynomial. Another example of a polynomial, nine a squared minus five. Even if I just have one number, even if I were to just write the number six, that can officially be considered a polynomial. If I were to write seven x squared minus three, let me do it in another variable, seven y squared minus three y plus pi, that too would be a polynomial. So these are examples of polynomials. What are examples of things that are not polynomials?"}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Even if I just have one number, even if I were to just write the number six, that can officially be considered a polynomial. If I were to write seven x squared minus three, let me do it in another variable, seven y squared minus three y plus pi, that too would be a polynomial. So these are examples of polynomials. What are examples of things that are not polynomials? Well, if I were to replace the seventh power right over here with the negative seventh power, so if I were to write 10x to the negative seventh power minus nine x squared plus 15x to the third power plus nine, this would not be a polynomial. So I think you might be sensing a rule here for what makes something a polynomial, that you have to have non-negative powers of your variable in each of the terms. And I just used that word terms, so let me explain it, because it'll help me explain what a polynomial is."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "What are examples of things that are not polynomials? Well, if I were to replace the seventh power right over here with the negative seventh power, so if I were to write 10x to the negative seventh power minus nine x squared plus 15x to the third power plus nine, this would not be a polynomial. So I think you might be sensing a rule here for what makes something a polynomial, that you have to have non-negative powers of your variable in each of the terms. And I just used that word terms, so let me explain it, because it'll help me explain what a polynomial is. A polynomial is something that is made up of a sum of terms. And so, for example, in this first polynomial, the first term is 10x to the seventh, the second term is negative nine x squared, the next term is 15x to the third, and then the last term, maybe you could say the fourth term, is nine. And you can see something, let me underline these."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And I just used that word terms, so let me explain it, because it'll help me explain what a polynomial is. A polynomial is something that is made up of a sum of terms. And so, for example, in this first polynomial, the first term is 10x to the seventh, the second term is negative nine x squared, the next term is 15x to the third, and then the last term, maybe you could say the fourth term, is nine. And you can see something, let me underline these. So these are all terms. This is a four-term polynomial right over here. And you could say, hey, wait, this thing you wrote in red, this also has four terms, but we have to put a few more rules for it to officially be a polynomial, especially a polynomial in one variable."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you can see something, let me underline these. So these are all terms. This is a four-term polynomial right over here. And you could say, hey, wait, this thing you wrote in red, this also has four terms, but we have to put a few more rules for it to officially be a polynomial, especially a polynomial in one variable. Each of those terms are going to be made up of a coefficient. This is the thing that multiplies the variable to some power. So in this first term, the coefficient is 10."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you could say, hey, wait, this thing you wrote in red, this also has four terms, but we have to put a few more rules for it to officially be a polynomial, especially a polynomial in one variable. Each of those terms are going to be made up of a coefficient. This is the thing that multiplies the variable to some power. So in this first term, the coefficient is 10. And let me write this word down, coefficient. It's another fancy word, but it's just a thing that's multiplied, in this case, times the variable, which is x to the seventh power. So the first coefficient is 10."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So in this first term, the coefficient is 10. And let me write this word down, coefficient. It's another fancy word, but it's just a thing that's multiplied, in this case, times the variable, which is x to the seventh power. So the first coefficient is 10. The next coefficient, and actually let me be careful here, because the second coefficient here is negative nine. So we are looking at coefficients. The third coefficient here is 15."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So the first coefficient is 10. The next coefficient, and actually let me be careful here, because the second coefficient here is negative nine. So we are looking at coefficients. The third coefficient here is 15. And you can view this fourth term or this fourth number as the coefficient, because this could be rewritten as, instead of just writing it as nine, you could write it as nine x to the zero power. And then it looks a little bit clearer like a coefficient. So in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter a, being multiplied by a variable, being raised to a non-negative integer power."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "The third coefficient here is 15. And you can view this fourth term or this fourth number as the coefficient, because this could be rewritten as, instead of just writing it as nine, you could write it as nine x to the zero power. And then it looks a little bit clearer like a coefficient. So in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter a, being multiplied by a variable, being raised to a non-negative integer power. So this right over here is a coefficient. It can be, if we're dealing, well, I don't wanna get too technical, it could be a positive, negative number, it could be any real number. We have our variable."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So in general, a polynomial is the sum of a finite number of terms where each term has a coefficient, which I could represent with the letter a, being multiplied by a variable, being raised to a non-negative integer power. So this right over here is a coefficient. It can be, if we're dealing, well, I don't wanna get too technical, it could be a positive, negative number, it could be any real number. We have our variable. And then the exponent here has to be non-negative, non-negative integer. So here, the reason why what I wrote in red is not a polynomial is because here, I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "We have our variable. And then the exponent here has to be non-negative, non-negative integer. So here, the reason why what I wrote in red is not a polynomial is because here, I have an exponent that is a negative integer. Let's give some other examples of things that are not polynomials. So if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the 1 1\u20442 power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer. It's 1 1\u20442. And this is the same thing as nine times the square root of a minus five."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's give some other examples of things that are not polynomials. So if I were to change the second one to, instead of nine a squared, if I wrote it as nine a to the 1 1\u20442 power minus five, this is not a polynomial because this exponent right over here, it is no longer an integer. It's 1 1\u20442. And this is the same thing as nine times the square root of a minus five. This also would not be a polynomial. Or if I were to write nine a to the a power minus five, a power minus five, also not a polynomial because here, the exponent is a variable. It's not a non-negative integer."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And this is the same thing as nine times the square root of a minus five. This also would not be a polynomial. Or if I were to write nine a to the a power minus five, a power minus five, also not a polynomial because here, the exponent is a variable. It's not a non-negative integer. So all of these are examples of polynomials. So there's a few more pieces of terminology that are valuable to know. Polynomials is the general, or a polynomial is a general term for one of these expressions that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "It's not a non-negative integer. So all of these are examples of polynomials. So there's a few more pieces of terminology that are valuable to know. Polynomials is the general, or a polynomial is a general term for one of these expressions that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form. But there's more specific terms for when you have only one term or two terms or three terms. So when you have one term, it's called a monomial. So this is a monomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Polynomials is the general, or a polynomial is a general term for one of these expressions that has multiple terms, a finite number, so not an infinite number, and each of the terms has this form. But there's more specific terms for when you have only one term or two terms or three terms. So when you have one term, it's called a monomial. So this is a monomial. This is an example of a monomial, which we could write as six x to the zero, but you could also, another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is a monomial. This is an example of a monomial, which we could write as six x to the zero, but you could also, another example of a monomial might be 10z to the 15th power. That's also a monomial. Your coefficient could be pi. Pi, whoops, it could be pi. So we could write pi times b to the fifth power. Any of these would be monomials."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Your coefficient could be pi. Pi, whoops, it could be pi. So we could write pi times b to the fifth power. Any of these would be monomials. So what's a binomial? Well, a binomial is where you have two terms. Monomial, mono for one, one term."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Any of these would be monomials. So what's a binomial? Well, a binomial is where you have two terms. Monomial, mono for one, one term. Binomial is you have two terms. So this right over here is a binomial. Binomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Monomial, mono for one, one term. Binomial is you have two terms. So this right over here is a binomial. Binomial. You have two terms. And all of these are polynomials, but these are subclassifications. So it's a binomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Binomial. You have two terms. And all of these are polynomials, but these are subclassifications. So it's a binomial. You have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So it's a binomial. You have one, two terms. Another example of a binomial would be three y to the third plus five y. Once again, you have two terms that have this form right over here. Now you'll also hear the term trinomial. Well, trinomial is when you have three terms. Trinomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Once again, you have two terms that have this form right over here. Now you'll also hear the term trinomial. Well, trinomial is when you have three terms. Trinomial. And this right over here is an example. This is the first term. This is the second term."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Trinomial. And this right over here is an example. This is the first term. This is the second term. And this is the third term. Now, the next word that you will hear often in the context with polynomials is the notion of the degree of a polynomial. And you might hear people say, what is the degree of a polynomial?"}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "This is the second term. And this is the third term. Now, the next word that you will hear often in the context with polynomials is the notion of the degree of a polynomial. And you might hear people say, what is the degree of a polynomial? Or what is the degree of a given term of a polynomial? So let's start with the degree of a given term. So let's go to this polynomial here."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you might hear people say, what is the degree of a polynomial? Or what is the degree of a given term of a polynomial? So let's start with the degree of a given term. So let's go to this polynomial here. We have this first term, 10x to the seventh. The degree is the power that we're raising the variable to. So this is a seventh degree term."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So let's go to this polynomial here. We have this first term, 10x to the seventh. The degree is the power that we're raising the variable to. So this is a seventh degree term. The second term is a second degree term. The third term is a third degree term. And you could view this constant term, which is really just nine, you could view that as, sometimes people will say the constant term."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "So this is a seventh degree term. The second term is a second degree term. The third term is a third degree term. And you could view this constant term, which is really just nine, you could view that as, sometimes people will say the constant term. Sometimes people will say the zeroth degree term. Now, if people are talking about the degree of the entire polynomial, they're gonna say, well, what is the degree of the highest term? What is the term with the highest degree?"}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "And you could view this constant term, which is really just nine, you could view that as, sometimes people will say the constant term. Sometimes people will say the zeroth degree term. Now, if people are talking about the degree of the entire polynomial, they're gonna say, well, what is the degree of the highest term? What is the term with the highest degree? That degree will be the degree of the entire polynomial. So this first polynomial, this is a seventh degree polynomial. This one right over here is a second degree polynomial because it has a second degree term, and that's the highest degree term."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "What is the term with the highest degree? That degree will be the degree of the entire polynomial. So this first polynomial, this is a seventh degree polynomial. This one right over here is a second degree polynomial because it has a second degree term, and that's the highest degree term. This right over here is a third degree. You could even say third degree binomial because its highest degree term has degree three. If this said five y to the seventh instead of five y, well, then it would be a seventh degree binomial."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "This one right over here is a second degree polynomial because it has a second degree term, and that's the highest degree term. This right over here is a third degree. You could even say third degree binomial because its highest degree term has degree three. If this said five y to the seventh instead of five y, well, then it would be a seventh degree binomial. This right over here is a 15th degree monomial. This is a second degree trinomial. Now, the last thing I will, or a few more things I will introduce you to is the idea of a leading term and a leading coefficient."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "If this said five y to the seventh instead of five y, well, then it would be a seventh degree binomial. This right over here is a 15th degree monomial. This is a second degree trinomial. Now, the last thing I will, or a few more things I will introduce you to is the idea of a leading term and a leading coefficient. So let me write this down, the notion of what it means to be leading. Well, it usually means, it can mean whatever is the first term or the coefficient. If you're saying leading term, it's the first term, and if you're saying leading coefficient, it's the coefficient in the first term, but it's oftentimes associated with a polynomial being written in standard form."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, the last thing I will, or a few more things I will introduce you to is the idea of a leading term and a leading coefficient. So let me write this down, the notion of what it means to be leading. Well, it usually means, it can mean whatever is the first term or the coefficient. If you're saying leading term, it's the first term, and if you're saying leading coefficient, it's the coefficient in the first term, but it's oftentimes associated with a polynomial being written in standard form. So standard form, standard form, is where you write the terms in degree order, starting with the highest degree term. So for example, what I have up here, this is not in standard form because I do have the highest degree term first, but then I should go to the next highest, which is the x to the third, but here I wrote x squared next, so this is not standard. If I wanted to write it in standard form, it would be 10x to the seventh power, which is the highest degree term, that's the degree seven, then 15x to the third, so plus 15x to the third, which is the next highest degree, then negative nine x squared is the next highest degree term, and then the lowest degree term here is plus nine or plus nine x to zero."}, {"video_title": "Polynomials intro Mathematics II High School Math Khan Academy.mp3", "Sentence": "If you're saying leading term, it's the first term, and if you're saying leading coefficient, it's the coefficient in the first term, but it's oftentimes associated with a polynomial being written in standard form. So standard form, standard form, is where you write the terms in degree order, starting with the highest degree term. So for example, what I have up here, this is not in standard form because I do have the highest degree term first, but then I should go to the next highest, which is the x to the third, but here I wrote x squared next, so this is not standard. If I wanted to write it in standard form, it would be 10x to the seventh power, which is the highest degree term, that's the degree seven, then 15x to the third, so plus 15x to the third, which is the next highest degree, then negative nine x squared is the next highest degree term, and then the lowest degree term here is plus nine or plus nine x to zero. Now this is in standard form. I've written the terms in order of decreasing degree with the highest degree first, and here it's clear that your leading term is 10x to the seventh, because it's the first one, and our leading coefficient here is the number 10. So there was a lot in that video, but hopefully the notion of a polynomial isn't seeming too intimidating at this point, and these are really useful words to be familiar with as you continue on on your math journey."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and think about it on your own before we work through it together. All right, so the fact that we have zeros at these values, that means that P of x, when x is equal to one of these values, is equal to zero. So P of negative four is equal to zero, P of three is equal to zero, and P of 1 8th is equal to zero. And before I even look at these choices, I could think about constructing a polynomial for which that is true. That's going to be true if I can express this polynomial as the product of expressions where each of these would make each of those expressions equal to zero. So what's an expression that would be zero when x is equal to negative four? Well, the expression x plus four, this is equal to zero when x is equal to negative four, so I like that."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And before I even look at these choices, I could think about constructing a polynomial for which that is true. That's going to be true if I can express this polynomial as the product of expressions where each of these would make each of those expressions equal to zero. So what's an expression that would be zero when x is equal to negative four? Well, the expression x plus four, this is equal to zero when x is equal to negative four, so I like that. What would be an expression that would be equal to zero when x is equal to three? Well, what about the expression x minus three? If x is equal to three, then this is going to be equal to zero, zero times anything is going to be equal to zero."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, the expression x plus four, this is equal to zero when x is equal to negative four, so I like that. What would be an expression that would be equal to zero when x is equal to three? Well, what about the expression x minus three? If x is equal to three, then this is going to be equal to zero, zero times anything is going to be equal to zero. So P of three would be zero in this case. And then what is an expression that would be equal to zero when x is equal to 1 8th? Well, that would be x minus 1 8th."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "If x is equal to three, then this is going to be equal to zero, zero times anything is going to be equal to zero. So P of three would be zero in this case. And then what is an expression that would be equal to zero when x is equal to 1 8th? Well, that would be x minus 1 8th. Now, though these aren't the only expressions, you could multiply them by constants, and still the principles that I just talked about would be true. But our polynomial would look something like this. You could try it out."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, that would be x minus 1 8th. Now, though these aren't the only expressions, you could multiply them by constants, and still the principles that I just talked about would be true. But our polynomial would look something like this. You could try it out. If x is equal to negative four, well, then this first expression is zero. Zero times something times something is zero. Same thing for x equals three."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "You could try it out. If x is equal to negative four, well, then this first expression is zero. Zero times something times something is zero. Same thing for x equals three. If this right over, if x equals three, then x minus three is equal to zero. And then zero times something times something is zero. And then if x is equal to 1 8th, this expression's going to be equal to zero."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Same thing for x equals three. If this right over, if x equals three, then x minus three is equal to zero. And then zero times something times something is zero. And then if x is equal to 1 8th, this expression's going to be equal to zero. Zero times something times something is going to be equal to zero. So which of these choices look like that? So let's see."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then if x is equal to 1 8th, this expression's going to be equal to zero. Zero times something times something is going to be equal to zero. So which of these choices look like that? So let's see. X plus four. I actually see that in choices B, and I see that in choices D. Choice C has x minus four there. So that would have a zero at x equals four."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let's see. X plus four. I actually see that in choices B, and I see that in choices D. Choice C has x minus four there. So that would have a zero at x equals four. If x equals four, this first expression, this first part of the expression would be equal to zero. But we care about that happening when x is equal to negative four. So I would actually rule out C, and then for the same reason I would rule out A."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So that would have a zero at x equals four. If x equals four, this first expression, this first part of the expression would be equal to zero. But we care about that happening when x is equal to negative four. So I would actually rule out C, and then for the same reason I would rule out A. So we're between B and D. And now let's see. Which of these have an x minus three in them? Well, I see an x minus three here."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So I would actually rule out C, and then for the same reason I would rule out A. So we're between B and D. And now let's see. Which of these have an x minus three in them? Well, I see an x minus three here. I see an x minus three there. So I like the, I still like B and D. I'll put another check mark right over there. And then last but not least, which of these would be equal to zero when x is equal to 1 8th?"}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, I see an x minus three here. I see an x minus three there. So I like the, I still like B and D. I'll put another check mark right over there. And then last but not least, which of these would be equal to zero when x is equal to 1 8th? Well, let's see. If I do 1 8th times 1 8th here, I'm gonna get 1 64th for this part of the expression. And so that's not going to be equal to zero."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then last but not least, which of these would be equal to zero when x is equal to 1 8th? Well, let's see. If I do 1 8th times 1 8th here, I'm gonna get 1 64th for this part of the expression. And so that's not going to be equal to zero. And these other two things aren't going to be equal to zero when x is equal to 1 8th. So this one is not looking so good. But let's verify this one."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And so that's not going to be equal to zero. And these other two things aren't going to be equal to zero when x is equal to 1 8th. So this one is not looking so good. But let's verify this one. This has, if x is equal to 1 8th, we have eight times 1 8th, which is one, minus one. That is going to be equal to zero. So this one checks out."}, {"video_title": "Zeros of polynomials matching equation to zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "But let's verify this one. This has, if x is equal to 1 8th, we have eight times 1 8th, which is one, minus one. That is going to be equal to zero. So this one checks out. And you might be thinking, hey, this last polynomial looks a little bit different than this polynomial that I wrote up here when I just tried to come up with a polynomial for which this would be true. And as I mentioned, you could take this and multiply it by constants and it would still be true. So if you just take this, and if we were to multiply it by eight, you would get p of x down here."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "What I wanna do in this video is use our knowledge of trigonometry and use our knowledge of triangles in order to figure out several things. So the first thing we wanna figure out is what's the measure of angle, what's the radian measure, what's the radian measure, measure of angle ABD, of angle ABD. Actually, let's just do that first, and then I'll talk about the other things that we need to think about. So I assume you've paused the video and tried to do this on your own. So let's think about what ABD would be. We know two of the angles of this triangle, so if we know two of the angles of this triangle, you should be able to figure out the third. Now, it might be a little bit unfamiliar."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So I assume you've paused the video and tried to do this on your own. So let's think about what ABD would be. We know two of the angles of this triangle, so if we know two of the angles of this triangle, you should be able to figure out the third. Now, it might be a little bit unfamiliar. We're used to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians, so we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians. So this angle plus that angle plus that angle are going to add up to pi. So let's just say that this right over here, let's just say measure of angle ABD in radians plus pi over four, plus pi over four, plus, this is a right angle, what would that be in radians?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, it might be a little bit unfamiliar. We're used to saying that the sum of the interior angles of a triangle add up to 180 degrees, but now we're thinking in terms of radians, so we could say that the sum of the angles of a triangle add up to, instead of saying 180 degrees, 180 degrees is the same thing as pi radians. So this angle plus that angle plus that angle are going to add up to pi. So let's just say that this right over here, let's just say measure of angle ABD in radians plus pi over four, plus pi over four, plus, this is a right angle, what would that be in radians? Well, a right angle in radians, a 90 degree angle in radians is pi over two radians. So plus pi over two. When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians, which is, of course, the same thing as 180 degrees."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So let's just say that this right over here, let's just say measure of angle ABD in radians plus pi over four, plus pi over four, plus, this is a right angle, what would that be in radians? Well, a right angle in radians, a 90 degree angle in radians is pi over two radians. So plus pi over two. When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians, which is, of course, the same thing as 180 degrees. And now we can solve for the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two minus pi over four. I just subtracted these two from both sides."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "When you take the sum of them, the interior angles of this triangle, they're going to add up to pi radians, which is, of course, the same thing as 180 degrees. And now we can solve for the measure of angle ABD. Measure of angle ABD is equal to pi minus pi over two minus pi over four. I just subtracted these two from both sides. So this is going to be equal to, as you could put a common denominator of four, then this is four pi over four. This is minus two pi over four. And this, of course, is minus pi over four."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "I just subtracted these two from both sides. So this is going to be equal to, as you could put a common denominator of four, then this is four pi over four. This is minus two pi over four. And this, of course, is minus pi over four. So four minus two minus one is gonna get us to one. So this is going to give us two pi over four. So the measure of angle ABD is actually the same as the measure of angle BAD."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And this, of course, is minus pi over four. So four minus two minus one is gonna get us to one. So this is going to give us two pi over four. So the measure of angle ABD is actually the same as the measure of angle BAD. It is pi over four. So that angle right over there is pi over four. Now what does that help us with?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So the measure of angle ABD is actually the same as the measure of angle BAD. It is pi over four. So that angle right over there is pi over four. Now what does that help us with? So if we know that this is pi over four, and that is pi over four radians, and once again, we know this is a unit circle. So we know that the length of segment AB, which is a radius of the circle, or is the radius of the circle, is length one. What else do we know about this triangle?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now what does that help us with? So if we know that this is pi over four, and that is pi over four radians, and once again, we know this is a unit circle. So we know that the length of segment AB, which is a radius of the circle, or is the radius of the circle, is length one. What else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well, sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "What else do we know about this triangle? Can we figure out the lengths of segment AD and the length of segment DB? Well, sure, because we have two base angles that have the same measure, that means that the corresponding sides are also going to have the same measure. That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, if we were to, not completely flip it over, but if we were to make it look like this, so the triangle, we could make it look like, let me make it look a little bit more like this. Actually, I want to make it look like a right angle, though."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "That means that this side is going to be congruent to that side. I can reorient it in a way that might make it a little easier to realize. If we were to flip it over, if we were to, not completely flip it over, but if we were to make it look like this, so the triangle, we could make it look like, let me make it look a little bit more like this. Actually, I want to make it look like a right angle, though. So my triangle, let me make it look like, there you go. So if this is D, if this is D, this is B, this is A, this is our right angle. Now, this is pi over four radians, and this is also pi over four radians."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Actually, I want to make it look like a right angle, though. So my triangle, let me make it look like, there you go. So if this is D, if this is D, this is B, this is A, this is our right angle. Now, this is pi over four radians, and this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle, an isosceles, but because they're not all the same, you know it's not equilateral. If all of the angles were the same, this would be equilateral, but this is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, this is pi over four radians, and this is also pi over four radians. When your two base angles are the same, you know you're dealing with an isosceles triangle, an isosceles, but because they're not all the same, you know it's not equilateral. If all of the angles were the same, this would be equilateral, but this is an isosceles non-equilateral triangle. So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same. This is an isosceles triangle. And so how does that help us figure out the lengths of the sides?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So if your base angles are the same, then you also know that the corresponding sides are going to be the same. These two sides are the same. This is an isosceles triangle. And so how does that help us figure out the lengths of the sides? Well, if you say that this side has length x, that means that this side has length x. If this side has length x, then this side has length x. And now we can use the Pythagorean theorem."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And so how does that help us figure out the lengths of the sides? Well, if you say that this side has length x, that means that this side has length x. If this side has length x, then this side has length x. And now we can use the Pythagorean theorem. We could say that x squared, this x squared plus this x squared, is equal to the hypotenuse squared, is equal to one squared. Or we could write that two x squared is equal to one, or that x squared is equal to one over two, or just taking the principal root of both sides, we get x is equal to one over the square root of two. And a lot of folks don't like having a radical in the denominator."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And now we can use the Pythagorean theorem. We could say that x squared, this x squared plus this x squared, is equal to the hypotenuse squared, is equal to one squared. Or we could write that two x squared is equal to one, or that x squared is equal to one over two, or just taking the principal root of both sides, we get x is equal to one over the square root of two. And a lot of folks don't like having a radical in the denominator. They don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, you'll see the numerator will have the square root of two, and in the denominator, we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And a lot of folks don't like having a radical in the denominator. They don't like having a rational number in the denominator. So we could rationalize the denominator by multiplying by the square root of two over the square root of two, which would be, you'll see the numerator will have the square root of two, and in the denominator, we're just going to have square root of two times square root of two is just two. So we've already been able to figure out several interesting things. We were able to figure out a measure of angle ABD in radians, we were able to figure out the lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work that we have done. So let's first think about what is the sine."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So we've already been able to figure out several interesting things. We were able to figure out a measure of angle ABD in radians, we were able to figure out the lengths of segment AD and the length of segment BD. Now what I want to do is figure out what are the sine, cosine, and tangent of pi over four radians, given all of the work that we have done. So let's first think about what is the sine. Let me do this in a color, do it in orange. Given all the work we've done, what is the sine of pi over four radians? And I encourage you, once again, pause the video, think about the unit circle definition of trig functions, and think about what this is."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So let's first think about what is the sine. Let me do this in a color, do it in orange. Given all the work we've done, what is the sine of pi over four radians? And I encourage you, once again, pause the video, think about the unit circle definition of trig functions, and think about what this is. Well, the unit circle definition of trig functions, and this angle, this pi over four radians, is forming an angle with a positive x-axis, and where its terminal ray intersects the unit circle, the x and y coordinates of this point are what specify the cosine and sine. So the coordinates of this point are going to be the cosine of pi over four radians is the x coordinate, and the sine of pi over four radians is the y coordinate, sine of pi over four. So what's the y coordinate going to be if we want the sine of pi over four?"}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And I encourage you, once again, pause the video, think about the unit circle definition of trig functions, and think about what this is. Well, the unit circle definition of trig functions, and this angle, this pi over four radians, is forming an angle with a positive x-axis, and where its terminal ray intersects the unit circle, the x and y coordinates of this point are what specify the cosine and sine. So the coordinates of this point are going to be the cosine of pi over four radians is the x coordinate, and the sine of pi over four radians is the y coordinate, sine of pi over four. So what's the y coordinate going to be if we want the sine of pi over four? Well, it's going to be this length right over here, which is the same thing as this length, which is the length of x, which is square root of two over two. Now what is the cosine of pi over four? And once again, I encourage you to pause the video, think about it."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "So what's the y coordinate going to be if we want the sine of pi over four? Well, it's going to be this length right over here, which is the same thing as this length, which is the length of x, which is square root of two over two. Now what is the cosine of pi over four? And once again, I encourage you to pause the video, think about it. What's this x coordinate? What's the x coordinate? The x coordinate is this distance right over here, which is once again going to be x, which is square root of two over two."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "And once again, I encourage you to pause the video, think about it. What's this x coordinate? What's the x coordinate? The x coordinate is this distance right over here, which is once again going to be x, which is square root of two over two. Now, what is the tangent of pi over four going to be? Well, the tangent of pi over four is just sine of pi over four over cosine of pi over four. Now, both of these are this exact same thing."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "The x coordinate is this distance right over here, which is once again going to be x, which is square root of two over two. Now, what is the tangent of pi over four going to be? Well, the tangent of pi over four is just sine of pi over four over cosine of pi over four. Now, both of these are this exact same thing. They're both square root of two over two, so you're gonna have square root of two over two divided by square root of two over two. Well, that's just going to give you one. And that also makes sense, because remember, the tangent of this angle is the slope of this line."}, {"video_title": "Solving triangle in unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now, both of these are this exact same thing. They're both square root of two over two, so you're gonna have square root of two over two divided by square root of two over two. Well, that's just going to give you one. And that also makes sense, because remember, the tangent of this angle is the slope of this line. And we see the slope. For every x we move in the horizontal direction, we move x up. So our change in y over change in x is essentially x over x, which is equal to one."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to do this on your own before we work on this together. All right, now let's work on this. So they already give us a hint of how to solve it. They have the graph of y is equal to 3 halves to the x. They graph it right over here. And this gives us a hint, and especially because they want us to find an approximate solution, that maybe we can solve this equation or approximate a solution to this equation through graphing. And the way we could do that is we could take each side of this equation and set them up as a function."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "They have the graph of y is equal to 3 halves to the x. They graph it right over here. And this gives us a hint, and especially because they want us to find an approximate solution, that maybe we can solve this equation or approximate a solution to this equation through graphing. And the way we could do that is we could take each side of this equation and set them up as a function. We could set y equals to each side of it. So if we set y equals to the left-hand side, we get y is equal to 3 halves to the x power, which is what they originally give us, the graph of that. And if we set y equal to the right-hand side, we get y is equal to five, and we can graph that."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And the way we could do that is we could take each side of this equation and set them up as a function. We could set y equals to each side of it. So if we set y equals to the left-hand side, we get y is equal to 3 halves to the x power, which is what they originally give us, the graph of that. And if we set y equal to the right-hand side, we get y is equal to five, and we can graph that. And what's interesting here is if we can find the x value that gives us the same y value on both of these equations, well, that means that those graphs are going to intersect. And if I'm getting the same y value for that x value in both of these, well, then that means that 3 halves to the x is going to be equal to five. And so we can look at where they intersect and get an approximate sense of what x value that is."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And if we set y equal to the right-hand side, we get y is equal to five, and we can graph that. And what's interesting here is if we can find the x value that gives us the same y value on both of these equations, well, that means that those graphs are going to intersect. And if I'm getting the same y value for that x value in both of these, well, then that means that 3 halves to the x is going to be equal to five. And so we can look at where they intersect and get an approximate sense of what x value that is. And we can see it, at least over here, it looks like x is roughly equal to four. So x is approximately equal to four. And if we wanted to, and we'd be done at that point, if you wanted to, you could try to test it out."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And so we can look at where they intersect and get an approximate sense of what x value that is. And we can see it, at least over here, it looks like x is roughly equal to four. So x is approximately equal to four. And if we wanted to, and we'd be done at that point, if you wanted to, you could try to test it out. You could say, hey, does that actually work out? 3 halves to the fourth power, is that equal to five? Let's see, three to the fourth is 81."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And if we wanted to, and we'd be done at that point, if you wanted to, you could try to test it out. You could say, hey, does that actually work out? 3 halves to the fourth power, is that equal to five? Let's see, three to the fourth is 81. Two to the fourth is 16. It gets us pretty close to five. 16 times five is 80."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "Let's see, three to the fourth is 81. Two to the fourth is 16. It gets us pretty close to five. 16 times five is 80. So it's not exact, but it gets us pretty close. And if you had a graphing calculator that could really zoom in and zoom in and zoom in, you would get a value, you would see that x is slightly different than x equals four. But let's do another example."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "16 times five is 80. So it's not exact, but it gets us pretty close. And if you had a graphing calculator that could really zoom in and zoom in and zoom in, you would get a value, you would see that x is slightly different than x equals four. But let's do another example. The key here is that we can approximate solutions to equations through graphing. So here we are told, this is the graph of y is equal to, we have this third degree polynomial right over here. Use the graph to answer the following questions."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "But let's do another example. The key here is that we can approximate solutions to equations through graphing. So here we are told, this is the graph of y is equal to, we have this third degree polynomial right over here. Use the graph to answer the following questions. How many solutions does the equation x to the third minus two x squared minus x plus one equals negative one have? Pause this video and try to think about that. Well, when we think about solutions to this, we could say, all right, well, let's imagine two functions."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "Use the graph to answer the following questions. How many solutions does the equation x to the third minus two x squared minus x plus one equals negative one have? Pause this video and try to think about that. Well, when we think about solutions to this, we could say, all right, well, let's imagine two functions. One is y is equal to x to the third minus two x squared minus x plus one, which we already have graphed here. And let's say that the other equation, or the other function, is y is equal to negative one. And then how many times do these intersect?"}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "Well, when we think about solutions to this, we could say, all right, well, let's imagine two functions. One is y is equal to x to the third minus two x squared minus x plus one, which we already have graphed here. And let's say that the other equation, or the other function, is y is equal to negative one. And then how many times do these intersect? That would tell us how many solutions we have. So that is y is equal to negative one. And so every time they intersect, that means we have a solution to our original equation."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And then how many times do these intersect? That would tell us how many solutions we have. So that is y is equal to negative one. And so every time they intersect, that means we have a solution to our original equation. And they intersect one, two, and three times. So this has three solutions. What about the second situation?"}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "And so every time they intersect, that means we have a solution to our original equation. And they intersect one, two, and three times. So this has three solutions. What about the second situation? How many solutions does the equation all of this business equal to have? Well, same drill. We could set y equals to x to the third minus two x squared minus x plus one."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "What about the second situation? How many solutions does the equation all of this business equal to have? Well, same drill. We could set y equals to x to the third minus two x squared minus x plus one. And then we could think about another function. What if y is equal to two? Well, y equals two would be up over there."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "We could set y equals to x to the third minus two x squared minus x plus one. And then we could think about another function. What if y is equal to two? Well, y equals two would be up over there. Y equals two. And we could see it only intersects y equals all of this business once. So this is only going to have one solution."}, {"video_title": "Solving equations by graphing intro Algebra 2 Khan Academy.mp3", "Sentence": "Well, y equals two would be up over there. Y equals two. And we could see it only intersects y equals all of this business once. So this is only going to have one solution. So the key here, and I'll just write it out, and these are screenshots from the exercise on Khan Academy where you'd have to type in one, or in the previous example, you would type in four. But these are examples where you can take an equation of one variable, set both sides of them independently equal to y, graph them, and then think about where they intersect. Because the x values where they intersect will be solutions to your original equation."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And to do that, I'll set up a little unit circle so that we can visualize what the tangent of various thetas are. So let's say that's a y-axis, this is my x-axis, that is my x-axis, and the unit circle would look something like this. And we already know, this is all a refresher of the unit circle definition of trig functions, that if I have an angle, an angle theta, where one side is the positive x-axis, and then the other side, so this is the other side, so the angle is formed like this, that where this ray intersects the unit circle, the coordinates of that, the x and y coordinates, are the sine of theta, sorry, the x coordinate is the cosine of theta, so it's the cosine of theta comma sine of theta. So the x coordinate here is cosine theta, the y coordinate there is the sine of theta. But we're concerned about tangent of theta. Well, we know that tangent of theta is the same thing as the sine of theta over the cosine of theta. Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the x coordinate here is cosine theta, the y coordinate there is the sine of theta. But we're concerned about tangent of theta. Well, we know that tangent of theta is the same thing as the sine of theta over the cosine of theta. Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line. It's going to be, this is going to be your change in y, change in y over change in x. This right over here is going to be the slope, the slope, I guess you could say, of this ray right over here. And so that's going to help us visualize what the tangents of different thetas are."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Or if you're starting from the origin, and you're going, and you're taking the value of essentially the y coordinate, the y coordinate over the x coordinate, it's essentially the slope of this line. It's going to be, this is going to be your change in y, change in y over change in x. This right over here is going to be the slope, the slope, I guess you could say, of this ray right over here. And so that's going to help us visualize what the tangents of different thetas are. So let me clean up my unit circle a little bit, just so that we can, all right, there we go. So now let's make a, let's make a table. So let's make a table."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so that's going to help us visualize what the tangents of different thetas are. So let me clean up my unit circle a little bit, just so that we can, all right, there we go. So now let's make a, let's make a table. So let's make a table. So if, so for various thetas, let's think about what tangent of theta is going to be. Tangent of theta. So maybe the easiest one, if theta is zero radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's make a table. So if, so for various thetas, let's think about what tangent of theta is going to be. Tangent of theta. So maybe the easiest one, if theta is zero radians. So if it's zero radians, what is the slope of this ray? Well, that ray is, the slope is zero. As x changes, y doesn't change at all."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So maybe the easiest one, if theta is zero radians. So if it's zero radians, what is the slope of this ray? Well, that ray is, the slope is zero. As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take, let's take pi over, pi over four radians. So this one right over here, theta is equal to pi over four."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "As x changes, y doesn't change at all. Now let's think about, and I'm just going to pick values that are very easy for us to think about what the tangent of those values are, and they'll help us form, they'll help us think about the shape of the graph of y is equal to tangent of theta. So let's take, let's take pi over, pi over four radians. So this one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes it's easier to think in degrees. That's a 45 degree angle."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this one right over here, theta is equal to pi over four. Now why is that interesting? And sometimes it's easier to think in degrees. That's a 45 degree angle. This here, your x coordinate and your y coordinate is the same. You might remember it's square root of two over two, but the important thing is, whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's a 45 degree angle. This here, your x coordinate and your y coordinate is the same. You might remember it's square root of two over two, but the important thing is, whatever you move in the x direction, you move the same in the y direction. So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one. Or sine of theta over cosine of theta, they're the same thing. So you're going to get one. So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the slope of this ray right over here is going to be equal to one, or another way of thinking about it, tangent of theta is going to be equal to one. Or sine of theta over cosine of theta, they're the same thing. So you're going to get one. So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle. So if I have theta is pi over four, then the tangent of theta is going to be equal to one. Now what if theta is equal to negative pi over four? So that is this right over here."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if you put, so let me just clean that up here, just because I'm going to keep reusing the same unit circle. So if I have theta is pi over four, then the tangent of theta is going to be equal to one. Now what if theta is equal to negative pi over four? So that is this right over here. So with x, so let me just draw a little triangle here. So when x, this x coordinate over here is square root of two over two, we know that, we've seen that multiple times, square root of two, actually let me label it a little bit better. So here our theta is equal to negative pi over four radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that is this right over here. So with x, so let me just draw a little triangle here. So when x, this x coordinate over here is square root of two over two, we know that, we've seen that multiple times, square root of two, actually let me label it a little bit better. So here our theta is equal to negative pi over four radians. Now, or you could, if you like to think in degrees, this would be negative 45 degrees. And now your sine and cosine of this angle are going to be the opposites of each other. The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So here our theta is equal to negative pi over four radians. Now, or you could, if you like to think in degrees, this would be negative 45 degrees. And now your sine and cosine of this angle are going to be the opposites of each other. The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two. The y coordinate here is negative square root of two. Negative square root of two over two. So what's the tangent?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The cosine is square root of two over two, the x coordinate of where this intersects is square root of two over two. The y coordinate here is negative square root of two. Negative square root of two over two. So what's the tangent? Well, it's going to be your sine over your cosine, which is going to be negative one. And you see that. For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's the tangent? Well, it's going to be your sine over your cosine, which is going to be negative one. And you see that. For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction. And so let me clean this up a little bit because I want to keep reusing my unit circle. So there you go. And so this is going to be negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "For however much you move in the x direction, you move the opposite of that, or you move the negative of that in the y direction. And so let me clean this up a little bit because I want to keep reusing my unit circle. So there you go. And so this is going to be negative one. This is going to be negative one. And so actually let's just start plotting a few of these points. So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so this is going to be negative one. This is going to be negative one. And so actually let's just start plotting a few of these points. So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one. Now let's think, right now if you just saw that, you might say, oh, maybe this is some type of a line, but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to pi over two? What happens to the slope of this line? So that is theta."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if we assume that this is the theta axis, if you can see that, that's the theta axis, and if this is the y axis, that's the y axis, we immediately see tangent of zero is zero, tangent of pi over four is one, I'm looking in radians, tangent of negative pi over four is negative one. Now let's think, right now if you just saw that, you might say, oh, maybe this is some type of a line, but we'll see very clearly it's not a line because what happens as our angle gets closer and closer to pi over two? What happens to the slope of this line? So that is theta. We're getting closer and closer to pi over two. Well, this ray, I guess I should say, is getting closer and closer to approaching the vertical. So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So that is theta. We're getting closer and closer to pi over two. Well, this ray, I guess I should say, is getting closer and closer to approaching the vertical. So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a, I'm going to draw essentially a vertical asymptote right over here at pi over two because it's not going to be, I guess one way to think about it is approaching infinity there. So this is going to be looking something like this. It's going to be looking something like this."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So its slope is getting more and more and more positive, and if you go all the way to pi over two, the slope at that point is really undefined, but it's approaching, one way to think about it is, it is approaching infinity. So as you get closer and closer to pi over two, so I'm going to make a, I'm going to draw essentially a vertical asymptote right over here at pi over two because it's not going to be, I guess one way to think about it is approaching infinity there. So this is going to be looking something like this. It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. And what happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's going to be looking something like this. The slope of the ray as you get closer and closer to pi over two is getting closer and closer to infinity. And what happens when the angle is getting closer and closer to negative pi over two? Is getting closer and closer to negative pi over two? Well, then the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Is getting closer and closer to negative pi over two? Well, then the slope is getting more and more and more negative. It's really approaching negative. It's approaching negative infinity. So let me draw that. So once again, not quite defined right over there. We have a vertical asymptote, and we are approaching negative infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's approaching negative infinity. So let me draw that. So once again, not quite defined right over there. We have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. So that's what the graph of tangent of theta looks, just over this section of, I guess we could say, the theta axis, but then we could keep going. There we could keep going."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We have a vertical asymptote, and we are approaching negative infinity. We are approaching negative infinity. So that's what the graph of tangent of theta looks, just over this section of, I guess we could say, the theta axis, but then we could keep going. There we could keep going. Because if our angle, right after we cross pi over two, so let's say we've just crossed pi over two. So we went right across it. Now what is the slope of this thing?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "There we could keep going. Because if our angle, right after we cross pi over two, so let's say we've just crossed pi over two. So we went right across it. Now what is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what is the slope of this thing? Well, the slope of this thing is hugely negative. It looks almost like what I just drew down here. It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative, and then as we increase our theta, as we increase our theta, it becomes less and less and less negative, all the way to when we go to, what is this, all the way until we go to, let me plot this, this angle right over here. Now what is this angle?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's hugely negative. So then the graph jumps back down here, and it's hugely negative again. It's hugely negative, and then as we increase our theta, as we increase our theta, it becomes less and less and less negative, all the way to when we go to, what is this, all the way until we go to, let me plot this, this angle right over here. Now what is this angle? This I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four?"}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now what is this angle? This I haven't told you yet. Let's say that this angle right over here is three pi over four. Now why did I pick three pi over four? Because that is pi over two, that is pi over two plus pi over four. Or you could say two pi's over four plus another pi over four is three pi over four. And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now why did I pick three pi over four? Because that is pi over two, that is pi over two plus pi over four. Or you could say two pi's over four plus another pi over four is three pi over four. And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive. So the slope here is going to be the slope at the same slope as we had for negative pi over four radians. We're going to have a slope of negative one. So at three pi over four, we have a slope of negative one."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the reason why this is interesting is because it is another, it's forming another, I guess you could say, pi over four, pi over four, pi over two triangle, or 45, 45, 90 triangle, where the x and y coordinates, or the x and y distances have the same magnitude, but now the x is going to be negative and the y is positive. So the slope here is going to be the slope at the same slope as we had for negative pi over four radians. We're going to have a slope of negative one. So at three pi over four, we have a slope of negative one. Then we increase our angle all the way to pi. Now our slope is back to zero. Our slope is back to zero."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So at three pi over four, we have a slope of negative one. Then we increase our angle all the way to pi. Now our slope is back to zero. Our slope is back to zero. And then as we go beyond that, as we go to, as we increase by another, as we increase by another pi over four, our slope goes back to being positive one. Our slope goes back to being positive one. And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Our slope is back to zero. And then as we go beyond that, as we go to, as we increase by another, as we increase by another pi over four, our slope goes back to being positive one. Our slope goes back to being positive one. And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity. The slope, notice if you move a little bit in the x direction, you're moving a lot up in the y direction. So once again, so now the graph is going to look like this. Let me do it in a color that you can actually see."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then once again, as we approach three pi over two, our slope is becoming more and more and more positive, getting, approaching positive, approaching positive infinity. The slope, notice if you move a little bit in the x direction, you're moving a lot up in the y direction. So once again, so now the graph is going to look like this. Let me do it in a color that you can actually see. The graph is going to look something, something like, something like this. And it will just continue to do this. It will just continue to do this every, every pi radians."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me do it in a color that you can actually see. The graph is going to look something, something like, something like this. And it will just continue to do this. It will just continue to do this every, every pi radians. Every, actually maybe we do that as a dotted line. Every pi radians, over and over and over again. So let me go back pi, and I can draw these asymptotes."}, {"video_title": "Tangent graph Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It will just continue to do this every, every pi radians. Every, actually maybe we do that as a dotted line. Every pi radians, over and over and over again. So let me go back pi, and I can draw these asymptotes. I can draw these asymptotes. And so we draw that and that. And so the graph of tangent, the graph of tangent of theta is going to look, is going to look something, something like this."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "Rewrite P of X as a product of linear factors. Pause this video and see if you can have a go at that. All right, now let's work on this together. Because they give us one of the factors, what we can do is say, hey, what happens if I divide X plus six into P of X? What do I have left over? It looks like I'm still going to have a quadratic, and then I'll probably have to factor that somehow to get a product of linear factors. So let's get going."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "Because they give us one of the factors, what we can do is say, hey, what happens if I divide X plus six into P of X? What do I have left over? It looks like I'm still going to have a quadratic, and then I'll probably have to factor that somehow to get a product of linear factors. So let's get going. So if I were to try to figure out what X plus six divided into X to the third plus nine X squared, and now we're gonna have to be careful. You might be tempted to just write minus 108 there, but then this gets tricky because you have your third degree column, your second degree column, you need your first degree column, but you just put your zero degree, your constant column here. So to make sure we have good hygiene, we could write plus zero X, and I encourage you to actually always do this if you're writing out a polynomial so that you don't skip that place, so to speak, minus 108."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "So let's get going. So if I were to try to figure out what X plus six divided into X to the third plus nine X squared, and now we're gonna have to be careful. You might be tempted to just write minus 108 there, but then this gets tricky because you have your third degree column, your second degree column, you need your first degree column, but you just put your zero degree, your constant column here. So to make sure we have good hygiene, we could write plus zero X, and I encourage you to actually always do this if you're writing out a polynomial so that you don't skip that place, so to speak, minus 108. And so then you say, all right, let's look at the highest degree terms. X goes into X to the third X squared times. X squared times six is six X squared."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "So to make sure we have good hygiene, we could write plus zero X, and I encourage you to actually always do this if you're writing out a polynomial so that you don't skip that place, so to speak, minus 108. And so then you say, all right, let's look at the highest degree terms. X goes into X to the third X squared times. X squared times six is six X squared. X squared times X is X to the third. We wanna subtract. We've done this multiple times, so I'm going a little bit faster than normal."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "X squared times six is six X squared. X squared times X is X to the third. We wanna subtract. We've done this multiple times, so I'm going a little bit faster than normal. Those cancel out. Nine X squared minus six X squared is three X squared. Bring down that zero X."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "We've done this multiple times, so I'm going a little bit faster than normal. Those cancel out. Nine X squared minus six X squared is three X squared. Bring down that zero X. And then how many times does X go into three X squared? Well, it goes three X times, and we would write it in this column. And notice, if we didn't keep this column for our first degree terms, we'd be kinda confused where to write that three X right about now."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "Bring down that zero X. And then how many times does X go into three X squared? Well, it goes three X times, and we would write it in this column. And notice, if we didn't keep this column for our first degree terms, we'd be kinda confused where to write that three X right about now. And so three X plus times six, I should say, is 18 X. Three X times X is three X squared. We wanna subtract what we have in that, I guess that color is mauve, light purple, not sure."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "And notice, if we didn't keep this column for our first degree terms, we'd be kinda confused where to write that three X right about now. And so three X plus times six, I should say, is 18 X. Three X times X is three X squared. We wanna subtract what we have in that, I guess that color is mauve, light purple, not sure. And so we get three X squareds cancel out, and then zero X minus 18 X is negative 18 X. Bring down that negative 108. And so then we have X goes into negative 18 X, negative 18 times."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "We wanna subtract what we have in that, I guess that color is mauve, light purple, not sure. And so we get three X squareds cancel out, and then zero X minus 18 X is negative 18 X. Bring down that negative 108. And so then we have X goes into negative 18 X, negative 18 times. Negative 18 times six is negative 108. That was working out nicely. Negative 18 times X is negative 18 X."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "And so then we have X goes into negative 18 X, negative 18 times. Negative 18 times six is negative 108. That was working out nicely. Negative 18 times X is negative 18 X. And then we wanna subtract what we have in this not so pleasant brown color. And so I will multiply them both by a negative. And so I am left with zero."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "Negative 18 times X is negative 18 X. And then we wanna subtract what we have in this not so pleasant brown color. And so I will multiply them both by a negative. And so I am left with zero. Everything just cancels out. And so I can rewrite P of X. I can rewrite P of X as being equal to X plus six times X squared plus three X minus 18. But I'm not done yet because this is not a linear factor."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "And so I am left with zero. Everything just cancels out. And so I can rewrite P of X. I can rewrite P of X as being equal to X plus six times X squared plus three X minus 18. But I'm not done yet because this is not a linear factor. This is still quadratic. So let's see, can I think of two numbers that add up to three? And then when I multiply, I get negative 18."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "But I'm not done yet because this is not a linear factor. This is still quadratic. So let's see, can I think of two numbers that add up to three? And then when I multiply, I get negative 18. So I'll need different signs. And then the obvious one is positive six and negative three. And if that, what I just did seems like voodoo to you."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "And then when I multiply, I get negative 18. So I'll need different signs. And then the obvious one is positive six and negative three. And if that, what I just did seems like voodoo to you. I encourage you to review factoring polynomials. But this I can rewrite because negative six plus, or actually I should say positive six plus negative three is equal to three. And then positive six times negative three is equal to negative 18."}, {"video_title": "Factoring using polynomial division missing term Algebra 2 Khan Academy.mp3", "Sentence": "And if that, what I just did seems like voodoo to you. I encourage you to review factoring polynomials. But this I can rewrite because negative six plus, or actually I should say positive six plus negative three is equal to three. And then positive six times negative three is equal to negative 18. So I can rewrite this as X plus six times X plus six times X minus three. And so there we have it. We have a product of linear factors."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "And what I want you to do, like always, pause the video and see if you can rewrite each of these logarithmic expressions in a simpler way. And I'll give you a hint in case you haven't started yet. The hint is that if you think about how you might be able to change the base of the logarithms or the logarithmic expressions, you might be able to simplify this a good bit. And I'll give you even a further hint. When I'm talking about change of base, I'm saying that if I have the log base, and I'll color code it, log base A of B, log base A of B, this is going to be equal to log of B, log of B over log of A, over log of A. Now you might be saying, wait, wait, we wrote a logarithm here, but you didn't write what the base is. Well, this is going to be true regardless of which base you choose, as long as you pick the same base."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I'll give you even a further hint. When I'm talking about change of base, I'm saying that if I have the log base, and I'll color code it, log base A of B, log base A of B, this is going to be equal to log of B, log of B over log of A, over log of A. Now you might be saying, wait, wait, we wrote a logarithm here, but you didn't write what the base is. Well, this is going to be true regardless of which base you choose, as long as you pick the same base. This could be base nine, base nine in either case. Now typically, people choose base 10. So 10 is the most typical one to use."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, this is going to be true regardless of which base you choose, as long as you pick the same base. This could be base nine, base nine in either case. Now typically, people choose base 10. So 10 is the most typical one to use. And that's because most people's calculators, or there might be logarithmic tables for base 10. So here, you're saying the exponent that I have to raise A to to get to B is equal to the exponent I have to raise 10 to to get to B, divided by the exponent I have to raise 10 to to get to A. This is a really, really useful thing to know if you are dealing with logarithms."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "So 10 is the most typical one to use. And that's because most people's calculators, or there might be logarithmic tables for base 10. So here, you're saying the exponent that I have to raise A to to get to B is equal to the exponent I have to raise 10 to to get to B, divided by the exponent I have to raise 10 to to get to A. This is a really, really useful thing to know if you are dealing with logarithms. And we prove it in another video. But now let's see if we can apply it. So going back to this yellow expression, this, once again, is the same thing as one divided by this right over here."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "This is a really, really useful thing to know if you are dealing with logarithms. And we prove it in another video. But now let's see if we can apply it. So going back to this yellow expression, this, once again, is the same thing as one divided by this right over here. So let me write it that way, actually. This is one divided by log base B of four. Well, let's use what we just said over here to rewrite it."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "So going back to this yellow expression, this, once again, is the same thing as one divided by this right over here. So let me write it that way, actually. This is one divided by log base B of four. Well, let's use what we just said over here to rewrite it. So this is going to be equal to, this is going to be equal to one divided by, instead of writing it log base B of four, we could write it as log of four. And if I don't write the base there, we can assume that it is base 10. Log of four over log of B."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, let's use what we just said over here to rewrite it. So this is going to be equal to, this is going to be equal to one divided by, instead of writing it log base B of four, we could write it as log of four. And if I don't write the base there, we can assume that it is base 10. Log of four over log of B. Now, if I divide by some fraction or some rational expression, that's the same thing as multiplying by the reciprocal. So this is going to be equal to one times the reciprocal of this. Log of B over log of four, which of course is just going to be log of B over log of four."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Log of four over log of B. Now, if I divide by some fraction or some rational expression, that's the same thing as multiplying by the reciprocal. So this is going to be equal to one times the reciprocal of this. Log of B over log of four, which of course is just going to be log of B over log of four. I just multiplied it by one. And so we can go in the other direction now using this little tool we established at the beginning of the video. This is the same thing as log base four of B. Log base four of B."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Log of B over log of four, which of course is just going to be log of B over log of four. I just multiplied it by one. And so we can go in the other direction now using this little tool we established at the beginning of the video. This is the same thing as log base four of B. Log base four of B. So we have a pretty neat result that actually came out here. We didn't prove it for any values, although we have a pretty general B here. If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "This is the same thing as log base four of B. Log base four of B. So we have a pretty neat result that actually came out here. We didn't prove it for any values, although we have a pretty general B here. If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases. This is log base B. What exponent do I have to raise B to to get to four? And then here I have what exponent do I have to raise four to to get to B?"}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "If I take the reciprocal of a logarithmic expression, I essentially have swapped the bases. This is log base B. What exponent do I have to raise B to to get to four? And then here I have what exponent do I have to raise four to to get to B? Now, it might seem a little bit magical until you actually put some tangible numbers here. Then it starts to make sense, especially relative to fractional exponents. For example, we know that four to the third power is equal to 64."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "And then here I have what exponent do I have to raise four to to get to B? Now, it might seem a little bit magical until you actually put some tangible numbers here. Then it starts to make sense, especially relative to fractional exponents. For example, we know that four to the third power is equal to 64. So if I had log base four of 64, that's going to be equal to three. And if I were to say log base 64 of four, well now I'm going to have to raise that to the 1 3rd power. So notice, they are the reciprocal of each other."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "For example, we know that four to the third power is equal to 64. So if I had log base four of 64, that's going to be equal to three. And if I were to say log base 64 of four, well now I'm going to have to raise that to the 1 3rd power. So notice, they are the reciprocal of each other. So actually not so magical after all, but it's nice to see how everything fits together. Now let's try to tackle this one over here. So I have log base C of 16 times log base two of C. All right."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "So notice, they are the reciprocal of each other. So actually not so magical after all, but it's nice to see how everything fits together. Now let's try to tackle this one over here. So I have log base C of 16 times log base two of C. All right. So this one, once again, it might be nice to rewrite each of these as a rational expression dealing with log base 10. So this first one, I could write this as log base 10 of 16. Remember, if I don't write the base, you could assume it's 10."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I have log base C of 16 times log base two of C. All right. So this one, once again, it might be nice to rewrite each of these as a rational expression dealing with log base 10. So this first one, I could write this as log base 10 of 16. Remember, if I don't write the base, you could assume it's 10. Over log, over log base 10 of C. And we're going to be multiplying this by, now this is going to be, we could write this as log base 10 of C, log base 10 of C over, over log base 10 of two. Log base 10 of two. Once again, I could have these little 10s here if it makes you comfortable."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Remember, if I don't write the base, you could assume it's 10. Over log, over log base 10 of C. And we're going to be multiplying this by, now this is going to be, we could write this as log base 10 of C, log base 10 of C over, over log base 10 of two. Log base 10 of two. Once again, I could have these little 10s here if it makes you comfortable. I could do something like that, but I don't have to. And now this is interesting, because if I'm multiplying by log of C and dividing by log of C, both of them base 10, well, those are going to cancel out, and I'm going to be left with log base 16, sorry, log base 10 of 16 over, over log base 10 of two. And we know how to go the other direction here."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "Once again, I could have these little 10s here if it makes you comfortable. I could do something like that, but I don't have to. And now this is interesting, because if I'm multiplying by log of C and dividing by log of C, both of them base 10, well, those are going to cancel out, and I'm going to be left with log base 16, sorry, log base 10 of 16 over, over log base 10 of two. And we know how to go the other direction here. This is going to be, this is going to be the logarithm, log base two of 16. Log base two of 16, and we're not done yet, because all this is is what power do I need to raise two to to get to 16? I have to raise two to the, I have to raise two to the fourth power."}, {"video_title": "Using the logarithm change of base rule Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we know how to go the other direction here. This is going to be, this is going to be the logarithm, log base two of 16. Log base two of 16, and we're not done yet, because all this is is what power do I need to raise two to to get to 16? I have to raise two to the, I have to raise two to the fourth power. We do the blue color. I have to raise two to the fourth power to get to 16. So this is kind of a cool thing, because in the beginning I started with this variable C. It looked like we were going to have to deal with a pretty abstract thing, but you can actually evaluate this, this kind of crazy looking expression right over here, evaluates to the number four."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "The hottest day of the year in Santiago, Chile, on average, is January 7th, when the average high temperature is 29 degrees Celsius. The coolest day of the year has an average high temperature of 14 degrees Celsius. Use a trigonometric function to model the temperature in Santiago, Chile, using 365 days as the length of a year. Remember that January 7th is the summer in Santiago. How many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Remember that January 7th is the summer in Santiago. How many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So let's do this in two parts. So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So first, let's try to figure out a trigonometric function that models the temperature in Santiago, Chile. So we'll have temperature as a function of days, where days are the number of days after January 7th. And once we have that trigonometric function to model that, then we can answer the second part, I guess the essential question, which is how many days after January 7th is the first spring day when the temperature reaches 20 degrees Celsius? So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So to think about this, let's graph it. Let's graph it, and it should become pretty apparent why they are suggesting that we use a trigonometric function to model this, because our seasonal variations, they're cyclical, they go up and down. And actually, if you look at the average temperature for any city over the course of the year, it really does look like a trigonometric function. So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's, so let's, this axis right over here, this is the passage of the days. Let's do D for days, and that's going to be in days after January 7th. So this right over here would be January 7th. And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And the vertical axis, this is the horizontal, the vertical axis is going to be in terms of Celsius degrees. So this, let's see, the high is 29, and I could write 29 degrees Celsius, and then, or the highest average day. And then if this is zero, then 14, which is the lowest average day, 14 degrees Celsius. And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And so our temperature will vary between these two extremes. Our temperature is going to vary, the highest average day, which they already told us is January 7th, we get to 29 degrees Celsius. And then the coldest day of the year, on average, you get to an average high of 14 degrees Celsius. So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So it looks like this. We're talking about average highs on a given day. And the reason why a trigonometric function is a good idea is because it's cyclical. So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So if this is January 7th, if you go 365 days in the future, you're back at January 7th. So if the average high temperature is 29 degrees Celsius on that day, the average high temperature is going to be 29 degrees Celsius on that day. Now, we're using a trigonometric function, so we're gonna hit our low point exactly halfway in between. So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there. And this is a high point."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we're gonna hit our low point exactly halfway in between, something right like that. And so our function is going to look like this. Our function, so let me see, let me just draw the low point right over there. And this is a high point. It's a high point right over there. That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And this is a high point. It's a high point right over there. That looks pretty good. And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days. If we go 365 days later, we're at the same point in the cycle."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And then I have the high point right over here, and then I just need to connect them, and there you go. I've drawn one period of our trigonometric function. And our period is 365 days. If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we go 365 days later, we're at the same point in the cycle. We are at the same point in the year. We're at the same point in the year. So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So what I want you to do right now is given what I've just drawn, try to model this right. So this right over here is, let's write this, this is capital T as a function of D. Try to figure out an expression for capital T as a function of D. And remember, it's going to be some trigonometric function. So I'm assuming you've given a go at it, and you might say, well, this looks like a cosine curve. Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Maybe it could be a sine curve. Which one should I use? And you could actually use either one, but I always like to go with the simpler one. You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "You just think about, well, if this were angles, either actually degrees or radians, which trigonometric function starts at your maximum point? Well, cosine of zero is one. The cosine starts at your maximum point. Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "Sine of zero is zero. So I'm going to use cosine here. I'm going to use a cosine function. So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it. So let's think about how we would do that."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So temperature as a function of days is going to be some amplitude times our cosine function. And we're going to have some argument to our cosine function. And then I'm probably going to have to shift it. So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So let's think about how we would do that. Well, what's the midline here? Well, the midline's the halfway point between our high and our low. So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So our middle point, if we were to visualize it, looks just like that is our midline right over there. And what value is this? Well, what's the average of 29 and 14? 29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "29 plus 14 is 43 divided by 2 is 21.5 degrees Celsius. So that's our midline. So essentially, we've shifted up our function by that amount. If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5. That's how much we've shifted it up."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "If we just had a regular cosine function, our midline would be at 0. But now we're at 21.5 degrees Celsius. So I'll just write plus 21.5. That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "That's how much we've shifted it up. Now, what's the amplitude? Well, our amplitude is how much we diverge from the midline. So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline. So minus 7.5."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So over here, we're 7.5 above the midline. So that's plus 7.5. Here, we're 7.5 below the midline. So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5. So that's our amplitude."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So minus 7.5. So our amplitude is 7.5. So the maximum amount we go away from the midline is 7.5. So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days. And what do we want?"}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So that's our amplitude. And now let's think about our argument to the cosine function right over here. So it's going to be a function of the days. And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "And what do we want? When 365 days have gone by, we want this entire argument to be 2 pi. So when d is 365, we want this whole thing to evaluate to 2 pi. So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them. That's why I always try to reason through them again."}, {"video_title": "Modeling annual temperature variation with trigonometry Trigonometry Khan Academy.mp3", "Sentence": "So we could put 2 pi over 365 in here. And you might remember your formulas. I always forget them. That's why I always try to reason through them again. The formula is, oh, you want 2 pi divided by your period and all the rest. But I just like to think, OK, look, after one period, which is 365 days, I want the whole argument over here to be 2 pi. I want to go around the unit circle once."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "Solve the equation for t and express your answer in terms of base 10 logarithms. And this equation says 10 to the 2t minus 3 is equal to 7. We want to solve for t in terms of base 10 logarithms. So let me get my little scratch pad out and I've copied and pasted the same problem. So I'm just going to rewrite it. So they have 10 to the 2t minus 3 is equal to 7. And actually let me color code this a little bit."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let me get my little scratch pad out and I've copied and pasted the same problem. So I'm just going to rewrite it. So they have 10 to the 2t minus 3 is equal to 7. And actually let me color code this a little bit. So 10 to the 2t minus 3 is equal to 7. So this is clearly an exponential form right over here. If we want to write it in logarithmic form, that'll essentially allow us to solve for the exponent."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "And actually let me color code this a little bit. So 10 to the 2t minus 3 is equal to 7. So this is clearly an exponential form right over here. If we want to write it in logarithmic form, that'll essentially allow us to solve for the exponent. So we could say this is the exact same truth about the universe as saying that the log base 10 of 7 is equal to 2t minus 3. Let's just make sure that makes sense. This is saying 10 to the 2t minus 3 is equal to 7."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we want to write it in logarithmic form, that'll essentially allow us to solve for the exponent. So we could say this is the exact same truth about the universe as saying that the log base 10 of 7 is equal to 2t minus 3. Let's just make sure that makes sense. This is saying 10 to the 2t minus 3 is equal to 7. This is saying that the power that I need to raise 10 to to get to 7 is 2t minus 3. Or 10 to the 2t minus 3 power is equal to 7. So these are equivalent statements."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is saying 10 to the 2t minus 3 is equal to 7. This is saying that the power that I need to raise 10 to to get to 7 is 2t minus 3. Or 10 to the 2t minus 3 power is equal to 7. So these are equivalent statements. What this form does is it starts to put into a form that's easier to solve for t. Now if we want to solve for t, we can add 3 to both sides. So if we add 3 to both sides, we are going to get log base 10 of 7 plus 3. And this plus 3 is, of course, outside of the logarithm to make it clear."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "So these are equivalent statements. What this form does is it starts to put into a form that's easier to solve for t. Now if we want to solve for t, we can add 3 to both sides. So if we add 3 to both sides, we are going to get log base 10 of 7 plus 3. And this plus 3 is, of course, outside of the logarithm to make it clear. It's just like that. And on the right-hand side, this is going to be equal to just 2t. And now to solve for t, we just divide both sides by 2."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "And this plus 3 is, of course, outside of the logarithm to make it clear. It's just like that. And on the right-hand side, this is going to be equal to just 2t. And now to solve for t, we just divide both sides by 2. So if we divide both sides by 2, we get t is equal to all of this business. Log base 10 of 7 plus 3, all of that over 2. So let me see if I can remember this and write it in the actual box that they gave us."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "And now to solve for t, we just divide both sides by 2. So if we divide both sides by 2, we get t is equal to all of this business. Log base 10 of 7 plus 3, all of that over 2. So let me see if I can remember this and write it in the actual box that they gave us. So we want to do log base 10 of 7 and then that plus 3. Yep, it's formatting it right. Divided by 2."}, {"video_title": "Solving exponential equation with logarithm Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let me see if I can remember this and write it in the actual box that they gave us. So we want to do log base 10 of 7 and then that plus 3. Yep, it's formatting it right. Divided by 2. So that's what I'm claiming that t is equal to in terms of base 10 logarithms. So let me check my answer. And I got it right."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So for example, if we look right over here, x is 1.585. b to the 1.585 is 3. So this is telling us that b to the 1.585 is equal to 3. Similarly, I can never say that, it's telling us that b to the 2.322 is 5. b to the 2.807 is 7. b to the 2.169 is 9. Now this table over here is telling us for any given value of y, what is going to be log base b of y? So this tells us that log base b of a is 0. Log base b of 2 is 1. Log base b of 2c is 1.585."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now this table over here is telling us for any given value of y, what is going to be log base b of y? So this tells us that log base b of a is 0. Log base b of 2 is 1. Log base b of 2c is 1.585. Log base b of 10d. So this is literally telling us that log base b of 10d is equal to 2.322. That's what this last column tells us."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Log base b of 2c is 1.585. Log base b of 10d. So this is literally telling us that log base b of 10d is equal to 2.322. That's what this last column tells us. Now what I challenge you to do is pause this video. And using just the information here, and you don't need a calculator. In fact, you can't use a calculator."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's what this last column tells us. Now what I challenge you to do is pause this video. And using just the information here, and you don't need a calculator. In fact, you can't use a calculator. I forbid you. Try to figure out what a, b, c, and d are, just using your powers of reasoning. No calculator involved."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "In fact, you can't use a calculator. I forbid you. Try to figure out what a, b, c, and d are, just using your powers of reasoning. No calculator involved. Just use your powers of reasoning. Can you figure out what a, b, c, and d are? So I'm assuming you've given a go at it."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "No calculator involved. Just use your powers of reasoning. Can you figure out what a, b, c, and d are? So I'm assuming you've given a go at it. So let's see what we can deduce from this. So here we have just a bunch of numbers. We need to figure out what b is."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it. So let's see what we can deduce from this. So here we have just a bunch of numbers. We need to figure out what b is. So these are all kind of b to the 1.585 power is 3. I don't really know what to make sense of this stuff here. Maybe this table will help us."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "We need to figure out what b is. So these are all kind of b to the 1.585 power is 3. I don't really know what to make sense of this stuff here. Maybe this table will help us. So this first, let me do this in different colors. This first column right over here tells us that log base b of a, so now y is equal to a, that that is equal to 0. Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Maybe this table will help us. So this first, let me do this in different colors. This first column right over here tells us that log base b of a, so now y is equal to a, that that is equal to 0. Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? Well, you raise it to the 0 power."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now this is an equivalent statement to saying that b to the a power is equal to, oh sorry, not b to the a power. This is an equivalent statement to saying b to the 0 power is equal to a. This is saying what exponent do I need to raise b to to get a? Well, you raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, you raise it to the 0 power. This is saying b to the 0 power is equal to a. Now what is anything to the 0 power, assuming that it's not 0? If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value. So since we know that b is not 0, anything to the 0 power is going to be 1. So this tells us that a is equal to 1. So we got 1 figured out."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we're assuming that we're dealing with that b is not 0, if we're assuming that b is not 0, then so we're going to assume that, and we can't assume, and I think that's a safe assumption because we're raising b to all of these other powers, we're getting a non-zero value. So since we know that b is not 0, anything to the 0 power is going to be 1. So this tells us that a is equal to 1. So we got 1 figured out. a is equal to 1. Now let's look at this next piece of information right over here. What does that tell us?"}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we got 1 figured out. a is equal to 1. Now let's look at this next piece of information right over here. What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I need to raise b to to get to 2 is 1. Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "What does that tell us? That tells us that log base b of 2 is equal to 1. This is equivalent to saying the power that I need to raise b to to get to 2 is 1. Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2. So I'm raising something to the first power, and I'm getting 2. What is this thing? Well, that means that b must be 2."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or if I want to write it in exponential form, I could write this as saying that b to the first power is equal to 2. So I'm raising something to the first power, and I'm getting 2. What is this thing? Well, that means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that means that b must be 2. 2 to the first power is 2. So b is equal to 2. b to the first power is equal to 2. You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. So we've been able to figure that out."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "You could say b to the first is equal to 2 to the first. That's also equal to 2. So b must be equal to 2. So we've been able to figure that out. So this is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we've been able to figure that out. So this is a 2 right over here. It actually makes sense. 2 to the 1.585 power, yeah, that feels right. That's about 3. Now let's see what else we can do. Let's see if we can figure out c. So let's look at this column."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 to the 1.585 power, yeah, that feels right. That's about 3. Now let's see what else we can do. Let's see if we can figure out c. So let's look at this column. Let's see what this column is telling us. That column, we could read it as log base b. Now our y is 2c."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's see if we can figure out c. So let's look at this column. Let's see what this column is telling us. That column, we could read it as log base b. Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write an exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585?"}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now our y is 2c. Log base b of 2c is equal to 1.585. Or we could read this as b, if we write an exponential form, b to the 1.585 is equal to 2c. Now what's b to the 1.585? Well, they tell us right over here that b to the 1.585 is 3. So this right over here is equal to 3. So we get 2c is equal to 3."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now what's b to the 1.585? Well, they tell us right over here that b to the 1.585 is 3. So this right over here is equal to 3. So we get 2c is equal to 3. Or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. And now we have this last column, which I will circle in purple."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we get 2c is equal to 3. Or divide both sides by 2, we would get c is equal to 1.5. This is working out pretty well. And now we have this last column, which I will circle in purple. And we can write this as log base b of 10d is equal to 2.322. So this is saying the power I need to raise b to get to 10d is 2.322. Or in exponential form, b to the 2.322 is equal to 10d."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "And now we have this last column, which I will circle in purple. And we can write this as log base b of 10d is equal to 2.322. So this is saying the power I need to raise b to get to 10d is 2.322. Or in exponential form, b to the 2.322 is equal to 10d. Now what is b to the 2.322? Well, they tell us over here b to the 2.322 is 5. So this is equal to 5."}, {"video_title": "Deducing with logarithmic and exponential form Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or in exponential form, b to the 2.322 is equal to 10d. Now what is b to the 2.322? Well, they tell us over here b to the 2.322 is 5. So this is equal to 5. So we could write 10d is equal to 5. Or divide both sides by 10, d is equal to 0.5. And we're done."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And as we'll see when we're adding complex numbers, you can only add the real parts to each other and you can only add the imaginary parts to each other. So let's add the real parts. So we have a 5 plus a 3. And then the imaginary parts, we have a 2i, so plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3, that's pretty straightforward."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And then the imaginary parts, we have a 2i, so plus 2i. And then we have a negative 7i, or we're subtracting 7i. So minus 7i right over here. And 5 plus 3, that's pretty straightforward. That's just going to be 8. And then if I have 2 of something, and from that I subtract 7 of that something, and in this case the something is the imaginary unit, the number i. If I have 2i's and I take away 7i's, then I have negative 5i's."}, {"video_title": "Adding complex numbers Imaginary and complex numbers Algebra II Khan Academy.mp3", "Sentence": "And 5 plus 3, that's pretty straightforward. That's just going to be 8. And then if I have 2 of something, and from that I subtract 7 of that something, and in this case the something is the imaginary unit, the number i. If I have 2i's and I take away 7i's, then I have negative 5i's. 2 minus 7 is negative 5. So then I have negative 5i. So when you add these two complex numbers, you get 8 minus 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We're asked to multiply the complex number 1 minus 3i times the complex number 2 plus 5i. And the general idea here is you can multiply these complex numbers like you would have multiplied any traditional binomial. You just have to remember that this isn't a variable. This is the imaginary unit i, or it's just i. But we could do that in two ways. We could just do the distributive property twice, which I like a little bit more just because it actually, you're doing it from a fundamental principle. There's nothing new."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is the imaginary unit i, or it's just i. But we could do that in two ways. We could just do the distributive property twice, which I like a little bit more just because it actually, you're doing it from a fundamental principle. There's nothing new. Or you could use FOIL, which you also use when you first multiply binomials. And I'll do it both ways. So you could view, this is just a number, 1 minus 3i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "There's nothing new. Or you could use FOIL, which you also use when you first multiply binomials. And I'll do it both ways. So you could view, this is just a number, 1 minus 3i. And so we can distribute it over the two numbers inside of this expression. So when we're multiplying it times this entire expression, we can multiply 1 minus 3i times 2 and 1 minus 3i times 5i. So let's do that."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So you could view, this is just a number, 1 minus 3i. And so we can distribute it over the two numbers inside of this expression. So when we're multiplying it times this entire expression, we can multiply 1 minus 3i times 2 and 1 minus 3i times 5i. So let's do that. So this can be rewritten as 1 minus 3i times 2. I'll write the 2 out front. Plus 1 minus 3i times 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So let's do that. So this can be rewritten as 1 minus 3i times 2. I'll write the 2 out front. Plus 1 minus 3i times 5i. All I did is a distributive property here. All I said is, look, if I have a times b plus c, this is the same thing as ab plus ac. I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Plus 1 minus 3i times 5i. All I did is a distributive property here. All I said is, look, if I have a times b plus c, this is the same thing as ab plus ac. I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i. And then I can do it again. I have a 2 now times 1 minus 3i. I can distribute it."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I just distributed the a on the b or the c. I distributed the 1 minus 3i on the 2 and the 5i. And then I can do it again. I have a 2 now times 1 minus 3i. I can distribute it. 2 times 1 is 2. 2 times negative 3i is negative 6i. And over here, I'll do it again."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I can distribute it. 2 times 1 is 2. 2 times negative 3i is negative 6i. And over here, I'll do it again. 5i times 1. So it's plus 5i times 1 is 5i. And then 5i times negative 3i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And over here, I'll do it again. 5i times 1. So it's plus 5i times 1 is 5i. And then 5i times negative 3i. So let's be careful here. 5 times negative 3 is negative 15. And then I have an i times an i. I'm multiplying 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then 5i times negative 3i. So let's be careful here. 5 times negative 3 is negative 15. And then I have an i times an i. I'm multiplying 5i. Let me do this over here. 5i times negative 3i. This is the same thing as 5 times negative 3 times i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then I have an i times an i. I'm multiplying 5i. Let me do this over here. 5i times negative 3i. This is the same thing as 5 times negative 3 times i. So the 5 times negative 3 is negative 15. And then we have i times i, which is i squared. Now, we know what i squared is by definition."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is the same thing as 5 times negative 3 times i. So the 5 times negative 3 is negative 15. And then we have i times i, which is i squared. Now, we know what i squared is by definition. i squared is negative 1. So you have negative 15 times negative 1. Well, that's the same thing as positive 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, we know what i squared is by definition. i squared is negative 1. So you have negative 15 times negative 1. Well, that's the same thing as positive 15. So this can be rewritten as 2 minus 6i plus 5i. Negative 15 times negative 1 is positive 15. Now, we can add the real parts."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, that's the same thing as positive 15. So this can be rewritten as 2 minus 6i plus 5i. Negative 15 times negative 1 is positive 15. Now, we can add the real parts. We have a 2 and we have a positive 15. So 2 plus 15. And we can add the imaginary parts."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, we can add the real parts. We have a 2 and we have a positive 15. So 2 plus 15. And we can add the imaginary parts. We have a negative 6. So we have a negative 6i, I should say. And then we have plus 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And we can add the imaginary parts. We have a negative 6. So we have a negative 6i, I should say. And then we have plus 5i. So plus 5i. And 2 plus 15 is 17. And if I have negative 6 of something plus 5 of that something, what do I have?"}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then we have plus 5i. So plus 5i. And 2 plus 15 is 17. And if I have negative 6 of something plus 5 of that something, what do I have? If I have 5 of that something and I take 6 of that something away, then I have negative 1 of that something. Negative 6i plus 5i is negative 1i. Or I could just say minus i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And if I have negative 6 of something plus 5 of that something, what do I have? If I have 5 of that something and I take 6 of that something away, then I have negative 1 of that something. Negative 6i plus 5i is negative 1i. Or I could just say minus i. So in this way, I just multiplied these two expressions, or these two complex numbers, really. I multiplied them just using the distributive property twice. You could also do it using FOIL."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Or I could just say minus i. So in this way, I just multiplied these two expressions, or these two complex numbers, really. I multiplied them just using the distributive property twice. You could also do it using FOIL. And I'll do that right now really fast. It is a little bit faster, but it's a little bit mechanical. So you might forget why you're doing it in the first place."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "You could also do it using FOIL. And I'll do that right now really fast. It is a little bit faster, but it's a little bit mechanical. So you might forget why you're doing it in the first place. But at the end of the day, you are doing the same thing here. You're essentially multiplying every term of this first number, or every part of this first number, times every part of the second number. And FOIL just makes sure that we're doing it."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So you might forget why you're doing it in the first place. But at the end of the day, you are doing the same thing here. You're essentially multiplying every term of this first number, or every part of this first number, times every part of the second number. And FOIL just makes sure that we're doing it. And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it, just in case that's the way you're learning it. So FOIL says, let's do the first numbers. Let's multiply the first numbers."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And FOIL just makes sure that we're doing it. And let me just write FOIL out here, which I'm not a huge fan of, but I'll do it, just in case that's the way you're learning it. So FOIL says, let's do the first numbers. Let's multiply the first numbers. So that's going to be the 1 times the 2. So 1 times the 2. That is the F in FOIL."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let's multiply the first numbers. So that's going to be the 1 times the 2. So 1 times the 2. That is the F in FOIL. Then it says, let's multiply the outer numbers times each other. So that's 1 times 5i. So plus 1 times 5i."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "That is the F in FOIL. Then it says, let's multiply the outer numbers times each other. So that's 1 times 5i. So plus 1 times 5i. This is the O in FOIL, the outer numbers. Then we do the inner numbers. Negative 3i times 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So plus 1 times 5i. This is the O in FOIL, the outer numbers. Then we do the inner numbers. Negative 3i times 2. So this is negative 3i times 2. Those are the inner numbers. And then we do the last numbers."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Negative 3i times 2. So this is negative 3i times 2. Those are the inner numbers. And then we do the last numbers. Negative 3i times 5i. So negative 3i times 5i. These are the last numbers."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then we do the last numbers. Negative 3i times 5i. So negative 3i times 5i. These are the last numbers. And that's all that FOIL is telling us. It's just making sure we're multiplying every part of this number times every part of that number. And then when we simplify it, 1 times 2 is 2."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "These are the last numbers. And that's all that FOIL is telling us. It's just making sure we're multiplying every part of this number times every part of that number. And then when we simplify it, 1 times 2 is 2. 1 times 5i is 5i. Negative 3i times 2 is negative 6i. And negative 3i times 5i, well, we already figured out what that was."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then when we simplify it, 1 times 2 is 2. 1 times 5i is 5i. Negative 3i times 2 is negative 6i. And negative 3i times 5i, well, we already figured out what that was. Negative 3i times 5i turns out to be 15. Negative 3 times 5 is negative 15. But i times i is negative 1."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And negative 3i times 5i, well, we already figured out what that was. Negative 3i times 5i turns out to be 15. Negative 3 times 5 is negative 15. But i times i is negative 1. Negative 15 times negative 1 is positive 15. Add the real parts. 2 plus 15."}, {"video_title": "Multiplying complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But i times i is negative 1. Negative 15 times negative 1 is positive 15. Add the real parts. 2 plus 15. You get 17. Add the imaginary parts. You have 5i minus 6i."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "What I hope to do in this video is get some practice simplifying some fairly hairy exponential expressions. So let's get started. Let's say that I have the expression 10 times 9 to the t over 2 plus 2 power times 5 to the 3t. And what I want to do is simplify this as much as possible and preferably get it in the form of a times b to the t. And like always, I encourage you to pause this video and see if you can do this on your own using exponent properties, your knowledge, your deep knowledge of exponent properties. All right, so let's work through this together. And it's really just about breaking the pieces up. So 10, I'll just leave that as 10 for now."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "And what I want to do is simplify this as much as possible and preferably get it in the form of a times b to the t. And like always, I encourage you to pause this video and see if you can do this on your own using exponent properties, your knowledge, your deep knowledge of exponent properties. All right, so let's work through this together. And it's really just about breaking the pieces up. So 10, I'll just leave that as 10 for now. There doesn't seem to be much to do there. But there's all sorts of interesting things going on here. So 9 to the t over 2 plus 2, so this right over here, I could break this up using the fact that, and I'll just write the properties over here."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So 10, I'll just leave that as 10 for now. There doesn't seem to be much to do there. But there's all sorts of interesting things going on here. So 9 to the t over 2 plus 2, so this right over here, I could break this up using the fact that, and I'll just write the properties over here. If I have 9 to the a plus b power, this is the same thing as 9 to the a times 9 to the b power. And over here, I have 9 to the t over 2 plus 2. So I could rewrite this as 9 to the t over 2 power times 9 squared."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So 9 to the t over 2 plus 2, so this right over here, I could break this up using the fact that, and I'll just write the properties over here. If I have 9 to the a plus b power, this is the same thing as 9 to the a times 9 to the b power. And over here, I have 9 to the t over 2 plus 2. So I could rewrite this as 9 to the t over 2 power times 9 squared. All right, now let's move over to 5 to the 3t. Well, if I have a to the bc, so you could view this as 5 to the 3 times t. This is the same thing as a to the b and then that to the c power. So I could write this as, this is going to be the same thing as 5 to the 3rd and then that to the t power."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So I could rewrite this as 9 to the t over 2 power times 9 squared. All right, now let's move over to 5 to the 3t. Well, if I have a to the bc, so you could view this as 5 to the 3 times t. This is the same thing as a to the b and then that to the c power. So I could write this as, this is going to be the same thing as 5 to the 3rd and then that to the t power. And the whole reason I did that is, well, this is just going to be a number and then I'm going to have some number to the t power. I want to get as many things just raised to the t power as possible just to see if I can simplify this thing. So this character right over here is going to be 81."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So I could write this as, this is going to be the same thing as 5 to the 3rd and then that to the t power. And the whole reason I did that is, well, this is just going to be a number and then I'm going to have some number to the t power. I want to get as many things just raised to the t power as possible just to see if I can simplify this thing. So this character right over here is going to be 81. 9 squared is 81. 5 to the 3rd power, let's see, 25 times 5, that's 125. So we're making good progress."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So this character right over here is going to be 81. 9 squared is 81. 5 to the 3rd power, let's see, 25 times 5, that's 125. So we're making good progress. And so the only thing we really have to simplify at this point is 9 to the t over 2. And actually, let me do that over here. 9 to the t over 2."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So we're making good progress. And so the only thing we really have to simplify at this point is 9 to the t over 2. And actually, let me do that over here. 9 to the t over 2. Well, that's the same thing. That's the same thing as 9 to the 1 half times t. And by this property right over here, that's the same thing as 9 to the 1, whoops, 9 to the 1 half times, 9 to the 1 half and then that to the t power. So what's 9 to the 1 half?"}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "9 to the t over 2. Well, that's the same thing. That's the same thing as 9 to the 1 half times t. And by this property right over here, that's the same thing as 9 to the 1, whoops, 9 to the 1 half times, 9 to the 1 half and then that to the t power. So what's 9 to the 1 half? Well, that's 3. So this is going to be equal to 3 to the t power. So this right over here is 3 to the t power."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So what's 9 to the 1 half? Well, that's 3. So this is going to be equal to 3 to the t power. So this right over here is 3 to the t power. So now this is getting interesting. So I have the 10 out front, I have the 10 out front times 3 to the t, 3 to the t, actually let me write the 81 first. 10 times 81 times 3 to the t, so all I did is I just swapped the order that I'm multiplying, times 125 to the t, times 125 to the t power."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "So this right over here is 3 to the t power. So now this is getting interesting. So I have the 10 out front, I have the 10 out front times 3 to the t, 3 to the t, actually let me write the 81 first. 10 times 81 times 3 to the t, so all I did is I just swapped the order that I'm multiplying, times 125 to the t, times 125 to the t power. Now, the 10 times 81, I can just multiply that out, that's going to be 810. And then what's 3 to the t times 125 to the t? Well, this is another exponent property at play here."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "10 times 81 times 3 to the t, so all I did is I just swapped the order that I'm multiplying, times 125 to the t, times 125 to the t power. Now, the 10 times 81, I can just multiply that out, that's going to be 810. And then what's 3 to the t times 125 to the t? Well, this is another exponent property at play here. Because if I have a times b to the t power, that's a to the t times b to the t. Or another way to think about it, if I have a to the t times b to the t, that's the same thing as a times b to the t power. And so over here I have 3 to the t times 125 to the t. So it's going to be the same thing as 3 times 125 to the t power. So this part of it, right over here, I could rewrite it as 3 times 125, and I'm going to raise that whole thing to the t power."}, {"video_title": "Simplifying an exponential expression Algebra II Khan Academy.mp3", "Sentence": "Well, this is another exponent property at play here. Because if I have a times b to the t power, that's a to the t times b to the t. Or another way to think about it, if I have a to the t times b to the t, that's the same thing as a times b to the t power. And so over here I have 3 to the t times 125 to the t. So it's going to be the same thing as 3 times 125 to the t power. So this part of it, right over here, I could rewrite it as 3 times 125, and I'm going to raise that whole thing to the t power. So I'm in the home stretch. This is going to be 810 times, 3 times 125 is 375, times 375 to the t power. And that's about as simplified as we can get, and we did indeed write it in the form that we hoped to write it in, that a times b to the t. Where if we actually match this form right over here, this right over here would be our a, and then our b would be the 375."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "And someone were to come up to you and say, hey look, this is an interesting function, but I'm curious, I like the number 1,111, and I'm curious at what point, for what t value, will my y be equal to 1,111? And so I encourage you to pause this video and think about it on your own. At what t value will this, will y be equal to, or roughly equal to 1,111? And if you see the need, you might want to use a calculator. So I'm assuming you've given a go at it, let's work through this together. So we wanna say, when does five times two to the t power equal 1,111? So let's write that down."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "And if you see the need, you might want to use a calculator. So I'm assuming you've given a go at it, let's work through this together. So we wanna say, when does five times two to the t power equal 1,111? So let's write that down. So when does five times two to the t power equal 1,111? So whenever we're doing anything algebraically, it's always a little bit useful to see if we can isolate the variable that we're trying to solve for. We're trying to find what t value will make this equal that right over there."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "So let's write that down. So when does five times two to the t power equal 1,111? So whenever we're doing anything algebraically, it's always a little bit useful to see if we can isolate the variable that we're trying to solve for. We're trying to find what t value will make this equal that right over there. So a good first step would maybe try to get this five out of the left hand side. So let's divide the left by five, but if we wanna keep this to being an equality, we have to do the same thing to both sides. So we get two to the t power is equal to 1,111 over five."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "We're trying to find what t value will make this equal that right over there. So a good first step would maybe try to get this five out of the left hand side. So let's divide the left by five, but if we wanna keep this to being an equality, we have to do the same thing to both sides. So we get two to the t power is equal to 1,111 over five. So how do we solve for t here? Well, what function is essentially the inverse of the exponential function? Well, it would be the logarithm."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "So we get two to the t power is equal to 1,111 over five. So how do we solve for t here? Well, what function is essentially the inverse of the exponential function? Well, it would be the logarithm. If we say that a to the b power is equal to c, then that means that log base c, sorry, log base a of c is equal to b. A to the b power is equal to c. Log base a of c says what power do I need to raise a to to get to c? Well, I need to raise a to the b power to get to c. A to the b power is equal to c. So these two are actually equivalent statements."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Well, it would be the logarithm. If we say that a to the b power is equal to c, then that means that log base c, sorry, log base a of c is equal to b. A to the b power is equal to c. Log base a of c says what power do I need to raise a to to get to c? Well, I need to raise a to the b power to get to c. A to the b power is equal to c. So these two are actually equivalent statements. So let's take log base two of both sides of this equation. So on the left hand side, on the left hand side, you have log base two of two to the t power, two to the t power. And on the right hand side, you have log base two of 1,111 over five."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Well, I need to raise a to the b power to get to c. A to the b power is equal to c. So these two are actually equivalent statements. So let's take log base two of both sides of this equation. So on the left hand side, on the left hand side, you have log base two of two to the t power, two to the t power. And on the right hand side, you have log base two of 1,111 over five. Now why is this useful right over here? So this is what power do we have to raise two to to get to the two to the t power? Well, to get to two to the t power, we have to raise two to the t power."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "And on the right hand side, you have log base two of 1,111 over five. Now why is this useful right over here? So this is what power do we have to raise two to to get to the two to the t power? Well, to get to two to the t power, we have to raise two to the t power. So this thing right over here, this thing right over here just simplifies to, this just simplifies to t. That just simplifies to t. And on the right hand side, we have log base two, we have all of this business right over here. So I'll just write it over. T is equal to log base two of 1,111 over five."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "Well, to get to two to the t power, we have to raise two to the t power. So this thing right over here, this thing right over here just simplifies to, this just simplifies to t. That just simplifies to t. And on the right hand side, we have log base two, we have all of this business right over here. So I'll just write it over. T is equal to log base two of 1,111 over five. 1,111 over five. So this is an expression that gives us our t value. But then the next question is, well, how do we actually figure out what this is?"}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "T is equal to log base two of 1,111 over five. 1,111 over five. So this is an expression that gives us our t value. But then the next question is, well, how do we actually figure out what this is? And if you take out your calculator, you'll quickly notice that there is no, there is no log base two button. So how do we actually compute it? And here we just have to apply a very useful property of exponents."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "But then the next question is, well, how do we actually figure out what this is? And if you take out your calculator, you'll quickly notice that there is no, there is no log base two button. So how do we actually compute it? And here we just have to apply a very useful property of exponents. If we have, if we have log, if we have log base two of, well, really anything, well, let me write it this way. If we have log base a of c, we can compute this as log base anything of c over log base that same anything of a. This anything has to be the same thing."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "And here we just have to apply a very useful property of exponents. If we have, if we have log, if we have log base two of, well, really anything, well, let me write it this way. If we have log base a of c, we can compute this as log base anything of c over log base that same anything of a. This anything has to be the same thing. And our calculator is useful because it has a log, the log, when you just press log, it's log base 10. If you press ln, it's natural log or log base e. I like to just use the log base 10. So this is going to be the same thing as log base 10 of 1,111 over five over log base 10 of two."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "This anything has to be the same thing. And our calculator is useful because it has a log, the log, when you just press log, it's log base 10. If you press ln, it's natural log or log base e. I like to just use the log base 10. So this is going to be the same thing as log base 10 of 1,111 over five over log base 10 of two. So we can get our calculator out. And we could have done log base e if we wanted. That would have been natural log, but I'll just use a log button."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "So this is going to be the same thing as log base 10 of 1,111 over five over log base 10 of two. So we can get our calculator out. And we could have done log base e if we wanted. That would have been natural log, but I'll just use a log button. So this is logarithm of 1,111 over five. So that's this part right over here. This is log base 10 implicitly."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "That would have been natural log, but I'll just use a log button. So this is logarithm of 1,111 over five. So that's this part right over here. This is log base 10 implicitly. That's what the log button is, divided by log base 10 of two. And then that gives us seven, well, it just keeps on going. This is approximately equal to 7.796."}, {"video_title": "Solving exponential equation Exponential and logarithmic functions Algebra II Khan Academy.mp3", "Sentence": "This is log base 10 implicitly. That's what the log button is, divided by log base 10 of two. And then that gives us seven, well, it just keeps on going. This is approximately equal to 7.796. 7.796, I'll just, 7.796. So this is approximately equal to 7.796. So when t is roughly equal to that, you're going to have y equaling 1,111."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "At time t increased, as time t increased, the temperature c of the coffee began to decrease exponentially and approach room temperature of 20 degrees Celsius. Which of the following graphs could model this relationship? So we're starting at 90 degrees Celsius. It looks like all of the graphs start at 90 degrees Celsius at t equals zero. And we're going to approach the room temperature of 20 degrees Celsius. So this first one does approach the room temperature of 20 degrees Celsius as t increases. Now this one, when t is 70, I'm assuming this is in minutes, when t is 70, it looks like it has the temperature going to zero degrees Celsius."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "It looks like all of the graphs start at 90 degrees Celsius at t equals zero. And we're going to approach the room temperature of 20 degrees Celsius. So this first one does approach the room temperature of 20 degrees Celsius as t increases. Now this one, when t is 70, I'm assuming this is in minutes, when t is 70, it looks like it has the temperature going to zero degrees Celsius. So that cup of coffee is going to start freezing. So I think I could rule out b. Also, this looks like a linear model, not an exponential one."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now this one, when t is 70, I'm assuming this is in minutes, when t is 70, it looks like it has the temperature going to zero degrees Celsius. So that cup of coffee is going to start freezing. So I think I could rule out b. Also, this looks like a linear model, not an exponential one. c, it does get us to this end state that stays at 20 degrees, but it doesn't look like an exponential model. It looks like it's linearly decreasing, and then it stops linearly decreasing after 50 minutes, and then it just stays constant at that temperature of 20 degrees. So even though it gets us to the right place, it does not look like an exponential decay."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "Also, this looks like a linear model, not an exponential one. c, it does get us to this end state that stays at 20 degrees, but it doesn't look like an exponential model. It looks like it's linearly decreasing, and then it stops linearly decreasing after 50 minutes, and then it just stays constant at that temperature of 20 degrees. So even though it gets us to the right place, it does not look like an exponential decay. So I would rule choice c out as well. So a is looking good. d, we are starting at 90."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So even though it gets us to the right place, it does not look like an exponential decay. So I would rule choice c out as well. So a is looking good. d, we are starting at 90. It does look like an exponential function. We have exponential decay right over here. And we are approaching something, but it's not the room temperature of 20 degrees Celsius."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "d, we are starting at 90. It does look like an exponential function. We have exponential decay right over here. And we are approaching something, but it's not the room temperature of 20 degrees Celsius. We're approaching 30 degrees Celsius here. So I'd also rule out d. So a is looking good. It's an exponential, it's decreasing exponentially, starting at 90 degrees Celsius, and it's approaching the room temperature of 20 degrees Celsius."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we are approaching something, but it's not the room temperature of 20 degrees Celsius. We're approaching 30 degrees Celsius here. So I'd also rule out d. So a is looking good. It's an exponential, it's decreasing exponentially, starting at 90 degrees Celsius, and it's approaching the room temperature of 20 degrees Celsius. Let's do another one of these. So it says, let me scroll up a little bit. So it says, after the closing of the mills, the population of the town starts decreasing exponentially."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's an exponential, it's decreasing exponentially, starting at 90 degrees Celsius, and it's approaching the room temperature of 20 degrees Celsius. Let's do another one of these. So it says, let me scroll up a little bit. So it says, after the closing of the mills, the population of the town starts decreasing exponentially. The graph below represents the population, p, in thousands of the town, t years, after the closing of the mill. All right, so it looks like the population starts at 40,000. It's decreasing exponentially."}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "So it says, after the closing of the mills, the population of the town starts decreasing exponentially. The graph below represents the population, p, in thousands of the town, t years, after the closing of the mill. All right, so it looks like the population starts at 40,000. It's decreasing exponentially. It looks like over time, the population is approaching 20,000 people. So what is the question here? Based on the graph with the mill closed, what does the population of the town approach as time increases?"}, {"video_title": "End behavior of algebraic models Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's decreasing exponentially. It looks like over time, the population is approaching 20,000 people. So what is the question here? Based on the graph with the mill closed, what does the population of the town approach as time increases? Well, we just said it. As time increases, it looks like it's coming close to, it's approaching 20,000. It's approaching 20,000."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "What we're going to do in this video is use what we know about polynomials and how to manipulate them, and what we've talked about of whether two polynomials are equal to each other for all values of the variable that they're written in, so whether we're dealing with a polynomial identity, and we're gonna use those skills in order to prove some properties of relationships between numbers. So if I were to list out some integers, I could go zero, I could go one, I could go two, three, four, five, and if I were to list out the squares of these, if I were to create a sequence of integer squares, well, zero squared would be zero, one squared would be one, two squared is four, three squared is nine, four squared is 16, five squared is 25, and we could, of course, keep going in either case. But the first thing I want you to think about before you even write down a polynomial or try to construct one is look at this sequence of integer squares, and do you see any pattern in terms of the difference between successive terms of this sequence of integer squares? All right, now let's think about this together. So to go from zero to one, you add one. Or to go from one to four, you add three. To go from four to nine, you add five."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's think about this together. So to go from zero to one, you add one. Or to go from one to four, you add three. To go from four to nine, you add five. To go from nine to 16, you add seven. It seems like a pattern here. As we go to successive terms of this sequence of integer squares, we're adding increasing odd numbers."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "To go from four to nine, you add five. To go from nine to 16, you add seven. It seems like a pattern here. As we go to successive terms of this sequence of integer squares, we're adding increasing odd numbers. So I'm guessing that if I add nine here, which is the next odd number, I'm gonna get to 25, and that indeed is the case. And you could test that out. Well, if I add 11, which would be the next odd number, what do I get to?"}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "As we go to successive terms of this sequence of integer squares, we're adding increasing odd numbers. So I'm guessing that if I add nine here, which is the next odd number, I'm gonna get to 25, and that indeed is the case. And you could test that out. Well, if I add 11, which would be the next odd number, what do I get to? I get to 36, which is the square of six. But how can we feel good that this always is true, that this never breaks down? Well, one way to do it is to think a little bit more generally, and that's where our algebra and our knowledge of polynomials are going to be useful."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "Well, if I add 11, which would be the next odd number, what do I get to? I get to 36, which is the square of six. But how can we feel good that this always is true, that this never breaks down? Well, one way to do it is to think a little bit more generally, and that's where our algebra and our knowledge of polynomials are going to be useful. So let's say we go all the way, and we're just speaking generally now. So we have the number n, and then with the next number after that is going to be n plus one. And then if we think about what the corresponding terms in the sequence of integer squares would be, well, that would be when we square it, when we get to n, we would get n squared."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "Well, one way to do it is to think a little bit more generally, and that's where our algebra and our knowledge of polynomials are going to be useful. So let's say we go all the way, and we're just speaking generally now. So we have the number n, and then with the next number after that is going to be n plus one. And then if we think about what the corresponding terms in the sequence of integer squares would be, well, that would be when we square it, when we get to n, we would get n squared. And when we get to n plus one, we would have n plus one squared. And let's see if we could think about what the difference between these two things are. The difference between 25 and 16 is nine."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "And then if we think about what the corresponding terms in the sequence of integer squares would be, well, that would be when we square it, when we get to n, we would get n squared. And when we get to n plus one, we would have n plus one squared. And let's see if we could think about what the difference between these two things are. The difference between 25 and 16 is nine. The difference between 16 and nine is seven. So let's think about what the difference between n plus one squared and n squared is. And how do we write that as a polynomial?"}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "The difference between 25 and 16 is nine. The difference between 16 and nine is seven. So let's think about what the difference between n plus one squared and n squared is. And how do we write that as a polynomial? Well, it'll just be n plus one squared minus n squared. And now let's see if we can rewrite this, algebraically manipulate this, so we can set up a polynomial identity that describes this pattern that we just saw. So what I'll do is I'm just going to expand out n plus one squared right over there."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "And how do we write that as a polynomial? Well, it'll just be n plus one squared minus n squared. And now let's see if we can rewrite this, algebraically manipulate this, so we can set up a polynomial identity that describes this pattern that we just saw. So what I'll do is I'm just going to expand out n plus one squared right over there. So that is going to be n squared plus two n plus one. And then we have this minus n squared here, so minus n squared. And so we see that n squared minus n squared cancels out."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "So what I'll do is I'm just going to expand out n plus one squared right over there. So that is going to be n squared plus two n plus one. And then we have this minus n squared here, so minus n squared. And so we see that n squared minus n squared cancels out. And so we can rewrite everything we have here as n plus one squared minus n squared. So this is really the difference between successive terms in our sequence of integer squares is going to be equal to two n plus one for any integer n. Well, for any integer n, what is two n plus one going to be? And especially here we're dealing with the positive integers."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "And so we see that n squared minus n squared cancels out. And so we can rewrite everything we have here as n plus one squared minus n squared. So this is really the difference between successive terms in our sequence of integer squares is going to be equal to two n plus one for any integer n. Well, for any integer n, what is two n plus one going to be? And especially here we're dealing with the positive integers. Well, for any integer n, this is going to be an odd integer. If you take any integer, you multiply it by two, this part is going to be even. But then you add one to that, you're going to get an odd integer."}, {"video_title": "Describing numerical relationships with polynomial identities Algebra 2 Khan Academy.mp3", "Sentence": "And especially here we're dealing with the positive integers. Well, for any integer n, this is going to be an odd integer. If you take any integer, you multiply it by two, this part is going to be even. But then you add one to that, you're going to get an odd integer. And you can also see that this increases by two as n increases. So when you go from one odd integer, you go add two to the next odd integer. You add two to the next odd integer, which is exactly what is described there."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "We're told that we want to factor the following expression and they ask us which pattern can we use to factor the expression and u and v are either constant integers or single variable expressions. So we'll do this one together and then we'll have a few more examples where I'll encourage you to pause the video. So when they're talking about patterns, they're really saying, hey, can we say that some of these can generally form a pattern that matches what we have here and then we can use that pattern to factor it into one of these forms. What do I mean by that? Well, let's just imagine something like u plus v squared. We've squared binomials in the past. This is going to be equal to u squared plus two times the product of these terms, so two u v, and then plus v squared."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "What do I mean by that? Well, let's just imagine something like u plus v squared. We've squared binomials in the past. This is going to be equal to u squared plus two times the product of these terms, so two u v, and then plus v squared. Now, when you look at this polynomial right over here, it actually has this form if you look at it carefully. How could it have this form? Well, if we view u squared as nine x to the eighth, then that means that u is, and let me write it as a capital U, u is equal to three x to the fourth because notice, if you square this, you're gonna get nine x to the eighth."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "This is going to be equal to u squared plus two times the product of these terms, so two u v, and then plus v squared. Now, when you look at this polynomial right over here, it actually has this form if you look at it carefully. How could it have this form? Well, if we view u squared as nine x to the eighth, then that means that u is, and let me write it as a capital U, u is equal to three x to the fourth because notice, if you square this, you're gonna get nine x to the eighth. So this right over here is u squared. And if we said that v squared is equal to y squared, so if this is capital V squared, then that means that v is equal to y. And then this would have to be two times u v, is it?"}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, if we view u squared as nine x to the eighth, then that means that u is, and let me write it as a capital U, u is equal to three x to the fourth because notice, if you square this, you're gonna get nine x to the eighth. So this right over here is u squared. And if we said that v squared is equal to y squared, so if this is capital V squared, then that means that v is equal to y. And then this would have to be two times u v, is it? Well, see, if I multiply u times v, I get three x to the fourth y, and then two times that is indeed six x to the fourth y, so this right over here is two u v. So notice, this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression, well, I would use a pattern for u plus v squared, so I would go with that choice right over there. Let's do a few more examples."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And then this would have to be two times u v, is it? Well, see, if I multiply u times v, I get three x to the fourth y, and then two times that is indeed six x to the fourth y, so this right over here is two u v. So notice, this polynomial, this higher-degree polynomial can be expressed in this pattern, which means it can be factored this way. So when they say which pattern can we use to factor this expression, well, I would use a pattern for u plus v squared, so I would go with that choice right over there. Let's do a few more examples. So here, once again, we're told the same thing. We're given a different expression, and they're asking us what pattern can we use to factor the expression. So I have these three terms here."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Let's do a few more examples. So here, once again, we're told the same thing. We're given a different expression, and they're asking us what pattern can we use to factor the expression. So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works. If this is u squared, if this is u squared, then that means that u is going to be equal to two x to the third power, and if this is v squared, then that means that v is equal to five."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So I have these three terms here. It looks like maybe I could use, I can see a perfect square here. Let's see if that works. If this is u squared, if this is u squared, then that means that u is going to be equal to two x to the third power, and if this is v squared, then that means that v is equal to five. Now, is this equal to two times u v? Well, let's see, two times u v would be equal to, well, you're not gonna have any y in it, so this is not going to be two u v, so this actually is not fitting the perfect square pattern, so we could rule this out, and both of these are perfect squares of some form. One just has a, I guess you'd say, adding v, the other one is subtracting v. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares, and we've seen this before."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "If this is u squared, if this is u squared, then that means that u is going to be equal to two x to the third power, and if this is v squared, then that means that v is equal to five. Now, is this equal to two times u v? Well, let's see, two times u v would be equal to, well, you're not gonna have any y in it, so this is not going to be two u v, so this actually is not fitting the perfect square pattern, so we could rule this out, and both of these are perfect squares of some form. One just has a, I guess you'd say, adding v, the other one is subtracting v. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares, and we've seen this before. This is u squared minus v squared, so you wouldn't have a three-term polynomial like that, so we could rule that one out, so I would pick that we can't use any of the patterns. Let's do yet another example, and I encourage you, pause the video and see if you can work this one out on your own. So, same idea."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "One just has a, I guess you'd say, adding v, the other one is subtracting v. This right over here, if I were to multiply this out, this is going to be equal to, this is a difference of squares, and we've seen this before. This is u squared minus v squared, so you wouldn't have a three-term polynomial like that, so we could rule that one out, so I would pick that we can't use any of the patterns. Let's do yet another example, and I encourage you, pause the video and see if you can work this one out on your own. So, same idea. They wanna factor the following expression, and this one essentially has two terms. We have a term here, and we have a term here. They are both, they both look like they are the square of something, and we have a difference of squares, so this is making me feel pretty good about this pattern, but let's see if that works out."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So, same idea. They wanna factor the following expression, and this one essentially has two terms. We have a term here, and we have a term here. They are both, they both look like they are the square of something, and we have a difference of squares, so this is making me feel pretty good about this pattern, but let's see if that works out. Remember, u plus v times u minus v is equal to u squared minus v squared, so if this is equal to u squared, then that means that capital U is equal to six x squared. That works, and if this is equal to v squared, well, that means that v is equal to y plus three, so this is fitting this pattern right over here, and they're just asking us to say, what pattern can we use to factor the expression? They're not asking us to actually factor it, so we'll just pick this choice, but once you identify the pattern, it's actually pretty straightforward to factor it, because if you say this is just going to factor into u plus v times u minus v, well, u plus v is going to be six x squared plus v plus y plus three times u minus v. U is six x squared."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "They are both, they both look like they are the square of something, and we have a difference of squares, so this is making me feel pretty good about this pattern, but let's see if that works out. Remember, u plus v times u minus v is equal to u squared minus v squared, so if this is equal to u squared, then that means that capital U is equal to six x squared. That works, and if this is equal to v squared, well, that means that v is equal to y plus three, so this is fitting this pattern right over here, and they're just asking us to say, what pattern can we use to factor the expression? They're not asking us to actually factor it, so we'll just pick this choice, but once you identify the pattern, it's actually pretty straightforward to factor it, because if you say this is just going to factor into u plus v times u minus v, well, u plus v is going to be six x squared plus v plus y plus three times u minus v. U is six x squared. Minus v is minus, we could write minus y plus three, or we could distribute the negative sign, but either way, this might make it a little bit clearer what we just did. We used the pattern to factor this higher degree polynomial, which is essentially just the difference of squares. Let's do one last example."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "They're not asking us to actually factor it, so we'll just pick this choice, but once you identify the pattern, it's actually pretty straightforward to factor it, because if you say this is just going to factor into u plus v times u minus v, well, u plus v is going to be six x squared plus v plus y plus three times u minus v. U is six x squared. Minus v is minus, we could write minus y plus three, or we could distribute the negative sign, but either way, this might make it a little bit clearer what we just did. We used the pattern to factor this higher degree polynomial, which is essentially just the difference of squares. Let's do one last example. So here, once again, we want to factor an expression. Which pattern can we use? Pause the video."}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Let's do one last example. So here, once again, we want to factor an expression. Which pattern can we use? Pause the video. All right, so we have two terms here, so it looks like it might be a difference of squares. If we set u is equal to seven, then this would be u u squared. But then what can we square to get 10x to the third power?"}, {"video_title": "Identifying quadratic patterns Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Pause the video. All right, so we have two terms here, so it looks like it might be a difference of squares. If we set u is equal to seven, then this would be u u squared. But then what can we square to get 10x to the third power? Remember, we want to have integer exponents here. And the square root of 10x to the third power, if I were to take the square root of 10x to the third power, it'd be something a little bit involved, like the square root of 10 times x times the square root of x to the third power, and I'm not going to get an integer exponent here. So it doesn't look like I can express this as v squared, so I would go with that we can't use any of the patterns."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "One of the solutions is x is equal to 0.5. Find the other solution. They say, hint, use a graphing calculator and round your answer to the nearest tenth. So pause this video and have a go at this if you like, and then we'll work on this together. And I encourage you to have a go at it. All right, now let's work on this together. Now the key here is to realize that we might be able to solve this by graphing, or at least approximate the solutions to this by graphing."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and have a go at this if you like, and then we'll work on this together. And I encourage you to have a go at it. All right, now let's work on this together. Now the key here is to realize that we might be able to solve this by graphing, or at least approximate the solutions to this by graphing. And the way we do that is if we have an equation, especially a hairy equation like this in one variable, we can set y equal to the left and then set y equal to the right, and then graph each of those functions, and then think about where they intersect, because they'll intersect at an x value that gives us the same y value, and that means that the two sides are the same. So what do I mean by that? Well, we could set y is equal to the negative natural log of two x, so we could have one equation or one function like that, and then we could have another equation or function that y is equal to two times the absolute value of x minus four minus seven, and let's see where they intersect."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "Now the key here is to realize that we might be able to solve this by graphing, or at least approximate the solutions to this by graphing. And the way we do that is if we have an equation, especially a hairy equation like this in one variable, we can set y equal to the left and then set y equal to the right, and then graph each of those functions, and then think about where they intersect, because they'll intersect at an x value that gives us the same y value, and that means that the two sides are the same. So what do I mean by that? Well, we could set y is equal to the negative natural log of two x, so we could have one equation or one function like that, and then we could have another equation or function that y is equal to two times the absolute value of x minus four minus seven, and let's see where they intersect. And the x values where they intersect, those are going to be solutions to that. So I'm going to use Desmos as my graphing calculator. So let's type in the two sides."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "Well, we could set y is equal to the negative natural log of two x, so we could have one equation or one function like that, and then we could have another equation or function that y is equal to two times the absolute value of x minus four minus seven, and let's see where they intersect. And the x values where they intersect, those are going to be solutions to that. So I'm going to use Desmos as my graphing calculator. So let's type in the two sides. So first I'll do the left side. So if y is equal to the negative natural log of two x, and actually let me make my color to be the same or as close as I can, so maybe closer to that bluer color. Okay, and then the next one, I want y is equal to two times, two times the absolute value."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "So let's type in the two sides. So first I'll do the left side. So if y is equal to the negative natural log of two x, and actually let me make my color to be the same or as close as I can, so maybe closer to that bluer color. Okay, and then the next one, I want y is equal to two times, two times the absolute value. Actually, I don't know whether Desmos prefers, I'll use that, actually that works, okay. X minus four, and then I will close my absolute value, and then I have minus seven, and I will do this in the red color so that we can keep track of things. Okay, so those are my two graphs, and now I just need to think about where they intersect."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "Okay, and then the next one, I want y is equal to two times, two times the absolute value. Actually, I don't know whether Desmos prefers, I'll use that, actually that works, okay. X minus four, and then I will close my absolute value, and then I have minus seven, and I will do this in the red color so that we can keep track of things. Okay, so those are my two graphs, and now I just need to think about where they intersect. And one of the solutions is x equal to 0.5. That's not the one they want. They want the other solution, so to speak."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "Okay, so those are my two graphs, and now I just need to think about where they intersect. And one of the solutions is x equal to 0.5. That's not the one they want. They want the other solution, so to speak. So let's see. So let's see, we have one solution. Actually, let me zoom in a little bit."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "They want the other solution, so to speak. So let's see. So let's see, we have one solution. Actually, let me zoom in a little bit. So when x is equal to 0.5, that's this solution, that's this solution right over here. It looks like y is equal to zero there. But then the other point of intersection seems to be right over here."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "Actually, let me zoom in a little bit. So when x is equal to 0.5, that's this solution, that's this solution right over here. It looks like y is equal to zero there. But then the other point of intersection seems to be right over here. And actually, Desmos has a nice little feature where it'll tell us that point right over there. But you could even approximate it. You can see that x is over six, and that each of these, let's see, one, two, three, four, five, each of those is 0.2."}, {"video_title": "Solving equations by graphing graphing calculator Algebra 2 Khan Academy.mp3", "Sentence": "But then the other point of intersection seems to be right over here. And actually, Desmos has a nice little feature where it'll tell us that point right over there. But you could even approximate it. You can see that x is over six, and that each of these, let's see, one, two, three, four, five, each of those is 0.2. So it's gonna be 6.2 something is what I would do, and they want us to round to the nearest 10th anyway, so you don't even need to use that feature. But you can see very clearly that when x is equal to approximately 6.238, that we get a y is equal to negative 2.54. Or another way to think about it is when x is approximately equal to, when x is approximately equal to 6.2, that the two sides of this equation are going to be equal to each other, or approximately equal to each other."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We are asked, are the following functions even, odd, or neither? So pause this video and try to work that out on your own before we work through it together. All right, now let's just remind ourselves of a definition for even and odd functions. One way to think about it is what happens when you take f of negative x? If f of negative x is equal to the function again, then we're dealing with an even function. If we evaluate f of negative x, instead of getting the function, we get the negative of the function, then we're dealing with an odd function. And if neither of these are true, it is neither."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "One way to think about it is what happens when you take f of negative x? If f of negative x is equal to the function again, then we're dealing with an even function. If we evaluate f of negative x, instead of getting the function, we get the negative of the function, then we're dealing with an odd function. And if neither of these are true, it is neither. So let's go to this first one right over here. F of x is equal to five over three minus x to the fourth. And the best way I can think about tackling this is let's just evaluate what f of negative x would be equal to."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And if neither of these are true, it is neither. So let's go to this first one right over here. F of x is equal to five over three minus x to the fourth. And the best way I can think about tackling this is let's just evaluate what f of negative x would be equal to. That would be equal to five over three minus, and everywhere we see an x, we're going to replace that with a negative x to the fourth power. Now what is negative x to the fourth power? Well, if you multiply a negative times a negative times a negative, how many times did I do that?"}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And the best way I can think about tackling this is let's just evaluate what f of negative x would be equal to. That would be equal to five over three minus, and everywhere we see an x, we're going to replace that with a negative x to the fourth power. Now what is negative x to the fourth power? Well, if you multiply a negative times a negative times a negative, how many times did I do that? If you take a negative to the fourth power, you're going to get a positive. So that's going to be equal to five over three minus x to the fourth, which is once again equal to f of x. And so this first one right over here, f of negative x is equal to f of x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, if you multiply a negative times a negative times a negative, how many times did I do that? If you take a negative to the fourth power, you're going to get a positive. So that's going to be equal to five over three minus x to the fourth, which is once again equal to f of x. And so this first one right over here, f of negative x is equal to f of x. It is clearly even. Let's do another example. So this one right over here, g of x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And so this first one right over here, f of negative x is equal to f of x. It is clearly even. Let's do another example. So this one right over here, g of x. Let's just evaluate g of negative x. And at any point you feel inspired and you didn't figure it out the first time, pause the video again and try to work it out on your own. Well, g of negative x is equal to one over negative x plus the cube root of negative x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this one right over here, g of x. Let's just evaluate g of negative x. And at any point you feel inspired and you didn't figure it out the first time, pause the video again and try to work it out on your own. Well, g of negative x is equal to one over negative x plus the cube root of negative x. And let's see, can we simplify this any? Well, we could rewrite this as the negative of one over x. And then I could view negative x as the same thing as negative one times x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, g of negative x is equal to one over negative x plus the cube root of negative x. And let's see, can we simplify this any? Well, we could rewrite this as the negative of one over x. And then I could view negative x as the same thing as negative one times x. And so we can factor out, or I should say we could take the negative one out of the radical. What is the cube root of negative one? Well, it's negative one."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then I could view negative x as the same thing as negative one times x. And so we can factor out, or I should say we could take the negative one out of the radical. What is the cube root of negative one? Well, it's negative one. So we could say minus, we could say minus one times the cube root, or we could just say the negative of the cube root of x. And then we can factor out a negative. So this is going to be equal to negative of one over x plus the cube root of x, which is equal to the negative of g of x, which is equal to the negative of g of x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, it's negative one. So we could say minus, we could say minus one times the cube root, or we could just say the negative of the cube root of x. And then we can factor out a negative. So this is going to be equal to negative of one over x plus the cube root of x, which is equal to the negative of g of x, which is equal to the negative of g of x. And so this is odd. F of negative x is equal to the negative of f of x. Well, in this case, it's g of x. G of negative x is equal to the negative of g of x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So this is going to be equal to negative of one over x plus the cube root of x, which is equal to the negative of g of x, which is equal to the negative of g of x. And so this is odd. F of negative x is equal to the negative of f of x. Well, in this case, it's g of x. G of negative x is equal to the negative of g of x. Let's do the third one. So here we've got h of x. And let's just evaluate h of negative x. H of negative x is equal to two to the negative x plus two to the negative of negative x, which would be two to the positive x."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, in this case, it's g of x. G of negative x is equal to the negative of g of x. Let's do the third one. So here we've got h of x. And let's just evaluate h of negative x. H of negative x is equal to two to the negative x plus two to the negative of negative x, which would be two to the positive x. Well, this is the same thing as our original h of x. This is just equal to h of x. You just swap these two terms."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And let's just evaluate h of negative x. H of negative x is equal to two to the negative x plus two to the negative of negative x, which would be two to the positive x. Well, this is the same thing as our original h of x. This is just equal to h of x. You just swap these two terms. And so this is clearly even. And then last but not least, we have j of x. So let's evaluate j of, I'm gonna write y."}, {"video_title": "Even and odd functions Equations Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You just swap these two terms. And so this is clearly even. And then last but not least, we have j of x. So let's evaluate j of, I'm gonna write y. Let's evaluate j of negative x is equal to negative x over one minus negative x, which is equal to negative x over one plus x. And let's see. There's no clear way of factoring out a negative or doing something interesting where I get either back to j of x or I get to negative j of x."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We're asked, what are the domain and range of the sine function? So to think about that, let's actually draw the sine function out. And what I have here, on the left-hand side right over here, I've got a unit circle. And I can, let me truncate this a little bit. I don't need that space right there. So let me clear that out. So I have a unit circle on the left-hand side right over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And I can, let me truncate this a little bit. I don't need that space right there. So let me clear that out. So I have a unit circle on the left-hand side right over here. And I'm going to use that to figure out the values of sine of theta for a given theta. So in the unit circle, this is x and this is y. Or you could even view this as the, well, we could just use x or y."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So I have a unit circle on the left-hand side right over here. And I'm going to use that to figure out the values of sine of theta for a given theta. So in the unit circle, this is x and this is y. Or you could even view this as the, well, we could just use x or y. And so for a given theta, we can see where that angle entered, the terminal side of the angle intersects the unit circle. And then the y-coordinate of that point is going to be sine of theta. And over here, I'm going to graph still y in the vertical axis."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Or you could even view this as the, well, we could just use x or y. And so for a given theta, we can see where that angle entered, the terminal side of the angle intersects the unit circle. And then the y-coordinate of that point is going to be sine of theta. And over here, I'm going to graph still y in the vertical axis. But I'm going to graph the graph of y is equal to sine of theta. y is equal to sine of theta. And on the horizontal axis, I'm not going to graph x."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And over here, I'm going to graph still y in the vertical axis. But I'm going to graph the graph of y is equal to sine of theta. y is equal to sine of theta. And on the horizontal axis, I'm not going to graph x. But I'm going to graph theta. You can view theta as the independent variable here. And it's going to be theta is going to be in radians."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And on the horizontal axis, I'm not going to graph x. But I'm going to graph theta. You can view theta as the independent variable here. And it's going to be theta is going to be in radians. So we're essentially going to pick a bunch of thetas and then come up with what sine of theta is and then graph it. So let's set up a little bit of a table here. Let's set up a little bit of a table."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And it's going to be theta is going to be in radians. So we're essentially going to pick a bunch of thetas and then come up with what sine of theta is and then graph it. So let's set up a little bit of a table here. Let's set up a little bit of a table. And so over here, I have theta. And over here, we're going to figure out what sine of theta is. And we could do a bunch of theta values."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Let's set up a little bit of a table. And so over here, I have theta. And over here, we're going to figure out what sine of theta is. And we could do a bunch of theta values. We could start, let's say we start at 0. Let's say we start at theta is equal to 0. What is sine of theta going to be?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And we could do a bunch of theta values. We could start, let's say we start at 0. Let's say we start at theta is equal to 0. What is sine of theta going to be? Well, when the angle is 0, we intersect the unit circle right over there. The y-coordinate of this is still 0. This is the point 1, 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What is sine of theta going to be? Well, when the angle is 0, we intersect the unit circle right over there. The y-coordinate of this is still 0. This is the point 1, 0. The y-coordinate is 0. So sine of theta is 0. Or we could say sine of 0 is equal to 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "This is the point 1, 0. The y-coordinate is 0. So sine of theta is 0. Or we could say sine of 0 is equal to 0. Sine of 0 is equal to 0. Now let's try theta is equal to pi over 2. Theta is equal to pi over 2."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Or we could say sine of 0 is equal to 0. Sine of 0 is equal to 0. Now let's try theta is equal to pi over 2. Theta is equal to pi over 2. I'm just doing the ones that are really easy to figure out. So if theta is equal to pi over 2, that's the same thing as a 90-degree angle. So the terminal side is going to be right along the y-axis, just like that."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Theta is equal to pi over 2. I'm just doing the ones that are really easy to figure out. So if theta is equal to pi over 2, that's the same thing as a 90-degree angle. So the terminal side is going to be right along the y-axis, just like that. And where it intersects the unit circle is right over here. And what point is that? Well, that's the point 0, 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So the terminal side is going to be right along the y-axis, just like that. And where it intersects the unit circle is right over here. And what point is that? Well, that's the point 0, 1. So what is sine of pi over 2? Well, sine of pi over 2 is just the y-coordinate right over here. It is 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, that's the point 0, 1. So what is sine of pi over 2? Well, sine of pi over 2 is just the y-coordinate right over here. It is 1. Sine of pi over 2 is 1. Let's keep going. You might see a little pattern here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "It is 1. Sine of pi over 2 is 1. Let's keep going. You might see a little pattern here. We're just going more and more around the circle. So let's think about what happens when theta is equal to pi. What is sine of pi?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "You might see a little pattern here. We're just going more and more around the circle. So let's think about what happens when theta is equal to pi. What is sine of pi? Well, we intersect the unit circle right over there. That coordinate is negative 1, 0. Sine is the y-coordinate."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What is sine of pi? Well, we intersect the unit circle right over there. That coordinate is negative 1, 0. Sine is the y-coordinate. So this right over here is sine of pi. Sine of pi is 0. Let's go to 3 pi over 2."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Sine is the y-coordinate. So this right over here is sine of pi. Sine of pi is 0. Let's go to 3 pi over 2. 3 pi over 2. Well, now we've gone 3 quarters of the way around the circle. We intersect."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Let's go to 3 pi over 2. 3 pi over 2. Well, now we've gone 3 quarters of the way around the circle. We intersect. The terminal side of the angle intersects the unit circle right over here. And so based on that, what is sine of 3 pi over 2? Well, this point right over here is the point negative."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We intersect. The terminal side of the angle intersects the unit circle right over here. And so based on that, what is sine of 3 pi over 2? Well, this point right over here is the point negative. Let me be careful. It is 0, negative 1. The sine of theta is the same thing as the y-coordinate."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, this point right over here is the point negative. Let me be careful. It is 0, negative 1. The sine of theta is the same thing as the y-coordinate. The y-coordinate is sine of theta. So when theta is 3 pi over 2, sine of theta is equal to negative 1. Now let's come full circle here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "The sine of theta is the same thing as the y-coordinate. The y-coordinate is sine of theta. So when theta is 3 pi over 2, sine of theta is equal to negative 1. Now let's come full circle here. So let's go all the way to theta equaling 2 pi. Let me do a color. Yeah, I'll just use the yellow here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Now let's come full circle here. So let's go all the way to theta equaling 2 pi. Let me do a color. Yeah, I'll just use the yellow here. What happens when theta is equal to 2 pi? Well, then we've gone all the way around the circle. And we are back to where we started."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Yeah, I'll just use the yellow here. What happens when theta is equal to 2 pi? Well, then we've gone all the way around the circle. And we are back to where we started. And the y-coordinate is 0. So sine of 2 pi is once again 0. And if we were to keep going around, we're going to start seeing, as we keep incrementing the angle, we're going to start seeing the same pattern emerge again."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And we are back to where we started. And the y-coordinate is 0. So sine of 2 pi is once again 0. And if we were to keep going around, we're going to start seeing, as we keep incrementing the angle, we're going to start seeing the same pattern emerge again. Well, let's try to graph this. So when theta is equal to 0, sine of theta is 0. When theta is equal to pi over 2, sine of theta is 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And if we were to keep going around, we're going to start seeing, as we keep incrementing the angle, we're going to start seeing the same pattern emerge again. Well, let's try to graph this. So when theta is equal to 0, sine of theta is 0. When theta is equal to pi over 2, sine of theta is 1. So we'll use the same scale. So sine of theta is equal to 1. I'll make this 1 on this axis and on that axis."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to pi over 2, sine of theta is 1. So we'll use the same scale. So sine of theta is equal to 1. I'll make this 1 on this axis and on that axis. So we can maybe see a little bit of a parallel here. When theta is equal to pi, sine of theta is 0. So when theta is equal to pi, sine of theta is 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I'll make this 1 on this axis and on that axis. So we can maybe see a little bit of a parallel here. When theta is equal to pi, sine of theta is 0. So when theta is equal to pi, sine of theta is 0. So we go back right over there. When theta is equal to 3 pi over 2, so that would be right over here, sine of theta is negative 1. So this is negative 1 over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So when theta is equal to pi, sine of theta is 0. So we go back right over there. When theta is equal to 3 pi over 2, so that would be right over here, sine of theta is negative 1. So this is negative 1 over here. I'll do the same scale over here. I'll make this negative. I'll make this negative 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So this is negative 1 over here. I'll do the same scale over here. I'll make this negative. I'll make this negative 1. And so sine of theta is negative 1. And then when theta is 2 pi, sine of theta is 0. So when theta is 2 pi, sine of theta is 0."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I'll make this negative 1. And so sine of theta is negative 1. And then when theta is 2 pi, sine of theta is 0. So when theta is 2 pi, sine of theta is 0. And so we can connect the dots. You can try other points in between, and you get a graph that looks something like this. My best attempt at drawing it freehand."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So when theta is 2 pi, sine of theta is 0. And so we can connect the dots. You can try other points in between, and you get a graph that looks something like this. My best attempt at drawing it freehand. It looks something like this. There's a reason why curves that look like this are called sinusoids, because they're the graph of a sine function. So just like this."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "My best attempt at drawing it freehand. It looks something like this. There's a reason why curves that look like this are called sinusoids, because they're the graph of a sine function. So just like this. But that's not the entire graph. We could keep going. We could add another pi over 2."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So just like this. But that's not the entire graph. We could keep going. We could add another pi over 2. If you added another pi over 2, so if you go to 2 pi, and then you add another pi over 2, so you could view this as 2 and 1 half pi, or however you want to think about it, then you're going to go back over here. So then you're going to get back to sine of theta being equal to 1. So you're going to go back to this point right over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We could add another pi over 2. If you added another pi over 2, so if you go to 2 pi, and then you add another pi over 2, so you could view this as 2 and 1 half pi, or however you want to think about it, then you're going to go back over here. So then you're going to get back to sine of theta being equal to 1. So you're going to go back to this point right over here. And you could keep going. You go another pi over 2, you're going to go back to this point, and you're going to be over here. And so the curve, or the function sine of theta, is really defined for any theta value, any real theta value that you choose."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So you're going to go back to this point right over here. And you could keep going. You go another pi over 2, you're going to go back to this point, and you're going to be over here. And so the curve, or the function sine of theta, is really defined for any theta value, any real theta value that you choose. So any theta value. You say, well, what about negatives? I mean, obviously I agree."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And so the curve, or the function sine of theta, is really defined for any theta value, any real theta value that you choose. So any theta value. You say, well, what about negatives? I mean, obviously I agree. As you keep increasing theta like this, we just keep going around and around the circle, and this pattern kind of emerges. But what happens when we go in the negative direction? Well, let's try it out."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "I mean, obviously I agree. As you keep increasing theta like this, we just keep going around and around the circle, and this pattern kind of emerges. But what happens when we go in the negative direction? Well, let's try it out. What happens if we were to take negative pi over 2? So let me do that. So negative pi over 2, well, that's going right over here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, let's try it out. What happens if we were to take negative pi over 2? So let me do that. So negative pi over 2, well, that's going right over here. And so we intersect the unit circle right over there. The y-coordinate is negative 1. So sine of negative pi over 2 is negative 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So negative pi over 2, well, that's going right over here. And so we intersect the unit circle right over there. The y-coordinate is negative 1. So sine of negative pi over 2 is negative 1. And we see that it just continues. It just continues. So sine of theta is defined for any positive, negative, any theta, positive or negative, non-negative, 0, anything."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So sine of negative pi over 2 is negative 1. And we see that it just continues. It just continues. So sine of theta is defined for any positive, negative, any theta, positive or negative, non-negative, 0, anything. So it's defined for anything. So let's go back to the question. So I could just keep drawing this function on and on and on."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So sine of theta is defined for any positive, negative, any theta, positive or negative, non-negative, 0, anything. So it's defined for anything. So let's go back to the question. So I could just keep drawing this function on and on and on. So let's go back to the question. What is the domain of the sine function? And just as a reminder, the domain are all of the inputs over which the function is defined, or all of the valid inputs into the function that the function will actually spit out a valid answer."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "So I could just keep drawing this function on and on and on. So let's go back to the question. What is the domain of the sine function? And just as a reminder, the domain are all of the inputs over which the function is defined, or all of the valid inputs into the function that the function will actually spit out a valid answer. So what is the domain of the sine function? Well, we already saw. We can put in any theta here."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "And just as a reminder, the domain are all of the inputs over which the function is defined, or all of the valid inputs into the function that the function will actually spit out a valid answer. So what is the domain of the sine function? Well, we already saw. We can put in any theta here. So you could say the domain is all real numbers. Now, what about the range? What about the range?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "We can put in any theta here. So you could say the domain is all real numbers. Now, what about the range? What about the range? Well, just as a review, the range is sometimes in more technical math classes called the image. It's the set of all the values that the function can actually take on. Well, what is that set?"}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "What about the range? Well, just as a review, the range is sometimes in more technical math classes called the image. It's the set of all the values that the function can actually take on. Well, what is that set? What is the range here? What is all the values that y equals sine of theta could actually take on? Well, we see that it keeps going between positive 1."}, {"video_title": "Example Graph, domain, and range of sine function Trigonometry Khan Academy.mp3", "Sentence": "Well, what is that set? What is the range here? What is all the values that y equals sine of theta could actually take on? Well, we see that it keeps going between positive 1. It keeps going between positive 1 and then to negative 1, and then back to positive 1, and then negative 1. It takes on all the values in between. So you see that sine of theta is always going to be less than or equal to 1, and it's always going to be greater than or equal to negative 1."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "We're told that Ben's home is X kilometers from an ice cream shop. Jerry's home is Y kilometers from the same shop. Then it tells us they each left their home at the same time and met at the ice cream shop at the same time. Ben walked an average speed, let me do this in a new color, average speed of five kilometers per hour, and Jerry rode his bicycle at an average speed of V kilometers per hour. Write an equation that relates X, Y, and V. Pause this video and see if you can do that. All right, so let's just remind ourselves how distance, speed, and time are related. You might be familiar with things like distance is equal to rate times time, or another way you could think about it is you could write that distance is equal to speed times time."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Ben walked an average speed, let me do this in a new color, average speed of five kilometers per hour, and Jerry rode his bicycle at an average speed of V kilometers per hour. Write an equation that relates X, Y, and V. Pause this video and see if you can do that. All right, so let's just remind ourselves how distance, speed, and time are related. You might be familiar with things like distance is equal to rate times time, or another way you could think about it is you could write that distance is equal to speed times time. Or if you wanna solve for time, you could divide both sides by speed. So you could get distance over speed, over speed is equal to time. Now, the reason why I set it up this way is that we know that Ben's time and Jerry's time is the same."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "You might be familiar with things like distance is equal to rate times time, or another way you could think about it is you could write that distance is equal to speed times time. Or if you wanna solve for time, you could divide both sides by speed. So you could get distance over speed, over speed is equal to time. Now, the reason why I set it up this way is that we know that Ben's time and Jerry's time is the same. They covered maybe different distances at maybe different speeds, but it took them the exact same amount of time. So Ben's distance divided by Ben's speed should be the same as Jerry's distance divided by Jerry's speed. So let me write that down."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Now, the reason why I set it up this way is that we know that Ben's time and Jerry's time is the same. They covered maybe different distances at maybe different speeds, but it took them the exact same amount of time. So Ben's distance divided by Ben's speed should be the same as Jerry's distance divided by Jerry's speed. So let me write that down. So Ben's distance, Ben's distance divided by Ben's speed, and let me write it in this color, Ben's, Ben's speed should be equal to Jerry's distance, Jerry's distance divided by Jerry's speed, Jerry's speed. Now, which of these do we know or do we already have variables defined? Well, we know that Ben's distance from the ice cream shop is x, so this is represented by x."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So let me write that down. So Ben's distance, Ben's distance divided by Ben's speed, and let me write it in this color, Ben's, Ben's speed should be equal to Jerry's distance, Jerry's distance divided by Jerry's speed, Jerry's speed. Now, which of these do we know or do we already have variables defined? Well, we know that Ben's distance from the ice cream shop is x, so this is represented by x. We know that Jerry's distance from the ice cream shop is represented by y, so this is y. We know that Ben's speed is five kilometers per hour, so we're assuming everything is in kilometers per hour, so this would be five. And then Jerry's speed is v kilometers per hour, so this is v right over there."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Well, we know that Ben's distance from the ice cream shop is x, so this is represented by x. We know that Jerry's distance from the ice cream shop is represented by y, so this is y. We know that Ben's speed is five kilometers per hour, so we're assuming everything is in kilometers per hour, so this would be five. And then Jerry's speed is v kilometers per hour, so this is v right over there. And so we could rewrite all of this as x over five is equal to y over v. And once again, the way that I've set this up, the left side is the amount of time Ben takes to get to the ice cream shop. This is, on the right-hand side, this is the amount of time Jerry takes to get to the ice cream shop, and they tell us it's the same amount of time. So there you have it."}, {"video_title": "Modeling with multiple variables Ice cream Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And then Jerry's speed is v kilometers per hour, so this is v right over there. And so we could rewrite all of this as x over five is equal to y over v. And once again, the way that I've set this up, the left side is the amount of time Ben takes to get to the ice cream shop. This is, on the right-hand side, this is the amount of time Jerry takes to get to the ice cream shop, and they tell us it's the same amount of time. So there you have it. We have an equation that relates x, y, and v, and they gave us the five. Now, it's completely possible that instead of the five, they gave us something else and Ben's speed was a variable. If they did that, then we would have a different given and maybe a different variable, but the structure of our equation would be the same, that Ben's distance divided by Ben's speed would need to be equal to Jerry's distance divided by Jerry's speed."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure this out on your own. All right, now let's do this together. So first, I always like to approach things intuitively. So let's say the first cylinder looks something like this, like this. And then the second cylinder, here it's 100 times taller. I would have trouble drawing something that's 100 times taller. But if it has the same volume, it's going to have to be a lot thinner."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So let's say the first cylinder looks something like this, like this. And then the second cylinder, here it's 100 times taller. I would have trouble drawing something that's 100 times taller. But if it has the same volume, it's going to have to be a lot thinner. So if you wanna maintain the volume, as you make the cylinder taller, and I'm not going anywhere close to 100 times as tall here, you're going to have to decrease the radius. So we would expect a radius to be a good bit less than 20 meters. So that's just the first intuition, just to make sure that we somehow don't get some number that's larger than 20 meters."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "But if it has the same volume, it's going to have to be a lot thinner. So if you wanna maintain the volume, as you make the cylinder taller, and I'm not going anywhere close to 100 times as tall here, you're going to have to decrease the radius. So we would expect a radius to be a good bit less than 20 meters. So that's just the first intuition, just to make sure that we somehow don't get some number that's larger than 20 meters. But how do we figure out what that could be? Well, now we can go back to the formula. And we know that Jill calculated that 20 meters is the radius."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So that's just the first intuition, just to make sure that we somehow don't get some number that's larger than 20 meters. But how do we figure out what that could be? Well, now we can go back to the formula. And we know that Jill calculated that 20 meters is the radius. So 20 is equal to the square root of V over pi h. And if this formula looks unfamiliar to you, just remember the volume of a cylinder is the area of either the top or the bottom. So pi r squared times the height. And if you were to just solve this for r, you would have this exact formula that Jill uses."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And we know that Jill calculated that 20 meters is the radius. So 20 is equal to the square root of V over pi h. And if this formula looks unfamiliar to you, just remember the volume of a cylinder is the area of either the top or the bottom. So pi r squared times the height. And if you were to just solve this for r, you would have this exact formula that Jill uses. So this isn't some new formula. This is probably something that you have seen already. So we know that 20 meters is equal to this."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And if you were to just solve this for r, you would have this exact formula that Jill uses. So this isn't some new formula. This is probably something that you have seen already. So we know that 20 meters is equal to this. And now we're talking about a situation where we're at a height that is 100 times taller. So this other cylinder is going to have a radius of square root of V is going to be the same. So let's just write that V there."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So we know that 20 meters is equal to this. And now we're talking about a situation where we're at a height that is 100 times taller. So this other cylinder is going to have a radius of square root of V is going to be the same. So let's just write that V there. Pi doesn't change. It's always going to be pi. And now instead of h, we have something that is 100 times taller."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So let's just write that V there. Pi doesn't change. It's always going to be pi. And now instead of h, we have something that is 100 times taller. So we could write that as 100h. And then what's another way to write this? Well, what I'm gonna do is try to bring out the 100."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And now instead of h, we have something that is 100 times taller. So we could write that as 100h. And then what's another way to write this? Well, what I'm gonna do is try to bring out the 100. So I still get the square root of V over pi h. So I could rewrite this as the square root of one over 100 times V over pi h, which I could write as the square root of one over 100, and I'm just using properties of radicals here, times the square root of V over pi h. Now we know what the square root of V over pi h is. We know that that is 20, and our units are meters. So this is 20."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "Well, what I'm gonna do is try to bring out the 100. So I still get the square root of V over pi h. So I could rewrite this as the square root of one over 100 times V over pi h, which I could write as the square root of one over 100, and I'm just using properties of radicals here, times the square root of V over pi h. Now we know what the square root of V over pi h is. We know that that is 20, and our units are meters. So this is 20. And then what's the square root of one over 100? Well, this is the same thing as one over the square root of 100. And of course, now it's gonna be times 20, times 20."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "So this is 20. And then what's the square root of one over 100? Well, this is the same thing as one over the square root of 100. And of course, now it's gonna be times 20, times 20. Well, the square root of 100, I should say the principal root of 100 is 10. So the radius of our new cylinder, of the second cylinder, is gonna be 1 1o of 20, which is equal to two meters. And we're done."}, {"video_title": "Interpreting expressions with multiple variables Cylinder Modeling Algebra 2 Khan Academy.mp3", "Sentence": "And of course, now it's gonna be times 20, times 20. Well, the square root of 100, I should say the principal root of 100 is 10. So the radius of our new cylinder, of the second cylinder, is gonna be 1 1o of 20, which is equal to two meters. And we're done. The second cylinder is going to have a radius of two meters, which meets our intuition. If we increase our height by a factor of 100, then our radius decreases by a factor of 10. The reason why is because you square the radius right over here."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "What we're going to do in this video is use our knowledge of the roots of this polynomial to think about intervals where this polynomial would be positive or negative. And the key realization is is that the sign of a polynomial stays the same between consecutive zeros. Let me just draw an arbitrary graph of a polynomial here to make you appreciate why that is true. So X axis, Y axis. And if I were to draw some arbitrary polynomial like that, you can see that between consecutive zeros, the sign is the same. Between this zero and this zero, the polynomial is positive. Between this zero and this zero, the polynomial is negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So X axis, Y axis. And if I were to draw some arbitrary polynomial like that, you can see that between consecutive zeros, the sign is the same. Between this zero and this zero, the polynomial is positive. Between this zero and this zero, the polynomial is negative. And that's almost intuitively true because if the sign did not stay the same, that means you would have to cross the X axis. So you would have a zero. But we're saying between consecutive zeros."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Between this zero and this zero, the polynomial is negative. And that's almost intuitively true because if the sign did not stay the same, that means you would have to cross the X axis. So you would have a zero. But we're saying between consecutive zeros. So between this zero and this zero, it is positive again. Then after that zero, it stays negative. Once again, the only way it wouldn't stay negative is if there were another zero."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "But we're saying between consecutive zeros. So between this zero and this zero, it is positive again. Then after that zero, it stays negative. Once again, the only way it wouldn't stay negative is if there were another zero. So now let's go back to this example here. And let me delete this because this is not the graph of P of X which I have just written down. Let's first think about its zeros."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Once again, the only way it wouldn't stay negative is if there were another zero. So now let's go back to this example here. And let me delete this because this is not the graph of P of X which I have just written down. Let's first think about its zeros. So the zeros are the X values that would either make X plus two equal zero, two X minus three equal zero, or X minus four equal zero. So first, we can think about, well, what X values would make X plus two, X plus two equal to zero? Well, that of course would be X equals negative two."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Let's first think about its zeros. So the zeros are the X values that would either make X plus two equal zero, two X minus three equal zero, or X minus four equal zero. So first, we can think about, well, what X values would make X plus two, X plus two equal to zero? Well, that of course would be X equals negative two. What X values would make two X minus three equal zero? Two X minus three equal to zero. Add three to both sides, you get two X equals three."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, that of course would be X equals negative two. What X values would make two X minus three equal zero? Two X minus three equal to zero. Add three to both sides, you get two X equals three. Divide both sides by two, you get X equals 3 1\u20442. And then last but not least, what X values would make X minus four equal to zero? Add four to both sides."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Add three to both sides, you get two X equals three. Divide both sides by two, you get X equals 3 1\u20442. And then last but not least, what X values would make X minus four equal to zero? Add four to both sides. You get X is equal to four. And so if we were to plot this, it would look something like this. So this is X equals negative two, X equals negative one, this is zero, this is one, two, three, and four."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Add four to both sides. You get X is equal to four. And so if we were to plot this, it would look something like this. So this is X equals negative two, X equals negative one, this is zero, this is one, two, three, and four. And let me draw the Y axis here. So the Y axis would look something like this. X and Y, we have a zero at X equals negative two, so our graph will intersect the X axis there."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So this is X equals negative two, X equals negative one, this is zero, this is one, two, three, and four. And let me draw the Y axis here. So the Y axis would look something like this. X and Y, we have a zero at X equals negative two, so our graph will intersect the X axis there. We have a zero at X is equal to 3 1\u20442, which is 1 1\u20442, which is right over there. And we have a zero at X equals four, which is right over there. And so we have several candidate intervals."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "X and Y, we have a zero at X equals negative two, so our graph will intersect the X axis there. We have a zero at X is equal to 3 1\u20442, which is 1 1\u20442, which is right over there. And we have a zero at X equals four, which is right over there. And so we have several candidate intervals. And actually, let me write this down in a table. So the intervals over which, and this would really be between consecutive zeros, intervals to consider. So I'll draw a little table here."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And so we have several candidate intervals. And actually, let me write this down in a table. So the intervals over which, and this would really be between consecutive zeros, intervals to consider. So I'll draw a little table here. So you have X is less than negative two. That's one interval. X is less than, actually, let me color code this."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So I'll draw a little table here. So you have X is less than negative two. That's one interval. X is less than, actually, let me color code this. So if I were to say the interval for X is less than negative two, so that's this yellow that I draw at the extreme left there, we could have an interval where X is between negative two and 3 1\u20442. So negative two is less than X, is less than positive 3 1\u20442. That would be this interval right over here."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "X is less than, actually, let me color code this. So if I were to say the interval for X is less than negative two, so that's this yellow that I draw at the extreme left there, we could have an interval where X is between negative two and 3 1\u20442. So negative two is less than X, is less than positive 3 1\u20442. That would be this interval right over here. You have the interval, I'm trying to use all my colors, between 3 1\u20442 and four, this interval here. So that would be 3 1\u20442 is less than X is less than four. And then last but not least, you have the interval where X is greater than four, that interval right over there."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "That would be this interval right over here. You have the interval, I'm trying to use all my colors, between 3 1\u20442 and four, this interval here. So that would be 3 1\u20442 is less than X is less than four. And then last but not least, you have the interval where X is greater than four, that interval right over there. So X is greater, greater than four. Now, there's a couple of ways of thinking about whether over that interval, our function is positive or negative. One method is to just evaluate our P, our function, at a point in the interval."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then last but not least, you have the interval where X is greater than four, that interval right over there. So X is greater, greater than four. Now, there's a couple of ways of thinking about whether over that interval, our function is positive or negative. One method is to just evaluate our P, our function, at a point in the interval. And if it's positive, well, that means that that whole interval is positive. If it's negative, well, that means that the whole interval is negative. And once again, it's intuitive, because if for whatever reason it were to switch, we would have another zero."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "One method is to just evaluate our P, our function, at a point in the interval. And if it's positive, well, that means that that whole interval is positive. If it's negative, well, that means that the whole interval is negative. And once again, it's intuitive, because if for whatever reason it were to switch, we would have another zero. I know I keep saying that. But another way to think about it is, over that interval, what is the behavior of X plus two, two X minus three, and X minus four? Think about whether they're positive or negative, and use our knowledge of multiplying positives and negatives together to figure out whether we're dealing with a positive or a negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And once again, it's intuitive, because if for whatever reason it were to switch, we would have another zero. I know I keep saying that. But another way to think about it is, over that interval, what is the behavior of X plus two, two X minus three, and X minus four? Think about whether they're positive or negative, and use our knowledge of multiplying positives and negatives together to figure out whether we're dealing with a positive or a negative. So let's do it, we could do it both ways. So let's think of this as our sample X, sample X value. And then let's see what we can intuit about, or deduce about whether over that interval we are positive or negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Think about whether they're positive or negative, and use our knowledge of multiplying positives and negatives together to figure out whether we're dealing with a positive or a negative. So let's do it, we could do it both ways. So let's think of this as our sample X, sample X value. And then let's see what we can intuit about, or deduce about whether over that interval we are positive or negative. So for X is less than negative two, maybe an easy one or an obvious one to use, it could be any value where X is less than negative two, but let's try X is equal to negative three. So you could try to evaluate P of negative three, you could just evaluate that, actually let's just do that. So that's going to be equal to negative one times, two times negative three is negative six, minus three is negative nine, negative nine, times negative three minus four, that is negative seven."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then let's see what we can intuit about, or deduce about whether over that interval we are positive or negative. So for X is less than negative two, maybe an easy one or an obvious one to use, it could be any value where X is less than negative two, but let's try X is equal to negative three. So you could try to evaluate P of negative three, you could just evaluate that, actually let's just do that. So that's going to be equal to negative one times, two times negative three is negative six, minus three is negative nine, negative nine, times negative three minus four, that is negative seven. So if you were to multiply all of this out, this would give you negative 63, which is clearly negative. So over this interval right over here, our polynomial is going to be negative. So, and then we can move on to the next one."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So that's going to be equal to negative one times, two times negative three is negative six, minus three is negative nine, negative nine, times negative three minus four, that is negative seven. So if you were to multiply all of this out, this would give you negative 63, which is clearly negative. So over this interval right over here, our polynomial is going to be negative. So, and then we can move on to the next one. And an interesting thing is we didn't even have to figure out the 63 part, we can just see that there's a negative times a negative times a negative, which is going to be a negative. And so let's just do that going forward. Let's just think about whether each of these are going to be positive or negative, and what would happen when you multiply those positive and negatives together."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So, and then we can move on to the next one. And an interesting thing is we didn't even have to figure out the 63 part, we can just see that there's a negative times a negative times a negative, which is going to be a negative. And so let's just do that going forward. Let's just think about whether each of these are going to be positive or negative, and what would happen when you multiply those positive and negatives together. Now in this second interval, between negative two and three halves, what is going to happen? Well, we could do a sample point. Let's say x is, let's say x is equal to zero."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Let's just think about whether each of these are going to be positive or negative, and what would happen when you multiply those positive and negatives together. Now in this second interval, between negative two and three halves, what is going to happen? Well, we could do a sample point. Let's say x is, let's say x is equal to zero. That might be pretty straightforward. Well, when x is equal to zero, we're going to be dealing with a positive times a negative times a negative. A positive times a negative times a negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Let's say x is, let's say x is equal to zero. That might be pretty straightforward. Well, when x is equal to zero, we're going to be dealing with a positive times a negative times a negative. A positive times a negative times a negative. And the reason why I did that, I just did it in my head. I said, okay, that's going to be a positive two times a negative three times a negative four. So a positive times a negative times a negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "A positive times a negative times a negative. And the reason why I did that, I just did it in my head. I said, okay, that's going to be a positive two times a negative three times a negative four. So a positive times a negative times a negative. So I could write it this way. It's going to be a positive times a negative times a negative. Well, a negative times a negative is a positive, and a positive times a positive is a positive."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So a positive times a negative times a negative. So I could write it this way. It's going to be a positive times a negative times a negative. Well, a negative times a negative is a positive, and a positive times a positive is a positive. So we are positive over that interval. And if you were to evaluate p of zero, you will get a positive value. Now what about this next interval?"}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Well, a negative times a negative is a positive, and a positive times a positive is a positive. So we are positive over that interval. And if you were to evaluate p of zero, you will get a positive value. Now what about this next interval? What about this next interval here between three halves and four? We could try x is equal to two. At when x is equal to two, we are going to get a positive times a positive times a two minus four is negative times a negative."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Now what about this next interval? What about this next interval here between three halves and four? We could try x is equal to two. At when x is equal to two, we are going to get a positive times a positive times a two minus four is negative times a negative. So this is going to be negative over that interval. And then last but not least, when x is greater than four, we could try x is equal to, let's say, five. We are going to have a positive, positive times a positive times a positive."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "At when x is equal to two, we are going to get a positive times a positive times a two minus four is negative times a negative. So this is going to be negative over that interval. And then last but not least, when x is greater than four, we could try x is equal to, let's say, five. We are going to have a positive, positive times a positive times a positive. So we are going to have a positive. And as I mentioned, you could also do it without the sample points. You could say, okay, when x is greater than four, you could say, okay, for any x greater than four, if you add two to it, that for sure is going to be positive."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We are going to have a positive, positive times a positive times a positive. So we are going to have a positive. And as I mentioned, you could also do it without the sample points. You could say, okay, when x is greater than four, you could say, okay, for any x greater than four, if you add two to it, that for sure is going to be positive. For any x greater than four, if you multiply it by two and subtract three, well, that's still going to be positive because two times something greater than four is definitely greater than three. And for any x greater than four, if you subtract four from it, you're still going to have a positive value. So that's another way to think about it even if you don't use a sample point."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "You could say, okay, when x is greater than four, you could say, okay, for any x greater than four, if you add two to it, that for sure is going to be positive. For any x greater than four, if you multiply it by two and subtract three, well, that's still going to be positive because two times something greater than four is definitely greater than three. And for any x greater than four, if you subtract four from it, you're still going to have a positive value. So that's another way to think about it even if you don't use a sample point. But there you have it. We've figured out the intervals over which the function is negative or positive. And we don't know exactly what the function looks like, but generally speaking, it's negative over this first interval, so it might look something like this."}, {"video_title": "Positive and negative intervals of polynomials Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So that's another way to think about it even if you don't use a sample point. But there you have it. We've figured out the intervals over which the function is negative or positive. And we don't know exactly what the function looks like, but generally speaking, it's negative over this first interval, so it might look something like this. It's positive over that next interval. And then it's negative over that third interval. And then it's positive over that last interval."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "we're asked to solve 2x squared plus 5 is equal to 6x. And so we have a quadratic equation here, but just to make it, put it into a form that we're more familiar with, let's try to put it into standard form. And standard form, of course, is the form ax squared plus bx plus c is equal to 0. And to do that, we essentially have to take the 6x and get rid of it from the right-hand side, so we just have a 0 on the right-hand side. And to do that, let's just subtract 6x from both sides of this equation. And so our left-hand side becomes 2x squared minus 6x plus 5 is equal to, and on our right-hand side, these two characters cancel out, and we just are left with 0. And there's many ways to solve this."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "And to do that, we essentially have to take the 6x and get rid of it from the right-hand side, so we just have a 0 on the right-hand side. And to do that, let's just subtract 6x from both sides of this equation. And so our left-hand side becomes 2x squared minus 6x plus 5 is equal to, and on our right-hand side, these two characters cancel out, and we just are left with 0. And there's many ways to solve this. We could try to factor it, and if I was trying to factor it, I would divide both sides by 2. If I divide both sides by 2, I would get integer coefficients on the x squared and the x term, but I would get 5 halves for the constant, so it's not one of these easy things to factor. We could complete the square, or we could apply the quadratic formula, which is really just a formula derived from completing the square, so let's do that in this scenario."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "And there's many ways to solve this. We could try to factor it, and if I was trying to factor it, I would divide both sides by 2. If I divide both sides by 2, I would get integer coefficients on the x squared and the x term, but I would get 5 halves for the constant, so it's not one of these easy things to factor. We could complete the square, or we could apply the quadratic formula, which is really just a formula derived from completing the square, so let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus, so that gives us two roots right over there, plus or minus square root of b squared minus 4ac over 2a. Let's apply that to this situation. Negative b, this right here is b, so negative b is negative negative 6, so that's going to be positive 6, plus or minus the square root of b squared."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "We could complete the square, or we could apply the quadratic formula, which is really just a formula derived from completing the square, so let's do that in this scenario. And the quadratic formula tells us that if we have something in standard form like this, that the roots of it are going to be negative b plus or minus, so that gives us two roots right over there, plus or minus square root of b squared minus 4ac over 2a. Let's apply that to this situation. Negative b, this right here is b, so negative b is negative negative 6, so that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a, which is 2, times 2 times c, which is 5, times 5, all of that over 2 times a. a is 2, so 2 times 2 is 4. So this is going to be equal to 6, plus or minus the square root of 36, so let me just figure this out, 36 minus, so this is 4 times 2 times 5, this is 40 over here, so 36 minus 40, and you already might be wondering what's going to happen here. All of that over 4, or this is equal to 6, plus or minus the square root of negative 4, 36 minus 40 is negative 4, over 4."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Negative b, this right here is b, so negative b is negative negative 6, so that's going to be positive 6, plus or minus the square root of b squared. Negative 6 squared is 36, minus 4 times a, which is 2, times 2 times c, which is 5, times 5, all of that over 2 times a. a is 2, so 2 times 2 is 4. So this is going to be equal to 6, plus or minus the square root of 36, so let me just figure this out, 36 minus, so this is 4 times 2 times 5, this is 40 over here, so 36 minus 40, and you already might be wondering what's going to happen here. All of that over 4, or this is equal to 6, plus or minus the square root of negative 4, 36 minus 40 is negative 4, over 4. You might say, hey wait Sal, negative 4, if I take a square root, I'm going to get an imaginary number. You would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number, so we're essentially going to get two complex numbers when we take the positive and negative version of this root."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "All of that over 4, or this is equal to 6, plus or minus the square root of negative 4, 36 minus 40 is negative 4, over 4. You might say, hey wait Sal, negative 4, if I take a square root, I'm going to get an imaginary number. You would be right. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number, so we're essentially going to get two complex numbers when we take the positive and negative version of this root. Let's do that. The square root of negative 4 is the same thing as 2i. We know that's the same thing as 2i, or if you want to think of it this way, the square root of negative 4 is the same thing as the square root of negative 1 times the square root of 4, which is the same, I could even do it one step."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number, so we're essentially going to get two complex numbers when we take the positive and negative version of this root. Let's do that. The square root of negative 4 is the same thing as 2i. We know that's the same thing as 2i, or if you want to think of it this way, the square root of negative 4 is the same thing as the square root of negative 1 times the square root of 4, which is the same, I could even do it one step. That's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. and the principal square root of negative 1 is i, times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "We know that's the same thing as 2i, or if you want to think of it this way, the square root of negative 4 is the same thing as the square root of negative 1 times the square root of 4, which is the same, I could even do it one step. That's the same thing as negative 1 times 4 under the radical, which is the same thing as the square root of negative 1 times the square root of 4. and the principal square root of negative 1 is i, times the principal square root of 4 is 2. So this is 2i, or i times 2. So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So this right over here is going to be 2i. So we are left with x is equal to 6 plus or minus 2i over 4. And if we were to simplify it, we could divide the numerator and the denominator by 2. And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3 halves plus 1 half i. That's if I take the positive version of the i there. Or we could view this as 3 halves minus 1 half i."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "And so that would be the same thing as 3 plus or minus i over 2. Or if you want to write them as two distinct complex numbers, you could write this as 3 plus i over 2, or 3 halves plus 1 half i. That's if I take the positive version of the i there. Or we could view this as 3 halves minus 1 half i. This and these two guys right here are equivalent. Those are the two roots. Now what I want to do is verify that these work, verify that these two roots."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Or we could view this as 3 halves minus 1 half i. This and these two guys right here are equivalent. Those are the two roots. Now what I want to do is verify that these work, verify that these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did, you can see that this is just dividing both of these by 2."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Now what I want to do is verify that these work, verify that these two roots. So this one I can rewrite as 3 plus i over 2. These are equivalent. All I did, you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1 half, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "All I did, you can see that this is just dividing both of these by 2. Or if you were to essentially factor out the 1 half, you could go either way on this expression. And this one over here is going to be 3 minus i over 2. Or you could go directly from this. This is 3 plus or minus i over 2, so 3 plus i over 2. Or 3 minus i over 2. This and this, or this and this, or this, these are all equal representations of both of the roots."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Or you could go directly from this. This is 3 plus or minus i over 2, so 3 plus i over 2. Or 3 minus i over 2. This and this, or this and this, or this, these are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "This and this, or this and this, or this, these are all equal representations of both of the roots. But let's see if they work. So I'm first going to try this character right over here. It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. So what we want to do is we want to take 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "It's going to get a little bit hairy, because we're going to have to square it and all the rest. But let's see if we can do it. So what we want to do is we want to take 2 times this quantity squared. So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity. As 6 times 3 plus i over 2. So what is 3 plus i squared?"}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So 2 times 3 plus i over 2 squared plus 5. And we want to verify that that's the same thing as 6 times this quantity. As 6 times 3 plus i over 2. So what is 3 plus i squared? So this is 2 times, let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of 3 and i. So 3 times i is 3i."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So what is 3 plus i squared? So this is 2 times, let me just square this. So 3 plus i, that's going to be 3 squared, which is 9, plus 2 times the product of 3 and i. So 3 times i is 3i. Times 2 is 6i, so plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out, either with the distributive property or FOIL it out. And you'll get the middle term will be, you'll get 3i twice, when you add them you get 6i."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So 3 times i is 3i. Times 2 is 6i, so plus 6i. And if that doesn't make sense to you, I encourage you to kind of multiply it out, either with the distributive property or FOIL it out. And you'll get the middle term will be, you'll get 3i twice, when you add them you get 6i. And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "And you'll get the middle term will be, you'll get 3i twice, when you add them you get 6i. And then plus i squared, and i squared is negative 1. Minus 1. All of that over 4. Plus 5 is equal to, if you divide the numerator and the denominator by 2, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it, just to save some screen real estate."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "All of that over 4. Plus 5 is equal to, if you divide the numerator and the denominator by 2, you get a 3 here and you get a 1 here. And 3 distributed on 3 plus i is equal to 9 plus 3i. And what we have over here, we can simplify it, just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "And what we have over here, we can simplify it, just to save some screen real estate. 9 minus 1 is 8. So if I get rid of this, this is just 8 plus 6i. We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2. And the denominator here is just going to be 2. This 2 and this 2 are going to cancel out."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "We can divide the numerator and the denominator right here by 2. So the numerator would become 4 plus 3i, if we divided it by 2. And the denominator here is just going to be 2. This 2 and this 2 are going to cancel out. So on the left-hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "This 2 and this 2 are going to cancel out. So on the left-hand side, we're left with 4 plus 3i plus 5. And this needs to be equal to 9 plus 3i. Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. Now let's try 3 minus i."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Well, you can see we have a 3i on both sides of this equation. And we have a 4 plus 5, which is exactly equal to 9. So this solution, 3 plus i, definitely works. Now let's try 3 minus i. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. So let me write it down over here. So we have, let me rewrite the original equation."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "Now let's try 3 minus i. So once again, just looking at the original equation, 2x squared plus 5 is equal to 6x. So let me write it down over here. So we have, let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So we have, let me rewrite the original equation. We have 2x squared plus 5 is equal to 6x. And now we're going to try this root, verify that it works. So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy, but as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared, 3 minus i squared, this 3 minus i times 3 minus i, which is, and you could get practice taking squares of two-termed expressions, or complex numbers in this case, actually, it's going to be 9, that's 3 squared, and then 3 times negative i is negative 3i, and then you're going to have two of those, so negative 6i, and then i squared, so negative i squared is also negative 1."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So we have 2 times 3 minus i over 2 squared plus 5 needs to be equal to 6 times this business. 6 times 3 minus i over 2. Once again, a little hairy, but as long as we do everything, we put our head down and focus on it, we should be able to get the right result. So 3 minus i squared, 3 minus i squared, this 3 minus i times 3 minus i, which is, and you could get practice taking squares of two-termed expressions, or complex numbers in this case, actually, it's going to be 9, that's 3 squared, and then 3 times negative i is negative 3i, and then you're going to have two of those, so negative 6i, and then i squared, so negative i squared is also negative 1. That's negative 1 times negative 1 times i times i, so that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root, not the principal square root, but one of the square roots of negative 1."}, {"video_title": "Example Complex roots for a quadratic Algebra II Khan Academy.mp3", "Sentence": "So 3 minus i squared, 3 minus i squared, this 3 minus i times 3 minus i, which is, and you could get practice taking squares of two-termed expressions, or complex numbers in this case, actually, it's going to be 9, that's 3 squared, and then 3 times negative i is negative 3i, and then you're going to have two of those, so negative 6i, and then i squared, so negative i squared is also negative 1. That's negative 1 times negative 1 times i times i, so that's also negative 1. Negative i squared is also equal to negative 1. Negative i is also another square root, not the principal square root, but one of the square roots of negative 1. So this is going to be, so now we're going to have a plus 1, oh, sorry, we're going to have a minus 1, because this is negative i squared, which is negative 1, negative 1, and all of that over 4, all of that over, that's 2 squared is 4, times 2 over here, times 2, plus 5, needs to be equal to, well, before I even multiply it out, we can divide the numerator and denominator by 2, so 6 divided by 2 is 3, 2 divided by 2 is 1, so 3 times 3 is 9, 3 times negative i is negative 3i, and if we simplify it a little bit more, 9 minus 1 is going to be, I'll do this in blue, 9 minus 1 is going to be 8, we have 8 minus 6i, and if we divide 8 minus 6i by 2 and 4 by 2, in the numerator we're going to get 4 minus 3i, and in the denominator over here, in the denominator over here, we're going to get a 2. We divided the numerator and the denominator by 2, then we have a 2 out here, and we have a 2 in the denominator, those two characters will cancel out, and so this expression right over here simplifies to 4 minus 3i, and we have a plus 5, needs to be equal to 9 minus 3i. We have a negative 3i on the left, negative 3i on the right, we have a 4 plus 5, we can evaluate it, this left-hand side is 9, 9 minus 3i, which is the exact same complex number as we have on the right-hand side, 9 minus 3i."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We see it right over here in gray. It's shown in the grid below. Graph the function g of x is equal to x minus two squared minus four in the interactive graph. And this is from the shifting functions exercise on Khan Academy. And we can see we can change, we can change the, we can change the graph of g of x. But let's see, we want to graph it properly. So let's see how they relate."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And this is from the shifting functions exercise on Khan Academy. And we can see we can change, we can change the, we can change the graph of g of x. But let's see, we want to graph it properly. So let's see how they relate. Well, let's think about a few things. Let's first just make g of x completely overlap. Well, actually, that's completely easier said than, okay, there you go."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's see how they relate. Well, let's think about a few things. Let's first just make g of x completely overlap. Well, actually, that's completely easier said than, okay, there you go. Now they're completely overlapping. And let's see how they're different. Well, g of x, if you look at what's going on here, instead of having an x squared, we have an x minus two squared."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, actually, that's completely easier said than, okay, there you go. Now they're completely overlapping. And let's see how they're different. Well, g of x, if you look at what's going on here, instead of having an x squared, we have an x minus two squared. So one way to think about it is, when x is zero, you have zero squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, g of x, if you look at what's going on here, instead of having an x squared, we have an x minus two squared. So one way to think about it is, when x is zero, you have zero squared is equal to zero. But how do you get zero here? Well, x has got to be equal to two. Two minus two squared is zero squared. If we don't look at the negative four just yet. And so we would want to shift this graph over two to the right."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, x has got to be equal to two. Two minus two squared is zero squared. If we don't look at the negative four just yet. And so we would want to shift this graph over two to the right. This is essentially how much do we shift to the right. It's sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well, okay, for the original graph, when it was just x squared, to get the zero squared, I just had to put x equals zero."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so we would want to shift this graph over two to the right. This is essentially how much do we shift to the right. It's sometimes a little bit counterintuitive that we have a negative there, because you might say, well, negative, that makes me think that I want to shift to the left. But you have to remind yourself is like, well, okay, for the original graph, when it was just x squared, to get the zero squared, I just had to put x equals zero. Now to get a zero squared, I have to put in a two. So this is actually shifting the graph to the right. And so what do we do with this negative four?"}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "But you have to remind yourself is like, well, okay, for the original graph, when it was just x squared, to get the zero squared, I just had to put x equals zero. Now to get a zero squared, I have to put in a two. So this is actually shifting the graph to the right. And so what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two squared, it's gonna shift it down by four."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so what do we do with this negative four? Well, this is a little bit more intuitive, or at least for me when I first learned it. This literally will just shift the graph down. Whatever your value is of x minus two squared, it's gonna shift it down by four. So what we want to do is just shift both of these points down by four. So this is gonna go from nine, this is gonna go from the coordinate five comma nine to five comma, if we go down four, five comma five. And this is gonna go from two comma zero to two comma negative four."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Whatever your value is of x minus two squared, it's gonna shift it down by four. So what we want to do is just shift both of these points down by four. So this is gonna go from nine, this is gonna go from the coordinate five comma nine to five comma, if we go down four, five comma five. And this is gonna go from two comma zero to two comma negative four. Two comma negative four. Did I do that right? I think that is, I think that's right."}, {"video_title": "Graphing shifted functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And this is gonna go from two comma zero to two comma negative four. Two comma negative four. Did I do that right? I think that is, I think that's right. What essentially what we have going on is g of x is f of x shifted two to the right and four down, two to the right and four down. And notice, if you look at the vertex here, we shifted two to the right and four down. And I shifted this one also, this one also I shifted two to the right and four down."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "When we first learned algebra together, we started factoring polynomials, especially quadratics. We recognized that an expression like x squared could be written as x times x. We also recognized that a polynomial like three x squared plus four x, that in this situation, both terms had the common factor of x, and you could factor that out. And so you could rewrite this as x times three x plus four. And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And so you could rewrite this as x times three x plus four. And we also learned to do fancier things. We learned to factor things like x squared plus seven x plus 12. We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We were able to say, hey, what two numbers would add up to seven, and if I were to multiply them, I'd get 12, and in those early videos, we show why that works, and say, well, three and four, so maybe this can be factored as x plus three times x plus four. If this is unfamiliar to you, I encourage you to go review that in some of the introductory factoring quadratics on Khan Academy. It should be a review at this point in your journey. We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x?"}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We also looked at things like differences of squares, x squared minus nine. Say, hey, that's x squared minus three squared, so we could factor that as x plus three times x minus three, and we looked at other types of quadratics. Now, as we go deeper into our algebra journeys, we're going to build on this to factor higher degree polynomials, third degree, fourth degree, fifth degree, which will be very useful in your mathematical careers, but we're going to start doing it by really looking at some of the structure, some of the patterns that we've seen in introductory algebra. For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "For example, let's say someone walks up to you on the street and says, can you factor x to the third plus seven x squared plus 12x? Well, at first you might say, oh, this is a third degree polynomial, that seems kind of intimidating, until you realize, hey, all of these terms have the common factor x, so if I factor that out, then it becomes x times x squared plus seven x plus 12, and then this is exactly what we saw over here, so we could rewrite all of this as x times x plus three times x plus four, so we're going to see that we might be able to do some simple factoring like this and even factoring multiple times. We might also start to appreciate structure that brings us back to some of what we saw in our introductory algebra, so for example, you might see something like this, where once again, someone walks up to you on the street and says, hey, you factor this, a to the fourth power plus seven a squared plus 12, and at first, you're like, wow, there's a fourth power here, what do I do, until you say, well, what if I were to rewrite this as a squared squared plus seven a squared plus 12, and now this a squared is looking an awful lot like this x over here. If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "If this were an x, then this would be x squared. If this were an x, then this would just be an x, and then these expressions would be the same, so when I factor it, everywhere I see an x, I could replace with an a squared, so I could factor this out, really looking at the same structure we have here as a squared plus three times a squared plus four. Now, I'm going really fast through this. This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "This is really the introductory video, the overview video. Don't worry if this is a little bit much too fast. This is really just to give you a sense of things. Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth."}, {"video_title": "Introduction to factoring higher degree polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Later in this unit, we're going to dig deeper into each of these cases, but just to give you a sense of where we're going, I'll give you another example that builds off of what you likely saw in your introductory algebra learning, so building off of the structure here, if someone were to walk up to you, again, a lot of people are walking up to you, and say factor four x to the sixth minus nine y to the fourth Well, at first, this looks quite intimidating until you realize that, hey, I could write both of these as squares. I could write this first one as two x to the third squared minus, and I could write this second term as three y squared squared, and now, this is just a difference of squares, so it'd be two x to the third plus three y squared times two x to the third minus three y squared. We'll also see things like this where we're going to be factoring multiple times, so once again, someone walks up to you on the street, and they say you're a very popular person. Someone walks up to you and says, up to you on the street and says, factor x to the fourth minus y to the fourth. Well, based on what we just saw, you could realize that this is the same thing as x squared squared minus y squared squared, and you say, okay, this is a difference of squares just like this was a difference of squares, so it's going to be the sum of x squared and y squared, x squared plus y squared, times the difference of them, x squared minus y squared. Now, this is fun because this is two a difference of squares, so we can rewrite this whole thing as I'll rewrite this first part, x squared, x squared plus y squared, and then we could factor this as a difference of squares just as we factored this up here, and we get x plus y times x minus y, so I'll leave you there. I've just bombarded you with a bunch of information, but this is really just to get you warmed up."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "Let's see if we can figure out what 256 to the 4 7ths power divided by two to the 4 7ths power is. And like always, pause the video and see if you can figure this out. All right, let's work through this together. And at first you might find this kind of daunting, especially when you see something like two to the 4 7ths power. Is that even, that's not going to be a whole number. How do I do this, especially without a calculator? And I should have said do this without a calculator."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And at first you might find this kind of daunting, especially when you see something like two to the 4 7ths power. Is that even, that's not going to be a whole number. How do I do this, especially without a calculator? And I should have said do this without a calculator. But then the key is to see that we can use our exponent properties to simplify this a little bit so that we can do this on paper. And the main property that might jump out at you is if I have something, if I have x to the a power over y to the a power, this is the same thing as x over y to the a power. And our situation right over here, 256 would be x, two would be y, and then a is 4 7ths."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And I should have said do this without a calculator. But then the key is to see that we can use our exponent properties to simplify this a little bit so that we can do this on paper. And the main property that might jump out at you is if I have something, if I have x to the a power over y to the a power, this is the same thing as x over y to the a power. And our situation right over here, 256 would be x, two would be y, and then a is 4 7ths. So we can rewrite this. This is going to be equal to, this is equal to 256 over two to the 4 7ths power. And so this is nice, we're already able to simplify this because we know 256 divided by two is 128."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And our situation right over here, 256 would be x, two would be y, and then a is 4 7ths. So we can rewrite this. This is going to be equal to, this is equal to 256 over two to the 4 7ths power. And so this is nice, we're already able to simplify this because we know 256 divided by two is 128. So this is 128 to the 4 7ths power. Now this might also seem a little bit difficult. How do I raise 128 to a fractional power?"}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And so this is nice, we're already able to simplify this because we know 256 divided by two is 128. So this is 128 to the 4 7ths power. Now this might also seem a little bit difficult. How do I raise 128 to a fractional power? But we just have to remind ourselves this is the same thing. This is the same thing as 128 to the 1 7th power then raised to the 4th power. We could also view it the other way around."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "How do I raise 128 to a fractional power? But we just have to remind ourselves this is the same thing. This is the same thing as 128 to the 1 7th power then raised to the 4th power. We could also view it the other way around. We could say that this is also 128 to the 4th power and then raise that to the 1 7th. But multiplying 128 four times, that's going to be very computationally intensive and then you have to find the 7th root of that. That seems pretty difficult so we don't want to go in that way."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "We could also view it the other way around. We could say that this is also 128 to the 4th power and then raise that to the 1 7th. But multiplying 128 four times, that's going to be very computationally intensive and then you have to find the 7th root of that. That seems pretty difficult so we don't want to go in that way. But if we can get the smaller number first, what is 128 to the 1 7th power, then that might be easier to raise to the 4th power. Now when you look at this, and knowing that probably the question writer in this case, I'm the person who presented it with you, is telling you that you're not going to use a calculator, it's a pretty good clue that all right, this is probably going to be something that I can figure out on my own. And you might recognize 128 as a power of two and maybe two to the 7th is 128 and we can verify that."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "That seems pretty difficult so we don't want to go in that way. But if we can get the smaller number first, what is 128 to the 1 7th power, then that might be easier to raise to the 4th power. Now when you look at this, and knowing that probably the question writer in this case, I'm the person who presented it with you, is telling you that you're not going to use a calculator, it's a pretty good clue that all right, this is probably going to be something that I can figure out on my own. And you might recognize 128 as a power of two and maybe two to the 7th is 128 and we can verify that. So let's see, two to the 1st is two, four, eight, 16, 32, 64, 128. Two times two is four, times two is eight, times two is 16, times two is 32, times two is 64, times two is 128. So two to the 7th power is equal to 128."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "And you might recognize 128 as a power of two and maybe two to the 7th is 128 and we can verify that. So let's see, two to the 1st is two, four, eight, 16, 32, 64, 128. Two times two is four, times two is eight, times two is 16, times two is 32, times two is 64, times two is 128. So two to the 7th power is equal to 128. Or another way of saying this exact same thing is that 128, 128 is equal to, or 128 to the 1 7th power is equal to two. Or you could even say that the 7th root, the 7th root of 128 is equal to two. So we can simplify this, this is two."}, {"video_title": "Evaluating quotient of fractional exponents Mathematics I High School Math Khan Academy.mp3", "Sentence": "So two to the 7th power is equal to 128. Or another way of saying this exact same thing is that 128, 128 is equal to, or 128 to the 1 7th power is equal to two. Or you could even say that the 7th root, the 7th root of 128 is equal to two. So we can simplify this, this is two. So our whole expression is now just two to the 4th power. Well that's just two times two, times two times two. So that's two to the 4th power, two to the 4th power which is just going to be equal to 16."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "The formula for the area of a triangle is A is equal to 1 half B times h. Where A is equal to area, B is equal to length of the base, and h is equal to the length of the height. So area is equal to 1 half times the length of the base times the length of the height. Solve this formula for the height. So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what B and h are. B would be the length of the base. So this distance right over here is B."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "So just to visualize this a little bit, let me draw a triangle here. Let me draw a triangle just so we know what B and h are. B would be the length of the base. So this distance right over here is B. And then this distance right here is our height. That is the height of the triangle. Let me do that in a lower case h because that's how we wrote it in the formula."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "So this distance right over here is B. And then this distance right here is our height. That is the height of the triangle. Let me do that in a lower case h because that's how we wrote it in the formula. Now they want us to solve this formula for the height. So the formula is area is equal to 1 half base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let me do that in a lower case h because that's how we wrote it in the formula. Now they want us to solve this formula for the height. So the formula is area is equal to 1 half base times height. And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right hand side. So let's get rid of everything else on the right hand side. I'll do it one step at a time."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "And we want to solve for h. We essentially want to isolate the h on one side of the equation. It's already on the right hand side. So let's get rid of everything else on the right hand side. I'll do it one step at a time. We can skip steps if we wanted to. But let's see if we can get rid of this 1 half. The best way to get rid of a 1 half, and it's being multiplied by h, is if we multiply both sides of the equation by its reciprocal."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "I'll do it one step at a time. We can skip steps if we wanted to. But let's see if we can get rid of this 1 half. The best way to get rid of a 1 half, and it's being multiplied by h, is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2 over 1 or by 2. So let's do that. Let's multiply."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "The best way to get rid of a 1 half, and it's being multiplied by h, is if we multiply both sides of the equation by its reciprocal. If we multiply both sides of the equation by 2 over 1 or by 2. So let's do that. Let's multiply. Remember, anything you do to one side of the equation, you also have to do to the other side of the equation. Now what did this do? Well the whole point behind multiplying by 2 is 2 times 1 half is 1."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let's multiply. Remember, anything you do to one side of the equation, you also have to do to the other side of the equation. Now what did this do? Well the whole point behind multiplying by 2 is 2 times 1 half is 1. So on the right hand side of the equation, we're just going to have a bh. And on the left hand side of the equation, we have a 2a. And we're almost there."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "Well the whole point behind multiplying by 2 is 2 times 1 half is 1. So on the right hand side of the equation, we're just going to have a bh. And on the left hand side of the equation, we have a 2a. And we're almost there. We have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "And we're almost there. We have a b multiplying by an h. If we want to just isolate the h, we could divide both sides of this equation by b. We're just dividing both sides. You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well the right hand side, the b's cancel out, and the left hand side we're just left with 2a over b. So we get h, and I'm just swapping the sides here, h is equal to 2a over b."}, {"video_title": "Rearrange formulas to isolate specific variables Linear equations Algebra I Khan Academy.mp3", "Sentence": "You can almost view b as the coefficient on the h. We're just dividing both sides by b. And then what do we get? Well the right hand side, the b's cancel out, and the left hand side we're just left with 2a over b. So we get h, and I'm just swapping the sides here, h is equal to 2a over b. 2a over b. And we're done. We have solved this formula for the height."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And it's zero is less than or equal to theta, which is less than or equal to two pi. So we're going to include zero and two pi in the possible values for theta. So to do this, I've set up a little chart for theta, cosine theta, and sine theta. And we can use this to, and the unit circle, to hopefully quickly graph what the graphs of y equals sine theta and y equals cosine theta are. And then we can think about how many times they intersect, and maybe where they actually intersect. So let's get started. So first of all, just to be clear, this is a unit circle."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we can use this to, and the unit circle, to hopefully quickly graph what the graphs of y equals sine theta and y equals cosine theta are. And then we can think about how many times they intersect, and maybe where they actually intersect. So let's get started. So first of all, just to be clear, this is a unit circle. This is the x-axis. This is the y-axis. Over here, we're going to graph these two graphs."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So first of all, just to be clear, this is a unit circle. This is the x-axis. This is the y-axis. Over here, we're going to graph these two graphs. So this is going to be the y-axis, and it's going to be a function of theta, not x, on the horizontal axis. So first, let's think about what happens when theta is equal to zero. So when theta is equal to zero, you're at this point right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Over here, we're going to graph these two graphs. So this is going to be the y-axis, and it's going to be a function of theta, not x, on the horizontal axis. So first, let's think about what happens when theta is equal to zero. So when theta is equal to zero, you're at this point right over here. Let me do it in a different color. You're at this point right over here on the unit circle. And what coordinate is that?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So when theta is equal to zero, you're at this point right over here. Let me do it in a different color. You're at this point right over here on the unit circle. And what coordinate is that? Well, that's the point one comma zero. And so based on that, what is cosine of theta when theta is equal to zero? Well, cosine of theta is one, and sine of theta is going to be zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And what coordinate is that? Well, that's the point one comma zero. And so based on that, what is cosine of theta when theta is equal to zero? Well, cosine of theta is one, and sine of theta is going to be zero. This is the x-axis at the point of intersection with the unit circle. This is the x-coordinate at the point of intersection with the unit circle. This is the y-coordinate."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, cosine of theta is one, and sine of theta is going to be zero. This is the x-axis at the point of intersection with the unit circle. This is the x-coordinate at the point of intersection with the unit circle. This is the y-coordinate. Let's keep going. What about pi over two? So pi over two, we are right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is the y-coordinate. Let's keep going. What about pi over two? So pi over two, we are right over here. What is that coordinate? Well, that's now x is zero, y is one. So based on that, cosine of theta is zero, and what is sine of theta?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So pi over two, we are right over here. What is that coordinate? Well, that's now x is zero, y is one. So based on that, cosine of theta is zero, and what is sine of theta? Well, that's going to be one. It's the y-coordinate right over here. Now let's go all the way to pi."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So based on that, cosine of theta is zero, and what is sine of theta? Well, that's going to be one. It's the y-coordinate right over here. Now let's go all the way to pi. We're at this point in the unit circle. What is the coordinate? Well, this is negative one comma zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's go all the way to pi. We're at this point in the unit circle. What is the coordinate? Well, this is negative one comma zero. So what is cosine of theta? Well, it's the x-coordinate here, which is negative one, and sine of theta is going to be the y-coordinate, which is zero. Now let's keep going."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, this is negative one comma zero. So what is cosine of theta? Well, it's the x-coordinate here, which is negative one, and sine of theta is going to be the y-coordinate, which is zero. Now let's keep going. Now we're down here at three pi over two. If we go all the way around to three pi over two, what is this coordinate? Well, this is zero, negative one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's keep going. Now we're down here at three pi over two. If we go all the way around to three pi over two, what is this coordinate? Well, this is zero, negative one. Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be? Well, it's going to be negative one. And then finally we go back to two pi, which is making a full revolution around the circle."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, this is zero, negative one. Cosine of theta is the x-coordinate here, so cosine of theta is going to be zero, and what is sine of theta going to be? Well, it's going to be negative one. And then finally we go back to two pi, which is making a full revolution around the circle. We went all the way around, and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals zero radians. And so what is cosine of theta?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then finally we go back to two pi, which is making a full revolution around the circle. We went all the way around, and we're back to this point right over here. So the coordinate is the exact same thing as when the angle equals zero radians. And so what is cosine of theta? Well, that's one, and sine of theta is zero. And from this we can make a rough sketch of the graph and think about where they might intersect. So first let's do cosine of theta."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so what is cosine of theta? Well, that's one, and sine of theta is zero. And from this we can make a rough sketch of the graph and think about where they might intersect. So first let's do cosine of theta. When theta is zero, and let me mark this off. So this is going to be when y is equal to one, and this is when y is equal to negative one. So y equals cosine of theta."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So first let's do cosine of theta. When theta is zero, and let me mark this off. So this is going to be when y is equal to one, and this is when y is equal to negative one. So y equals cosine of theta. I'm going to graph, let's see, theta equals zero, cosine of theta equals one. So cosine of theta is equal to one. When theta is equal to pi over two, cosine of theta is zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So y equals cosine of theta. I'm going to graph, let's see, theta equals zero, cosine of theta equals one. So cosine of theta is equal to one. When theta is equal to pi over two, cosine of theta is zero. When theta is equal to pi, cosine of theta is negative one. When theta is equal to three pi over two, cosine of theta is equal to zero. That's this right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to pi over two, cosine of theta is zero. When theta is equal to pi, cosine of theta is negative one. When theta is equal to three pi over two, cosine of theta is equal to zero. That's this right over here. And then finally when theta is two pi, cosine of theta is one again. And the curve will look something like this. My best attempt to draw it."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's this right over here. And then finally when theta is two pi, cosine of theta is one again. And the curve will look something like this. My best attempt to draw it. Make it a nice, smooth curve. So it's going to look something like this. The look of these curves should look somewhat familiar at this point."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "My best attempt to draw it. Make it a nice, smooth curve. So it's going to look something like this. The look of these curves should look somewhat familiar at this point. So it should look something like this. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The look of these curves should look somewhat familiar at this point. So it should look something like this. So this is the graph of y is equal to cosine of theta. Now let's do the same thing for sine theta. When theta is equal to zero, sine theta is zero. When theta is pi over two, sine of theta is one. When theta is equal to pi, sine of theta is zero."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's do the same thing for sine theta. When theta is equal to zero, sine theta is zero. When theta is pi over two, sine of theta is one. When theta is equal to pi, sine of theta is zero. When theta is equal to three pi over two, sine of theta is negative one. When theta is equal to two pi, sine of theta is equal to zero. And so the graph of sine of theta is going to look something like this."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "When theta is equal to pi, sine of theta is zero. When theta is equal to three pi over two, sine of theta is negative one. When theta is equal to two pi, sine of theta is equal to zero. And so the graph of sine of theta is going to look something like this. My best attempt at drawing it is going to look something like this. So just visually we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so the graph of sine of theta is going to look something like this. My best attempt at drawing it is going to look something like this. So just visually we can think about the question. At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? For theta being between zero and two pi, including those two points. Well, you just look at this graph. You see there's two points of intersection."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "At how many points do the graphs of y equals sine of theta and y equals cosine of theta intersect for this range for theta? For theta being between zero and two pi, including those two points. Well, you just look at this graph. You see there's two points of intersection. This point right over here and this point right over here. Just over the, between zero and two pi. These are cyclical graphs."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You see there's two points of intersection. This point right over here and this point right over here. Just over the, between zero and two pi. These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this two pi range for theta, you get two points of intersection. Now let's think about what they are."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "These are cyclical graphs. If we kept going, they would keep intersecting with each other. But just over this two pi range for theta, you get two points of intersection. Now let's think about what they are. Because they look to be pretty close between, right between zero and pi over two and right between pi and three pi over two. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over four."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about what they are. Because they look to be pretty close between, right between zero and pi over two and right between pi and three pi over two. So let's look at our unit circle if we can figure out what those values are. It looks like this is at pi over four. So let's verify that. So let's think about what these values are at pi over four. So pi over four is that angle, or that's the terminal side of it."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It looks like this is at pi over four. So let's verify that. So let's think about what these values are at pi over four. So pi over four is that angle, or that's the terminal side of it. So this is pi over four. Pi over four is the exact same thing as a 45 degree angle. So let's do pi over four right over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So pi over four is that angle, or that's the terminal side of it. So this is pi over four. Pi over four is the exact same thing as a 45 degree angle. So let's do pi over four right over here. So we have to figure out what this point is, what the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle?"}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's do pi over four right over here. So we have to figure out what this point is, what the coordinates are. So let's make this a right triangle. And so what do we know about this right triangle? And I'm going to draw it right over here to make it a little clearer. This is a very typical type of right triangle, so it's good to get some familiarity with it. Let me draw my best attempt."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so what do we know about this right triangle? And I'm going to draw it right over here to make it a little clearer. This is a very typical type of right triangle, so it's good to get some familiarity with it. Let me draw my best attempt. All right. So we know it's a right triangle. We know that this is 45 degrees."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let me draw my best attempt. All right. So we know it's a right triangle. We know that this is 45 degrees. What is the length of the hypotenuse? Well, this is a unit circle. It has radius one, so the length of the hypotenuse here is one."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We know that this is 45 degrees. What is the length of the hypotenuse? Well, this is a unit circle. It has radius one, so the length of the hypotenuse here is one. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It has radius one, so the length of the hypotenuse here is one. And what do we know about this angle right over here? Well, we know that it too must be 45 degrees because all of these angles have to add up to 180. And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean theorem to think about the length of those sides. So using the Pythagorean theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, let's call these, if this has length a, well, then this also has length a."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And since these two angles are the same, we know that these two sides are going to be the same. And then we could use the Pythagorean theorem to think about the length of those sides. So using the Pythagorean theorem, knowing that these two sides are equal, what do we get for the length of those sides? Well, let's call these, if this has length a, well, then this also has length a. And we can use the Pythagorean theorem, and we could say a squared plus a squared is equal to the hypotenuse squared is equal to one. Or, 2 a squared is equal to one, a squared is equal to one half. Take the principal root of both sides, a is equal to the square root of one half, which is the square root of one, which is just one, over the square root of two."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, let's call these, if this has length a, well, then this also has length a. And we can use the Pythagorean theorem, and we could say a squared plus a squared is equal to the hypotenuse squared is equal to one. Or, 2 a squared is equal to one, a squared is equal to one half. Take the principal root of both sides, a is equal to the square root of one half, which is the square root of one, which is just one, over the square root of two. We can rationalize the denominator here by multiplying by square root of two over square root of two, which gives us a is equal to, in the numerators, square root of two. And in the denominator, square root of two times square root of two is two. 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Take the principal root of both sides, a is equal to the square root of one half, which is the square root of one, which is just one, over the square root of two. We can rationalize the denominator here by multiplying by square root of two over square root of two, which gives us a is equal to, in the numerators, square root of two. And in the denominator, square root of two times square root of two is two. 2. So this length is square root of 2 over 2. And this length is the same thing. So this length right over here is square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "2. So this length is square root of 2 over 2. And this length is the same thing. So this length right over here is square root of 2 over 2. And this height right over here is also square root of 2 over 2. So based on that, what is this coordinate point? Well, it's square root of 2 over 2 to the right in the positive direction."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this length right over here is square root of 2 over 2. And this height right over here is also square root of 2 over 2. So based on that, what is this coordinate point? Well, it's square root of 2 over 2 to the right in the positive direction. So x is equal to square root of 2 over 2. And y is square root of 2 over 2 in the upwards direction, the vertical direction, the positive vertical direction. So it's also square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, it's square root of 2 over 2 to the right in the positive direction. So x is equal to square root of 2 over 2. And y is square root of 2 over 2 in the upwards direction, the vertical direction, the positive vertical direction. So it's also square root of 2 over 2. Cosine of theta is just the x-coordinate. So it's square root of 2 over 2. Sine of theta is just the y-coordinate."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's also square root of 2 over 2. Cosine of theta is just the x-coordinate. So it's square root of 2 over 2. Sine of theta is just the y-coordinate. So you see immediately that they are indeed equal at that point. So at this point, they are both equal to square root of 2. They're both equal to square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Sine of theta is just the y-coordinate. So you see immediately that they are indeed equal at that point. So at this point, they are both equal to square root of 2. They're both equal to square root of 2 over 2. Now what about this point right over here, which looks right in between pi and 3 pi over 2? So that's going to be, so this is pi. This is 3 pi over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "They're both equal to square root of 2 over 2. Now what about this point right over here, which looks right in between pi and 3 pi over 2? So that's going to be, so this is pi. This is 3 pi over 2. It is right over here. So it's another pi over 4 plus pi. So pi plus pi over 4 is the same thing as 4 pi over 4 plus pi over 4."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is 3 pi over 2. It is right over here. So it's another pi over 4 plus pi. So pi plus pi over 4 is the same thing as 4 pi over 4 plus pi over 4. So this is the angle 5 pi over 4. So this is 5 pi over 4. So this is equal to 5 pi over 4."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So pi plus pi over 4 is the same thing as 4 pi over 4 plus pi over 4. So this is the angle 5 pi over 4. So this is 5 pi over 4. So this is equal to 5 pi over 4. So that's what we're trying to figure out. What are the value of these functions at theta equal 5 pi over 4? Well, there's multiple ways to think about it."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to 5 pi over 4. So that's what we're trying to figure out. What are the value of these functions at theta equal 5 pi over 4? Well, there's multiple ways to think about it. You can even use a little bit of geometry to say, well, if this is a 45 degree angle, then this right over here is also a 45 degree angle. You could say that the reference angle in terms of degrees is 45 degrees. And we could do a very similar thing."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, there's multiple ways to think about it. You can even use a little bit of geometry to say, well, if this is a 45 degree angle, then this right over here is also a 45 degree angle. You could say that the reference angle in terms of degrees is 45 degrees. And we could do a very similar thing. We can draw a right triangle. We know the hypotenuse is 1. We know that if this is a right angle, this is 45 degrees."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we could do a very similar thing. We can draw a right triangle. We know the hypotenuse is 1. We know that if this is a right angle, this is 45 degrees. If that's 45 degrees, then this is also 45 degrees. And we have a triangle that's very similar. They're actually congruent triangles."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We know that if this is a right angle, this is 45 degrees. If that's 45 degrees, then this is also 45 degrees. And we have a triangle that's very similar. They're actually congruent triangles. So hypotenuse is 1, 45, 45, 90. We then know that the length of this side is square root of 2 over 2. And the length of this side is square root of 2 over 2, the exact same logic we used over here."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "They're actually congruent triangles. So hypotenuse is 1, 45, 45, 90. We then know that the length of this side is square root of 2 over 2. And the length of this side is square root of 2 over 2, the exact same logic we used over here. So based on that, what is the coordinate of that point? Well, let's think about the x value. It's square root of 2 over 2 in the negative direction."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the length of this side is square root of 2 over 2, the exact same logic we used over here. So based on that, what is the coordinate of that point? Well, let's think about the x value. It's square root of 2 over 2 in the negative direction. We have to go square root of 2 over 2 to the left of the origin. So it's negative square root of 2 over 2. This point on the x-axis is negative square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's square root of 2 over 2 in the negative direction. We have to go square root of 2 over 2 to the left of the origin. So it's negative square root of 2 over 2. This point on the x-axis is negative square root of 2 over 2. What about the y value? Well, we have to go square root of 2 over 2 down in the downward direction from the origin. So it's also negative square root of 2 over 2."}, {"video_title": "Example Intersection of sine and cosine Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This point on the x-axis is negative square root of 2 over 2. What about the y value? Well, we have to go square root of 2 over 2 down in the downward direction from the origin. So it's also negative square root of 2 over 2. So cosine of theta is negative square root of 2 over 2. And sine of theta is also negative square root of 2 over 2. And so we see that we do indeed have the same value for cosine of theta and sine of theta right there."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "And you could solve this by doing algebraic long division, by taking all of this business and dividing it by x minus three and figuring out if you have a remainder. If you do end up with a remainder, then this is not a factor of this. But if you don't have a remainder, then that means that this divides fully into this right over here without a remainder, which means it is a factor. So if the remainder is equal to zero, the remainder is equal to zero if and only if it's a factor. It is a factor. And we know a very fast way of calculating the remainder of when you take some polynomial and you divide it by a first degree expression like this. I guess you could say when you divide it by a first degree polynomial like this."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "So if the remainder is equal to zero, the remainder is equal to zero if and only if it's a factor. It is a factor. And we know a very fast way of calculating the remainder of when you take some polynomial and you divide it by a first degree expression like this. I guess you could say when you divide it by a first degree polynomial like this. The polynomial remainder theorem, the polynomial remainder theorem tells us that if we take some, let me do this, some polynomial, P of x, and we were to divide it by some x minus a, then the remainder is just going to be equal to our polynomial evaluated at, our polynomial evaluated at a. So let's just see what's a in this case. Well, in this case, our a is positive three."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "I guess you could say when you divide it by a first degree polynomial like this. The polynomial remainder theorem, the polynomial remainder theorem tells us that if we take some, let me do this, some polynomial, P of x, and we were to divide it by some x minus a, then the remainder is just going to be equal to our polynomial evaluated at, our polynomial evaluated at a. So let's just see what's a in this case. Well, in this case, our a is positive three. So let's just evaluate our polynomial at x equals three. If what we get is equal to zero, that means our remainder is zero, and that means that x minus three is a factor. If we get some other remainder, that means, well, we have a non-zero remainder, and this isn't a factor."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "Well, in this case, our a is positive three. So let's just evaluate our polynomial at x equals three. If what we get is equal to zero, that means our remainder is zero, and that means that x minus three is a factor. If we get some other remainder, that means, well, we have a non-zero remainder, and this isn't a factor. So let's try it out. So we're gonna have, so I'm just gonna do it, I'll do it all in magenta. It might be a little computationally intensive."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "If we get some other remainder, that means, well, we have a non-zero remainder, and this isn't a factor. So let's try it out. So we're gonna have, so I'm just gonna do it, I'll do it all in magenta. It might be a little computationally intensive. So it's gonna be two times three to the fourth power. Three to the fourth, three to the third is 81. 81 minus 11, yeah, this is gonna get a little computationally intensive, but let's see if we can power through it."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "It might be a little computationally intensive. So it's gonna be two times three to the fourth power. Three to the fourth, three to the third is 81. 81 minus 11, yeah, this is gonna get a little computationally intensive, but let's see if we can power through it. 11 times 27, I probably should have picked a simpler example, but let's just keep going. Plus 15 times nine, plus four times three is 12, minus 12, so lucky for us, at least those last two terms cancel out. And so this is going to be, the rest from here is arithmetic."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "81 minus 11, yeah, this is gonna get a little computationally intensive, but let's see if we can power through it. 11 times 27, I probably should have picked a simpler example, but let's just keep going. Plus 15 times nine, plus four times three is 12, minus 12, so lucky for us, at least those last two terms cancel out. And so this is going to be, the rest from here is arithmetic. Two times 81 is 162. Now let's think about what 27 times 11 is. So let's see, 27 times 10 is going to be 270."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "And so this is going to be, the rest from here is arithmetic. Two times 81 is 162. Now let's think about what 27 times 11 is. So let's see, 27 times 10 is going to be 270. 270 plus another 27 is minus 297. 297, did I do that yet? 270, so 27 times 10 is 270, plus 27, 297, yep, that's right."}, {"video_title": "Polynomial remainder theorem to test factor Algebra II Khan Academy.mp3", "Sentence": "So let's see, 27 times 10 is going to be 270. 270 plus another 27 is minus 297. 297, did I do that yet? 270, so 27 times 10 is 270, plus 27, 297, yep, that's right. And then we have, I'm prone to make Carol's Errors here. Let's see, 90 plus 45 is 135, so plus 135. And let's see, if I were to take 162 and 135, that's going to give me 297 minus 297, minus 200, let me do that green color, minus 297, and we do indeed equal zero."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "The form of your answer should either be just a clean polynomial or some polynomial plus some constant over x plus two where p of x is a polynomial and k is an integer. Fair enough. And if we were doing this on Khan Academy, this is a screenshot from Khan Academy, it, we would have to type this in, but we're just going to do it by hand. And like always, pause this video and try to do it on your own before we work through it together. All right, now let's work through it together. And what we're trying to do is divide x plus two into three x to the third power plus four x squared minus three x plus seven. And so like always, we focus on the highest degree terms first."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "And like always, pause this video and try to do it on your own before we work through it together. All right, now let's work through it together. And what we're trying to do is divide x plus two into three x to the third power plus four x squared minus three x plus seven. And so like always, we focus on the highest degree terms first. X goes into three x to the third power how many times? Well, three x squared times. We'd want to put that in the second degree column, three x squared."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "And so like always, we focus on the highest degree terms first. X goes into three x to the third power how many times? Well, three x squared times. We'd want to put that in the second degree column, three x squared. Three x squared times two is six x squared. Three x squared times x is three x to the third power. There's something very meditative about algebraic long division."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "We'd want to put that in the second degree column, three x squared. Three x squared times two is six x squared. Three x squared times x is three x to the third power. There's something very meditative about algebraic long division. Anyway, we'd want to subtract what we just wrote from what we have up here. So let's subtract. And these characters cancel out."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "There's something very meditative about algebraic long division. Anyway, we'd want to subtract what we just wrote from what we have up here. So let's subtract. And these characters cancel out. And then four x squared minus six x squared is negative two x squared. Bring down that negative three x. And now we would want to say, hey, how many times does x go into negative two x squared?"}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "And these characters cancel out. And then four x squared minus six x squared is negative two x squared. Bring down that negative three x. And now we would want to say, hey, how many times does x go into negative two x squared? Well, it would go negative two x times. Put that in our first degree column. Negative two x times two is negative four x."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "And now we would want to say, hey, how many times does x go into negative two x squared? Well, it would go negative two x times. Put that in our first degree column. Negative two x times two is negative four x. Negative two x times x is negative two x squared. Now we want to subtract what we have here in orange from what we have up here in teal. So we either put a negative around the whole thing or we distribute that negative and that becomes a positive."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "Negative two x times two is negative four x. Negative two x times x is negative two x squared. Now we want to subtract what we have here in orange from what we have up here in teal. So we either put a negative around the whole thing or we distribute that negative and that becomes a positive. That becomes a positive. And so this is equal to the x squared terms cancel out. Negative three x plus four x is just going to be a straight up x."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "So we either put a negative around the whole thing or we distribute that negative and that becomes a positive. That becomes a positive. And so this is equal to the x squared terms cancel out. Negative three x plus four x is just going to be a straight up x. Bring down that seven, x plus seven. How many times does x go into x? Well, one time."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "Negative three x plus four x is just going to be a straight up x. Bring down that seven, x plus seven. How many times does x go into x? Well, one time. Actually, let me use a new color here. So how many times does x go into x? It goes one time."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "Well, one time. Actually, let me use a new color here. So how many times does x go into x? It goes one time. Put that in the constant column. One times two is two. One times x is x."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "It goes one time. Put that in the constant column. One times two is two. One times x is x. We want to subtract these characters. And we're left with seven minus two is five. And so we can rewrite this whole thing as, we deserve, I guess, a little bit of a drum roll."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "One times x is x. We want to subtract these characters. And we're left with seven minus two is five. And so we can rewrite this whole thing as, we deserve, I guess, a little bit of a drum roll. Three x squared minus two x plus one plus the remainder, five, over x plus two. One way to think about it is, hey, I have this remainder. I'd have to keep dividing it by x plus two if I really want to figure out exactly what this is."}, {"video_title": "Dividing polynomials by linear expressions Algebra 2 Khan Academy.mp3", "Sentence": "And so we can rewrite this whole thing as, we deserve, I guess, a little bit of a drum roll. Three x squared minus two x plus one plus the remainder, five, over x plus two. One way to think about it is, hey, I have this remainder. I'd have to keep dividing it by x plus two if I really want to figure out exactly what this is. Now, if I wanted these expressions to be completely identical, I would put a condition on the domain that x cannot be equal to negative two, because if x was equal to negative two, we'd be dividing by zero here. But for the purposes of this exercise, you just have to input, you just have to input this part right over here. You'd have to type it in, which I guess isn't the easiest thing to do in the world, but it's worth doing."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And this is really just a fancy way of seeing whether an expression that involves a polynomial is equal to another expression. So for example, you're familiar with x squared plus two x plus one. We've seen polynomials like this multiple times. This is a quadratic. And you might recognize that this would be equal to x plus one squared. That for any value of x, x squared plus two x plus one is the same thing as adding one to that x and then squaring the whole thing. And we saw this when we first learned how to multiply binomials, and we took the square of binomials."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "This is a quadratic. And you might recognize that this would be equal to x plus one squared. That for any value of x, x squared plus two x plus one is the same thing as adding one to that x and then squaring the whole thing. And we saw this when we first learned how to multiply binomials, and we took the square of binomials. But now we're going to do this with slightly more complicated expressions, things that aren't just simple quadratics, or that might not be as obvious as this. And the way that we're going to prove whether they're true or not is just with a little bit of algebraic manipulation. So for example, if someone walked up to you on the street and said, all right, m to the third minus one, is it equal to m minus one times one plus m plus m squared?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And we saw this when we first learned how to multiply binomials, and we took the square of binomials. But now we're going to do this with slightly more complicated expressions, things that aren't just simple quadratics, or that might not be as obvious as this. And the way that we're going to prove whether they're true or not is just with a little bit of algebraic manipulation. So for example, if someone walked up to you on the street and said, all right, m to the third minus one, is it equal to m minus one times one plus m plus m squared? Pause this video and see what you would tell that person, whether you could prove whether it is or is not a true polynomial identity. Okay, let's do it together. And the way I would tackle this is I would expand out, I would multiply out what we have on the right-hand side."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "So for example, if someone walked up to you on the street and said, all right, m to the third minus one, is it equal to m minus one times one plus m plus m squared? Pause this video and see what you would tell that person, whether you could prove whether it is or is not a true polynomial identity. Okay, let's do it together. And the way I would tackle this is I would expand out, I would multiply out what we have on the right-hand side. So this is going to be equal to, so first I could take this m and then multiply it times every term in this second expression. So m times one is m. M times m is m squared. And then m times m squared is m to the third power."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And the way I would tackle this is I would expand out, I would multiply out what we have on the right-hand side. So this is going to be equal to, so first I could take this m and then multiply it times every term in this second expression. So m times one is m. M times m is m squared. And then m times m squared is m to the third power. And then I would take this negative one and then multiply and then distribute that times every term in that other expression. So negative one times one is negative one. Negative one times m is negative m. And negative one times m squared is negative m squared."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And then m times m squared is m to the third power. And then I would take this negative one and then multiply and then distribute that times every term in that other expression. So negative one times one is negative one. Negative one times m is negative m. And negative one times m squared is negative m squared. And now let's see if we can simplify this. We have an m and a negative m, so those are going to cancel out. We have an m squared and a negative m squared, so those cancel out."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "Negative one times m is negative m. And negative one times m squared is negative m squared. And now let's see if we can simplify this. We have an m and a negative m, so those are going to cancel out. We have an m squared and a negative m squared, so those cancel out. And so we are going to be left with m to the third power minus one. Now clearly, m to the third power minus one is going to be equal to m to the third power minus one for any value of m. These are identical expressions. So this is indeed a polynomial identity."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "We have an m squared and a negative m squared, so those cancel out. And so we are going to be left with m to the third power minus one. Now clearly, m to the third power minus one is going to be equal to m to the third power minus one for any value of m. These are identical expressions. So this is indeed a polynomial identity. Let's do another example. Let's say someone were to walk up to you on the street and said, quick, n plus three squared plus two n. Is that equal to eight n plus 13? Is this a polynomial identity?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "So this is indeed a polynomial identity. Let's do another example. Let's say someone were to walk up to you on the street and said, quick, n plus three squared plus two n. Is that equal to eight n plus 13? Is this a polynomial identity? Pause this video and see if you can figure that out. All right, now we're gonna work on that together. And I would do it the exact same way."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "Is this a polynomial identity? Pause this video and see if you can figure that out. All right, now we're gonna work on that together. And I would do it the exact same way. I would try to simplify with a little bit of algebra. Maybe the easiest thing to do first, and you could do this in multiple ways, is I have these n terms, two n's here, eight n's over here. Well, what if I were to get these two n's out of the left-hand side?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And I would do it the exact same way. I would try to simplify with a little bit of algebra. Maybe the easiest thing to do first, and you could do this in multiple ways, is I have these n terms, two n's here, eight n's over here. Well, what if I were to get these two n's out of the left-hand side? So if I were to just subtract two n from both sides of this equation, I am going to get on the left-hand side n plus three squared. And on the right-hand side, I am going to get six n, eight n minus two n, plus 13. Now, what's n plus three squared?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "Well, what if I were to get these two n's out of the left-hand side? So if I were to just subtract two n from both sides of this equation, I am going to get on the left-hand side n plus three squared. And on the right-hand side, I am going to get six n, eight n minus two n, plus 13. Now, what's n plus three squared? Well, that's going to be n squared plus two times three times n. If what I just did does not seem familiar to you, I encourage you to look at the videos about squaring binomials. But this is going to be plus six n plus three squared, which is nine. And is this going to be equal to six n plus 13?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "Now, what's n plus three squared? Well, that's going to be n squared plus two times three times n. If what I just did does not seem familiar to you, I encourage you to look at the videos about squaring binomials. But this is going to be plus six n plus three squared, which is nine. And is this going to be equal to six n plus 13? Well, already this is starting to look a little bit sketchy. But let's just keep going with the algebra. So let's see, if we subtract six n from both sides, what do you get?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "And is this going to be equal to six n plus 13? Well, already this is starting to look a little bit sketchy. But let's just keep going with the algebra. So let's see, if we subtract six n from both sides, what do you get? Well, on the left-hand side, you're just going to have n squared plus nine. And on the right-hand side, you're going to get 13. Now, are there values of n for which this is not always true?"}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "So let's see, if we subtract six n from both sides, what do you get? Well, on the left-hand side, you're just going to have n squared plus nine. And on the right-hand side, you're going to get 13. Now, are there values of n for which this is not always true? Well, sure, I can find a lot of values of n for which this is not always true. If n is zero, this is not going to be true. If n is one, this is not going to be true."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "Now, are there values of n for which this is not always true? Well, sure, I can find a lot of values of n for which this is not always true. If n is zero, this is not going to be true. If n is one, this is not going to be true. If n is two, this actually would be true. But if n is three, this is not going to be true. If n is four, five, et cetera."}, {"video_title": "Polynomial identities introduction Algebra 2 Khan Academy.mp3", "Sentence": "If n is one, this is not going to be true. If n is two, this actually would be true. But if n is three, this is not going to be true. If n is four, five, et cetera. So for actually most values of n, this is not going to be true. So in order for it to be a polynomial identity, it has to be true for all of the values that are legitimate values that you can evaluate for the variable in question. So this one right over here is not a polynomial, a polynomial identity."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "Rewrite the formula so it solves for Fahrenheit. So just like I mentioned, the C is Celsius, is the Celsius temperature, and the F is the Fahrenheit temperature. So right now it's expressed in terms of, we've solved for C in terms of F, now we want to solve for F in terms of C. So let's see how we're going to do that. So I'll just rewrite it. C is equal to 5 ninths times F minus 32. And we want to solve for F. So the first useful thing to do might be to get rid of this 5 ninths from both sides, or at least from the right side of the equation. So we just have an F minus 32."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "So I'll just rewrite it. C is equal to 5 ninths times F minus 32. And we want to solve for F. So the first useful thing to do might be to get rid of this 5 ninths from both sides, or at least from the right side of the equation. So we just have an F minus 32. And the easiest way to do that is to multiply by the inverse of 5 ninths. So if we multiply both sides of this equation by 9 fifths, and I could put it here, but it'll squeeze it a little tighter. Let me try it."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "So we just have an F minus 32. And the easiest way to do that is to multiply by the inverse of 5 ninths. So if we multiply both sides of this equation by 9 fifths, and I could put it here, but it'll squeeze it a little tighter. Let me try it. 9 fifths in there. If I do it to the right hand side of the equation, I also have to do it to the left hand side of the equation. And what do I get?"}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "Let me try it. 9 fifths in there. If I do it to the right hand side of the equation, I also have to do it to the left hand side of the equation. And what do I get? The left hand side becomes 9 fifths Celsius, or the Celsius temperature, 9 fifths times the Celsius temperature, is equal to, the whole reason we wanted to multiply the right side by 9 fifths is these 9's cancel out, this 5 cancel out, and this just becomes a 1. So 9 fifths times the Celsius temperature is going to be equal to this expression. The Fahrenheit temperature minus 32."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "And what do I get? The left hand side becomes 9 fifths Celsius, or the Celsius temperature, 9 fifths times the Celsius temperature, is equal to, the whole reason we wanted to multiply the right side by 9 fifths is these 9's cancel out, this 5 cancel out, and this just becomes a 1. So 9 fifths times the Celsius temperature is going to be equal to this expression. The Fahrenheit temperature minus 32. We don't need the parentheses anymore because this is the only expression on the right hand side. Now, we want to solve for the Fahrenheit temperature. So let's see if we can get rid of this 32 from the right hand side."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "The Fahrenheit temperature minus 32. We don't need the parentheses anymore because this is the only expression on the right hand side. Now, we want to solve for the Fahrenheit temperature. So let's see if we can get rid of this 32 from the right hand side. Easiest way I know how to do that is to add 32. Add 32 to both sides of this equation. Add 32 to both sides."}, {"video_title": "Solving for F in terms of C Linear equations Algebra I Khan Academy.mp3", "Sentence": "So let's see if we can get rid of this 32 from the right hand side. Easiest way I know how to do that is to add 32. Add 32 to both sides of this equation. Add 32 to both sides. The left hand side now becomes 9 fifths times the Celsius temperature plus 32. 9 fifths times the Celsius temperature plus 32, and that's going to be equal to, this negative 32 and positive 32 cancel out, that was the whole point behind adding 32, so it's just going to be equal to F. And we're done. We have rewritten this formula up here so it solves for Fahrenheit."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "The three points plotted below are on the graph of y is equal to b to the x power. Based only on these three points, plot the three corresponding points that must be on the graph of y is equal to log base b of x by clicking on the graph. So I've actually copied and pasted this problem on my little scratch pad so I can mark it up a little bit. So what is this first function? This first function is telling us x and this is y is equal to b to the x power. So when x is equal to zero, y is equal to one. When x is equal to zero, y is equal to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So what is this first function? This first function is telling us x and this is y is equal to b to the x power. So when x is equal to zero, y is equal to one. When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the first power is equal to four. y is equal to four."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "When x is equal to zero, y is equal to one. That's this point right over here. When x is equal to one, b to the first power is equal to four. y is equal to four. So another way of thinking of this, y or four is equal to b to the first power. And actually we can deduce then that b must be four. So that's this point right over there."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "y is equal to four. So another way of thinking of this, y or four is equal to b to the first power. And actually we can deduce then that b must be four. So that's this point right over there. And then this point is telling us that b to the second power is equal to sixteen. So when x is equal to two, b to the second power, y is equal to sixteen. Now we want to plot the three corresponding points on this function."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's this point right over there. And then this point is telling us that b to the second power is equal to sixteen. So when x is equal to two, b to the second power, y is equal to sixteen. Now we want to plot the three corresponding points on this function. So let me draw another table here. So now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. And so what are the possibilities here?"}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Now we want to plot the three corresponding points on this function. So let me draw another table here. So now it's essentially the inverse function where this is going to be x and we want to calculate y is equal to log base b of x. And so what are the possibilities here? So what I want to do is think, let's take these values because these are the ones, these are essentially inverse functions. Log is the inverse of exponents. So if we take the points one, four, and sixteen."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "And so what are the possibilities here? So what I want to do is think, let's take these values because these are the ones, these are essentially inverse functions. Log is the inverse of exponents. So if we take the points one, four, and sixteen. So what is y going to be here? y is going to be log base b of one. So this is saying what power I need to raise b to to get to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if we take the points one, four, and sixteen. So what is y going to be here? y is going to be log base b of one. So this is saying what power I need to raise b to to get to one. Well, if we assume that b is non-zero, and that's a reasonable assumption because b to different powers are non-zero, this is going to be zero for any non-zero b. So this is going to be zero right over here. So we have the point one, zero."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is saying what power I need to raise b to to get to one. Well, if we assume that b is non-zero, and that's a reasonable assumption because b to different powers are non-zero, this is going to be zero for any non-zero b. So this is going to be zero right over here. So we have the point one, zero. So it's that point over there. And notice, this point corresponds to this point. We've essentially swapped the x's and y's."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we have the point one, zero. So it's that point over there. And notice, this point corresponds to this point. We've essentially swapped the x's and y's. And in general, when you're taking an inverse, you're going to reflect over the line y is equal to x. And this is clearly a reflection over that line. So let's look over here."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "We've essentially swapped the x's and y's. And in general, when you're taking an inverse, you're going to reflect over the line y is equal to x. And this is clearly a reflection over that line. So let's look over here. When x is equal to four, what is log base b of four? What is the power I need to raise b to to get to four? Well, we see right over here, b to the first power is equal to four."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's look over here. When x is equal to four, what is log base b of four? What is the power I need to raise b to to get to four? Well, we see right over here, b to the first power is equal to four. We already figured that out. When I take b to the first power is equal to four. So this right over here is going to be equal to one."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, we see right over here, b to the first power is equal to four. We already figured that out. When I take b to the first power is equal to four. So this right over here is going to be equal to one. So when x is equal to four, y is equal to one. And notice, once again, it is a reflection over the line y is equal to x. So when x is equal to 16, then y is equal to log base b of 16."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this right over here is going to be equal to one. So when x is equal to four, y is equal to one. And notice, once again, it is a reflection over the line y is equal to x. So when x is equal to 16, then y is equal to log base b of 16. The power I need to raise b to to get to 16. Well, we already know. If we take b squared, we get to 16."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "So when x is equal to 16, then y is equal to log base b of 16. The power I need to raise b to to get to 16. Well, we already know. If we take b squared, we get to 16. So this is equal to two. So when x is equal to 16, y is equal to two. Notice, we've essentially just swapped the x and y values for each of these points."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "If we take b squared, we get to 16. So this is equal to two. So when x is equal to 16, y is equal to two. Notice, we've essentially just swapped the x and y values for each of these points. This is y, this is a reflection over the line y is equal to x. Now let's actually do that on the actual interface. And the whole reason is to give you this appreciation that these are inverse functions of each other."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "Notice, we've essentially just swapped the x and y values for each of these points. This is y, this is a reflection over the line y is equal to x. Now let's actually do that on the actual interface. And the whole reason is to give you this appreciation that these are inverse functions of each other. So let's plot the points. So that point corresponding to that point. So x zero, y one corresponds to x one, y zero."}, {"video_title": "Plotting points of logarithmic function Logarithms Algebra II Khan Academy.mp3", "Sentence": "And the whole reason is to give you this appreciation that these are inverse functions of each other. So let's plot the points. So that point corresponding to that point. So x zero, y one corresponds to x one, y zero. Here x is one, y is four. That corresponds to x four, y one. Here x is two, y is 16."}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "We are told that Jade is making pancakes using flour, eggs, and milk. This table gives the cost per kilogram of each ingredient and the amount in kilograms that Jade uses, all right? The total amount Jade spends on ingredients is $6. Write an equation that relates A, B, C, and D. Pause this video and see if you can have a go at this. All right, now let's go through this together. So I like to do these in real time so that you can see my thought process. So I'm here with you right now."}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "Write an equation that relates A, B, C, and D. Pause this video and see if you can have a go at this. All right, now let's go through this together. So I like to do these in real time so that you can see my thought process. So I'm here with you right now. So let's see, the total Jade spends, and this is going to be on flour, eggs, and milk, is going to be $6. So one way to think about it is, and I'll do this in different colors, the flour dollars plus the eggs, eggs dollars, or the amount that Jade's gonna spend on eggs, plus the amount that Jade spends on milk, I'll call that the milk dollars, is going to be equal to the total amount, is going to be equal to $6. And so what's the total amount that Jade is going to spend on flour?"}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "So I'm here with you right now. So let's see, the total Jade spends, and this is going to be on flour, eggs, and milk, is going to be $6. So one way to think about it is, and I'll do this in different colors, the flour dollars plus the eggs, eggs dollars, or the amount that Jade's gonna spend on eggs, plus the amount that Jade spends on milk, I'll call that the milk dollars, is going to be equal to the total amount, is going to be equal to $6. And so what's the total amount that Jade is going to spend on flour? Well, we can just look right over here, $0.9 per kilogram times eight kilograms. So Jade's going to spend $0.9 A. I'm just going to multiply these two things to figure out how much Jade spends on flour. And so this is going to be equal to, and I'll lose the dollar symbol just so we can focus on the numbers and the variables, 0.9 A, 0.9 A."}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "And so what's the total amount that Jade is going to spend on flour? Well, we can just look right over here, $0.9 per kilogram times eight kilograms. So Jade's going to spend $0.9 A. I'm just going to multiply these two things to figure out how much Jade spends on flour. And so this is going to be equal to, and I'll lose the dollar symbol just so we can focus on the numbers and the variables, 0.9 A, 0.9 A. And this is, of course, is going to be in dollars, which is important. I'm gonna add dollars plus dollars plus dollars to get dollars. Now, what about eggs?"}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "And so this is going to be equal to, and I'll lose the dollar symbol just so we can focus on the numbers and the variables, 0.9 A, 0.9 A. And this is, of course, is going to be in dollars, which is important. I'm gonna add dollars plus dollars plus dollars to get dollars. Now, what about eggs? Well, the same notion, 0.2 kilograms times B dollars per kilogram. If I take the product of these two, I'm gonna get 0.2 B dollars, or I could just think of it as 0.2 B if I don't write the dollar symbol. And then last but not least, on milk, let me look at the product of these two things, of this and of this."}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "Now, what about eggs? Well, the same notion, 0.2 kilograms times B dollars per kilogram. If I take the product of these two, I'm gonna get 0.2 B dollars, or I could just think of it as 0.2 B if I don't write the dollar symbol. And then last but not least, on milk, let me look at the product of these two things, of this and of this. So D kilograms, it's C dollars per kilogram. That's just going to be CD dollars. So plus CD."}, {"video_title": "Modeling with multiple variables Pancakes Modeling Algebra Khan Academy.mp3", "Sentence": "And then last but not least, on milk, let me look at the product of these two things, of this and of this. So D kilograms, it's C dollars per kilogram. That's just going to be CD dollars. So plus CD. And all of that is going to be equal to $6. And we're done. We wrote an equation that relates A, B, C, and D."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So $8 each day on using the oven and $1.50 on ingredients for each pizza. One day the price for the ingredients increased from $1.50 to $2 per pizza. Dominique made some calculations and found that she should bake eight pizzas more each day. So the expenses for a single pizza would remain the same. And I assume they're saying the total expenses for a single pizza because clearly the ingredients cost is not the same. We're talking about the total ingredients. So if we were to spread the cost of the oven across all of the pizzas, write an equation to find, or the total cost for the oven per day to spread that across all the pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So the expenses for a single pizza would remain the same. And I assume they're saying the total expenses for a single pizza because clearly the ingredients cost is not the same. We're talking about the total ingredients. So if we were to spread the cost of the oven across all of the pizzas, write an equation to find, or the total cost for the oven per day to spread that across all the pizzas. Write an equation to find out how many pizzas Dominique baked each day before the change in price. Use P to represent the number of pizzas. So let's just think about her total cost per pizza before and then her total cost per pizza after if she bakes eight more pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if we were to spread the cost of the oven across all of the pizzas, write an equation to find, or the total cost for the oven per day to spread that across all the pizzas. Write an equation to find out how many pizzas Dominique baked each day before the change in price. Use P to represent the number of pizzas. So let's just think about her total cost per pizza before and then her total cost per pizza after if she bakes eight more pizzas. So before, we're gonna use P to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients for each pizza. So 1.5 or $1.50 times the number of pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So let's just think about her total cost per pizza before and then her total cost per pizza after if she bakes eight more pizzas. So before, we're gonna use P to say that's the number of pizzas she baked per day before the change in price. So before the change in price, on a given day, she would spend $8 on the oven and then $1.50 on ingredients for each pizza. So 1.5 or $1.50 times the number of pizzas. This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So 1.5 or $1.50 times the number of pizzas. This would be her total cost on all the pizzas in that day. It's the oven cost plus it's the ingredients cost. So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if you wanted this on a per pizza basis, you would just divide by the number of pizzas. Now let's think about what happens after the change in price. After the change in price, her cost per day for the oven is still $8. But now she has to spend $2 per pizza on ingredients. So $2 per pizza. And instead of saying that she's baking P pizzas, let's say that she's now baking eight more pizzas each day. So it's going to be P plus eight."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "But now she has to spend $2 per pizza on ingredients. So $2 per pizza. And instead of saying that she's baking P pizzas, let's say that she's now baking eight more pizzas each day. So it's going to be P plus eight. And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making P plus eight pizzas, you would divide by P plus eight. And the problem tells us that these two things are equivalent."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So it's going to be P plus eight. And so this is going to be her total cost for all of the pizzas she's now baking. And so if you want it on a per pizza basis, well, she's now making P plus eight pizzas, you would divide by P plus eight. And the problem tells us that these two things are equivalent. Here you had a higher cost of ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what P has to be. P has to be some number, some number of pizzas, so that these two expressions are equal."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And the problem tells us that these two things are equivalent. Here you had a higher cost of ingredients per pizzas, but since you are now baking more pizzas, you're spreading the oven cost amongst more and more pizzas. So let's think about what P has to be. P has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made P, is going to be the same as her total cost per pizza when she's making P plus eight pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "P has to be some number, some number of pizzas, so that these two expressions are equal. Her total cost per pizza before, when she only made P, is going to be the same as her total cost per pizza when she's making P plus eight pizzas. So these two things need to be equal. So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked each day before the change in price, and we used P to represent the number of pizzas. But now for fun, let's actually just solve for P. So let's just simplify things a little bit. And actually, so this part right over here, actually, let's just cross multiply this on both sides, or another way of thinking is, multiply both sides times P plus eight, and multiply both sides times P. So if we multiply by P plus eight, and we multiply by P, we multiply by P plus eight, and we multiply by P, that cancels with that, that cancels with that."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we did that first part, or we did what they asked us. We wrote an equation to find out how many pizzas Dominique baked each day before the change in price, and we used P to represent the number of pizzas. But now for fun, let's actually just solve for P. So let's just simplify things a little bit. And actually, so this part right over here, actually, let's just cross multiply this on both sides, or another way of thinking is, multiply both sides times P plus eight, and multiply both sides times P. So if we multiply by P plus eight, and we multiply by P, we multiply by P plus eight, and we multiply by P, that cancels with that, that cancels with that. On the left-hand side, so let's see, we have to just do the distributive property twice right over here. What is P times eight plus 1.5P? Well, that's going to be eight P, I'm just multiplying the P times this stuff first, plus 1.5P squared, and now let's multiply, now let's multiply the eight times both of these terms."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And actually, so this part right over here, actually, let's just cross multiply this on both sides, or another way of thinking is, multiply both sides times P plus eight, and multiply both sides times P. So if we multiply by P plus eight, and we multiply by P, we multiply by P plus eight, and we multiply by P, that cancels with that, that cancels with that. On the left-hand side, so let's see, we have to just do the distributive property twice right over here. What is P times eight plus 1.5P? Well, that's going to be eight P, I'm just multiplying the P times this stuff first, plus 1.5P squared, and now let's multiply, now let's multiply the eight times both of these terms. So plus 64, plus eight times 1.5, that is 12, plus 12P, and that's going to be equal to, let's see, let's multiply P times all of this business. So that's going to be equal to eight P, eight times P is eight P, and let's see, I could distribute these terms and then multiply by P. So two times P is two P, times P is two P squared, plus two P squared, and then two times eight is 16, times P is 16P. So now we have, well, we essentially end up with a quadratic equation, but let's simplify it a little bit so that we can either factor it or apply the quadratic formula."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Well, that's going to be eight P, I'm just multiplying the P times this stuff first, plus 1.5P squared, and now let's multiply, now let's multiply the eight times both of these terms. So plus 64, plus eight times 1.5, that is 12, plus 12P, and that's going to be equal to, let's see, let's multiply P times all of this business. So that's going to be equal to eight P, eight times P is eight P, and let's see, I could distribute these terms and then multiply by P. So two times P is two P, times P is two P squared, plus two P squared, and then two times eight is 16, times P is 16P. So now we have, well, we essentially end up with a quadratic equation, but let's simplify it a little bit so that we can either factor it or apply the quadratic formula. So let's see, let's throw, let's subtract two P squared from, actually, let's subtract 1.5P squared from both sides. So subtract 1.5P squared, subtract 1.5P squared. Actually, let me just put everything on the left-hand side just because that's, might be a little bit more intuitive."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So now we have, well, we essentially end up with a quadratic equation, but let's simplify it a little bit so that we can either factor it or apply the quadratic formula. So let's see, let's throw, let's subtract two P squared from, actually, let's subtract 1.5P squared from both sides. So subtract 1.5P squared, subtract 1.5P squared. Actually, let me just put everything on the left-hand side just because that's, might be a little bit more intuitive. So let's subtract two P squared from both sides, subtract two P squared from both sides. Let's subtract 16P from both sides. We have an eight P and a 12P, and then we're gonna subtract a 16P from both sides."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Actually, let me just put everything on the left-hand side just because that's, might be a little bit more intuitive. So let's subtract two P squared from both sides, subtract two P squared from both sides. Let's subtract 16P from both sides. We have an eight P and a 12P, and then we're gonna subtract a 16P from both sides. And then, oh, actually, let's subtract an eight P as well from both sides. We have a 16P and an eight P. So that actually works out quite well. So now we've subtracted eight P from both sides, 16P from both sides."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "We have an eight P and a 12P, and then we're gonna subtract a 16P from both sides. And then, oh, actually, let's subtract an eight P as well from both sides. We have a 16P and an eight P. So that actually works out quite well. So now we've subtracted eight P from both sides, 16P from both sides. We've essentially subtracted all of this stuff from both sides, and we are left with, let's see, 1.5, I'll do it in degree order. 1.5P squared minus two P squared is negative, negative 0.5P squared. Now let's see, these cancel out."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So now we've subtracted eight P from both sides, 16P from both sides. We've essentially subtracted all of this stuff from both sides, and we are left with, let's see, 1.5, I'll do it in degree order. 1.5P squared minus two P squared is negative, negative 0.5P squared. Now let's see, these cancel out. 12P minus 16P is minus four P. And then we have plus 64, plus 64, and then that is going to be equal to zero. That is going to be equal to zero. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative two."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now let's see, these cancel out. 12P minus 16P is minus four P. And then we have plus 64, plus 64, and then that is going to be equal to zero. That is going to be equal to zero. And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative two. By negative two. I want the coefficient over here to be one. So then we get P squared plus eight P, P squared plus eight P is going to be equal to, is going to be equal to, let's see, negative, oh, negative times negative two."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "And just to simplify this a little bit, or just to make this a little bit cleaner, let's multiply both sides of this equation by negative two. By negative two. I want the coefficient over here to be one. So then we get P squared plus eight P, P squared plus eight P is going to be equal to, is going to be equal to, let's see, negative, oh, negative times negative two. So minus 128 is going to be equal to zero. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128, and if we were to, and if we were to add them together, we get positive eight."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So then we get P squared plus eight P, P squared plus eight P is going to be equal to, is going to be equal to, let's see, negative, oh, negative times negative two. So minus 128 is going to be equal to zero. So let's see if we can factor this. Can we think of two numbers where if we take their product, we get negative 128, and if we were to, and if we were to add them together, we get positive eight. So they're gonna have different signs right over here. So let's see, if we say 12 times, let's see, 12 times, well, let's see, what numbers could this be? So if we were to think about 128 is the same thing as, let's see, 16, let's see, 16 goes into 128."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Can we think of two numbers where if we take their product, we get negative 128, and if we were to, and if we were to add them together, we get positive eight. So they're gonna have different signs right over here. So let's see, if we say 12 times, let's see, 12 times, well, let's see, what numbers could this be? So if we were to think about 128 is the same thing as, let's see, 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128. Does it go eight times?"}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So if we were to think about 128 is the same thing as, let's see, 16, let's see, 16 goes into 128. Let me work through this. 16 goes into 128. Does it go eight times? Eight times six is, eight times six is 48. Eight times 10 is 80 plus 40 is 128. Yep, it goes eight times."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Does it go eight times? Eight times six is, eight times six is 48. Eight times 10 is 80 plus 40 is 128. Yep, it goes eight times. So 16 and eight seem to work. So if you have positive 16 and negative eight, their product would be negative 128. So we can factor this out as P plus 16 times P minus eight is equal to zero."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Yep, it goes eight times. So 16 and eight seem to work. So if you have positive 16 and negative eight, their product would be negative 128. So we can factor this out as P plus 16 times P minus eight is equal to zero. Now this is going to be equal to zero if one of these two, at least one of these is going to be equal to zero. So we have two solutions. Either P plus 16 is going to be equal to zero or P minus eight is equal to zero."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "So we can factor this out as P plus 16 times P minus eight is equal to zero. Now this is going to be equal to zero if one of these two, at least one of these is going to be equal to zero. So we have two solutions. Either P plus 16 is going to be equal to zero or P minus eight is equal to zero. This one right over here, subtract 16 from both sides, you get P is equal to negative 16. Here you get P is equal to eight if you add eight to both sides. Now we're talking about a number of pizzas made."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Either P plus 16 is going to be equal to zero or P minus eight is equal to zero. This one right over here, subtract 16 from both sides, you get P is equal to negative 16. Here you get P is equal to eight if you add eight to both sides. Now we're talking about a number of pizzas made. So we're not, this one doesn't apply. This would be like her, Dominique eating 16 pizzas or somehow destroying 16 pizzas a day. We're not interested in that solution."}, {"video_title": "Making more pizzas to spread cost per pizza Functions and their graphs Algebra II Khan Academy.mp3", "Sentence": "Now we're talking about a number of pizzas made. So we're not, this one doesn't apply. This would be like her, Dominique eating 16 pizzas or somehow destroying 16 pizzas a day. We're not interested in that solution. So if we want the solution to the original question, the number of pizzas she made before the increase in price, she made eight pizzas, she made eight pizzas per day. So P right over here needs to be equal to eight. So before the change in price, she made eight pizzas a day."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "So let's say that, I guess again, someone walks up to you in the street and says, what is x squared plus one divided by x plus two? So pause this video and have a go at that, and I'll give you a little bit of a warning. This one's a little bit more involved than you might expect. All right, so there is two ways to approach this. Either we can try to re-express the numerator where it involves an x plus two somehow, or we could try to do algebraic long division. So let me do it the first way. So x squared plus one, it's not obvious that you can factor it out, but can you write something that has x plus two as a factor and interestingly enough, has no first degree terms, because we don't want some first degree, weird first degree terms sitting up there."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "All right, so there is two ways to approach this. Either we can try to re-express the numerator where it involves an x plus two somehow, or we could try to do algebraic long division. So let me do it the first way. So x squared plus one, it's not obvious that you can factor it out, but can you write something that has x plus two as a factor and interestingly enough, has no first degree terms, because we don't want some first degree, weird first degree terms sitting up there. And the best thing that I could think of is constructing a difference of squares using x plus two. So we know that x plus two times x minus two is equal to x squared minus four. So what if we were to write x squared minus four up here, and then we would just have to add five to get to plus one?"}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "So x squared plus one, it's not obvious that you can factor it out, but can you write something that has x plus two as a factor and interestingly enough, has no first degree terms, because we don't want some first degree, weird first degree terms sitting up there. And the best thing that I could think of is constructing a difference of squares using x plus two. So we know that x plus two times x minus two is equal to x squared minus four. So what if we were to write x squared minus four up here, and then we would just have to add five to get to plus one? So what if we were to write x squared minus four, and then we write plus five? This expression and that expression up there, those are completely equivalent. But why did I do that?"}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "So what if we were to write x squared minus four up here, and then we would just have to add five to get to plus one? So what if we were to write x squared minus four, and then we write plus five? This expression and that expression up there, those are completely equivalent. But why did I do that? Well, now I can write x squared minus four as x plus two times x minus two. And so then I could rewrite this entire expression as x plus two times x minus two, all of that over x plus two, plus five, plus five, over x plus two. And now as long as x does not equal negative two, then we could divide the numerator and the denominator by x plus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "But why did I do that? Well, now I can write x squared minus four as x plus two times x minus two. And so then I could rewrite this entire expression as x plus two times x minus two, all of that over x plus two, plus five, plus five, over x plus two. And now as long as x does not equal negative two, then we could divide the numerator and the denominator by x plus two. And then we would be left with x minus two plus five over x plus two. And I'll put that little constraint if I wanna say that this expression is the same as that first expression for x not equaling negative two. And so here, we'd say, hey, x squared plus one divided by x plus two is x minus two."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "And now as long as x does not equal negative two, then we could divide the numerator and the denominator by x plus two. And then we would be left with x minus two plus five over x plus two. And I'll put that little constraint if I wanna say that this expression is the same as that first expression for x not equaling negative two. And so here, we'd say, hey, x squared plus one divided by x plus two is x minus two. And then we have a remainder of five, remainder of five. Now let's do the same question, or try to rewrite this using algebraic long division. We'll see that this is actually a little bit more straightforward."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "And so here, we'd say, hey, x squared plus one divided by x plus two is x minus two. And then we have a remainder of five, remainder of five. Now let's do the same question, or try to rewrite this using algebraic long division. We'll see that this is actually a little bit more straightforward. So we are going to divide x plus two into x squared plus one. Now when I write things out, I like to be very careful with my, I guess you could say, my different places for the different degrees. So x squared plus one has no first degree term."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "We'll see that this is actually a little bit more straightforward. So we are going to divide x plus two into x squared plus one. Now when I write things out, I like to be very careful with my, I guess you could say, my different places for the different degrees. So x squared plus one has no first degree term. So I'm gonna write the one out here. So second degree, no first degree term. And then we have a one, which you could view as our zero degree term or our constant term."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "So x squared plus one has no first degree term. So I'm gonna write the one out here. So second degree, no first degree term. And then we have a one, which you could view as our zero degree term or our constant term. And so we do the same drill. How many times does x go into x squared? We'll just look at the highest degree terms here."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "And then we have a one, which you could view as our zero degree term or our constant term. And so we do the same drill. How many times does x go into x squared? We'll just look at the highest degree terms here. X goes into x squared x times. That's first degree. So I put in the first degree column."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "We'll just look at the highest degree terms here. X goes into x squared x times. That's first degree. So I put in the first degree column. X times two is two x. X times x is x squared. And now we wanna subtract. And so what is this going to be equal to?"}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "So I put in the first degree column. X times two is two x. X times x is x squared. And now we wanna subtract. And so what is this going to be equal to? We know the x squareds cancel out. And then I'm gonna be subtracting negative two x from, you could view this as plus zero x up here, plus one. And so you're left with negative two x."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "And so what is this going to be equal to? We know the x squareds cancel out. And then I'm gonna be subtracting negative two x from, you could view this as plus zero x up here, plus one. And so you're left with negative two x. And then we bring down that one, plus one. X goes into negative two x, negative two times. Put that in the constant column."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "And so you're left with negative two x. And then we bring down that one, plus one. X goes into negative two x, negative two times. Put that in the constant column. Negative two times two is negative four. And then negative two times x is negative two x. Now we have to be very careful here because we wanna subtract the negative two x minus four from the negative two x plus one."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "Put that in the constant column. Negative two times two is negative four. And then negative two times x is negative two x. Now we have to be very careful here because we wanna subtract the negative two x minus four from the negative two x plus one. We could view it as this. Or we could just distribute the negative sign. And then this will be positive two x plus four."}, {"video_title": "Dividing quadratics by linear expressions with remainders missing x-term Algebra 2 Khan Academy.mp3", "Sentence": "Now we have to be very careful here because we wanna subtract the negative two x minus four from the negative two x plus one. We could view it as this. Or we could just distribute the negative sign. And then this will be positive two x plus four. And then the two x's, the two x and the negative two x cancels out. One plus four is five. And there's no obvious way of dividing x plus two into five so we would call that the remainder, exactly what we had before."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's do some work on logarithm properties. So let's just review real quick what a logarithm even is. So if I write, let's say I write log base x of a is equal to, I don't know, make up a letter, n. What does this mean? Well, this just means that x to the n equals a. I think we already know that. We've learned that in the logarithm video. And so it's very important to realize that when you evaluate a logarithm expression like log base x of a, the answer, or when you evaluate, what you get is an exponent, right? This n is really just an exponent."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, this just means that x to the n equals a. I think we already know that. We've learned that in the logarithm video. And so it's very important to realize that when you evaluate a logarithm expression like log base x of a, the answer, or when you evaluate, what you get is an exponent, right? This n is really just an exponent. This is equal to this thing, right? You could have written it just like this. You could have, because this n is equal to this, you could just write x, it's going to get a little messy, to the log base x of a is equal to a."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "This n is really just an exponent. This is equal to this thing, right? You could have written it just like this. You could have, because this n is equal to this, you could just write x, it's going to get a little messy, to the log base x of a is equal to a. All I did is I took this n and I replaced it with this term. And I wanted to write it this way because I want you to really get an intuitive understanding of the notion that a logarithm, when you evaluate it, really is an exponent. And we're going to take that notion and that's where really all of the logarithm properties come from."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "You could have, because this n is equal to this, you could just write x, it's going to get a little messy, to the log base x of a is equal to a. All I did is I took this n and I replaced it with this term. And I wanted to write it this way because I want you to really get an intuitive understanding of the notion that a logarithm, when you evaluate it, really is an exponent. And we're going to take that notion and that's where really all of the logarithm properties come from. So let me just do, what I actually want to do is I want to stumble upon the logarithm properties by playing around. And then later on I'll summarize them and clean it all up. But I wanted to show you maybe how people originally discovered this stuff."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we're going to take that notion and that's where really all of the logarithm properties come from. So let me just do, what I actually want to do is I want to stumble upon the logarithm properties by playing around. And then later on I'll summarize them and clean it all up. But I wanted to show you maybe how people originally discovered this stuff. So let's say that, I don't know, let's say that x, let me switch colors, I think that keeps things interesting. So let's say that x to the, I don't know, l is equal to a. Well if we write that as a logarithm, that same relationship as a logarithm, we could write that log base x of a is equal to l, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "But I wanted to show you maybe how people originally discovered this stuff. So let's say that, I don't know, let's say that x, let me switch colors, I think that keeps things interesting. So let's say that x to the, I don't know, l is equal to a. Well if we write that as a logarithm, that same relationship as a logarithm, we could write that log base x of a is equal to l, right? I just rewrote what I wrote on the top line. Now let me switch colors. If I were to say that x to the m is equal to b, same thing, I just switch letters."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well if we write that as a logarithm, that same relationship as a logarithm, we could write that log base x of a is equal to l, right? I just rewrote what I wrote on the top line. Now let me switch colors. If I were to say that x to the m is equal to b, same thing, I just switch letters. But that just means that log base x of b is equal to m. I just did the same thing that I did in this line, I just switched letters. So let's just keep going and see what happens. So let's say, let me get another color, I might have to run out of color, I have infinite color, I will never run out."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "If I were to say that x to the m is equal to b, same thing, I just switch letters. But that just means that log base x of b is equal to m. I just did the same thing that I did in this line, I just switched letters. So let's just keep going and see what happens. So let's say, let me get another color, I might have to run out of color, I have infinite color, I will never run out. So let's say I have x to the n, you're probably saying, Sal, where are you going with this? But you'll see, it's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's say, let me get another color, I might have to run out of color, I have infinite color, I will never run out. So let's say I have x to the n, you're probably saying, Sal, where are you going with this? But you'll see, it's pretty neat. x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Let's start with this right here."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "x to the n is equal to a times b. x to the n is equal to a times b. And that's just like saying that log base x is equal to a times b. So what can we do with all of this? Let's start with this right here. x to the n is equal to a times b. So how can we rewrite this? Well, a is this, and b is this, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's start with this right here. x to the n is equal to a times b. So how can we rewrite this? Well, a is this, and b is this, right? So let's rewrite that. So we know that x to the n is equal to a, a is this, x to the l, and what's b? Times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, a is this, and b is this, right? So let's rewrite that. So we know that x to the n is equal to a, a is this, x to the l, and what's b? Times b. Well, b is x to the m, right? Not doing anything fancy right now. Well, what's x to the l times x to the m?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Times b. Well, b is x to the m, right? Not doing anything fancy right now. Well, what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, what's x to the l times x to the m? Well, we know from the exponents, when you multiply two expressions that have the same base and different exponents, you just add the exponents. So this is equal to, let me take a neutral color. I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors because I think that's useful. L plus m. It's kind of onerous to keep switching colors, but you get what I'm saying."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "I don't know if I said that verbally correct, but you get the point. When you have the same base and you're multiplying, you can just add the exponents. That equals x to the, I want to keep switching colors because I think that's useful. L plus m. It's kind of onerous to keep switching colors, but you get what I'm saying. So x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus m. So what do we know?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "L plus m. It's kind of onerous to keep switching colors, but you get what I'm saying. So x to the n is equal to x to the l plus m. Let me put the x here. Oh, I wanted that to be green. x to the l plus m. So what do we know? We know x to the n is equal to x to the l plus m, right? Well, we have the same base. These exponents must equal each other, right?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "x to the l plus m. So what do we know? We know x to the n is equal to x to the l plus m, right? Well, we have the same base. These exponents must equal each other, right? So we know that n is equal to l plus m. Well, what does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "These exponents must equal each other, right? So we know that n is equal to l plus m. Well, what does that do for us? I've kind of just been playing around with logarithms. Am I getting anywhere? I think you'll see that I am. Well, what's another way of writing n? So we said x to the n is equal to a times b. Oh, I actually skipped a step here."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "Am I getting anywhere? I think you'll see that I am. Well, what's another way of writing n? So we said x to the n is equal to a times b. Oh, I actually skipped a step here. So that means, so going back here, x to the n is equal to a times b. That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we said x to the n is equal to a times b. Oh, I actually skipped a step here. So that means, so going back here, x to the n is equal to a times b. That means that log base x of a times b is equal to n. You knew that. I hope you don't realize I'm not backtracking or anything. I just forgot to write this down when I first did it. But anyway, so what's n? What's another way of writing n?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "I hope you don't realize I'm not backtracking or anything. I just forgot to write this down when I first did it. But anyway, so what's n? What's another way of writing n? Well, another way of writing n is right here. Log base x of a times b. So now we know that if we just substitute n for that, we get log base x of a times b."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "What's another way of writing n? Well, another way of writing n is right here. Log base x of a times b. So now we know that if we just substitute n for that, we get log base x of a times b. And what does that equal? Well, that equals l. Another way to write l is right up here. It equals log base x of a plus m. And what's m?"}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "So now we know that if we just substitute n for that, we get log base x of a times b. And what does that equal? Well, that equals l. Another way to write l is right up here. It equals log base x of a plus m. And what's m? m is right here. It's log base x of b. And there we have our first logarithm property."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "It equals log base x of a plus m. And what's m? m is right here. It's log base x of b. And there we have our first logarithm property. The log base x of a times b, well, that just equals the log of base x of a plus the log base x of b. And this is, hopefully, this proves that to you. And if you want the intuition of why this works out, it falls from the fact that logarithms are nothing but exponents."}, {"video_title": "Proof log a + log b = log ab Logarithms Algebra II Khan Academy.mp3", "Sentence": "And there we have our first logarithm property. The log base x of a times b, well, that just equals the log of base x of a plus the log base x of b. And this is, hopefully, this proves that to you. And if you want the intuition of why this works out, it falls from the fact that logarithms are nothing but exponents. So with that, I'll leave you in this video. In the next video, I will prove another logarithm property. I'll see you soon."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and try to figure that out before we work on that together. All right, now let's work on it together. And just as a reminder of what a function and an inverse even does, if this is the domain of a function, and that's the set of all values that you could input into the function for x and get a valid output, and so let's say you have an x here, it's a member of the domain, and if I were to apply the function to it, if I were to input that x into that function, then the function is going to output a value in the range of the function, and we call that value f of x. Now an inverse, that goes the other way. If you were to input the f of x value into the function, that's going to take us back to x. So that's exactly what f inverse does. Now how do we actually figure out the inverse of a function, especially a function that's defined with a rational expression like this?"}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Now an inverse, that goes the other way. If you were to input the f of x value into the function, that's going to take us back to x. So that's exactly what f inverse does. Now how do we actually figure out the inverse of a function, especially a function that's defined with a rational expression like this? Well, the way that I think about it is, let's say that y is equal to our function of x, or y is a function of x. So we could say that y is equal to two x plus five over four minus three x. For our inverse, the relationship between x and y is going to be swapped, and so in our inverse, it's going to be true that x is going to be equal to two y plus five over four minus three y."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "Now how do we actually figure out the inverse of a function, especially a function that's defined with a rational expression like this? Well, the way that I think about it is, let's say that y is equal to our function of x, or y is a function of x. So we could say that y is equal to two x plus five over four minus three x. For our inverse, the relationship between x and y is going to be swapped, and so in our inverse, it's going to be true that x is going to be equal to two y plus five over four minus three y. And then to be able to express this as a function of x, to say that what is y is a function of x for our inverse, we now have to solve for y. So it's just a little bit of algebra here. So let's see if we can do that."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "For our inverse, the relationship between x and y is going to be swapped, and so in our inverse, it's going to be true that x is going to be equal to two y plus five over four minus three y. And then to be able to express this as a function of x, to say that what is y is a function of x for our inverse, we now have to solve for y. So it's just a little bit of algebra here. So let's see if we can do that. So the first thing that I would do is multiply both sides of this equation by four minus three y. If we do that, on the left-hand side, we are going to get x times each of these terms, so we're going to get four x minus three y x, and then that's going to be equal to, on the right-hand side, since we multiplied by the denominator here, we're just going to be left with the numerator, it's going to be equal to two y plus five. And this could be a little bit intimidating because we're seeing an x's and y's, what are we trying to do?"}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's see if we can do that. So the first thing that I would do is multiply both sides of this equation by four minus three y. If we do that, on the left-hand side, we are going to get x times each of these terms, so we're going to get four x minus three y x, and then that's going to be equal to, on the right-hand side, since we multiplied by the denominator here, we're just going to be left with the numerator, it's going to be equal to two y plus five. And this could be a little bit intimidating because we're seeing an x's and y's, what are we trying to do? Remember, we're trying to solve for y. So let's gather all the y terms on one side and all the non-y terms on the other side. So let's get rid of this two y here."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "And this could be a little bit intimidating because we're seeing an x's and y's, what are we trying to do? Remember, we're trying to solve for y. So let's gather all the y terms on one side and all the non-y terms on the other side. So let's get rid of this two y here. Actually, well, I could go either way. Let's get rid of this two y here. So let's subtract two y from both sides."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's get rid of this two y here. Actually, well, I could go either way. Let's get rid of this two y here. So let's subtract two y from both sides. And let's get rid of this four x from the left-hand side. So let's subtract four x from both sides. And then what are we going to be left with?"}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's subtract two y from both sides. And let's get rid of this four x from the left-hand side. So let's subtract four x from both sides. And then what are we going to be left with? On the left-hand side, we're left with minus, or negative five, or actually, it would be this way. It would be negative three y x minus two y, and you might say, hey, where is this going? But I'll show you in a second."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "And then what are we going to be left with? On the left-hand side, we're left with minus, or negative five, or actually, it would be this way. It would be negative three y x minus two y, and you might say, hey, where is this going? But I'll show you in a second. Is equal to, those cancel out, and we're going to have five minus four x. Now, once again, we are trying to solve for y. So let's factor out a y here, and then we are going to have y times negative three x minus two is equal to five minus four x."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "But I'll show you in a second. Is equal to, those cancel out, and we're going to have five minus four x. Now, once again, we are trying to solve for y. So let's factor out a y here, and then we are going to have y times negative three x minus two is equal to five minus four x. And now this is the home stretch. We can just divide both sides of this equation by negative three x minus two, and we're going to get y is equal to five minus four x over negative three x minus two. Now, another way that you could express this, you could multiply both the numerator and the denominator by a negative one."}, {"video_title": "Finding inverses of rational functions Equations Algebra 2 Khan Academy.mp3", "Sentence": "So let's factor out a y here, and then we are going to have y times negative three x minus two is equal to five minus four x. And now this is the home stretch. We can just divide both sides of this equation by negative three x minus two, and we're going to get y is equal to five minus four x over negative three x minus two. Now, another way that you could express this, you could multiply both the numerator and the denominator by a negative one. That won't change the value. And then you would get, you would get in the numerator four x minus five, and in the denominator, you would get a three x plus two. So there you have it."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Let's see if we can simplify 6 to the 1 half power times the 5th root of 6 and all of that to the 3rd power. I encourage you to pause this video and try it on your own. So let me actually color code these exponents just so we can keep track of them a little better. So that's the 1 half power in blue. This is the 5th root here in magenta. And let's see in green, let's think about this 3rd power. So one way to think about this 5th root is that this is the exact same thing as raising this 6 to the 1 5th power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So that's the 1 half power in blue. This is the 5th root here in magenta. And let's see in green, let's think about this 3rd power. So one way to think about this 5th root is that this is the exact same thing as raising this 6 to the 1 5th power. So let's write it like that. So this part right over here we could rewrite as 6 to the 1 5th power. And then that whole thing gets raised to the 3rd power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So one way to think about this 5th root is that this is the exact same thing as raising this 6 to the 1 5th power. So let's write it like that. So this part right over here we could rewrite as 6 to the 1 5th power. And then that whole thing gets raised to the 3rd power. And of course we have this 6 to the 1 half power out here. 6 to the 1 half power times all of this business right over here. Now what happens if we raise something to an exponent and then raise that whole thing to another exponent?"}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And then that whole thing gets raised to the 3rd power. And of course we have this 6 to the 1 half power out here. 6 to the 1 half power times all of this business right over here. Now what happens if we raise something to an exponent and then raise that whole thing to another exponent? Well we've already seen in our exponent properties that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the 3 5th power. And of course we're multiplying that times 6 to the 1 half power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now what happens if we raise something to an exponent and then raise that whole thing to another exponent? Well we've already seen in our exponent properties that's the equivalent of raising this to the product of these two exponents. So this part right over here could be rewritten as 6 to the 3 5th power. And of course we're multiplying that times 6 to the 1 half power. And if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way because these all equal each other. This is the same thing as 6 being raised to the 1 half plus 3 5th power."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And of course we're multiplying that times 6 to the 1 half power. And if you're multiplying some base to this exponent and then the same base again to another exponent, we know that this is going to be the same thing. And actually we could put these equal signs the whole way because these all equal each other. This is the same thing as 6 being raised to the 1 half plus 3 5th power. 1 half plus 3 over 5. Now what's 1 half plus 3 over 5? Well we could find a common denominator."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "This is the same thing as 6 being raised to the 1 half plus 3 5th power. 1 half plus 3 over 5. Now what's 1 half plus 3 over 5? Well we could find a common denominator. It would be 10. So that's the same thing as 6 to the 1 half. We can write it as 5 over 10 plus 3 5th is the same thing as 6 over 10."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Well we could find a common denominator. It would be 10. So that's the same thing as 6 to the 1 half. We can write it as 5 over 10 plus 3 5th is the same thing as 6 over 10. 6 over 10 power which is the same thing. And we deserve a little bit of a drum roll here. This wasn't that long of a problem."}, {"video_title": "Simplifying with exponent properties Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We can write it as 5 over 10 plus 3 5th is the same thing as 6 over 10. 6 over 10 power which is the same thing. And we deserve a little bit of a drum roll here. This wasn't that long of a problem. 6 to the 11 10th power. And so that looks pretty simplified to me. I guess we're done."}, {"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": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "But there's something about this that might pop out at you. The thing that pops out at me at least is that 25 is a perfect square, x to the fourth is a perfect square, so 25x to the fourth is a perfect square, and 9 is also a perfect square. So maybe this is the square of some binomial, and to confirm it, this center term has to be 2 times the product of the terms that you're squaring on either end. Let me explain that a little bit better. So 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Let me explain that a little bit better. So 25x to the fourth, that is the same thing as 5x squared squared, right? So it's a perfect square. 9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now what is 30x squared? Well, what happens if we take 5 times plus or minus 3?"}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "9 is the exact same thing as, well, it could be plus or minus 3 squared. It could be either one. Now what is 30x squared? Well, what happens if we take 5 times plus or minus 3? So remember, this needs to be 2 times the product of what's inside the square, the square root of this and the square root of that. So if we take, given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Well, what happens if we take 5 times plus or minus 3? So remember, this needs to be 2 times the product of what's inside the square, the square root of this and the square root of that. So if we take, given that there's a negative sign here and 5 is positive, we want to take the negative 3, right? That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "That's the only way we're going to get a negative over there, so let's just try it with negative 3. So what is 2 times 5x squared times negative 3? What is this? Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square, so we can just rewrite this as this is equal to 5x squared. We do it the same color."}, {"video_title": "Example 5 Factoring a fourth degree polynomial using the perfect square\u201d form Khan Academy.mp3", "Sentence": "Well, 2 times 5x squared is 10x squared times negative 3. It is equal to negative 30x squared. We know that this is a perfect square, so we can just rewrite this as this is equal to 5x squared. We do it the same color. 5x squared minus 3 times 5x squared minus 3. And we saw in the last video why this works, and if you want to verify it for yourself, multiply this out. You will get 25x to the fourth minus 30x squared plus 9."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "All right. Complete the following sentence about the half-life of the bacteria culture. The number of bacteria is halved every blank seconds. So, t is being given to us in seconds. So, let's think about this a little bit. Let's think about, I'll draw a little table here. I'll draw a little table here."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So, t is being given to us in seconds. So, let's think about this a little bit. Let's think about, I'll draw a little table here. I'll draw a little table here. So, if we have, this is t, and this is n of t, and I'll start with a straightforward t. At time equals zero, right when we start, this whole thing, if this t is zero, then 1 1 2 of the zero over 5.5, that's just 1 1 2 of the zero, that's all gonna be one, and you're just gonna be left with 1,000, 1,000 bacteria in the Petri dish. Now, at what point do we get to multiply by 1 1 2? At what point do we get to say 1,000 times 1 1 2?"}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "I'll draw a little table here. So, if we have, this is t, and this is n of t, and I'll start with a straightforward t. At time equals zero, right when we start, this whole thing, if this t is zero, then 1 1 2 of the zero over 5.5, that's just 1 1 2 of the zero, that's all gonna be one, and you're just gonna be left with 1,000, 1,000 bacteria in the Petri dish. Now, at what point do we get to multiply by 1 1 2? At what point do we get to say 1,000 times 1 1 2? Well, in order to say 1,000 times 1 1 2, the exponent here has to be one. So, at what time is the exponent here going to be one? Well, the exponent here is going to be one, this whole exponent's going to be one when t is equal to 5.5 seconds."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "At what point do we get to say 1,000 times 1 1 2? Well, in order to say 1,000 times 1 1 2, the exponent here has to be one. So, at what time is the exponent here going to be one? Well, the exponent here is going to be one, this whole exponent's going to be one when t is equal to 5.5 seconds. So, t is 5.5 seconds. And likewise, we wait another 5.5 seconds, so if we go to 11 seconds, then this is going to be 1,000 times, 11 divided by 5.5 is two, so times 1 1 2 to the second power, so times 1 1 2, times 1 1 2. So, every 5 1 2 seconds, we will essentially have half of the bacteria that we had 5 1 2 seconds ago."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, the exponent here is going to be one, this whole exponent's going to be one when t is equal to 5.5 seconds. So, t is 5.5 seconds. And likewise, we wait another 5.5 seconds, so if we go to 11 seconds, then this is going to be 1,000 times, 11 divided by 5.5 is two, so times 1 1 2 to the second power, so times 1 1 2, times 1 1 2. So, every 5 1 2 seconds, we will essentially have half of the bacteria that we had 5 1 2 seconds ago. So, the number of bacteria is halved every 5.5 seconds. And you see it in the formula, in the function definition right over there, but it's nice to reason it through and to really digest why it makes sense. Let's do a few more of these."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So, every 5 1 2 seconds, we will essentially have half of the bacteria that we had 5 1 2 seconds ago. So, the number of bacteria is halved every 5.5 seconds. And you see it in the formula, in the function definition right over there, but it's nice to reason it through and to really digest why it makes sense. Let's do a few more of these. The chemical element einsteinium-253 naturally loses its mass over time. A sample of einsteinium-253 had an initial mass of 320 grams when we measured it. The relationship between the elapsed time t in days and the mass m of t in grams left in the sample is modeled by the following function."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Let's do a few more of these. The chemical element einsteinium-253 naturally loses its mass over time. A sample of einsteinium-253 had an initial mass of 320 grams when we measured it. The relationship between the elapsed time t in days and the mass m of t in grams left in the sample is modeled by the following function. Complete the following sentence about the rate of change in the mass sample. The sample loses 87.5% of its mass every blank days. So, this one, instead of saying how much we grew or shrunk by, we're saying a percent change."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "The relationship between the elapsed time t in days and the mass m of t in grams left in the sample is modeled by the following function. Complete the following sentence about the rate of change in the mass sample. The sample loses 87.5% of its mass every blank days. So, this one, instead of saying how much we grew or shrunk by, we're saying a percent change. So, if you lose 87.5%, that means that you are left with 12.5%, which is the same thing, is that you have, is the same way of saying that you have 0.125% of your mass. So, another way of thinking about it is the sample is 0.125 of its mass, of its original mass, or how long does it take for the sample to be 0.125 of its mass? And there we could do a similar idea."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "So, this one, instead of saying how much we grew or shrunk by, we're saying a percent change. So, if you lose 87.5%, that means that you are left with 12.5%, which is the same thing, is that you have, is the same way of saying that you have 0.125% of your mass. So, another way of thinking about it is the sample is 0.125 of its mass, of its original mass, or how long does it take for the sample to be 0.125 of its mass? And there we could do a similar idea. You see the 0.125 right over here. And so, I could draw a table if you like, although I think you might guess where this is going. But let me draw a little table here."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "And there we could do a similar idea. You see the 0.125 right over here. And so, I could draw a table if you like, although I think you might guess where this is going. But let me draw a little table here. So, t and m of t. When t is zero, m of t is 320. And so, at what point is t, at what time is m of t going to be 320 times 0.125? Because this, going from this to this, that is losing 87.5% of your mass."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "But let me draw a little table here. So, t and m of t. When t is zero, m of t is 320. And so, at what point is t, at what time is m of t going to be 320 times 0.125? Because this, going from this to this, that is losing 87.5% of your mass. Losing 87, I'll use, so let me just write it this way. So, this is minus 87.5% of the mass. You've lost 0.875 to get to 0.125."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Because this, going from this to this, that is losing 87.5% of your mass. Losing 87, I'll use, so let me just write it this way. So, this is minus 87.5% of the mass. You've lost 0.875 to get to 0.125. So, this, you could just use 0.125 to the first power. So, what t do you have to make this exponent equal one? Well, t has to be 61.4."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "You've lost 0.875 to get to 0.125. So, this, you could just use 0.125 to the first power. So, what t do you have to make this exponent equal one? Well, t has to be 61.4. 61.4, and where t is in days. 61.4 days. Now, you might be tempted to always just pattern match, so whatever's in the denominator here, but I really encourage you to think about it, because that's the whole point of these problems."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Well, t has to be 61.4. 61.4, and where t is in days. 61.4 days. Now, you might be tempted to always just pattern match, so whatever's in the denominator here, but I really encourage you to think about it, because that's the whole point of these problems. If you just are pattern matching these, well, I don't know how helpful that's going to be for you. Let's do one more of these. Howard started studying how the number of branches on his tree grows over time."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Now, you might be tempted to always just pattern match, so whatever's in the denominator here, but I really encourage you to think about it, because that's the whole point of these problems. If you just are pattern matching these, well, I don't know how helpful that's going to be for you. Let's do one more of these. Howard started studying how the number of branches on his tree grows over time. The relationship between the elapsed time t in years since Howard started studying the tree and the number of its branches, n of t, is modeled by the following function. Complete the following sentence about the rate of change in the number of branches. Howard's tree gains 4 5ths more branches every blank years."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Howard started studying how the number of branches on his tree grows over time. The relationship between the elapsed time t in years since Howard started studying the tree and the number of its branches, n of t, is modeled by the following function. Complete the following sentence about the rate of change in the number of branches. Howard's tree gains 4 5ths more branches every blank years. So, gaining 4 5ths is equivalent to multiplying, multiplying by, and remember, you're gaining 4 5ths of what you already are. You're not just gaining the number 4 5ths. You're getting 4 5ths of what you already are."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "Howard's tree gains 4 5ths more branches every blank years. So, gaining 4 5ths is equivalent to multiplying, multiplying by, and remember, you're gaining 4 5ths of what you already are. You're not just gaining the number 4 5ths. You're getting 4 5ths of what you already are. So, that's the equivalent of multiplying by 1 plus 4 5ths, or 9 5ths. So, gaining 4 5ths is the same thing as multiplying by 9 5ths. If I'm five years old, and if I gain 4 5ths of my age, I would gain five, I would gain four years to get to be nine years old, which means I've multiplied my age by 9 5ths."}, {"video_title": "Interpreting time in exponential models Mathematics II High School Math Khan Academy.mp3", "Sentence": "You're getting 4 5ths of what you already are. So, that's the equivalent of multiplying by 1 plus 4 5ths, or 9 5ths. So, gaining 4 5ths is the same thing as multiplying by 9 5ths. If I'm five years old, and if I gain 4 5ths of my age, I would gain five, I would gain four years to get to be nine years old, which means I've multiplied my age by 9 5ths. So, Howard's tree, you could say, grows by a factor of 9 5ths every how many years? Well, you can see over here, the common ratio is 9 5ths, and so you're going to grow by 9 5ths every time t is a multiple of 7.3, or I guess you could say every time t increases by 7.3, you're going to, then this exponent is going to increase by a whole, and so you could view that as multiplying again by 9 5ths. So, Howard's tree gains 4 5ths more branches every 7.3 years."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "One way is we can just multiply the height of this big rectangle times the width of this big rectangle. So what's its height? Well, from here to here, that distance is going to be y squared, and then from there to there, that distance is going to be negative six y. And I know what you're thinking, how can my distance be negative six y? Isn't a distance always positive? Well, even negative six y can be positive if y is negative. So it's completely reasonable to say, well, this distance could be negative six y."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "And I know what you're thinking, how can my distance be negative six y? Isn't a distance always positive? Well, even negative six y can be positive if y is negative. So it's completely reasonable to say, well, this distance could be negative six y. So this entire, the entire height right over here is going to be, it's going to be y squared minus six y. Or you could view it as y squared plus this distance, which is negative six y. Y squared plus negative six y, which is the same thing as y squared minus six y. So that's the height of this big rectangle."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "So it's completely reasonable to say, well, this distance could be negative six y. So this entire, the entire height right over here is going to be, it's going to be y squared minus six y. Or you could view it as y squared plus this distance, which is negative six y. Y squared plus negative six y, which is the same thing as y squared minus six y. So that's the height of this big rectangle. What's its width? Well, the width is going to be, the width of this purple rectangle, it's going to be three y squared plus the width of this yellow rectangle, which is negative two y, and that can have a negative out here, the same logic why this could have a negative, why the negative six y could have a negative. And then plus the width of this blue rectangle."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "So that's the height of this big rectangle. What's its width? Well, the width is going to be, the width of this purple rectangle, it's going to be three y squared plus the width of this yellow rectangle, which is negative two y, and that can have a negative out here, the same logic why this could have a negative, why the negative six y could have a negative. And then plus the width of this blue rectangle. And so if you add them all together, the width of the entire rectangle is going to be three y squared minus two y, minus two y, plus one. And just like that, this expression that I just wrote down will give us the area for the entire, the area for the entire big rectangle. Now there's another way to do it, and a big clue was that we subdivided the big rectangle into these six smaller rectangles, and we have the dimensions for the six smaller rectangles."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "And then plus the width of this blue rectangle. And so if you add them all together, the width of the entire rectangle is going to be three y squared minus two y, minus two y, plus one. And just like that, this expression that I just wrote down will give us the area for the entire, the area for the entire big rectangle. Now there's another way to do it, and a big clue was that we subdivided the big rectangle into these six smaller rectangles, and we have the dimensions for the six smaller rectangles. And so we could find the area for each of these, and then we could add them all together. So let's look at this first one. Height times width."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "Now there's another way to do it, and a big clue was that we subdivided the big rectangle into these six smaller rectangles, and we have the dimensions for the six smaller rectangles. And so we could find the area for each of these, and then we could add them all together. So let's look at this first one. Height times width. The area of this purple rectangle is gonna be the height, y squared, times the width, which is three y squared, which is going to be equal to, it's gonna be three, and then y squared times y squared is y to the fourth power. What's the area of this yellow rectangle? Height is y squared."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "Height times width. The area of this purple rectangle is gonna be the height, y squared, times the width, which is three y squared, which is going to be equal to, it's gonna be three, and then y squared times y squared is y to the fourth power. What's the area of this yellow rectangle? Height is y squared. It's gonna be y squared times the width, times negative two y, which is going to give us negative two y to the third power. What about the blue one? Well, height times width is gonna be y squared times one, which of course is just going to be equal to y squared."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "Height is y squared. It's gonna be y squared times the width, times negative two y, which is going to give us negative two y to the third power. What about the blue one? Well, height times width is gonna be y squared times one, which of course is just going to be equal to y squared. Now this green one, it's gonna be the height, which is now negative six y times the width, which is three y squared, which is going to be equal to, see, negative six times three is negative 18, and then y times y squared is y to the third power. Now the area of this gray rectangle is gonna be the height, which is negative six y, times the width, which is negative two y, which gets us negative six times negative two is positive 12. Y times y is y squared."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "Well, height times width is gonna be y squared times one, which of course is just going to be equal to y squared. Now this green one, it's gonna be the height, which is now negative six y times the width, which is three y squared, which is going to be equal to, see, negative six times three is negative 18, and then y times y squared is y to the third power. Now the area of this gray rectangle is gonna be the height, which is negative six y, times the width, which is negative two y, which gets us negative six times negative two is positive 12. Y times y is y squared. And then finally, the area of this rectangle right over here is gonna be the height, which is negative six y, times the width, which is just one, which is equal to negative six y. And so if we want the area of this entire rectangle, we can just add up the areas of the smaller ones. So it's going to be equal to the three, it's going to be equal to the three y to the fourth, three y to the fourth, plus negative two y to the third power, negative, let me write this in a color that corresponds to that, negative two y to the third power, plus y squared, plus y squared, minus 18 y to the third power, so minus 18 y to the third power, plus 12 y squared, let's write that in black, so plus 12 y squared, and then last but not least, we have the minus six y, minus six y."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "Y times y is y squared. And then finally, the area of this rectangle right over here is gonna be the height, which is negative six y, times the width, which is just one, which is equal to negative six y. And so if we want the area of this entire rectangle, we can just add up the areas of the smaller ones. So it's going to be equal to the three, it's going to be equal to the three y to the fourth, three y to the fourth, plus negative two y to the third power, negative, let me write this in a color that corresponds to that, negative two y to the third power, plus y squared, plus y squared, minus 18 y to the third power, so minus 18 y to the third power, plus 12 y squared, let's write that in black, so plus 12 y squared, and then last but not least, we have the minus six y, minus six y. So this is an expression for the area of the entire thing, but we can simplify it more. So let's see, we have, we only have one fourth degree term, so I'll just rewrite that. So we have one fourth degree term, so I'm just gonna rewrite that, three y to the fourth power."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "So it's going to be equal to the three, it's going to be equal to the three y to the fourth, three y to the fourth, plus negative two y to the third power, negative, let me write this in a color that corresponds to that, negative two y to the third power, plus y squared, plus y squared, minus 18 y to the third power, so minus 18 y to the third power, plus 12 y squared, let's write that in black, so plus 12 y squared, and then last but not least, we have the minus six y, minus six y. So this is an expression for the area of the entire thing, but we can simplify it more. So let's see, we have, we only have one fourth degree term, so I'll just rewrite that. So we have one fourth degree term, so I'm just gonna rewrite that, three y to the fourth power. Now how many third degree terms do we have? We have negative two y to the third power, we have negative 18 y to the third power. So if we add these two together, how many y to the third powers do we have?"}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "So we have one fourth degree term, so I'm just gonna rewrite that, three y to the fourth power. Now how many third degree terms do we have? We have negative two y to the third power, we have negative 18 y to the third power. So if we add these two together, how many y to the third powers do we have? Well negative two plus negative 18 is negative 20. Negative 20 y to the third power. And then how many second degree terms do we have?"}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "So if we add these two together, how many y to the third powers do we have? Well negative two plus negative 18 is negative 20. Negative 20 y to the third power. And then how many second degree terms do we have? Well we have one y squared right over here, and then we have another 12 y squareds, you add those together, you're gonna have 13 y squareds. And then finally we still have this, we still need to subtract this six y. And there you have it, another expression for the area of the entire rectangle."}, {"video_title": "Multiplying binomials by polynomials area model Algebra II High School Math Khan Academy.mp3", "Sentence": "And then how many second degree terms do we have? Well we have one y squared right over here, and then we have another 12 y squareds, you add those together, you're gonna have 13 y squareds. And then finally we still have this, we still need to subtract this six y. And there you have it, another expression for the area of the entire rectangle. And the whole point of doing this is to realize that this up here and this down here are equivalent, and that the way that we multiply this actually corresponds to exactly how we found the areas of the smaller rectangles right over here. You would say y squared times three y squared is three y fourth, y squared times negative two y is negative two y to the third power, y squared plus one, y squared times one is y squared, which is exactly what we did when we found the area of these rectangles in this, I guess you could say in this top row. And then you would take the negative six, and you would say negative six times three y squared is negative 18 y to the third, negative six times negative two y is positive 12 y squared, negative six times one is, or negative six y times one is negative six y."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And the reason why it says interactive graph is this is a screenshot from this type of exercise on Khan Academy, and on Khan Academy, you'd be able to click on points here and they'd put little dots, and you can either delete them and put them someplace else, so it would be an interactive graph. But this is just a screenshot, so I'm just going to draw on top of this. But the main goal is, what are the zeros of this polynomial, and then you just have to plot it on this graph. So pause this video and have a go at it. All right, now to figure out the zeros of a polynomial, you essentially have to figure out the X values that would make the polynomial equal to zero. Or another way to think about it is the X values that would make this equation true. X to the third plus X squared minus nine X minus nine is equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and have a go at it. All right, now to figure out the zeros of a polynomial, you essentially have to figure out the X values that would make the polynomial equal to zero. Or another way to think about it is the X values that would make this equation true. X to the third plus X squared minus nine X minus nine is equal to zero. Now the best way to do that is to try to factor this expression. Now this is a third-degree polynomial, which isn't always so easy to factor, so let's see how we might approach it. The first thing I look for is, are there any common factors to all of these terms?"}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "X to the third plus X squared minus nine X minus nine is equal to zero. Now the best way to do that is to try to factor this expression. Now this is a third-degree polynomial, which isn't always so easy to factor, so let's see how we might approach it. The first thing I look for is, are there any common factors to all of these terms? And it doesn't look like there is. The next thing I could look for, I could think about whether factoring by grouping could work here. And when I think about factoring by grouping, I would look at the first two terms and I would look at the last two terms."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "The first thing I look for is, are there any common factors to all of these terms? And it doesn't look like there is. The next thing I could look for, I could think about whether factoring by grouping could work here. And when I think about factoring by grouping, I would look at the first two terms and I would look at the last two terms. And I would say, is there anything I could factor out of these first two terms that would, or what's the most that I could factor out of these first two terms? And then what's the most that I could factor out of these last two terms? And then would it leave something similar once I've done that factoring?"}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And when I think about factoring by grouping, I would look at the first two terms and I would look at the last two terms. And I would say, is there anything I could factor out of these first two terms that would, or what's the most that I could factor out of these first two terms? And then what's the most that I could factor out of these last two terms? And then would it leave something similar once I've done that factoring? Now what I mean is, for these first two, we have a common factor of X squared. So let's factor out an X squared and these first two terms become X squared times X plus one. And then for these second two terms, I can factor out a negative nine."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then would it leave something similar once I've done that factoring? Now what I mean is, for these first two, we have a common factor of X squared. So let's factor out an X squared and these first two terms become X squared times X plus one. And then for these second two terms, I can factor out a negative nine. So I could rewrite it as negative nine times X plus one. Now that all worked out quite nicely because now we see, if we view, if we view this is now our first term and this is our second term, we can see that X plus one is a factor of both of them. And so we can factor that out."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then for these second two terms, I can factor out a negative nine. So I could rewrite it as negative nine times X plus one. Now that all worked out quite nicely because now we see, if we view, if we view this is now our first term and this is our second term, we can see that X plus one is a factor of both of them. And so we can factor that out. We can factor out the X plus one. And I'll do that in this light blue color. Actually, let me do it in a slightly darker blue color."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And so we can factor that out. We can factor out the X plus one. And I'll do that in this light blue color. Actually, let me do it in a slightly darker blue color. And so if you factor out the X plus one, you're left with X plus one times X squared, X squared minus nine, minus nine. And that is going to be equal to zero. Now we're not done factoring yet because now we have a difference of squares."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Actually, let me do it in a slightly darker blue color. And so if you factor out the X plus one, you're left with X plus one times X squared, X squared minus nine, minus nine. And that is going to be equal to zero. Now we're not done factoring yet because now we have a difference of squares. X squared minus nine, this is going to be equal to, and let me just write it all out. So I have this X plus one here. So I have X plus one."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "Now we're not done factoring yet because now we have a difference of squares. X squared minus nine, this is going to be equal to, and let me just write it all out. So I have this X plus one here. So I have X plus one. And then the X squared minus nine, I can write as X plus three times X minus three. If any of what I'm doing feels unfamiliar to you, if that first factoring feels unfamiliar, I encourage you to review factoring by grouping. And if what I just did looks unfamiliar, I encourage you to look at factoring differences of squares."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So I have X plus one. And then the X squared minus nine, I can write as X plus three times X minus three. If any of what I'm doing feels unfamiliar to you, if that first factoring feels unfamiliar, I encourage you to review factoring by grouping. And if what I just did looks unfamiliar, I encourage you to look at factoring differences of squares. But anyway, all of that would be equal to zero. Now if I have the product of several things equaling zero, if any one of those things is equal to zero, that would make the whole expression equal to zero. So we have a situation where one solution would be the solution that makes X plus one equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And if what I just did looks unfamiliar, I encourage you to look at factoring differences of squares. But anyway, all of that would be equal to zero. Now if I have the product of several things equaling zero, if any one of those things is equal to zero, that would make the whole expression equal to zero. So we have a situation where one solution would be the solution that makes X plus one equal to zero. And once again, I'm gonna do that darker color. X plus one equal zero. And that of course is X is equal to negative one."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So we have a situation where one solution would be the solution that makes X plus one equal to zero. And once again, I'm gonna do that darker color. X plus one equal zero. And that of course is X is equal to negative one. Another solution is what would make X plus three equal to zero. And that of course is X is equal to negative three, subtract three from both sides. And then another solution is going to be whatever X value makes X minus three equal to zero."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And that of course is X is equal to negative one. Another solution is what would make X plus three equal to zero. And that of course is X is equal to negative three, subtract three from both sides. And then another solution is going to be whatever X value makes X minus three equal to zero. Add three to both sides, you get X is equal to three. So there you have it. We have our three zeros."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then another solution is going to be whatever X value makes X minus three equal to zero. Add three to both sides, you get X is equal to three. So there you have it. We have our three zeros. Our polynomial evaluated at any of these X values will be equal to zero. So we can plot it here on this interactive graph. I'm just gonna draw on it."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We have our three zeros. Our polynomial evaluated at any of these X values will be equal to zero. So we can plot it here on this interactive graph. I'm just gonna draw on it. So we have X equals negative one, which is right over there. X equals negative three, which is right over there. And X equals three, which is right over there."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "I'm just gonna draw on it. So we have X equals negative one, which is right over there. X equals negative three, which is right over there. And X equals three, which is right over there. And the reason why you might wanna do this type of thing, this exercise just asks us to do this and we're done. But the reason why this is useful is this can help inform what the graph looks like. This tells us where our graph intersects the X axis."}, {"video_title": "Zeros of polynomials (with factoring) grouping Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And X equals three, which is right over there. And the reason why you might wanna do this type of thing, this exercise just asks us to do this and we're done. But the reason why this is useful is this can help inform what the graph looks like. This tells us where our graph intersects the X axis. So our graph might do something like this. Or it might do something like this. And we would have to look at other information to think about what that might be."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Pause this video and see if you can work through that. All right, now let's work through it together. So there's a couple of ways that we can conceptualize average rate of change of a function. One way to think about it is it's our change in the value of our function divided by our change in x, or it's our change in the value of our function per x on average. So you can view it as change in the value of function divided by your change in x. If you say that y is equal to f of x, you could also express it as change in y over change in x. On average, how much does a function change per unit change in x on average?"}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "One way to think about it is it's our change in the value of our function divided by our change in x, or it's our change in the value of our function per x on average. So you can view it as change in the value of function divided by your change in x. If you say that y is equal to f of x, you could also express it as change in y over change in x. On average, how much does a function change per unit change in x on average? Now we could do this with a table, or we could try to conceptualize it visually. But let's just do this one with a table, and then we'll try to connect the dots a little bit with a visual. So if we have x here, and then if we have y is equal to f of x right over here, when x is equal to negative two, what is y going to be equal to, or what is f of negative two?"}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "On average, how much does a function change per unit change in x on average? Now we could do this with a table, or we could try to conceptualize it visually. But let's just do this one with a table, and then we'll try to connect the dots a little bit with a visual. So if we have x here, and then if we have y is equal to f of x right over here, when x is equal to negative two, what is y going to be equal to, or what is f of negative two? Well, let's see, f of, so y is equal to f of negative two, which is going to be equal to negative eight. That's negative two to the third power, minus four times negative two, so that's minus negative eight, so that's plus eight, that equals zero. And then when x is equal to three, I'm going to the right end of that interval, well now y is equal to f of three, which is equal to 27, three to the third power, minus four times three, minus 12, which is equal to 15."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So if we have x here, and then if we have y is equal to f of x right over here, when x is equal to negative two, what is y going to be equal to, or what is f of negative two? Well, let's see, f of, so y is equal to f of negative two, which is going to be equal to negative eight. That's negative two to the third power, minus four times negative two, so that's minus negative eight, so that's plus eight, that equals zero. And then when x is equal to three, I'm going to the right end of that interval, well now y is equal to f of three, which is equal to 27, three to the third power, minus four times three, minus 12, which is equal to 15. So what is our change in y over our change in x over this interval? Well, our y went from zero to 15, so we have a increase of 15 in y. And what was our change in x?"}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And then when x is equal to three, I'm going to the right end of that interval, well now y is equal to f of three, which is equal to 27, three to the third power, minus four times three, minus 12, which is equal to 15. So what is our change in y over our change in x over this interval? Well, our y went from zero to 15, so we have a increase of 15 in y. And what was our change in x? Well, we went from negative two to positive three, so we had a plus five change in x, so our change in x is plus five. And so our average rate of change of y with respect to x, or the rate of change of our function with respect to x over the interval is going to be equal to three. If you wanted to think about this visually, I could try to sketch this."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And what was our change in x? Well, we went from negative two to positive three, so we had a plus five change in x, so our change in x is plus five. And so our average rate of change of y with respect to x, or the rate of change of our function with respect to x over the interval is going to be equal to three. If you wanted to think about this visually, I could try to sketch this. So this is the x-axis, the y-axis, and our function does something like this. So at x equals negative two, f of x is zero. And then it goes up, and then it comes back down, and then it does something like this."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "If you wanted to think about this visually, I could try to sketch this. So this is the x-axis, the y-axis, and our function does something like this. So at x equals negative two, f of x is zero. And then it goes up, and then it comes back down, and then it does something like this. It does something like this, and it does, and it was going before this. And so the interval that we care about, we're going from negative two to three, which is right about there. So that's x equals negative two to x equals three."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "And then it goes up, and then it comes back down, and then it does something like this. It does something like this, and it does, and it was going before this. And so the interval that we care about, we're going from negative two to three, which is right about there. So that's x equals negative two to x equals three. And so what we wanna do at the left end of the interval, our function is equal to zero, so we're at this point right over here. I'll do this in a new color. We're at this point right over there."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "So that's x equals negative two to x equals three. And so what we wanna do at the left end of the interval, our function is equal to zero, so we're at this point right over here. I'll do this in a new color. We're at this point right over there. And at the right end of our function, f of three is 15. So we are up here someplace. Let me connect the curve a little bit."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "We're at this point right over there. And at the right end of our function, f of three is 15. So we are up here someplace. Let me connect the curve a little bit. We are going to be up there. And so when we're thinking about the average rate of change over this interval, we're really thinking about the slope of the line that connects these two points. So the line that connects these two points looks something like this."}, {"video_title": "Finding average rate of change of polynomials Algebra 2 Khan Academy.mp3", "Sentence": "Let me connect the curve a little bit. We are going to be up there. And so when we're thinking about the average rate of change over this interval, we're really thinking about the slope of the line that connects these two points. So the line that connects these two points looks something like this. And we're just calculating what is our change in y, which is going to be this, our change in y. And we see that the value of our function increased by 15 divided by our change in x. So this right over here is our change in x, which we see we went from negative two to three."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "In this video, I want to introduce you to the number i, which is sometimes called the imaginary unit. And what you're going to see here, and it might be a little bit difficult to fully appreciate right from the get-go, it's a more bizarre number than some of the other wacky numbers we learn in mathematics like pi or e. And it's more bizarre because it doesn't have a tangible value in the sense that we're normally used to defining numbers. i is defined as the number whose square is equal to negative 1. This is the definition of i. And it leads to all sorts of interesting things. Now some places you will see i defined this way, i as being equal to the principal square root of negative 1. I wanted to just point out to you that this is not wrong, and it might make sense to you."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This is the definition of i. And it leads to all sorts of interesting things. Now some places you will see i defined this way, i as being equal to the principal square root of negative 1. I wanted to just point out to you that this is not wrong, and it might make sense to you. If something squared is negative 1, then maybe it's the principal square root of negative 1. And so these seem to be almost the same statement. But I just want to make you a little careful."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I wanted to just point out to you that this is not wrong, and it might make sense to you. If something squared is negative 1, then maybe it's the principal square root of negative 1. And so these seem to be almost the same statement. But I just want to make you a little careful. When you do this, some people will even go as far as saying this is wrong. And it actually turns out that they are wrong to say that this is wrong. But when you do this, you have to be a little bit careful about what it means to take a principal square root of a negative number and it being defined for imaginary, and as we'll learn in the future, complex numbers."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But I just want to make you a little careful. When you do this, some people will even go as far as saying this is wrong. And it actually turns out that they are wrong to say that this is wrong. But when you do this, you have to be a little bit careful about what it means to take a principal square root of a negative number and it being defined for imaginary, and as we'll learn in the future, complex numbers. But for your understanding right now, you don't have to differentiate them. You don't have to split hairs between any of these definitions. Now, with this definition, let's just think about what the different powers of i are."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "But when you do this, you have to be a little bit careful about what it means to take a principal square root of a negative number and it being defined for imaginary, and as we'll learn in the future, complex numbers. But for your understanding right now, you don't have to differentiate them. You don't have to split hairs between any of these definitions. Now, with this definition, let's just think about what the different powers of i are. Because you can imagine, if something squared is negative 1, if I take it to all sorts of powers, maybe that should give us all sorts of weird things. And what we'll see is that the powers of i are kind of neat, because they kind of cycle, or they do cycle, through a set of values. So I could start with i to the 0th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now, with this definition, let's just think about what the different powers of i are. Because you can imagine, if something squared is negative 1, if I take it to all sorts of powers, maybe that should give us all sorts of weird things. And what we'll see is that the powers of i are kind of neat, because they kind of cycle, or they do cycle, through a set of values. So I could start with i to the 0th power. And so you might say, hey, look, anything to the 0th power is 1. So i to the 0th power is 1. And that is true."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So I could start with i to the 0th power. And so you might say, hey, look, anything to the 0th power is 1. So i to the 0th power is 1. And that is true. And you could actually derive that even from this definition. But this is pretty straightforward. Anything to the 0th power, including i, is 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And that is true. And you could actually derive that even from this definition. But this is pretty straightforward. Anything to the 0th power, including i, is 1. Then you say, OK, what is i to the 1st power? Well, anything to the 1st power is just that number times itself once. So that's just going to be i, really by the definition of what it means to take an exponent."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Anything to the 0th power, including i, is 1. Then you say, OK, what is i to the 1st power? Well, anything to the 1st power is just that number times itself once. So that's just going to be i, really by the definition of what it means to take an exponent. So that completely makes sense. And then you have i to the 2nd power. Well, by definition, i to the 2nd power is equal to negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that's just going to be i, really by the definition of what it means to take an exponent. So that completely makes sense. And then you have i to the 2nd power. Well, by definition, i to the 2nd power is equal to negative 1. Let's try i to the 3rd power. I'll do this in a color I haven't used. i to the 3rd power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, by definition, i to the 2nd power is equal to negative 1. Let's try i to the 3rd power. I'll do this in a color I haven't used. i to the 3rd power. Well, that's going to be i to the 2nd power times i. And we know that i to the 2nd power is negative 1. So it's negative 1 times i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "i to the 3rd power. Well, that's going to be i to the 2nd power times i. And we know that i to the 2nd power is negative 1. So it's negative 1 times i. Let me make it clear. This is the same thing as this, which is the same thing as that. i squared is negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it's negative 1 times i. Let me make it clear. This is the same thing as this, which is the same thing as that. i squared is negative 1. So when you multiply it out, negative 1 times i we'll write as negative i. Now, what happens if we take i to the 4th power? I'll do it up here."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "i squared is negative 1. So when you multiply it out, negative 1 times i we'll write as negative i. Now, what happens if we take i to the 4th power? I'll do it up here. i to the 4th power. Well, once again, this is going to be i times i to the 3rd power. So that's i times i to the 3rd power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "I'll do it up here. i to the 4th power. Well, once again, this is going to be i times i to the 3rd power. So that's i times i to the 3rd power. But what was i to the 3rd power? i to the 3rd power was negative i. This over here is negative i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that's i times i to the 3rd power. But what was i to the 3rd power? i to the 3rd power was negative i. This over here is negative i. And so i times i would give you negative 1. But you have a negative out here. So it's i times i is negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "This over here is negative i. And so i times i would give you negative 1. But you have a negative out here. So it's i times i is negative 1. And then you have a negative. That gives you positive 1. Let me write it down."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So it's i times i is negative 1. And then you have a negative. That gives you positive 1. Let me write it down. So this is the same thing as i times negative i, which is the same thing as negative 1 times, remember, multiplication is commutative. If we're just multiplying a bunch of numbers, we can switch the order. So this is the same thing as negative 1 times i times i. i times i, by definition, is negative 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let me write it down. So this is the same thing as i times negative i, which is the same thing as negative 1 times, remember, multiplication is commutative. If we're just multiplying a bunch of numbers, we can switch the order. So this is the same thing as negative 1 times i times i. i times i, by definition, is negative 1. Negative 1 times negative 1 is equal to 1. So i to the 4th is the same thing as i to the 0th power. Now let's try i to the 5th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So this is the same thing as negative 1 times i times i. i times i, by definition, is negative 1. Negative 1 times negative 1 is equal to 1. So i to the 4th is the same thing as i to the 0th power. Now let's try i to the 5th power. Well, that's just going to be i to the 4th times i. And we know what i to the 4th is. It is 1."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Now let's try i to the 5th power. Well, that's just going to be i to the 4th times i. And we know what i to the 4th is. It is 1. So it's 1 times i, or it is just i again. So once again, it is exactly the same thing as i to the 1st power. Let's try, and just to see the pattern keep going, let's try i to the 6th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "It is 1. So it's 1 times i, or it is just i again. So once again, it is exactly the same thing as i to the 1st power. Let's try, and just to see the pattern keep going, let's try i to the 6th power. Well, that's i times i to the 5th power. That's i times i to the 5th. i to the 5th, we already established, is just i."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let's try, and just to see the pattern keep going, let's try i to the 6th power. Well, that's i times i to the 5th power. That's i times i to the 5th. i to the 5th, we already established, is just i. So it's i times i. It is equal to, by definition, i times i is negative 1. And then let's finish up."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "i to the 5th, we already established, is just i. So it's i times i. It is equal to, by definition, i times i is negative 1. And then let's finish up. Well, we could keep going this way. We can keep putting higher and higher powers of i here. And we'll see that it keeps cycling back."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then let's finish up. Well, we could keep going this way. We can keep putting higher and higher powers of i here. And we'll see that it keeps cycling back. And in the next video, I'll teach you how, taking an arbitrarily high power of i, how you can figure out what that's going to be. But let's just verify this cycle keeps going. i to the 7th power is equal to i times i to the 6th power."}, {"video_title": "Introduction to i and imaginary numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And we'll see that it keeps cycling back. And in the next video, I'll teach you how, taking an arbitrarily high power of i, how you can figure out what that's going to be. But let's just verify this cycle keeps going. i to the 7th power is equal to i times i to the 6th power. i to the 6th power is negative 1. i times negative 1 is negative i. And if you take i to the 8th, once again, it'll be 1. i to the 9th will be i again. So on and so forth."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "And because we know that 4 times 4 times 4, or 4 to the 3rd power, is equal to 64, if we're looking for the cube root of 64, we're looking for a number that that number times that number times that same number is going to be equal to 64. Well, we know that number is 4, so this thing right over here is going to be 4. Now we're going to think of slightly more complex fractional exponents. The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2 3rd power is. And here I'm going to use a property of exponents that we'll study more later on."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "The one we see here has a 1 in the numerator. Now we're going to see something different. So what I want to do is think about what 64 to the 2 3rd power is. And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that, let's say with a simple number, if I raise something to the 3rd power, and then I were to raise that to, say, the 4th power, this is going to be the same thing as raising it to the 2 to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the 4th power, and then the 3rd power. All this is saying is if I raise something to a power, and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "And here I'm going to use a property of exponents that we'll study more later on. But this property of exponents is the idea that, let's say with a simple number, if I raise something to the 3rd power, and then I were to raise that to, say, the 4th power, this is going to be the same thing as raising it to the 2 to the 3 times 4 power, or 2 to the 12th power, which you could also write as raising it to the 4th power, and then the 3rd power. All this is saying is if I raise something to a power, and then raise that whole thing to a power, it's the same thing as multiplying the two exponents. This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2 3rds is the same thing as 1 3rd times 2. So we could go in the other direction. We could say, hey, look, well, this is going to be the same thing as 64 to the 1 3rd power, and then that thing squared."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "This is the same thing as 2 to the 12th. So we could use that property here to say, well, 2 3rds is the same thing as 1 3rd times 2. So we could go in the other direction. We could say, hey, look, well, this is going to be the same thing as 64 to the 1 3rd power, and then that thing squared. Notice, I'm raising something to a power, and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2 3rds power. Now, why did I do this?"}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "We could say, hey, look, well, this is going to be the same thing as 64 to the 1 3rd power, and then that thing squared. Notice, I'm raising something to a power, and then raising that to a power. If I were to multiply these two things, I would get 64 to the 2 3rds power. Now, why did I do this? Well, we already know what 64 to the 1 3rd power is. We just calculated it. That's equal to 4."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Now, why did I do this? Well, we already know what 64 to the 1 3rd power is. We just calculated it. That's equal to 4. So we could say that this is equal to, and I'll write in that same yellow color, this is equal to 4 squared. This is equal to 4 squared, which is equal to 16. Which is equal to 16."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "That's equal to 4. So we could say that this is equal to, and I'll write in that same yellow color, this is equal to 4 squared. This is equal to 4 squared, which is equal to 16. Which is equal to 16. So 64 to 2 3rds power is equal to 16. The way I think of it, let me find the cube root of 64, which is 4, and then let me square it. And that is gonna get me to 16."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Which is equal to 16. So 64 to 2 3rds power is equal to 16. The way I think of it, let me find the cube root of 64, which is 4, and then let me square it. And that is gonna get me to 16. Now I'll give you an even hairier problem, and I encourage you to try this one on your own before I work through it. So we're gonna work with 8 over 27. And we're gonna raise this thing to the negative, to the negative, and I'll try to color code it, negative 2 over 3 power, to the negative 2 3rds power."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "And that is gonna get me to 16. Now I'll give you an even hairier problem, and I encourage you to try this one on your own before I work through it. So we're gonna work with 8 over 27. And we're gonna raise this thing to the negative, to the negative, and I'll try to color code it, negative 2 over 3 power, to the negative 2 3rds power. I encourage you to pause this and try this on your own. Well, the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says take the reciprocal of this to the positive exponent."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "And we're gonna raise this thing to the negative, to the negative, and I'll try to color code it, negative 2 over 3 power, to the negative 2 3rds power. I encourage you to pause this and try this on your own. Well, the first thing I do whenever I see a negative exponent is to say, well, how can I get rid of that negative exponent? And I just remind myself, well, the negative exponent really just says take the reciprocal of this to the positive exponent. So this is going to be equal to, the reciprocal of this is 27, I'm using a different color. Let me use that light mauve color. So this is going to be equal to 27 over 8, over 8, I just took the reciprocal of this right over here."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "And I just remind myself, well, the negative exponent really just says take the reciprocal of this to the positive exponent. So this is going to be equal to, the reciprocal of this is 27, I'm using a different color. Let me use that light mauve color. So this is going to be equal to 27 over 8, over 8, I just took the reciprocal of this right over here. It's equal to 27 over 8 to the positive 2 3rds power, to the positive 2 3rds power. So notice, all I did, got rid of the exponent and took the reciprocal of the base right over here. 8 over 27 is the base, negative 2 3rds is the exponent."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "So this is going to be equal to 27 over 8, over 8, I just took the reciprocal of this right over here. It's equal to 27 over 8 to the positive 2 3rds power, to the positive 2 3rds power. So notice, all I did, got rid of the exponent and took the reciprocal of the base right over here. 8 over 27 is the base, negative 2 3rds is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over denominator some power, and this is another very powerful exponent property, this is going to be the exact same thing, this is going to be the exact same thing as raising 27 to the 2 3rds power, to the 2 over 3 power, over, over 8 to the 2 3rds power, 8 to the 2 3rds power. This is another very powerful exponent property."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "8 over 27 is the base, negative 2 3rds is the exponent. Now, how can we handle this? Well, we've already seen that if I have a numerator to some power over denominator some power, and this is another very powerful exponent property, this is going to be the exact same thing, this is going to be the exact same thing as raising 27 to the 2 3rds power, to the 2 over 3 power, over, over 8 to the 2 3rds power, 8 to the 2 3rds power. This is another very powerful exponent property. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator of that power over the denominator raised to that power. Now, let's think about what this is. Well, just like we saw before, this is going to be the same thing, this is going to be the same thing as 27 to the 1 3rd power, 27 to the 1 3rd power, and then that squared, and then that squared, because 1 3rd times 2 is 2 3rds, so I'm gonna raise 27 to the 1 3rd power and then square whatever that is, and that is going to be over, oh, this color coding is making this, just switch a lot of colors, this is going to be over 8 to the 1 3rd power, 8 to the 1 3rd power, to the, and then that's gonna be raised to the 2nd power, same thing we're doing in the denominator, we raise 8 to the 1 3rd and then square that."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "This is another very powerful exponent property. Notice, if I have something divided by something and I'm raising the whole thing to a power, I can essentially raise the numerator of that power over the denominator raised to that power. Now, let's think about what this is. Well, just like we saw before, this is going to be the same thing, this is going to be the same thing as 27 to the 1 3rd power, 27 to the 1 3rd power, and then that squared, and then that squared, because 1 3rd times 2 is 2 3rds, so I'm gonna raise 27 to the 1 3rd power and then square whatever that is, and that is going to be over, oh, this color coding is making this, just switch a lot of colors, this is going to be over 8 to the 1 3rd power, 8 to the 1 3rd power, to the, and then that's gonna be raised to the 2nd power, same thing we're doing in the denominator, we raise 8 to the 1 3rd and then square that. So what's this going to be? Well, 27 to the 1 3rd power, 27 to the 1 3rd power, is the cube root of 27, it's some number, that number times that same number is gonna be equal to 27. Well, it might jump out at you already that 3 to the 3rd is equal to 27, or that 27 to the 1 3rd is equal to 3."}, {"video_title": "Fractional exponents with numerators other than 1 Algebra I Khan Academy.mp3", "Sentence": "Well, just like we saw before, this is going to be the same thing, this is going to be the same thing as 27 to the 1 3rd power, 27 to the 1 3rd power, and then that squared, and then that squared, because 1 3rd times 2 is 2 3rds, so I'm gonna raise 27 to the 1 3rd power and then square whatever that is, and that is going to be over, oh, this color coding is making this, just switch a lot of colors, this is going to be over 8 to the 1 3rd power, 8 to the 1 3rd power, to the, and then that's gonna be raised to the 2nd power, same thing we're doing in the denominator, we raise 8 to the 1 3rd and then square that. So what's this going to be? Well, 27 to the 1 3rd power, 27 to the 1 3rd power, is the cube root of 27, it's some number, that number times that same number is gonna be equal to 27. Well, it might jump out at you already that 3 to the 3rd is equal to 27, or that 27 to the 1 3rd is equal to 3. So the numerator, we're gonna end up with 3 squared, and then in the denominator, we are going to end up with, well, what's 8 to the 1 3rd power? Well, 2 times 2 times 2 is 8, so this is 8 to the 1 3rd is 2, and then we are going to, and then let me do that same orange color, 8 to the 1 3rd is 2, and then we're going to square that. So this is going to simplify to 3 squared over 2 squared, which is just going to be equal to 9 over 4."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And you could do this, you could figure this out with algebraic long division, but I'll give you a hint. It is much simpler and much less computational intensive and takes much less space on your paper if you use the polynomial remainder theorem. And if that's unfamiliar to you, there's other videos that actually cover that. So why don't you have a go at it? All right, so now let's work through this together. The polynomial remainder theorem tells us that the remainder, when I take a polynomial, if I take a polynomial, if I take a polynomial, p of x, and if I were to divide it by, if I were to divide it by an x minus a, the remainder, the remainder, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of, p of a."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So why don't you have a go at it? All right, so now let's work through this together. The polynomial remainder theorem tells us that the remainder, when I take a polynomial, if I take a polynomial, if I take a polynomial, p of x, and if I were to divide it by, if I were to divide it by an x minus a, the remainder, the remainder, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of, p of a. So in this case, our p of x is this. What is our a? Well, our a is going to be positive two."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Is just going to be equal to p of, p of a. So in this case, our p of x is this. What is our a? Well, our a is going to be positive two. Remember, it's x minus a. So let me do this. Our a is equal to positive two."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Well, our a is going to be positive two. Remember, it's x minus a. So let me do this. Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder, the remainder in this case is going to be equal to p of two, p of two, which is equal to, so let's see, it's going to be, I'll just do it all in magenta, negative three times eight, minus, let's see, minus four times four."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder, the remainder in this case is going to be equal to p of two, p of two, which is equal to, so let's see, it's going to be, I'll just do it all in magenta, negative three times eight, minus, let's see, minus four times four. Plus four times four, plus 20, minus seven. So let's see, this is negative 24 minus 16, plus 20, plus 20, minus seven. And so that gives us, let's see, negative 24 minus 16, this is negative 40."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So the remainder, the remainder in this case is going to be equal to p of two, p of two, which is equal to, so let's see, it's going to be, I'll just do it all in magenta, negative three times eight, minus, let's see, minus four times four. Plus four times four, plus 20, minus seven. So let's see, this is negative 24 minus 16, plus 20, plus 20, minus seven. And so that gives us, let's see, negative 24 minus 16, this is negative 40. All right, I'm just going to do it step by step. This is equal to negative, actually I can do this in my head. All right, here we go."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And so that gives us, let's see, negative 24 minus 16, this is negative 40. All right, I'm just going to do it step by step. This is equal to negative, actually I can do this in my head. All right, here we go. So this is negative 40, plus 20 is negative 20, minus seven is negative 27. Negative 27. And that was pretty neat, because if we did, if we tried, if we attempted to do this without the polynomial remainder theorem, we would have had to do a bunch of algebraic long division."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "All right, here we go. So this is negative 40, plus 20 is negative 20, minus seven is negative 27. Negative 27. And that was pretty neat, because if we did, if we tried, if we attempted to do this without the polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the quotient and all of that, but we don't need, we don't need the quotient. We don't need to know, so if we did all the algebraic long division, we would have taken our P of X, and then we would have divided the X minus A into it, X minus A into it, and we would have gotten, we would have gotten a quotient here, Q of X, and we would have done all this business down here, all this algebraic long division, probably wouldn't have even fit on the page. But eventually, we would have gotten to a point where we got an expression that has a lower degree than this, and it would have to be a constant, because this is a first degree, so it would have to be essentially a zero degree."}, {"video_title": "Polynomial remainder theorem example Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And that was pretty neat, because if we did, if we tried, if we attempted to do this without the polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the quotient and all of that, but we don't need, we don't need the quotient. We don't need to know, so if we did all the algebraic long division, we would have taken our P of X, and then we would have divided the X minus A into it, X minus A into it, and we would have gotten, we would have gotten a quotient here, Q of X, and we would have done all this business down here, all this algebraic long division, probably wouldn't have even fit on the page. But eventually, we would have gotten to a point where we got an expression that has a lower degree than this, and it would have to be a constant, because this is a first degree, so it would have to be essentially a zero degree. So we would have eventually gotten to our negative 27. But this was much, much, much, much easier than having to go through this entire exercise. Hopefully you appreciated that."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So I have a function, well they've already used f, g, and h, so I'll use j. So a function j is odd if you evaluate j at some value, so let's say j of a, and if you evaluate that j at the negative of that value, and if these two things are the negative of each other, then my function is odd. If these two things were the same, if you didn't have this negative here, then it would be an even function. So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point, so let's say when x is equal to 2. So we get f of 2, f of 2 is equal to 2."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So let's see which of these meet the criteria of being odd. So let's look at f of x. So we could pick a particular point, so let's say when x is equal to 2. So we get f of 2, f of 2 is equal to 2. Now what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So we get f of 2, f of 2 is equal to 2. Now what is f of negative 2? f of negative 2 looks like it is 6. f of negative 2 is equal to 6. So these aren't the negative of each other. In order for this to be odd, f of negative 2 would have had to be equal to the negative of this, would have had to be equal to negative 2. So f of x is definitely not odd. So all I had to do was find even one case that violated this constraint to be odd, and so I can say it's definitely not odd."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So these aren't the negative of each other. In order for this to be odd, f of negative 2 would have had to be equal to the negative of this, would have had to be equal to negative 2. So f of x is definitely not odd. So all I had to do was find even one case that violated this constraint to be odd, and so I can say it's definitely not odd. Now let's look at g of x. So I could use the same, let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So all I had to do was find even one case that violated this constraint to be odd, and so I can say it's definitely not odd. Now let's look at g of x. So I could use the same, let's see, when x is equal to 2, we get g of 2 is equal to negative 7. Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation, and it looks like that's the case for any x we pick, that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Now let's look at when g is negative 2. So we get g of negative 2 is also equal to negative 7. So here we have a situation, and it looks like that's the case for any x we pick, that g of x is going to be equal to g of negative x. So g of x is equal to g of negative x. It's symmetric around the y, or I should say the vertical axis right over here. So g of x is even, not odd. So which of these functions is odd?"}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So g of x is equal to g of negative x. It's symmetric around the y, or I should say the vertical axis right over here. So g of x is even, not odd. So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So which of these functions is odd? Definitely not g of x. So our last hope is h of x. Let's see if h of x seems to meet the criteria. So if we, I'll do it in this green color. So if we take h of 1, and we can look at it even visually. So h of 1 gets us right over here."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Let's see if h of x seems to meet the criteria. So if we, I'll do it in this green color. So if we take h of 1, and we can look at it even visually. So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1, for 2. Well, 2 is at the x-axis, but that's definitely h of 2 is 0, h of negative 2 is 0, but those are the negatives of each other."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So h of 1 gets us right over here. h of negative 1 seems to get us an equal amount, an equal distance, negative. So it seems to fit for 1, for 2. Well, 2 is at the x-axis, but that's definitely h of 2 is 0, h of negative 2 is 0, but those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number, and h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "Well, 2 is at the x-axis, but that's definitely h of 2 is 0, h of negative 2 is 0, but those are the negatives of each other. 0 is equal to negative 0. If we go to, say, h of 4, h of 4 is this negative number, and h of negative 4 seems to be a positive number of the same magnitude. So once again, this is the negative of this. So it looks like this is indeed an odd function. And another way to visually spot an odd function is a function, it's going to go through the origin, and you could essentially flip it over on both axes. So if you flip this, the right half, over the left half, and then flip that over the horizontal axis, you are going to get this right over here."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So once again, this is the negative of this. So it looks like this is indeed an odd function. And another way to visually spot an odd function is a function, it's going to go through the origin, and you could essentially flip it over on both axes. So if you flip this, the right half, over the left half, and then flip that over the horizontal axis, you are going to get this right over here. So you see here we're going up and to the right, here we're going to go down and to the left, and then you curve right over there, you curve up just like that. But the easiest way to test it is just to do what we did. Look at a given x."}, {"video_title": "Recognizing features of functions (example 2) Algebra II Khan Academy.mp3", "Sentence": "So if you flip this, the right half, over the left half, and then flip that over the horizontal axis, you are going to get this right over here. So you see here we're going up and to the right, here we're going to go down and to the left, and then you curve right over there, you curve up just like that. But the easiest way to test it is just to do what we did. Look at a given x. So for example, when x is equal to 8, h of 8 looks like this number right around 8. h of negative 8 looks like it's pretty close to negative 8. So they seem to be the negative of each other. It sounds like a car crash just happened outside."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "We're told we want to find the zeros of this polynomial, and they give us the polynomial right over here, and it's in factored form. And they say plot all the zeros, or the x-intercepts, of the polynomial in the interactive graph. And so this is a screenshot from Khan Academy. If you were doing it on Khan Academy, you would click where the zeros are to plot the zeros, but I'm just going to draw it in. So pause this video and see if you can have a go at this before we work on this together. All right, now let's work on this together. So the zeros are the x-values that make our polynomial equal to zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "If you were doing it on Khan Academy, you would click where the zeros are to plot the zeros, but I'm just going to draw it in. So pause this video and see if you can have a go at this before we work on this together. All right, now let's work on this together. So the zeros are the x-values that make our polynomial equal to zero. So another way to think about it is, for what x-values are p of x equal to zero? Those would be the zeros. So essentially we have to say, hey, what x-values would make two x times two x plus three times x minus two, because this is p of x, what x-values would make this equal to zero?"}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So the zeros are the x-values that make our polynomial equal to zero. So another way to think about it is, for what x-values are p of x equal to zero? Those would be the zeros. So essentially we have to say, hey, what x-values would make two x times two x plus three times x minus two, because this is p of x, what x-values would make this equal to zero? Well, as we've talked about in previous videos, if you take the product of things and that equal's zero, if any one of those things equals zero, at least one of those things equals zero, it would make the whole product equal to zero. So for example, if two x is equal to zero, it would make the whole thing zero. So two x could be equal to zero, and if two x is equal to zero that means x is equal to zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So essentially we have to say, hey, what x-values would make two x times two x plus three times x minus two, because this is p of x, what x-values would make this equal to zero? Well, as we've talked about in previous videos, if you take the product of things and that equal's zero, if any one of those things equals zero, at least one of those things equals zero, it would make the whole product equal to zero. So for example, if two x is equal to zero, it would make the whole thing zero. So two x could be equal to zero, and if two x is equal to zero that means x is equal to zero. And you could try that out, if x is equal to zero this part right over here is going to be equal to zero, it doesn't matter what these other two things are. Zero times something times something is going to be equal to zero. And then you could say, well, well maybe two x plus three is equal to zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So two x could be equal to zero, and if two x is equal to zero that means x is equal to zero. And you could try that out, if x is equal to zero this part right over here is going to be equal to zero, it doesn't matter what these other two things are. Zero times something times something is going to be equal to zero. And then you could say, well, well maybe two x plus three is equal to zero. So we could just write that, two x plus three is equal to zero and if that were true, what would x, or what would x have to be in order to make that true? Subtract three from both sides, two x would have to be equal to negative three or x would be equal to negative 3 1\u20442. So this is another x value that would make the whole thing zero because if x is equal to negative 3 1\u20442, then two x plus three is equal to zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then you could say, well, well maybe two x plus three is equal to zero. So we could just write that, two x plus three is equal to zero and if that were true, what would x, or what would x have to be in order to make that true? Subtract three from both sides, two x would have to be equal to negative three or x would be equal to negative 3 1\u20442. So this is another x value that would make the whole thing zero because if x is equal to negative 3 1\u20442, then two x plus three is equal to zero. You take a zero times whatever this is and whatever that is, you're gonna get zero. And then last but not least, x minus two could be equal to zero. That would make the whole product equal to zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So this is another x value that would make the whole thing zero because if x is equal to negative 3 1\u20442, then two x plus three is equal to zero. You take a zero times whatever this is and whatever that is, you're gonna get zero. And then last but not least, x minus two could be equal to zero. That would make the whole product equal to zero. So what x value makes x minus two equal zero? Well, add two to both sides and you would get x is equal to two. If x equals two, that equals zero."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "That would make the whole product equal to zero. So what x value makes x minus two equal zero? Well, add two to both sides and you would get x is equal to two. If x equals two, that equals zero. Doesn't matter what these other two things are. Zero times something times something is going to be equal to zero. So just like that, we have the zeros of our polynomial."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "If x equals two, that equals zero. Doesn't matter what these other two things are. Zero times something times something is going to be equal to zero. So just like that, we have the zeros of our polynomial. And the reason why they have x intercepts in parentheses here is that's where the graph of P of x, if you say y equals P of x, that's where it would intersect the x-axis. And that's because that's where our polynomial is equal to zero. So let's see, we have x equals zero, which is right over there."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So just like that, we have the zeros of our polynomial. And the reason why they have x intercepts in parentheses here is that's where the graph of P of x, if you say y equals P of x, that's where it would intersect the x-axis. And that's because that's where our polynomial is equal to zero. So let's see, we have x equals zero, which is right over there. Once again, if you were doing this on Khan Academy, you would just click right over there and it would put a little dot there. We have x is equal to negative 3 1\u20442, which is the same thing as negative 1 1\u20442, so that's right over there. And then we have x equals two, which is right over there."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "So let's see, we have x equals zero, which is right over there. Once again, if you were doing this on Khan Academy, you would just click right over there and it would put a little dot there. We have x is equal to negative 3 1\u20442, which is the same thing as negative 1 1\u20442, so that's right over there. And then we have x equals two, which is right over there. So those are the x intercepts, or the zeros of that polynomial. Now, this is useful in life because you could use it to graph a function. I don't know exactly what this function looks like."}, {"video_title": "Zeros of polynomials plotting zeros Polynomial graphs Algebra 2 Khan Academy.mp3", "Sentence": "And then we have x equals two, which is right over there. So those are the x intercepts, or the zeros of that polynomial. Now, this is useful in life because you could use it to graph a function. I don't know exactly what this function looks like. Maybe it looks something like this. Maybe it looks something like this. We would have to try out a few other values to get a sense of that, but we at least know where it's intersecting the x-axis."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Which of the following is the graph of y is equal to two times the square root of negative x minus one and they give us some choices here and so I encourage you to pause this video and try to figure it out on your own before we work through this together. All right, now let's work through this together and the way that I'm going to do it is I'm actually going to try to draw what the graph of two times the square root of negative x minus one should look like and then I'll just look at which of the choices is closest to what I drew. And the way that I'm going to do that is I'm going to do it step by step. So we already see what y equals the square root of x looks like. But let's say we just want to build up. So let's say we want to now figure out what is the graph of y is equal to the square root of, instead of an x under the radical sign, let me put a negative x under the radical sign. What would that do to it?"}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So we already see what y equals the square root of x looks like. But let's say we just want to build up. So let's say we want to now figure out what is the graph of y is equal to the square root of, instead of an x under the radical sign, let me put a negative x under the radical sign. What would that do to it? Well, whatever was happening at a certain value of x will now happen at the negative of that value of x. So square root of x is not defined for negative numbers. Now this one won't be defined for positive numbers."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "What would that do to it? Well, whatever was happening at a certain value of x will now happen at the negative of that value of x. So square root of x is not defined for negative numbers. Now this one won't be defined for positive numbers. And the behavior that you saw at x equals two, you would now see at x equals negative two. The behavior that you saw at x equals four, you will now see at x equals negative four. And so on and so forth."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Now this one won't be defined for positive numbers. And the behavior that you saw at x equals two, you would now see at x equals negative two. The behavior that you saw at x equals four, you will now see at x equals negative four. And so on and so forth. So the y equals the square root of negative x is going to look like this. You've essentially flipped it over the y, we have flipped it over the y-axis. All right, so we've done this part."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "And so on and so forth. So the y equals the square root of negative x is going to look like this. You've essentially flipped it over the y, we have flipped it over the y-axis. All right, so we've done this part. Now let's scale that. Now let's multiply that by two. So what would y is equal to two times the square root of negative x look like?"}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "All right, so we've done this part. Now let's scale that. Now let's multiply that by two. So what would y is equal to two times the square root of negative x look like? Well, it would look like this red curve, but at any given x value, we're gonna get twice as high. So at x equals negative four, instead of getting to two, we're now going to get to four. At x equals negative nine, instead of getting to three, we're now going to get to six."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So what would y is equal to two times the square root of negative x look like? Well, it would look like this red curve, but at any given x value, we're gonna get twice as high. So at x equals negative four, instead of getting to two, we're now going to get to four. At x equals negative nine, instead of getting to three, we're now going to get to six. Now at x equals zero, we're still going to be at zero because two times zero is zero. So it's going to look like that. Something like that."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "At x equals negative nine, instead of getting to three, we're now going to get to six. Now at x equals zero, we're still going to be at zero because two times zero is zero. So it's going to look like that. Something like that. So that's y equals two times the square root of negative x. And then last but not least, what will y, let me do that in a different color, what will y equals two times the square root of negative x minus one look like? Well, whatever y value we were getting before, we're now just going to shift everything down by one."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Something like that. So that's y equals two times the square root of negative x. And then last but not least, what will y, let me do that in a different color, what will y equals two times the square root of negative x minus one look like? Well, whatever y value we were getting before, we're now just going to shift everything down by one. So if we were at six before, we're going to be at five now. If we were at four before, we're now going to be at three. If we were at zero before, we're now going to be at negative one."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Well, whatever y value we were getting before, we're now just going to shift everything down by one. So if we were at six before, we're going to be at five now. If we were at four before, we're now going to be at three. If we were at zero before, we're now going to be at negative one. And so our curve is going to look something like that. So let's look for, let's see which choices match that. So let me scroll down here."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "If we were at zero before, we're now going to be at negative one. And so our curve is going to look something like that. So let's look for, let's see which choices match that. So let me scroll down here. And both C and D kind of look right, but notice, right at zero, we wanted to be at negative one. So D is exactly what we had drawn. And at nine, we're at five, or at negative nine, we're at five."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So let me scroll down here. And both C and D kind of look right, but notice, right at zero, we wanted to be at negative one. So D is exactly what we had drawn. And at nine, we're at five, or at negative nine, we're at five. At negative four, we're at three. And at zero, we're at negative one, exactly what we had drawn. Let's do another example."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "And at nine, we're at five, or at negative nine, we're at five. At negative four, we're at three. And at zero, we're at negative one, exactly what we had drawn. Let's do another example. So here, this is a similar question. Now they graphed the cube root of x. Y is equal to the cube root of x. And then they say, which of the following is the graph of this business?"}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Let's do another example. So here, this is a similar question. Now they graphed the cube root of x. Y is equal to the cube root of x. And then they say, which of the following is the graph of this business? And they give us choices again. So once again, pause this video and try to work it out on your own before we do this together. All right, now let's work on this together."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "And then they say, which of the following is the graph of this business? And they give us choices again. So once again, pause this video and try to work it out on your own before we do this together. All right, now let's work on this together. And I'm gonna do the same technique. I'm just gonna build it up piece by piece. So this is already y is equal to the cube root of x."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "All right, now let's work on this together. And I'm gonna do the same technique. I'm just gonna build it up piece by piece. So this is already y is equal to the cube root of x. So now let's build up on that. Let's say we wanna now have an x plus two under the radical sign. So let's graph y is equal to the cube root of x plus two."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So this is already y is equal to the cube root of x. So now let's build up on that. Let's say we wanna now have an x plus two under the radical sign. So let's graph y is equal to the cube root of x plus two. Well, what this does is it shifts the curve two to the left. And we've gone over this in multiple videos before. So we are now here."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So let's graph y is equal to the cube root of x plus two. Well, what this does is it shifts the curve two to the left. And we've gone over this in multiple videos before. So we are now here. And you can even try some values out to verify that. At x equals zero, at x equals zero, or actually, let me put it this way. At x equals negative two, you're gonna take the cube root of zero, which is right over there."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So we are now here. And you can even try some values out to verify that. At x equals zero, at x equals zero, or actually, let me put it this way. At x equals negative two, you're gonna take the cube root of zero, which is right over there. So we have now shifted two to the left to look something like this. And now let's build up on that. Let's multiply this times the negative."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "At x equals negative two, you're gonna take the cube root of zero, which is right over there. So we have now shifted two to the left to look something like this. And now let's build up on that. Let's multiply this times the negative. So y is equal to the negative of the cube root of x plus two. What would that look like? Well, if you multiply your whole expression, or in this case, the whole graph or the whole function by a negative, you're gonna flip it over the horizontal axis."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Let's multiply this times the negative. So y is equal to the negative of the cube root of x plus two. What would that look like? Well, if you multiply your whole expression, or in this case, the whole graph or the whole function by a negative, you're gonna flip it over the horizontal axis. And so it is now going to look like this. Whatever y value we're going to get before for a given x, you're now getting the opposite, the negative of it. So it's going to look like that, something like that."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Well, if you multiply your whole expression, or in this case, the whole graph or the whole function by a negative, you're gonna flip it over the horizontal axis. And so it is now going to look like this. Whatever y value we're going to get before for a given x, you're now getting the opposite, the negative of it. So it's going to look like that, something like that. So that is y equal to the negative of the cube root of x plus two. And then last but not least, we are going to think about, and I'm searching for an appropriate color. I haven't used orange yet."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So it's going to look like that, something like that. So that is y equal to the negative of the cube root of x plus two. And then last but not least, we are going to think about, and I'm searching for an appropriate color. I haven't used orange yet. Y is equal to the negative of the cube root of x plus two. And I'm going to add five. So all that's going to do is take this last graph and shift it up by five."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "I haven't used orange yet. Y is equal to the negative of the cube root of x plus two. And I'm going to add five. So all that's going to do is take this last graph and shift it up by five. Whatever y value is going to get before, I'm now going to get five higher. So five higher, let's see, I was at zero here. So now I'm going to be at five here."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So all that's going to do is take this last graph and shift it up by five. Whatever y value is going to get before, I'm now going to get five higher. So five higher, let's see, I was at zero here. So now I'm going to be at five here. So it's going to look, it's going to look something, something like, something like that. I know I'm not drawing it perfectly, but you get the general idea. Now let's look at the choices."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So now I'm going to be at five here. So it's going to look, it's going to look something, something like, something like that. I know I'm not drawing it perfectly, but you get the general idea. Now let's look at the choices. And I think the key point to look at is this point right over here that in our original graph was at zero, zero. Now it is going to be at negative two comma five. So let's look for it."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "Now let's look at the choices. And I think the key point to look at is this point right over here that in our original graph was at zero, zero. Now it is going to be at negative two comma five. So let's look for it. And it also should be flipped. So on the left-hand side, we have the top part. And on the right-hand side, we have the part that goes lower."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "So let's look for it. And it also should be flipped. So on the left-hand side, we have the top part. And on the right-hand side, we have the part that goes lower. So let's see. So A, C, and B all have the left-hand side as the higher part, and then the right-hand side being the lower part. But we wanted this point to be at negative two comma five."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "And on the right-hand side, we have the part that goes lower. So let's see. So A, C, and B all have the left-hand side as the higher part, and then the right-hand side being the lower part. But we wanted this point to be at negative two comma five. A doesn't have it there. B doesn't have it there. D, we already said, goes in the wrong direction."}, {"video_title": "Graphing square and cube root functions Algebra 2 Khan academy.mp3", "Sentence": "But we wanted this point to be at negative two comma five. A doesn't have it there. B doesn't have it there. D, we already said, goes in the wrong direction. It's increasing. So let's see. Negative two comma five, it's indeed what we expected."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "This is a screenshot from an exercise on Khan Academy. It says the intergraphic, the interactive graph below contains the graph of y is equal to log base two of x as a dashed curve, and you can see it down there as that dashed curve, with the points one comma zero and two comma one highlighted. Adjust the movable graph to draw y is equal to four times log base two of x plus six minus seven. And so if you happen to have this exercise in front of you, I encourage you to do that, or if you're just thinking about it in your head, think about how you would approach this, and I'll give you a hint. To go from our original y is equal to log base two of x to all of this, it's really going to be a series of transformations, and on this tool right over here, what we can do is we can move this vertical asymptote around, so that's one thing we can move, and then we can also move two of these points. So where we're starting is right, we are starting right over there, and so let's see, and that was just the graph of y is equal to log base two of x. So let's just do these transformations one at a time."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "And so if you happen to have this exercise in front of you, I encourage you to do that, or if you're just thinking about it in your head, think about how you would approach this, and I'll give you a hint. To go from our original y is equal to log base two of x to all of this, it's really going to be a series of transformations, and on this tool right over here, what we can do is we can move this vertical asymptote around, so that's one thing we can move, and then we can also move two of these points. So where we're starting is right, we are starting right over there, and so let's see, and that was just the graph of y is equal to log base two of x. So let's just do these transformations one at a time. So the first thing I am going to do, instead of just doing log base two of x, let's do log base two of x plus six. So if you replace your x with an x plus six, what is it going to do? Well, it's going to shift everything six to the left, and if that doesn't make intuitive sense to you, I encourage you to watch some of the introductory videos on shifting transformations."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "So let's just do these transformations one at a time. So the first thing I am going to do, instead of just doing log base two of x, let's do log base two of x plus six. So if you replace your x with an x plus six, what is it going to do? Well, it's going to shift everything six to the left, and if that doesn't make intuitive sense to you, I encourage you to watch some of the introductory videos on shifting transformations. So everything is going to shift six to the left. So this vertical asymptote is going to shift six to the left, it's gonna be, instead of being at x equals zero, it's going to go all the way to x equals negative six. This point right over here, which was at one comma zero, it's going to go six to the left."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "Well, it's going to shift everything six to the left, and if that doesn't make intuitive sense to you, I encourage you to watch some of the introductory videos on shifting transformations. So everything is going to shift six to the left. So this vertical asymptote is going to shift six to the left, it's gonna be, instead of being at x equals zero, it's going to go all the way to x equals negative six. This point right over here, which was at one comma zero, it's going to go six to the left. One, two, three, four, five, six, and this point, which was at two comma one, it's going to go six to the left. One, two, three, four, five, and six. So so far what we have graphed is log base two of x plus six."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "This point right over here, which was at one comma zero, it's going to go six to the left. One, two, three, four, five, six, and this point, which was at two comma one, it's going to go six to the left. One, two, three, four, five, and six. So so far what we have graphed is log base two of x plus six. So the next thing we might want to do is, what is four times log base two of x plus six? And the way to think about it is, whatever y value we were getting before, we're now going to get four times that. So when x is equal to negative five, we're getting a y value of zero, but four times zero is still zero, so that point will stay the same."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "So so far what we have graphed is log base two of x plus six. So the next thing we might want to do is, what is four times log base two of x plus six? And the way to think about it is, whatever y value we were getting before, we're now going to get four times that. So when x is equal to negative five, we're getting a y value of zero, but four times zero is still zero, so that point will stay the same. But when x is equal to negative four, we're getting a y value of one. But now that's going to be four times higher, because we're putting that four out front, so instead of being at four, instead of being at one, it's going to be at four. So this right over here is the graph of y is equal to log base two of x plus six."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "So when x is equal to negative five, we're getting a y value of zero, but four times zero is still zero, so that point will stay the same. But when x is equal to negative four, we're getting a y value of one. But now that's going to be four times higher, because we're putting that four out front, so instead of being at four, instead of being at one, it's going to be at four. So this right over here is the graph of y is equal to log base two of x plus six. And then the last thing we have to consider is, well, we're gonna take all of that, and then we're going to subtract seven to get to our target graph. So whatever points we are here, we are now going to subtract seven. So this is at y equals zero, but now we're gonna subtract seven, so we're gonna go down one, two, three, four, five, six, seven."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "So this right over here is the graph of y is equal to log base two of x plus six. And then the last thing we have to consider is, well, we're gonna take all of that, and then we're going to subtract seven to get to our target graph. So whatever points we are here, we are now going to subtract seven. So this is at y equals zero, but now we're gonna subtract seven, so we're gonna go down one, two, three, four, five, six, seven. I went off the screen a little bit, but let me see if I can scroll down a little bit so that you can see that. See, almost, there you go. Now you can see, I moved this down from zero to negative seven, and then this one I have to move down seven, one, two, three, four, five, six, and seven."}, {"video_title": "Graphing logarithmic functions (example 2) Algebra 2 Khan Academy.mp3", "Sentence": "So this is at y equals zero, but now we're gonna subtract seven, so we're gonna go down one, two, three, four, five, six, seven. I went off the screen a little bit, but let me see if I can scroll down a little bit so that you can see that. See, almost, there you go. Now you can see, I moved this down from zero to negative seven, and then this one I have to move down seven, one, two, three, four, five, six, and seven. And we're done. There you have it. That is the graph of y is equal to four times log base two of x plus six minus seven."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And they say that we can factor the expression as u plus v squared, where u and v are either constant integers or single variable expressions. What are u and v? And then they ask us to actually factor the expression. So pause this video and see if you can work on that. All right, so let's go with the first part of it. So they say they can factor the expression as u plus v squared. So how do we see this expression in terms of u plus v squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and see if you can work on that. All right, so let's go with the first part of it. So they say they can factor the expression as u plus v squared. So how do we see this expression in terms of u plus v squared? Well, one way is to just remind ourselves what u plus v squared even is. U plus v squared, this is just going to be the square of a binomial, and we've seen this in many, many other videos. This is going to be u squared plus two times the product of these two terms, so two uv plus v squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So how do we see this expression in terms of u plus v squared? Well, one way is to just remind ourselves what u plus v squared even is. U plus v squared, this is just going to be the square of a binomial, and we've seen this in many, many other videos. This is going to be u squared plus two times the product of these two terms, so two uv plus v squared. If you've never seen this before and you're not sure where this came from, I encourage you to watch some of those early videos where we explain this out. But does this match this pattern? Well, can we express this term as u squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "This is going to be u squared plus two times the product of these two terms, so two uv plus v squared. If you've never seen this before and you're not sure where this came from, I encourage you to watch some of those early videos where we explain this out. But does this match this pattern? Well, can we express this term as u squared? Well, if this is u squared, then u would have to be equal to x plus seven. And when I say, actually, let me be a little careful. Can we express this entire thing right over here as u squared?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, can we express this term as u squared? Well, if this is u squared, then u would have to be equal to x plus seven. And when I say, actually, let me be a little careful. Can we express this entire thing right over here as u squared? If u squared is equal to x plus seven squared, that means that u is going to be equal to x plus seven. And then this right over here would have to be v squared. If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Can we express this entire thing right over here as u squared? If u squared is equal to x plus seven squared, that means that u is going to be equal to x plus seven. And then this right over here would have to be v squared. If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth. So v is equal to y squared. Now, they already told us that this can be factored as the expression u plus v squared, but let's make sure that this actually works. Is this middle term right over here, is this truly equal to two times u times v, two uv?"}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "If this is v squared, then that means that v is equal to y squared, because y squared squared is equal to y to the fourth. So v is equal to y squared. Now, they already told us that this can be factored as the expression u plus v squared, but let's make sure that this actually works. Is this middle term right over here, is this truly equal to two times u times v, two uv? Well, let's see. Two times u would be two times x plus seven times v times y squared. And that's exactly what we have right over here."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Is this middle term right over here, is this truly equal to two times u times v, two uv? Well, let's see. Two times u would be two times x plus seven times v times y squared. And that's exactly what we have right over here. It's two y squared times x plus seven. So this kind of hairy-looking expression actually does fit this pattern right over here. So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And that's exactly what we have right over here. It's two y squared times x plus seven. So this kind of hairy-looking expression actually does fit this pattern right over here. So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared. Now, using that, we can now actually factor the expression. We can write this thing as being equal to u plus v squared. And we know what u and v are."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So you can view it as u plus v squared, where u is equal to x plus seven, and v is equal to y squared. Now, using that, we can now actually factor the expression. We can write this thing as being equal to u plus v squared. And we know what u and v are. So this whole expression is going to be equal to u, which is x plus seven, and I'll put it in parentheses just so you see it very clearly, plus v plus y squared squared, because that's exactly what we wrote over there. And of course, you don't have to write these parentheses. You could rewrite this as x plus seven plus y squared squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "And we know what u and v are. So this whole expression is going to be equal to u, which is x plus seven, and I'll put it in parentheses just so you see it very clearly, plus v plus y squared squared, because that's exactly what we wrote over there. And of course, you don't have to write these parentheses. You could rewrite this as x plus seven plus y squared squared. Let's do another example. So here, once again, we are told that we want to factor the following expression, and they're saying that we can factor the expression as u plus v times u minus v, where u and v are either constant integers or single-variable expressions. So pause this video and try to figure out what u and v are, and then actually factor the expression."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "You could rewrite this as x plus seven plus y squared squared. Let's do another example. So here, once again, we are told that we want to factor the following expression, and they're saying that we can factor the expression as u plus v times u minus v, where u and v are either constant integers or single-variable expressions. So pause this video and try to figure out what u and v are, and then actually factor the expression. All right, well, let's just remind ourselves in general what u plus v times u minus v is equal to. Well, if this is unfamiliar to you, I encourage you to watch the videos on difference of squares. But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video and try to figure out what u and v are, and then actually factor the expression. All right, well, let's just remind ourselves in general what u plus v times u minus v is equal to. Well, if this is unfamiliar to you, I encourage you to watch the videos on difference of squares. But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared. If you actually take the trouble of multiplying this out, you're going to see that that middle term, that middle third terms, or the middle terms, I should say, cancel out, so you're just left with the u squared minus the v squared. And so does this fit this pattern? Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "But when you multiply this all out, this is going to give you a difference of squares, u squared minus v squared. If you actually take the trouble of multiplying this out, you're going to see that that middle term, that middle third terms, or the middle terms, I should say, cancel out, so you're just left with the u squared minus the v squared. And so does this fit this pattern? Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x. Notice, u squared would be two x squared, which is four x squared, and then v would have to be equal to the square root of nine y to the sixth, which the square root of nine is three, and the square root of y to the sixth is going to be y to the third power. And then we could use that to factor the expression, because we could say, hey, this right over here is the same thing as u squared minus v squared, so it's going to be equal to, we can factor it out as, or factor it as, u minus or u plus v times u minus v. So what's that going to be equal to? So u plus v is going to be equal to two x plus three y to the third, and then u minus v is going to be equal to two x, which is our u right over here, minus our v, minus three y to the third."}, {"video_title": "Factorization with substitution Polynomial factorization Algebra 2 Khan Academy.mp3", "Sentence": "Well, in order for this to be u squared and for this to be v squared, that means u squared is equal to four x squared, so that means that u would have to be equal to the square root of that, which would be two times x. Notice, u squared would be two x squared, which is four x squared, and then v would have to be equal to the square root of nine y to the sixth, which the square root of nine is three, and the square root of y to the sixth is going to be y to the third power. And then we could use that to factor the expression, because we could say, hey, this right over here is the same thing as u squared minus v squared, so it's going to be equal to, we can factor it out as, or factor it as, u minus or u plus v times u minus v. So what's that going to be equal to? So u plus v is going to be equal to two x plus three y to the third, and then u minus v is going to be equal to two x, which is our u right over here, minus our v, minus three y to the third. So there you had it. We factored the expression. You might want to write it down here, but we just did it right up there, and we're done."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "We see it's an exponential model here. How many bacteria will make up the culture after 120 minutes? So, really they just want to say, well, what is B of 120 going to be? And so it's going to be 10 times two to the 120 divided by 12th power. So this is going to be equal to 10 times two to the 120 divided by 12 is 10th power. So this is going to be equal to 10 times, two to the 10th power is 1,024. If you want to verify that, you could say, well, two to the fifth is equal to 32."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "And so it's going to be 10 times two to the 120 divided by 12th power. So this is going to be equal to 10 times two to the 120 divided by 12 is 10th power. So this is going to be equal to 10 times, two to the 10th power is 1,024. If you want to verify that, you could say, well, two to the fifth is equal to 32. And so two to the 10th is going to be two to the fifth times two to the fifth. And 32 times 32 is, let's see, 64, 0, 6. So let's see, we're going to have six, sorry."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "If you want to verify that, you could say, well, two to the fifth is equal to 32. And so two to the 10th is going to be two to the fifth times two to the fifth. And 32 times 32 is, let's see, 64, 0, 6. So let's see, we're going to have six, sorry. Three times 32 is 96. And so you have a four, 12, 10, 24. So this is going to be 1024."}, {"video_title": "Exponential model word problem bacteria growth High School Math Khan Academy.mp3", "Sentence": "So let's see, we're going to have six, sorry. Three times 32 is 96. And so you have a four, 12, 10, 24. So this is going to be 1024. 10 times that is going to be equal to 1,024, 0. So 10,200, 10,240 bacteria. And we're done."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's work through this together. And to start, let's just remind ourselves what an average rate of change even is. You can view it as the change in your value of the function for a given change in the underlying variable, for a given change in x. We could also view this as, if we wanna figure out the interval, we could say our x final minus our x initial. And in the numerator, it would be the value of our function at the x final minus the value of the function at our x initial. Now, they aren't asking us to calculate this for all of these different intervals. They're just asking us whether it is positive."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "We could also view this as, if we wanna figure out the interval, we could say our x final minus our x initial. And in the numerator, it would be the value of our function at the x final minus the value of the function at our x initial. Now, they aren't asking us to calculate this for all of these different intervals. They're just asking us whether it is positive. And if you look over here, as long as our x final is greater than x initial, in order to have a positive average rate of change, we just need to figure out whether h at x final is greater than h at x initial. If the value of the function at the higher endpoint is larger than the value of the function at the lower endpoint, then we have a positive average rate of change. So let's see if that's happening for any of these choices."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "They're just asking us whether it is positive. And if you look over here, as long as our x final is greater than x initial, in order to have a positive average rate of change, we just need to figure out whether h at x final is greater than h at x initial. If the value of the function at the higher endpoint is larger than the value of the function at the lower endpoint, then we have a positive average rate of change. So let's see if that's happening for any of these choices. So let's see, h of zero, this endpoint, is going to be equal to zero. If I just say 1 1 8 times zero minus zero. And h of two is equal to 1 1 8 times two to the third power is eight."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So let's see if that's happening for any of these choices. So let's see, h of zero, this endpoint, is going to be equal to zero. If I just say 1 1 8 times zero minus zero. And h of two is equal to 1 1 8 times two to the third power is eight. So 1 1 8 times eight is one minus four. So that's going to be, this is negative three. And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And h of two is equal to 1 1 8 times two to the third power is eight. So 1 1 8 times eight is one minus four. So that's going to be, this is negative three. And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger. This is a negative average rate of change. So I'll rule this one out. And actually, just to help us visualize this, I did go to Desmos and graph this function."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And so we don't have a situation where h at our endpoint, at our higher endpoint, is actually larger. This is a negative average rate of change. So I'll rule this one out. And actually, just to help us visualize this, I did go to Desmos and graph this function. And we can visually see that we have a negative average rate of change from x equals zero to x equals two. At x equals zero, this is where our function is. At x equals two, this is where our function is."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And actually, just to help us visualize this, I did go to Desmos and graph this function. And we can visually see that we have a negative average rate of change from x equals zero to x equals two. At x equals zero, this is where our function is. At x equals two, this is where our function is. And so you can see, at x equals two, our function has a lower value. You could also think of the average rate of change as the slope of the line that connects the two endpoints on the function. And so you can see it has a negative slope."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "At x equals two, this is where our function is. And so you can see, at x equals two, our function has a lower value. You could also think of the average rate of change as the slope of the line that connects the two endpoints on the function. And so you can see it has a negative slope. So we have a negative average rate of change between those two points. Now what about between these two? So h of zero, we already calculated as zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And so you can see it has a negative slope. So we have a negative average rate of change between those two points. Now what about between these two? So h of zero, we already calculated as zero. And what is h of eight? Well, let's see, that's 1 8th times eight to the third power. Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So h of zero, we already calculated as zero. And what is h of eight? Well, let's see, that's 1 8th times eight to the third power. Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power. So that's going to be 64. Minus eight to the second power, minus 64. So that's equal to zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "Well, if I do eight to the third power, but then divide by eight, that's the same thing as eight to the second power. So that's going to be 64. Minus eight to the second power, minus 64. So that's equal to zero. So here, we have a zero average rate of change, because this numerator's going to be zero, so we can rule that out. And you see it right over here. When x is equal to zero, our function's there."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So that's equal to zero. So here, we have a zero average rate of change, because this numerator's going to be zero, so we can rule that out. And you see it right over here. When x is equal to zero, our function's there. When x is equal to eight, our function is there. And you can see that the slope of the line that connects those two points is zero. So you have zero average rate of change between those two points."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "When x is equal to zero, our function's there. When x is equal to eight, our function is there. And you can see that the slope of the line that connects those two points is zero. So you have zero average rate of change between those two points. Now what about choice C? So let's see, h of six is going to be equal to 1 8th times six to the third power. So let's see, 36 times six is 180 plus 36."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So you have zero average rate of change between those two points. Now what about choice C? So let's see, h of six is going to be equal to 1 8th times six to the third power. So let's see, 36 times six is 180 plus 36. So that is going to be 216. 216 minus 36. 216 is six times six times six."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So let's see, 36 times six is 180 plus 36. So that is going to be 216. 216 minus 36. 216 is six times six times six. And then if we divide that by eight, that is going to be the same thing as. This is 36. And then we have 6 8ths of 36."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "216 is six times six times six. And then if we divide that by eight, that is going to be the same thing as. This is 36. And then we have 6 8ths of 36. So this is going to be, this is going to simplify to 3 4ths times 36 minus 36, which is going to be equal to negative nine. You could have done it with a calculator or done some long division. But hopefully what I just did makes some sense."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And then we have 6 8ths of 36. So this is going to be, this is going to simplify to 3 4ths times 36 minus 36, which is going to be equal to negative nine. You could have done it with a calculator or done some long division. But hopefully what I just did makes some sense. It's a little bit of arithmetic. And so h of six, we have our function is negative nine. And then h of eight, I'll draw a line here so we don't make it too messy."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "But hopefully what I just did makes some sense. It's a little bit of arithmetic. And so h of six, we have our function is negative nine. And then h of eight, I'll draw a line here so we don't make it too messy. H of eight, we already know, is equal to zero. So our function at this end point is higher than the value of our function at this end point. So we do have a positive average rate of change."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "And then h of eight, I'll draw a line here so we don't make it too messy. H of eight, we already know, is equal to zero. So our function at this end point is higher than the value of our function at this end point. So we do have a positive average rate of change. So I would pick that choice right over there. And you could see it visually. H of six, when x is equal to six, our value of our function is negative nine."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So we do have a positive average rate of change. So I would pick that choice right over there. And you could see it visually. H of six, when x is equal to six, our value of our function is negative nine. And when x is equal to eight, our value of our function is zero. And so the line that connects those two points definitely has a positive slope. So we have a positive average rate of change over that interval."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "H of six, when x is equal to six, our value of our function is negative nine. And when x is equal to eight, our value of our function is zero. And so the line that connects those two points definitely has a positive slope. So we have a positive average rate of change over that interval. Now, if we were just doing this on our own, we'd be done. But we could just check this one right over here. If we compare h of zero, we already know is zero."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "So we have a positive average rate of change over that interval. Now, if we were just doing this on our own, we'd be done. But we could just check this one right over here. If we compare h of zero, we already know is zero. And h of six, we already know, is equal to negative nine. So this is a negative average rate of change because at the higher end point right over here, we have a lower value of our function. So we'd rule this out."}, {"video_title": "Sign of average rate of change of polynomial Algebra 2 Khan Academy.mp3", "Sentence": "If we compare h of zero, we already know is zero. And h of six, we already know, is equal to negative nine. So this is a negative average rate of change because at the higher end point right over here, we have a lower value of our function. So we'd rule this out. And you see it right over here. If you go from x equals zero where the function is to x equals six where the function is, it looks something, it looks something like that. Clearly, that line has a negative slope."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "I encourage you to pause the video and see if you can solve for x before we work through it together. All right, so one thing we could do is we could try to isolate each of the radicals on either side of the equation. So let's subtract this one from both sides so I can get it onto the right-hand side, or a version of it on the right-hand side. So I'm subtracting it from the left-hand side and from the right-hand side. And so this is going to get us, that is going to get us. On the left-hand side, I just have square root, these cancel out, so I'm just left with the square root of three x minus seven is going to be equal to this, the negative of the square root of two x minus one. So now we can square both sides."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I'm subtracting it from the left-hand side and from the right-hand side. And so this is going to get us, that is going to get us. On the left-hand side, I just have square root, these cancel out, so I'm just left with the square root of three x minus seven is going to be equal to this, the negative of the square root of two x minus one. So now we can square both sides. And we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here, we're going to get the same value. So the solution we might get might be the version when we're solving for the positive square root, not when we take the negative of it. So we have to test our solutions at the end to make sure that they're actually valid for our original equation."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So now we can square both sides. And we always have to be careful when we're doing that because whether we're squaring the positive or the negative square root here, we're going to get the same value. So the solution we might get might be the version when we're solving for the positive square root, not when we take the negative of it. So we have to test our solutions at the end to make sure that they're actually valid for our original equation. But if you square both sides, on the left-hand side, we're going to get three x minus seven. And on the right-hand side, a negative square is just a positive and the square root of two x minus one squared is going to be two x minus one. And I'll see, we can subtract two x from both sides to get all of our x's on one side."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we have to test our solutions at the end to make sure that they're actually valid for our original equation. But if you square both sides, on the left-hand side, we're going to get three x minus seven. And on the right-hand side, a negative square is just a positive and the square root of two x minus one squared is going to be two x minus one. And I'll see, we can subtract two x from both sides to get all of our x's on one side. So I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get rid of the negative seven. So add seven to both sides."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "And I'll see, we can subtract two x from both sides to get all of our x's on one side. So I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get rid of the negative seven. So add seven to both sides. And we are going to get, we are going to get three x minus two x is x is equal to negative one plus seven. X is equal to six. Now let's verify that this actually works."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So add seven to both sides. And we are going to get, we are going to get three x minus two x is x is equal to negative one plus seven. X is equal to six. Now let's verify that this actually works. So if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus the square root of two times six minus one needs to be equal to zero. So does this actually work out? Three times six minus seven."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now let's verify that this actually works. So if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus the square root of two times six minus one needs to be equal to zero. So does this actually work out? Three times six minus seven. So this is going to be the square root of 11 plus the square root of 11 needs to be equal to zero, which clearly is not going to be equal to zero. This is two square roots of 11, which does not equal zero. So this does not work."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Three times six minus seven. So this is going to be the square root of 11 plus the square root of 11 needs to be equal to zero, which clearly is not going to be equal to zero. This is two square roots of 11, which does not equal zero. So this does not work. And you might say, wait, how did this happen? I did all of this nice, neat algebra. I didn't make any mistakes."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "So this does not work. And you might say, wait, how did this happen? I did all of this nice, neat algebra. I didn't make any mistakes. But I got something that doesn't work. Well, this right here is an extraneous solution. Why is it an extraneous solution?"}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "I didn't make any mistakes. But I got something that doesn't work. Well, this right here is an extraneous solution. Why is it an extraneous solution? Because it's actually the solution to the equation. It's a solution to the equation, the square root of three x minus seven minus the square root of two x minus one is equal to zero. And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there?"}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "Why is it an extraneous solution? Because it's actually the solution to the equation. It's a solution to the equation, the square root of three x minus seven minus the square root of two x minus one is equal to zero. And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there? Well, the key is if when we added, when we took this onto the right-hand side and squared it, well, it all boiled down to this, regardless of what starting point you started with. If you did the exact same thing, you would have gotten to this point right over here. So the solution to this ended up being the solution to this starting point versus the one that we originally started with."}, {"video_title": "Solving square-root equations no solution Mathematics III High School Math Khan Academy.mp3", "Sentence": "And you might say, well, if it's the solution to that, if it's the solution to this thing right over here, how did I get the answer when I'm trying to do algebraic steps there? Well, the key is if when we added, when we took this onto the right-hand side and squared it, well, it all boiled down to this, regardless of what starting point you started with. If you did the exact same thing, you would have gotten to this point right over here. So the solution to this ended up being the solution to this starting point versus the one that we originally started with. So this one, interestingly, has no solutions. And it would actually be fun to think about why it has no solutions. We've shown to a certain degree the only solution you got by taking reasonable algebraic steps is an extraneous one."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Let's say that you are desperate for a dollar, and so you come to me, the local loan shark, and you say, hey, I need to borrow a dollar for a year. And I tell you, well, you know, I'm in a good mood, I'm willing to lend you that dollar that you need for a year, and I will lend it to you for the low interest of 100% per year. So 100%, 100% per year. So how much would you have to pay me in a year? Well, you're going to have to pay the original principal, what I lent you, plus 100% of that. So plus one other dollar, which is clearly going to be equal to $2. Now, you say, well, gee, that's a lot to have to pay, to have to pay back twice what I borrowed, and there's a possibility that I might have the money in six months."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So how much would you have to pay me in a year? Well, you're going to have to pay the original principal, what I lent you, plus 100% of that. So plus one other dollar, which is clearly going to be equal to $2. Now, you say, well, gee, that's a lot to have to pay, to have to pay back twice what I borrowed, and there's a possibility that I might have the money in six months. What kind of a deal can you get me for that, Mr. Loan Shark? And I say, oh, gee, if you're willing to pay back in six months, then I'll just charge you half the interest for half the time. So you borrow $1, and so in six months, I will charge you 50% interest over six months."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Now, you say, well, gee, that's a lot to have to pay, to have to pay back twice what I borrowed, and there's a possibility that I might have the money in six months. What kind of a deal can you get me for that, Mr. Loan Shark? And I say, oh, gee, if you're willing to pay back in six months, then I'll just charge you half the interest for half the time. So you borrow $1, and so in six months, I will charge you 50% interest over six months. This, of course, was one year. And so how much would you have to pay? Well, you would have to pay the original principal, what you borrowed, the $1, plus 50% of that $1."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So you borrow $1, and so in six months, I will charge you 50% interest over six months. This, of course, was one year. And so how much would you have to pay? Well, you would have to pay the original principal, what you borrowed, the $1, plus 50% of that $1. So plus 0.50, and that, of course, is equal to 1.5. So that is equal to $1.50. I'll just write it like this, $1.50."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Well, you would have to pay the original principal, what you borrowed, the $1, plus 50% of that $1. So plus 0.50, and that, of course, is equal to 1.5. So that is equal to $1.50. I'll just write it like this, $1.50. Now, you say, well, gee, that's, I guess, better, but what happens if I don't have the money then, if I still actually need a year? And I say, well, we actually have a system for that. What I'll do is just say that, okay, you don't have the money for me yet."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "I'll just write it like this, $1.50. Now, you say, well, gee, that's, I guess, better, but what happens if I don't have the money then, if I still actually need a year? And I say, well, we actually have a system for that. What I'll do is just say that, okay, you don't have the money for me yet. Essentially, we can think about it that I'll just lend that amount that you need for you for another six months. So we'll lend that out for another six months at the same interest rate, at 50%, for the next six months. And so then you'll owe me the principal, $1.50, plus 50% of the principal, plus 0.75."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "What I'll do is just say that, okay, you don't have the money for me yet. Essentially, we can think about it that I'll just lend that amount that you need for you for another six months. So we'll lend that out for another six months at the same interest rate, at 50%, for the next six months. And so then you'll owe me the principal, $1.50, plus 50% of the principal, plus 0.75. And that gets us to $2.25. So that equals $2.25. Or another way of thinking about it is to go from $1 over the first period, you just multiply that times 1.5."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And so then you'll owe me the principal, $1.50, plus 50% of the principal, plus 0.75. And that gets us to $2.25. So that equals $2.25. Or another way of thinking about it is to go from $1 over the first period, you just multiply that times 1.5. If you're going to grow something by 50%, you just multiply it times 1.5. And if you're going to grow it by another 50%, you can multiply by 1.5 again. So one way of thinking about it is that 50% interest is the same thing as multiplying by 1.5."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Or another way of thinking about it is to go from $1 over the first period, you just multiply that times 1.5. If you're going to grow something by 50%, you just multiply it times 1.5. And if you're going to grow it by another 50%, you can multiply by 1.5 again. So one way of thinking about it is that 50% interest is the same thing as multiplying by 1.5. So if you start with 1 and multiply by 1.5 twice, this is going to be the same thing. $2.25 is going to be 1 multiplied by 1.5 twice. Multiplied twice is the same thing as 1.5 squared."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So one way of thinking about it is that 50% interest is the same thing as multiplying by 1.5. So if you start with 1 and multiply by 1.5 twice, this is going to be the same thing. $2.25 is going to be 1 multiplied by 1.5 twice. Multiplied twice is the same thing as 1.5 squared. And you can see the same thing right over here. 100% is the same thing as multiplying by 2. That's literally multiplying by 1 plus 1."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Multiplied twice is the same thing as 1.5 squared. And you can see the same thing right over here. 100% is the same thing as multiplying by 2. That's literally multiplying by 1 plus 1. So this is multiplying by 2. So you could view this right over here as 1 times 1 to the first power. And you're only doing it over one period over that year."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "That's literally multiplying by 1 plus 1. So this is multiplying by 2. So you could view this right over here as 1 times 1 to the first power. And you're only doing it over one period over that year. And you say, once again, where is that 2? Well, if someone's asking for 100%, that means over the period, you're going to have to pay twice. You're going to have to pay the principal plus 100%."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And you're only doing it over one period over that year. And you say, once again, where is that 2? Well, if someone's asking for 100%, that means over the period, you're going to have to pay twice. You're going to have to pay the principal plus 100%. You're going to have to pay twice what you originally borrowed. If someone's charging you 50%, over every period, you're going to have to pay whatever you borrowed. So that's kind of the one part plus 50% of it."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "You're going to have to pay the principal plus 100%. You're going to have to pay twice what you originally borrowed. If someone's charging you 50%, over every period, you're going to have to pay whatever you borrowed. So that's kind of the one part plus 50% of it. So 1.5 times what you borrowed. So you multiply times 1.5 every time. If you wanted to see how this actually related to the interest, you could view this as, so this right over here is equal to 1 times the interest part is 1 plus 100% divided by one period to the first power."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So that's kind of the one part plus 50% of it. So 1.5 times what you borrowed. So you multiply times 1.5 every time. If you wanted to see how this actually related to the interest, you could view this as, so this right over here is equal to 1 times the interest part is 1 plus 100% divided by one period to the first power. And I know this seems like a crazy way of just rewriting what we just wrote over here, writing 1 plus 1. But you'll see that we can keep writing this as we compound over different periods. This one right over here, we can rewrite as 1 times 1 plus 100%."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "If you wanted to see how this actually related to the interest, you could view this as, so this right over here is equal to 1 times the interest part is 1 plus 100% divided by one period to the first power. And I know this seems like a crazy way of just rewriting what we just wrote over here, writing 1 plus 1. But you'll see that we can keep writing this as we compound over different periods. This one right over here, we can rewrite as 1 times 1 plus 100%. Here we took our 100% for the year and we divided it into two periods, two six-month periods, each of them at 50%. 1 plus 100% over 2 is the same thing as 1.5. And then we compounded it over two periods."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "This one right over here, we can rewrite as 1 times 1 plus 100%. Here we took our 100% for the year and we divided it into two periods, two six-month periods, each of them at 50%. 1 plus 100% over 2 is the same thing as 1.5. And then we compounded it over two periods. And actually, let me do that two periods into a different color. So this, the periods, let me do in this orange color right over here. And you might start to see a pattern forming."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And then we compounded it over two periods. And actually, let me do that two periods into a different color. So this, the periods, let me do in this orange color right over here. And you might start to see a pattern forming. So let's say, well gee, maybe I have the money back in, and you don't really like this because this is $2.25. That was more than the original $2. So you say, well, what if we do this over every 12 months?"}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And you might start to see a pattern forming. So let's say, well gee, maybe I have the money back in, and you don't really like this because this is $2.25. That was more than the original $2. So you say, well, what if we do this over every 12 months? I say, sure, we got a program for that. So after every 12 months, after every month, I should say, I'm just going to charge you 100% divided by 12 interest. So this is equal to 8 1 3rd%."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So you say, well, what if we do this over every 12 months? I say, sure, we got a program for that. So after every 12 months, after every month, I should say, I'm just going to charge you 100% divided by 12 interest. So this is equal to 8 1 3rd%. And taking, or having to pay back the principal plus 8 1 3rd%, that's the same thing as multiplying times 1.083 repeating. So after one month, you would have to pay 1.083 repeating. After two months, and this isn't to scale, that actually looks more like more than two months, but it's not completely at scale."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So this is equal to 8 1 3rd%. And taking, or having to pay back the principal plus 8 1 3rd%, that's the same thing as multiplying times 1.083 repeating. So after one month, you would have to pay 1.083 repeating. After two months, and this isn't to scale, that actually looks more like more than two months, but it's not completely at scale. After two months, you're going to have to multiply by this again. So times 1.083 repeating. And so that would get you to 1.083 repeating squared."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "After two months, and this isn't to scale, that actually looks more like more than two months, but it's not completely at scale. After two months, you're going to have to multiply by this again. So times 1.083 repeating. And so that would get you to 1.083 repeating squared. And if you went all the way down 12 months, so let me get myself some space here. So if you went all the way down 12 months, so let me just, so 12 months, or I should say from the beginning 12 months, so another 10 months. So what's the total interest you would have to pay over a year if you weren't able to keep coming up with the money, if you had to keep re-borrowing, and I kept compounding that interest?"}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And so that would get you to 1.083 repeating squared. And if you went all the way down 12 months, so let me get myself some space here. So if you went all the way down 12 months, so let me just, so 12 months, or I should say from the beginning 12 months, so another 10 months. So what's the total interest you would have to pay over a year if you weren't able to keep coming up with the money, if you had to keep re-borrowing, and I kept compounding that interest? Well, you're going to have to pay 1.083 to the, well, this is for one month, so you can view this as to the 1st power. This is for two months, so you're going to have to pay this to the 12th power. We have compounded over 12 periods, 8 1 3rd percent over 12 periods."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So what's the total interest you would have to pay over a year if you weren't able to keep coming up with the money, if you had to keep re-borrowing, and I kept compounding that interest? Well, you're going to have to pay 1.083 to the, well, this is for one month, so you can view this as to the 1st power. This is for two months, so you're going to have to pay this to the 12th power. We have compounded over 12 periods, 8 1 3rd percent over 12 periods. Or if you wanted to write it in this form right over here, this would be the same thing as our original principle, our original principle, times 1 plus 100 percent divided by 12. So now we've divided our 100 percent into 12 periods, and we're going to compound that 12 times. So we're going to take that to the 12th power."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "We have compounded over 12 periods, 8 1 3rd percent over 12 periods. Or if you wanted to write it in this form right over here, this would be the same thing as our original principle, our original principle, times 1 plus 100 percent divided by 12. So now we've divided our 100 percent into 12 periods, and we're going to compound that 12 times. So we're going to take that to the 12th power. So what is this going to equal to, this business right over here? So we can get a calculator out for that. So I'll get my TI-85 out."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So we're going to take that to the 12th power. So what is this going to equal to, this business right over here? So we can get a calculator out for that. So I'll get my TI-85 out. And so what is this going to be equal to? Well, we could do it a couple of ways, and this is 1.083. Let me actually, this is repeating right over here."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So I'll get my TI-85 out. And so what is this going to be equal to? Well, we could do it a couple of ways, and this is 1.083. Let me actually, this is repeating right over here. Let's get our calculator out. So we could do it a couple of ways. Actually, let me just write it this way."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Let me actually, this is repeating right over here. Let's get our calculator out. So we could do it a couple of ways. Actually, let me just write it this way. You're going to get the same value, and I don't have to rewrite this 1 here. I just did that there to kind of, hopefully you see the kind of structure in this expression. So 1 plus, 100 percent is the same thing as 1."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Actually, let me just write it this way. You're going to get the same value, and I don't have to rewrite this 1 here. I just did that there to kind of, hopefully you see the kind of structure in this expression. So 1 plus, 100 percent is the same thing as 1. So 1 divided by 12 to the 12th power, to the 12th power, 2.613. I'll just round, so approximately, approximately 2.613. Now you say, well, this is an interesting game."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So 1 plus, 100 percent is the same thing as 1. So 1 divided by 12 to the 12th power, to the 12th power, 2.613. I'll just round, so approximately, approximately 2.613. Now you say, well, this is an interesting game. You've almost forgot about your financial troubles, and you're just intrigued by what happens if we keep going this. If we, you know, here we compounded just, we have 100 percent over a year. Here we do 50 percent every six months."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Now you say, well, this is an interesting game. You've almost forgot about your financial troubles, and you're just intrigued by what happens if we keep going this. If we, you know, here we compounded just, we have 100 percent over a year. Here we do 50 percent every six months. Here we do a 12th of 100 percent, 8 and 1 third percent every 12 months, and we get to this number. Well, what happens if we did it every day? Every day."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Here we do 50 percent every six months. Here we do a 12th of 100 percent, 8 and 1 third percent every 12 months, and we get to this number. Well, what happens if we did it every day? Every day. So if I borrowed $1, and I said, well, gee, I'm just going to, each day, I'm just going to charge you, I'm just going to charge you 1,365th of 100 percent. So 100 percent divided by 365, and I'm going to compound that 365 times. And so you're curious mathematically, and you say, well, what do we get then?"}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Every day. So if I borrowed $1, and I said, well, gee, I'm just going to, each day, I'm just going to charge you, I'm just going to charge you 1,365th of 100 percent. So 100 percent divided by 365, and I'm going to compound that 365 times. And so you're curious mathematically, and you say, well, what do we get then? What do we get after a year? Well, you have your original principle. Let me scroll over a little bit more to the right so we have more space."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And so you're curious mathematically, and you say, well, what do we get then? What do we get after a year? Well, you have your original principle. Let me scroll over a little bit more to the right so we have more space. You're going to have your original principle times 1 plus 100 percent divided by not 12, now we've divided the 100 percent into 365 periods. But we're going to compound it. Remember, every time we have to multiply by 1 plus 100 percent over 365 every day that the loan is not paid."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Let me scroll over a little bit more to the right so we have more space. You're going to have your original principle times 1 plus 100 percent divided by not 12, now we've divided the 100 percent into 365 periods. But we're going to compound it. Remember, every time we have to multiply by 1 plus 100 percent over 365 every day that the loan is not paid. So to the 365th power. And you say, oh, gee, taking something to the 365th power, that's going to give me some huge number. But then you say, well, actually, maybe not so bad, because 100 percent divided by 365 is going to be a small number."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Remember, every time we have to multiply by 1 plus 100 percent over 365 every day that the loan is not paid. So to the 365th power. And you say, oh, gee, taking something to the 365th power, that's going to give me some huge number. But then you say, well, actually, maybe not so bad, because 100 percent divided by 365 is going to be a small number. This thing is going to be reasonably close to 1. And obviously, we could raise 1 to whatever power we want, and we don't get anything crazy. So let's see where this one goes."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "But then you say, well, actually, maybe not so bad, because 100 percent divided by 365 is going to be a small number. This thing is going to be reasonably close to 1. And obviously, we could raise 1 to whatever power we want, and we don't get anything crazy. So let's see where this one goes. So this is the same thing as 1 plus 100 percent is the same thing as 1 divided by 365 to the 365th power. And we get 2.71456. Let me put it over here."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "So let's see where this one goes. So this is the same thing as 1 plus 100 percent is the same thing as 1 divided by 365 to the 365th power. And we get 2.71456. Let me put it over here. So then we get, so this is approximately equal to 2. And this approximate is a very precise approximation. But 2.7, but my calculator's precision only goes so far."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "Let me put it over here. So then we get, so this is approximately equal to 2. And this approximate is a very precise approximation. But 2.7, but my calculator's precision only goes so far. 2.7145675. And it keeps going on and on. And this is really, really interesting."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "But 2.7, but my calculator's precision only goes so far. 2.7145675. And it keeps going on and on. And this is really, really interesting. Because as we take larger and larger numbers here, it doesn't just balloon into just some crazy ginormous number. It seems to be approaching some magical and mystical number. And it is, in fact, the case."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And this is really, really interesting. Because as we take larger and larger numbers here, it doesn't just balloon into just some crazy ginormous number. It seems to be approaching some magical and mystical number. And it is, in fact, the case. That if you were to just take larger and larger, if you were to take your 100 percent and divide by larger and larger numbers, but take it to that power, you're going to approach perhaps the most magical and mystical number of all, the number e. And you could see it right over here in your calculator. They have this e to the x, and I can do that. So e to the, I'll raise it to the first power so you can look at the calculator's internal representation of it."}, {"video_title": "e and compound interest Interest and debt Finance & Capital Markets Khan Academy.mp3", "Sentence": "And it is, in fact, the case. That if you were to just take larger and larger, if you were to take your 100 percent and divide by larger and larger numbers, but take it to that power, you're going to approach perhaps the most magical and mystical number of all, the number e. And you could see it right over here in your calculator. They have this e to the x, and I can do that. So e to the, I'll raise it to the first power so you can look at the calculator's internal representation of it. And you see, just already raising some, doing 1 plus 1 over 365 to the 365th power, we got pretty, we're starting to get really, really, really, really close to e. And I encourage you to try this with larger and larger numbers. And you're going to get closer and closer to this magical mystery. You almost wouldn't mind paying the loan shark e dollars because it's such a beautiful number."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And just to explain how this widget works, if you're trying to do it on Khan Academy, this dot right over here helps define the midline. You can move that up and down. And then this one right over here is a neighboring extreme point. So either a minimum or a maximum point. So there's a couple of ways that we could approach this. First of all, let's just think about what would cosine of pi x look like? And then we'll think about what the negative does and the plus 1.5."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So either a minimum or a maximum point. So there's a couple of ways that we could approach this. First of all, let's just think about what would cosine of pi x look like? And then we'll think about what the negative does and the plus 1.5. So cosine of pi x, when x is equal to zero, pi times zero is just going to be zero, cosine of zero is equal to one. And if we're just talking about cosine of pi x, that's going to be a maximum point when you hit one. Just cosine of pi x would oscillate between one and negative one."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And then we'll think about what the negative does and the plus 1.5. So cosine of pi x, when x is equal to zero, pi times zero is just going to be zero, cosine of zero is equal to one. And if we're just talking about cosine of pi x, that's going to be a maximum point when you hit one. Just cosine of pi x would oscillate between one and negative one. And then what would its period be if we're talking about cosine of pi x? Well, you might remember one way to think about the period is to take two pi and divide it by whatever the coefficient is on the x right over here. So two pi divided by pi would tell us that we have a period of two."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "Just cosine of pi x would oscillate between one and negative one. And then what would its period be if we're talking about cosine of pi x? Well, you might remember one way to think about the period is to take two pi and divide it by whatever the coefficient is on the x right over here. So two pi divided by pi would tell us that we have a period of two. And so how do we construct a period of two here? Well, that means that as we start here at x equals zero, we're at one, we want to get back to that maximum point by the time x is equal to two. So let me see how I can do that."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So two pi divided by pi would tell us that we have a period of two. And so how do we construct a period of two here? Well, that means that as we start here at x equals zero, we're at one, we want to get back to that maximum point by the time x is equal to two. So let me see how I can do that. If I were to squeeze it a little bit, that looks pretty good. And the reason why I worked on this midline point is I liked having this maximum point at one when x is equal to zero, because we said cosine of pi times zero should be equal to one. So that's why I'm just manipulating this other point in order to set the period right."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So let me see how I can do that. If I were to squeeze it a little bit, that looks pretty good. And the reason why I worked on this midline point is I liked having this maximum point at one when x is equal to zero, because we said cosine of pi times zero should be equal to one. So that's why I'm just manipulating this other point in order to set the period right. But this looks right. We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed two. So this is what the graph of cosine of pi x would look like."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So that's why I'm just manipulating this other point in order to set the period right. But this looks right. We're going from this maximum point, we're going all the way down and then back to that maximum point, and it looks like our period is indeed two. So this is what the graph of cosine of pi x would look like. Now, what about this negative sign? Well, the negative would essentially flip it around. So instead of whenever we're equaling one, we should be equal to negative one."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So this is what the graph of cosine of pi x would look like. Now, what about this negative sign? Well, the negative would essentially flip it around. So instead of whenever we're equaling one, we should be equal to negative one. And every time we're equal to negative one, we should be equal to one. So what I could do is I could just take that and then bring it down here. And there you have it, I flipped it around."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So instead of whenever we're equaling one, we should be equal to negative one. And every time we're equal to negative one, we should be equal to one. So what I could do is I could just take that and then bring it down here. And there you have it, I flipped it around. So this is the graph of y equals negative cosine of pi x. And then last but not least, we have this plus 1.5. So that's just going to shift everything up by 1.5."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "And there you have it, I flipped it around. So this is the graph of y equals negative cosine of pi x. And then last but not least, we have this plus 1.5. So that's just going to shift everything up by 1.5. So I'm just gonna shift everything up by, shift it up by 1.5, and shift it up by 1.5. And there you have it. That is the graph of negative cosine of pi x plus 1.5."}, {"video_title": "Example Graphing y=-cos(\u03c0\u22c5x)+1.5 Trigonometry Algebra 2 Khan Academy.mp3", "Sentence": "So that's just going to shift everything up by 1.5. So I'm just gonna shift everything up by, shift it up by 1.5, and shift it up by 1.5. And there you have it. That is the graph of negative cosine of pi x plus 1.5. And you can validate that that's our midline. We're still oscillating one above and one below. The negative sign, when cosine of pi times zero, that should be one, but then you take the negative of that, we get to negative one."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "So we have the graph here of y is equal to p of x. I could write it like this, y is equal to p of x. And they say what is the remainder when p of x is divided by x plus three? So pause this video and see if you can have a go at this. And they tell us your answer should be an integer. So as you might have assumed, this will involve the polynomial remainder theorem. And all that tells us is that, hey, if we were to take p of x and divide it by x plus three, whatever the remainder is here, so let's say the remainder is equal to k, that value k is what we would have gotten if we took our polynomial and we evaluated it at the value of x that would have made x plus three equal zero, or just what would happen if I evaluated our polynomial at x equals negative three. You have to be very careful there."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "And they tell us your answer should be an integer. So as you might have assumed, this will involve the polynomial remainder theorem. And all that tells us is that, hey, if we were to take p of x and divide it by x plus three, whatever the remainder is here, so let's say the remainder is equal to k, that value k is what we would have gotten if we took our polynomial and we evaluated it at the value of x that would have made x plus three equal zero, or just what would happen if I evaluated our polynomial at x equals negative three. You have to be very careful there. Sometimes people get confused. They see a positive three, and then they evaluate the polynomial at the positive three to figure out the remainder. No, if you saw a positive three there, you would evaluate the polynomial at negative three."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "You have to be very careful there. Sometimes people get confused. They see a positive three, and then they evaluate the polynomial at the positive three to figure out the remainder. No, if you saw a positive three there, you would evaluate the polynomial at negative three. But this should be equal to k as well. And so what is the remainder when p of x is divided by x plus three? Well, it's going to be equal to p of negative three."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "No, if you saw a positive three there, you would evaluate the polynomial at negative three. But this should be equal to k as well. And so what is the remainder when p of x is divided by x plus three? Well, it's going to be equal to p of negative three. P of negative three, it looks like it is equal to negative two, it is equal to negative two. So our remainder is equal to negative two in this situation. Let's do another example."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "Well, it's going to be equal to p of negative three. P of negative three, it looks like it is equal to negative two, it is equal to negative two. So our remainder is equal to negative two in this situation. Let's do another example. Actually, let's do several more examples. Here, we're told that p of x is equal to all of this business where k is an unknown integer. Very interesting."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "Let's do another example. Actually, let's do several more examples. Here, we're told that p of x is equal to all of this business where k is an unknown integer. Very interesting. P of x divided by x minus two has a remainder of one. What is the value of k? So pause this video again, see if you can work it out."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "Very interesting. P of x divided by x minus two has a remainder of one. What is the value of k? So pause this video again, see if you can work it out. All right, well, this second sentence that p of x divided by x minus two has a remainder of one, that tells us that p of, not of negative two, but p of positive two, whatever x value would make this expression equal zero, that p of two is equal to one. And then we could use this top information to figure out what p of two would be. It would be two to the fourth power minus two times two to the third power plus k times two squared, so times two squared, minus 11, and so all of that, that's p of two right over here, that's going to be equal to one."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "So pause this video again, see if you can work it out. All right, well, this second sentence that p of x divided by x minus two has a remainder of one, that tells us that p of, not of negative two, but p of positive two, whatever x value would make this expression equal zero, that p of two is equal to one. And then we could use this top information to figure out what p of two would be. It would be two to the fourth power minus two times two to the third power plus k times two squared, so times two squared, minus 11, and so all of that, that's p of two right over here, that's going to be equal to one. Two to the fourth is 16, and then two times two to the third, that's two to the fourth again, so it's minus 16, plus four k minus 11 is equal to one. These cancel out. And let's see, we can add 11 to both sides of this equation, and we get four k is equal to 12, divide both sides by four, and we get k is equal to three, and we're done."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "It would be two to the fourth power minus two times two to the third power plus k times two squared, so times two squared, minus 11, and so all of that, that's p of two right over here, that's going to be equal to one. Two to the fourth is 16, and then two times two to the third, that's two to the fourth again, so it's minus 16, plus four k minus 11 is equal to one. These cancel out. And let's see, we can add 11 to both sides of this equation, and we get four k is equal to 12, divide both sides by four, and we get k is equal to three, and we're done. Let's do another example. In fact, let's do two more, because we're having so much fun. So this next question tells us p of x is a polynomial, and they tell us what p of x divided by various things are, what the remainder would be when you divide p of x by these various expressions."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "And let's see, we can add 11 to both sides of this equation, and we get four k is equal to 12, divide both sides by four, and we get k is equal to three, and we're done. Let's do another example. In fact, let's do two more, because we're having so much fun. So this next question tells us p of x is a polynomial, and they tell us what p of x divided by various things are, what the remainder would be when you divide p of x by these various expressions. Find the following values of p of x, p of negative four and p of one. Pause this video and see if you can have a go at it. All right, so p of negative four, this is going to be equal to the remainder, remainder, when p of x divided by what?"}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "So this next question tells us p of x is a polynomial, and they tell us what p of x divided by various things are, what the remainder would be when you divide p of x by these various expressions. Find the following values of p of x, p of negative four and p of one. Pause this video and see if you can have a go at it. All right, so p of negative four, this is going to be equal to the remainder, remainder, when p of x divided by what? You might be tempted to say x minus four, but they're trying to trick you intentionally. This would be the remainder when p of x is divided by x plus four. And so they tell us right over here, p of x divided by x plus four has a remainder of three."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "All right, so p of negative four, this is going to be equal to the remainder, remainder, when p of x divided by what? You might be tempted to say x minus four, but they're trying to trick you intentionally. This would be the remainder when p of x is divided by x plus four. And so they tell us right over here, p of x divided by x plus four has a remainder of three. So it's going to be three right over there. And similarly, p of one, this is going to be the remainder, this is the remainder, when p of x divided by, not x plus one, but x minus one. So when p of x is divided by x minus one, the remainder is zero."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "And so they tell us right over here, p of x divided by x plus four has a remainder of three. So it's going to be three right over there. And similarly, p of one, this is going to be the remainder, this is the remainder, when p of x divided by, not x plus one, but x minus one. So when p of x is divided by x minus one, the remainder is zero. Let's do one last example. So once again, p of x is a polynomial. And then they give us a few values of p of x."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "So when p of x is divided by x minus one, the remainder is zero. Let's do one last example. So once again, p of x is a polynomial. And then they give us a few values of p of x. And they say what is the remainder when p of x is divided by x minus three? Pause the video and try to think about that. Well, we've gone over this multiple times."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "And then they give us a few values of p of x. And they say what is the remainder when p of x is divided by x minus three? Pause the video and try to think about that. Well, we've gone over this multiple times. The remainder when p of x is divided by x minus three, that would be p of, not negative three, p of positive three, whatever value makes, whatever value of x makes this entire expression equal zero. So p of positive three, p of positive three is equal to five. And similarly, what is the remainder?"}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "Well, we've gone over this multiple times. The remainder when p of x is divided by x minus three, that would be p of, not negative three, p of positive three, whatever value makes, whatever value of x makes this entire expression equal zero. So p of positive three, p of positive three is equal to five. And similarly, what is the remainder? Actually, no, not so similar. This is interesting. What is the remainder when p of x is divided by x?"}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "And similarly, what is the remainder? Actually, no, not so similar. This is interesting. What is the remainder when p of x is divided by x? I know what you're thinking. It's like, wait, what number am I dealing with? But if I were to rewrite this, instead of saying divided by x, if I were to say divided by x plus zero, then you'd be like, oh, now I get it."}, {"video_title": "Remainder theorem examples Polynomial Division Algebra 2 Khan Academy.mp3", "Sentence": "What is the remainder when p of x is divided by x? I know what you're thinking. It's like, wait, what number am I dealing with? But if I were to rewrite this, instead of saying divided by x, if I were to say divided by x plus zero, then you'd be like, oh, now I get it. Or if I wrote divided by x minus zero, you're like, oh, now I get it. This is going to be p of, and it doesn't matter whether I take a positive or a negative zero, it's going to be p of zero. And p of zero, they tell us, is negative one."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the first thing we have to ask ourselves is what does amplitude even refer to? Well, the amplitude of a periodic function is just half the difference between the minimum and maximum values it takes on. So if I were to draw a periodic function like this, and if we just go back and forth between two, let me draw it a little bit neater, it goes back and forth between two values like that. So between that value and that value, you take the difference between the two, and half of that is the amplitude. Another way of thinking about the amplitude is how much does it sway from its middle position? Right over here, we have y equals negative 1 half cosine of 3x. So what is going to be the amplitude of this?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So between that value and that value, you take the difference between the two, and half of that is the amplitude. Another way of thinking about the amplitude is how much does it sway from its middle position? Right over here, we have y equals negative 1 half cosine of 3x. So what is going to be the amplitude of this? Well, the easy way to think about it is just what is multiplying the cosine function? And you could do the same thing if it was a sine function. We have negative 1 half multiplying it."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what is going to be the amplitude of this? Well, the easy way to think about it is just what is multiplying the cosine function? And you could do the same thing if it was a sine function. We have negative 1 half multiplying it. So the amplitude in this situation is going to be the absolute value of negative 1 half, which is equal to 1 half. And you might say, well, why do I not care about the sine? Why do I take the absolute value of it?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "We have negative 1 half multiplying it. So the amplitude in this situation is going to be the absolute value of negative 1 half, which is equal to 1 half. And you might say, well, why do I not care about the sine? Why do I take the absolute value of it? Well, the negative just flips the function around. It's not going to change how much it sways between its minimum and maximum position. The other thing is, well, how is it just simply the absolute value of this thing?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Why do I take the absolute value of it? Well, the negative just flips the function around. It's not going to change how much it sways between its minimum and maximum position. The other thing is, well, how is it just simply the absolute value of this thing? And to realize the why, you just have to remember that a cosine function or a sine function varies between positive 1 and negative 1 if it's just a simple function. So this is just multiplying that positive 1 or negative 1. And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The other thing is, well, how is it just simply the absolute value of this thing? And to realize the why, you just have to remember that a cosine function or a sine function varies between positive 1 and negative 1 if it's just a simple function. So this is just multiplying that positive 1 or negative 1. And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1. Now you're changing it or you're multiplying it by this amount. So the amplitude is 1 half. Now let's think about the period."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so if normally the amplitude, if you didn't have any coefficient here, if the coefficient was positive or negative 1, the amplitude would just be 1. Now you're changing it or you're multiplying it by this amount. So the amplitude is 1 half. Now let's think about the period. So the first thing I want to ask you is what does the period of a cyclical function, even or a periodic function, I should say, what does the period of a periodic function even refer to? Well, let me draw some axes on this function right over here. Let's say that this right over here is the y-axis."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about the period. So the first thing I want to ask you is what does the period of a cyclical function, even or a periodic function, I should say, what does the period of a periodic function even refer to? Well, let me draw some axes on this function right over here. Let's say that this right over here is the y-axis. That's the y-axis. And let's just say, for the sake of argument, this is the x-axis right over here. So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Let's say that this right over here is the y-axis. That's the y-axis. And let's just say, for the sake of argument, this is the x-axis right over here. So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function. So what do they mean here? Well, what's repeating? So we go down and then up, just like that."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So the period of a periodic function is the length of the smallest interval that contains exactly one copy of the repeating pattern of that periodic function. So what do they mean here? Well, what's repeating? So we go down and then up, just like that. Then we go down and then we go up. So in this case, the length of the smallest interval that contains exactly one copy of the repeating pattern, this could be one of the smallest repeating patterns. And so this length between here and here would be one period."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we go down and then up, just like that. Then we go down and then we go up. So in this case, the length of the smallest interval that contains exactly one copy of the repeating pattern, this could be one of the smallest repeating patterns. And so this length between here and here would be one period. Then we could go between here and here is another period. And there's multiple. This isn't the only pattern that you could pick."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so this length between here and here would be one period. Then we could go between here and here is another period. And there's multiple. This isn't the only pattern that you could pick. You could say, well, I'm going to define my pattern starting here, going up, and then going down, like that. So you could say, that's my smallest length. And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This isn't the only pattern that you could pick. You could say, well, I'm going to define my pattern starting here, going up, and then going down, like that. So you could say, that's my smallest length. And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there. But either way, you're going to get the same length that it takes to repeat that pattern. So given that, what is the period of this function right over here? Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then you would see that, OK, well, if you go in the negative direction, the next repeating version of that pattern is right over there. But either way, you're going to get the same length that it takes to repeat that pattern. So given that, what is the period of this function right over here? Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here. So we divide it by the absolute value of 3, which is just 3. So we get 2 pi over 3. Now, we need to think about why does this work."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, to figure out the period, we just take 2 pi and divide it by the absolute value of the coefficient right over here. So we divide it by the absolute value of 3, which is just 3. So we get 2 pi over 3. Now, we need to think about why does this work. Well, if you think about just a traditional cosine function or a traditional sine function, it has a period of 2 pi. If you think about the unit circle, 2 pi, if you start at 0, 2 pi radians later, you're back to where you started. 2 pi radians, another 2 pi, you're back to where you started."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now, we need to think about why does this work. Well, if you think about just a traditional cosine function or a traditional sine function, it has a period of 2 pi. If you think about the unit circle, 2 pi, if you start at 0, 2 pi radians later, you're back to where you started. 2 pi radians, another 2 pi, you're back to where you started. If you go in the negative direction, you go negative 2 pi, you're back to where you started. For any angle here, if you go 2 pi, you're back to where you were before. You go negative 2 pi, you're back to where you were before."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "2 pi radians, another 2 pi, you're back to where you started. If you go in the negative direction, you go negative 2 pi, you're back to where you started. For any angle here, if you go 2 pi, you're back to where you were before. You go negative 2 pi, you're back to where you were before. So the periods for these are all 2 pi. And the reason why this makes sense is that this coefficient makes you get to 2 pi or negative, in this case, 2 pi, it's going to make you get to 2 pi all that much faster. And so your period is going to be a lower number."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You go negative 2 pi, you're back to where you were before. So the periods for these are all 2 pi. And the reason why this makes sense is that this coefficient makes you get to 2 pi or negative, in this case, 2 pi, it's going to make you get to 2 pi all that much faster. And so your period is going to be a lower number. It takes less length. You're going to get to 2 pi 3 times as fast. Now, you might say, well, why are you taking the absolute value here?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so your period is going to be a lower number. It takes less length. You're going to get to 2 pi 3 times as fast. Now, you might say, well, why are you taking the absolute value here? Well, if this was a negative number, it would get you to negative 2 pi all that much faster. But either way, you're going to be completing one cycle. So with that out of the way, let's visualize these two things."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now, you might say, well, why are you taking the absolute value here? Well, if this was a negative number, it would get you to negative 2 pi all that much faster. But either way, you're going to be completing one cycle. So with that out of the way, let's visualize these two things. Let's actually draw negative 1 half cosine of 3x. So let me draw my axes here, my best attempt. So this is my y-axis."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So with that out of the way, let's visualize these two things. Let's actually draw negative 1 half cosine of 3x. So let me draw my axes here, my best attempt. So this is my y-axis. This is my x-axis. And then let me draw some. So this is 0 right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is my y-axis. This is my x-axis. And then let me draw some. So this is 0 right over here. x is equal to 0. And let me draw x is equal to positive 1 half. I'll draw it right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is 0 right over here. x is equal to 0. And let me draw x is equal to positive 1 half. I'll draw it right over here. So x is equal to positive 1 half. And we haven't shifted this function up or down any. Then if we wanted to, we could add a constant out here, outside of the cosine function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "I'll draw it right over here. So x is equal to positive 1 half. And we haven't shifted this function up or down any. Then if we wanted to, we could add a constant out here, outside of the cosine function. But this is positive 1 half. Or we could just write that as 1 half. And then down here, let's say that this is negative 1 half."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Then if we wanted to, we could add a constant out here, outside of the cosine function. But this is positive 1 half. Or we could just write that as 1 half. And then down here, let's say that this is negative 1 half. And so let me draw that bound. I'm just drawing these dotted lines, so it'll become easy for me to draw. And what happens when this is 0?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then down here, let's say that this is negative 1 half. And so let me draw that bound. I'm just drawing these dotted lines, so it'll become easy for me to draw. And what happens when this is 0? Well, cosine of 0 is 1. But we're going to multiply it by negative 1 half. So it's going to be negative 1 half right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And what happens when this is 0? Well, cosine of 0 is 1. But we're going to multiply it by negative 1 half. So it's going to be negative 1 half right over here. And then it's going to start going up. It can only go in that direction. It's bounded."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it's going to be negative 1 half right over here. And then it's going to start going up. It can only go in that direction. It's bounded. It's going to start going up. Then it'll come back down. And then it will get back to that original point right over here."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's bounded. It's going to start going up. Then it'll come back down. And then it will get back to that original point right over here. And the question is, what is this distance? What is this length? What is this length going to be?"}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then it will get back to that original point right over here. And the question is, what is this distance? What is this length? What is this length going to be? Well, we know what its period is. It's 2 pi over 3. It's going to get to this point 3 times as fast as a traditional cosine function."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "What is this length going to be? Well, we know what its period is. It's 2 pi over 3. It's going to get to this point 3 times as fast as a traditional cosine function. So this is going to be 2 pi over 3. And then if you give it another 2 pi over 3, it's going to get back to that same point again. So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It's going to get to this point 3 times as fast as a traditional cosine function. So this is going to be 2 pi over 3. And then if you give it another 2 pi over 3, it's going to get back to that same point again. So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3. 4 pi over 3, you've completed another cycle. So that length right over there is the period. And you could also do the same thing in the negative direction."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So if you go another 2 pi over 3, so in this case, you've now gone 4 pi over 3. 4 pi over 3, you've completed another cycle. So that length right over there is the period. And you could also do the same thing in the negative direction. So this right over here would be negative 2 pi over 3. And to visualize the amplitude, you see that it can go 1 half. Well, there's two ways to think about it."}, {"video_title": "Example Amplitude and period Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And you could also do the same thing in the negative direction. So this right over here would be negative 2 pi over 3. And to visualize the amplitude, you see that it can go 1 half. Well, there's two ways to think about it. The difference between the maximum and the minimum point is 1. Half of that is 1 half. Or you could say that it's going 1 half in magnitude."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And what they ask us to do is write a formula for the function g in terms of f. So let's think about how to do it. And like always, pause the video and see if you can work through it on your own. All right, well, what I like to do is I like to focus on this minimum point because I think that's a very easy thing to look at because both of them have that minimum point right over there. And so we can think about how do we shift f, at least, especially this minimum point, how do we shift it to get to overlapping with g? Well, the first thing that might jump out at us is that we would want to shift to the left. And we'd want to shift to the left four. So let me do this in a new color."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And so we can think about how do we shift f, at least, especially this minimum point, how do we shift it to get to overlapping with g? Well, the first thing that might jump out at us is that we would want to shift to the left. And we'd want to shift to the left four. So let me do this in a new color. So I would want to shift to the left by four. So we have shifted to the left by four, or you could say we shifted by negative four. Either way, you could think about it."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let me do this in a new color. So I would want to shift to the left by four. So we have shifted to the left by four, or you could say we shifted by negative four. Either way, you could think about it. And then we need to shift down. So we need to shift, we need to go from y equals two to y is equal to negative five. So let me do that."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Either way, you could think about it. And then we need to shift down. So we need to shift, we need to go from y equals two to y is equal to negative five. So let me do that. So let's shift down. So we shift down by seven, or you could say we have a negative seven shift. So how do you express g of x if it's a version of f of x that's shifted to the left by four and shifted down by seven?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let me do that. So let's shift down. So we shift down by seven, or you could say we have a negative seven shift. So how do you express g of x if it's a version of f of x that's shifted to the left by four and shifted down by seven? Or you could say it had a negative four horizontal shift and had a negative seven vertical shift. Well, one way to think about it is g of x is going to be equal to f of, let me do it in a little darker color. It's going to be equal to f of x minus your horizontal shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So how do you express g of x if it's a version of f of x that's shifted to the left by four and shifted down by seven? Or you could say it had a negative four horizontal shift and had a negative seven vertical shift. Well, one way to think about it is g of x is going to be equal to f of, let me do it in a little darker color. It's going to be equal to f of x minus your horizontal shift. I'll write horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's going to be equal to f of x minus your horizontal shift. I'll write horizontal shift. So x minus your horizontal shift plus your vertical shift. So plus your vertical shift. Well, what is our horizontal shift here? Well, we're shifting to the left, so it was a negative shift. So our horizontal shift is negative four."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So plus your vertical shift. Well, what is our horizontal shift here? Well, we're shifting to the left, so it was a negative shift. So our horizontal shift is negative four. Now what's our vertical shift? Well, we went down, so our vertical shift is negative seven. So it's negative seven."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So our horizontal shift is negative four. Now what's our vertical shift? Well, we went down, so our vertical shift is negative seven. So it's negative seven. So there you have it. We get g of x, let me do it in that same color. We get g of x is equal to f of x minus negative four, or x plus four."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So it's negative seven. So there you have it. We get g of x, let me do it in that same color. We get g of x is equal to f of x minus negative four, or x plus four. And then we have plus negative seven, or you could just say minus seven. And we're done. And when I look at things like this, the negative seven is more intuitive to me."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We get g of x is equal to f of x minus negative four, or x plus four. And then we have plus negative seven, or you could just say minus seven. And we're done. And when I look at things like this, the negative seven is more intuitive to me. It's that I shifted it down, it makes sense that I have a negative seven. But at first when you work on these, you say, hey, wait, I shifted to the left. Why is it a plus four?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And when I look at things like this, the negative seven is more intuitive to me. It's that I shifted it down, it makes sense that I have a negative seven. But at first when you work on these, you say, hey, wait, I shifted to the left. Why is it a plus four? And the way I think about it is in order to get the same value out of the function, instead of inputting, so if you want to get the value of f of zero, you now have to put x equals negative four in, and then you get that same value. You still get to zero. So I don't know if that helps or hurts in terms of your understanding, but it often helps to try out some different values for x and seeing how it actually does shift the function."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Why is it a plus four? And the way I think about it is in order to get the same value out of the function, instead of inputting, so if you want to get the value of f of zero, you now have to put x equals negative four in, and then you get that same value. You still get to zero. So I don't know if that helps or hurts in terms of your understanding, but it often helps to try out some different values for x and seeing how it actually does shift the function. And if you're just trying to get your head around this piece, the horizontal shift, I recommend not even using this example. Use an example that only has a horizontal shift, and it'll become a little bit more intuitive. And we have many videos that go into much more depth that explain that."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So I don't know if that helps or hurts in terms of your understanding, but it often helps to try out some different values for x and seeing how it actually does shift the function. And if you're just trying to get your head around this piece, the horizontal shift, I recommend not even using this example. Use an example that only has a horizontal shift, and it'll become a little bit more intuitive. And we have many videos that go into much more depth that explain that. Let's do another example of this. So here we have y is equal to g of x in purple, and y is equal to f of x in blue. And they say given that f of x is equal to square root of x plus four minus two, write an expression for g of x in terms of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And we have many videos that go into much more depth that explain that. Let's do another example of this. So here we have y is equal to g of x in purple, and y is equal to f of x in blue. And they say given that f of x is equal to square root of x plus four minus two, write an expression for g of x in terms of x. So first, let me just write an expression for g of x in terms of f of x. We can see once again, it's just a shifted version of f of x. And remember, and I'll just write in general, so g of x is going to be equal to f of x minus your horizontal shift plus your vertical shift."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And they say given that f of x is equal to square root of x plus four minus two, write an expression for g of x in terms of x. So first, let me just write an expression for g of x in terms of f of x. We can see once again, it's just a shifted version of f of x. And remember, and I'll just write in general, so g of x is going to be equal to f of x minus your horizontal shift plus your vertical shift. Vertical shift. And so to go from f to g, what is your horizontal shift? Well, your horizontal shift is, if you take this point right over here, which should map to that point once we shift everything, your horizontal shift is two to the left."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "And remember, and I'll just write in general, so g of x is going to be equal to f of x minus your horizontal shift plus your vertical shift. Vertical shift. And so to go from f to g, what is your horizontal shift? Well, your horizontal shift is, if you take this point right over here, which should map to that point once we shift everything, your horizontal shift is two to the left. Or you could say it's a negative two horizontal shift, so that should be negative two. And then what is our vertical shift? Well, our vertical shift is we move, we go from y equals negative two to y equals three."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, your horizontal shift is, if you take this point right over here, which should map to that point once we shift everything, your horizontal shift is two to the left. Or you could say it's a negative two horizontal shift, so that should be negative two. And then what is our vertical shift? Well, our vertical shift is we move, we go from y equals negative two to y equals three. So we're shifting five up. So this is a vertical shift of positive five. So your vertical shift is five."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, our vertical shift is we move, we go from y equals negative two to y equals three. So we're shifting five up. So this is a vertical shift of positive five. So your vertical shift is five. So if we just wanted to write g of x in terms of f of x, like we just did in the previous example, we could say g of x is going to be equal to f of x minus negative two, which is x plus two, and then we have plus five. But that's not what they asked us to do. They asked us to write, whoops, they asked us to write an expression for g of x in terms of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So your vertical shift is five. So if we just wanted to write g of x in terms of f of x, like we just did in the previous example, we could say g of x is going to be equal to f of x minus negative two, which is x plus two, and then we have plus five. But that's not what they asked us to do. They asked us to write, whoops, they asked us to write an expression for g of x in terms of x. And so here we're actually going to use the definition of f of x. So let me make it clear. We know that f of x, f of x, is going to be equal to square root of x plus four minus two."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "They asked us to write, whoops, they asked us to write an expression for g of x in terms of x. And so here we're actually going to use the definition of f of x. So let me make it clear. We know that f of x, f of x, is going to be equal to square root of x plus four minus two. So given that, given that, what is f of x plus two? Well, f of x plus two is going to be equal to, everywhere we see an x, we're going to replace it with an x plus two. Square root of x plus two plus four minus two, which is equal to the square root of x plus six minus two."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "We know that f of x, f of x, is going to be equal to square root of x plus four minus two. So given that, given that, what is f of x plus two? Well, f of x plus two is going to be equal to, everywhere we see an x, we're going to replace it with an x plus two. Square root of x plus two plus four minus two, which is equal to the square root of x plus six minus two. Well, that's fair enough. That's just f of x plus two. Now what is f of x plus two plus five?"}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Square root of x plus two plus four minus two, which is equal to the square root of x plus six minus two. Well, that's fair enough. That's just f of x plus two. Now what is f of x plus two plus five? So f of x plus two plus five is going to be this thing right over here, plus five. So it's going to be equal to square root of x plus six minus two, and now we're going to add five. Let me do it in a different color."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Now what is f of x plus two plus five? So f of x plus two plus five is going to be this thing right over here, plus five. So it's going to be equal to square root of x plus six minus two, and now we're going to add five. Let me do it in a different color. So plus five, so plus five. And so what we end up with is going to be square root of x plus six minus two plus five is going to be plus three. So that is equal to g of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "Let me do it in a different color. So plus five, so plus five. And so what we end up with is going to be square root of x plus six minus two plus five is going to be plus three. So that is equal to g of x. Just as a reminder, what did we do here? First I expressed g of x in terms of f of x. I said, hey, to get from f of x to g of x, I shift two to the left, two to the left. It's a little counterintuitive that it's plus two makes it a shift of two to the left."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "So that is equal to g of x. Just as a reminder, what did we do here? First I expressed g of x in terms of f of x. I said, hey, to get from f of x to g of x, I shift two to the left, two to the left. It's a little counterintuitive that it's plus two makes it a shift of two to the left. If this was minus two, it would be a shift of two to the right, but like I just said in the previous example, it's good to try out some x's and to see why that makes sense. And then we shifted five up. So this was g of x in terms of f of x."}, {"video_title": "Shifting functions Mathematics III High School Math Khan Academy.mp3", "Sentence": "It's a little counterintuitive that it's plus two makes it a shift of two to the left. If this was minus two, it would be a shift of two to the right, but like I just said in the previous example, it's good to try out some x's and to see why that makes sense. And then we shifted five up. So this was g of x in terms of f of x. But then they told us what f of x actually is in terms of x. So I said, okay, well, what is f of x plus two? F of x plus two, we substituted x plus two for x and we got this, but g of x is f of x plus two plus five."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "We're told the graph of y equals two to the x is shown below, so that's the graph, it's an exponential function. Which of the following is the graph of y is equal to negative one times two to the x plus three plus four, and they give us four choices down here. And before we even look closely at those choices, let's just think about what this would look like if it was transformed into that. And you might notice that what we have here, this y that we wanna find the graph of, is a transformation of this original one. How do we transform it? Well, we replaced x with x plus three, then we multiplied that by negative one, and then we add four. So let's take it step by step."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "And you might notice that what we have here, this y that we wanna find the graph of, is a transformation of this original one. How do we transform it? Well, we replaced x with x plus three, then we multiplied that by negative one, and then we add four. So let's take it step by step. So this is y equals two to the x. What I wanna do next is, let's graph y is equal to two to the x plus three power. Well, if you replace x with x plus three, you're going to shift the graph to the left, to the left by three."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's take it step by step. So this is y equals two to the x. What I wanna do next is, let's graph y is equal to two to the x plus three power. Well, if you replace x with x plus three, you're going to shift the graph to the left, to the left by three. And that might be a little bit counterintuitive, but when we actually think about some points, it'll hopefully make some sense here. For example, over in our original graph, when x is equal to zero, y is equal to one. Well, how do we get y equal one for our new graph, for this thing right over here?"}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, if you replace x with x plus three, you're going to shift the graph to the left, to the left by three. And that might be a little bit counterintuitive, but when we actually think about some points, it'll hopefully make some sense here. For example, over in our original graph, when x is equal to zero, y is equal to one. Well, how do we get y equal one for our new graph, for this thing right over here? Well, to get y equals one here, the exponent here still has to be zero, so that's going to happen at x equals negative three. So that's going to happen at x equals negative three, y is equal to one. So notice, we shifted to the left by three."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, how do we get y equal one for our new graph, for this thing right over here? Well, to get y equals one here, the exponent here still has to be zero, so that's going to happen at x equals negative three. So that's going to happen at x equals negative three, y is equal to one. So notice, we shifted to the left by three. Likewise, in our original graph, when x is two, y is four. Well, how do we get y equals four in this thing right over here? Well, for y to be equal to four, this exponent here needs to be equal to two, and so for this exponent to be equal to two, because two squared is four, for this exponent to be equal to two, x is going to be equal to negative one."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So notice, we shifted to the left by three. Likewise, in our original graph, when x is two, y is four. Well, how do we get y equals four in this thing right over here? Well, for y to be equal to four, this exponent here needs to be equal to two, and so for this exponent to be equal to two, because two squared is four, for this exponent to be equal to two, x is going to be equal to negative one. So when x is equal to negative one, y is equal to four. When x is equal to negative one, y is equal to four. Notice, we shifted to the left by three."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Well, for y to be equal to four, this exponent here needs to be equal to two, and so for this exponent to be equal to two, because two squared is four, for this exponent to be equal to two, x is going to be equal to negative one. So when x is equal to negative one, y is equal to four. When x is equal to negative one, y is equal to four. Notice, we shifted to the left by three. And so this thing, which isn't our final graph that we're looking for, is going to look something like that, something like that, which shifted, y equals two to the x, shifted to the left by three. Now let's figure out what the graph of, now let's multiply this expression times negative one. Notice, we're slowly building up to our goal."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Notice, we shifted to the left by three. And so this thing, which isn't our final graph that we're looking for, is going to look something like that, something like that, which shifted, y equals two to the x, shifted to the left by three. Now let's figure out what the graph of, now let's multiply this expression times negative one. Notice, we're slowly building up to our goal. So now let's figure out the graph of y is equal to negative one times two to the x plus three. Well, here, when y equals two to the x plus three, if we multiply that times negative one, whatever y we had, we're going to have the negative of that. So instead of when x is equal to negative three having positive one, when x equals negative three, you're going to have negative one."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Notice, we're slowly building up to our goal. So now let's figure out the graph of y is equal to negative one times two to the x plus three. Well, here, when y equals two to the x plus three, if we multiply that times negative one, whatever y we had, we're going to have the negative of that. So instead of when x is equal to negative three having positive one, when x equals negative three, you're going to have negative one. We multiply by negative one. When x is equal to negative one, instead of having four, you're going to have negative four. So our graph is going to be flipped over, it's flipped over the x-axis, and it's going to look something, something, something like this."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So instead of when x is equal to negative three having positive one, when x equals negative three, you're going to have negative one. We multiply by negative one. When x is equal to negative one, instead of having four, you're going to have negative four. So our graph is going to be flipped over, it's flipped over the x-axis, and it's going to look something, something, something like this. And this is not a perfect drawing, but it'll give us a sense of things, and then we can look at which of these graphs match up to that. And then finally, we want to add that four there. So we want to figure out the graph of y equals negative one times two to the x plus three plus four."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So our graph is going to be flipped over, it's flipped over the x-axis, and it's going to look something, something, something like this. And this is not a perfect drawing, but it'll give us a sense of things, and then we can look at which of these graphs match up to that. And then finally, we want to add that four there. So we want to figure out the graph of y equals negative one times two to the x plus three plus four. So we want to take what we just had and shift it up by four. So instead of this being a negative one right over here, this is going to be a negative one plus four is three. Instead of this being a negative four, negative four plus four is zero."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So we want to figure out the graph of y equals negative one times two to the x plus three plus four. So we want to take what we just had and shift it up by four. So instead of this being a negative one right over here, this is going to be a negative one plus four is three. Instead of this being a negative four, negative four plus four is zero. Instead of our horizontal asymptote being at y equals zero, our horizontal asymptote is going to be at y equals four. So it's going to look like, let me draw, I can do a better job than that. Our horizontal asymptote is going to be right over there."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Instead of this being a negative four, negative four plus four is zero. Instead of our horizontal asymptote being at y equals zero, our horizontal asymptote is going to be at y equals four. So it's going to look like, let me draw, I can do a better job than that. Our horizontal asymptote is going to be right over there. So our graph is going to look something like, going to look something like, something like this. We just shifted that red graph up by four. Shifted it up by, shifted it up by four."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Our horizontal asymptote is going to be right over there. So our graph is going to look something like, going to look something like, something like this. We just shifted that red graph up by four. Shifted it up by, shifted it up by four. And we have a horizontal asymptote at y equals four. So let's look at which of these choices match that. So choice A right over here has a horizontal asymptote at y equals four, but it is shifted on the horizontal direction inappropriately."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "Shifted it up by, shifted it up by four. And we have a horizontal asymptote at y equals four. So let's look at which of these choices match that. So choice A right over here has a horizontal asymptote at y equals four, but it is shifted on the horizontal direction inappropriately. In fact, it looks like it might have not been shifted to the left. So we can rule this one out. So let's rule that one out."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So choice A right over here has a horizontal asymptote at y equals four, but it is shifted on the horizontal direction inappropriately. In fact, it looks like it might have not been shifted to the left. So we can rule this one out. So let's rule that one out. This one over here, well this one just, this one, this one approaches our asymptote as x increases, so that's not right. It should approach our asymptote as x decreases, so we rule that one out. Choice C looks like what we just graphed, horizontal asymptote at x equals four."}, {"video_title": "Transforming exponential graphs (example 2) Mathematics III High School Math Khan Academy.mp3", "Sentence": "So let's rule that one out. This one over here, well this one just, this one, this one approaches our asymptote as x increases, so that's not right. It should approach our asymptote as x decreases, so we rule that one out. Choice C looks like what we just graphed, horizontal asymptote at x equals four. When x equals negative three, y is equal to three, that's what we got. When x is equal to negative one, y is equal to zero. So this looks right."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Let's see if we can give ourselves a little bit of practice converting between radians and degrees and degrees and radians. And just as a review, let's just remind ourselves of the relationship. And I always do this before I have to convert between the two. If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "If I do one revolution of a circle, how many radians is that going to be? Well, we know one revolution of a circle is 2 pi radians. And how many degrees is that? If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "If I do one revolution around a circle, well, we know that that's 360. I can either write it with a little degree symbol right like that, or I could write it just like that. And this is really enough information for us to think about how to convert between radians and degrees. If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "If we want to simplify this a little bit, we can divide both sides by 2, and you could have pi radians are equal to 180 degrees, or another way to think about it, going halfway around a circle in radians is pi radians, or the arc that subtends that angle is pi radiuses. And that's also 180 degrees. And if you want to really think about, well, how many degrees are there per radian, you can divide both sides of this by pi. So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So if you divide both sides of this by pi, you get one radian. I have to go from plural to singular. One radian is equal to 180 over pi degrees. So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So all I did is I divided both sides by pi. And if you wanted to figure out how many radians are there per degree, you could divide both sides by 180. So you'd get pi over 180 radians is equal to 1 degree. So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So now I think we are ready to start converting. So let's convert 30 degrees to radians. So let's think about it. So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So I'm going to write it out. And actually, this might remind you of kind of unit analysis that you might do when you first did unit conversion, but it also works out here. So if I were to write 30 degrees, and this is how my brain likes to work with it, I like to write out the word degrees. And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And then I say, well, I want to convert to radians. So I really want to figure out how many radians are there per degree. So let me write this down. I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "I want to figure out how many radians do we have per degree. And I haven't filled out how many that is, but we see just the units will cancel out. If we have degrees times radians per degree, the degrees will cancel out and I'll be just left with radians. If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "If I multiply the number of degrees I have times the number of radians per degree, we're going to get radians. And hopefully that makes intuitive sense as well. And here we just have to think about, well, if I have pi radians, how many degrees is that? Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "Well, that's 180 degrees. It comes straight out of this right over here. Pi radians for every 180 degrees or pi over 180 radians per degree. And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And this is going to get us to 30 times pi over 180, which we'll simplify to 30 over 180 is 1 over 6. So this is equal to pi over 6. Actually, let me write the units out. This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "This is 30 radians, which is equal to pi over 6 radians. Now let's go the other way. Let's think about if we have pi over 3 radians, and I want to convert that to degrees. So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So what am I going to get if I convert that to degrees? Well, here we're going to want to figure out how many degrees are there per radian. And one way to think about it is, well, think about the pi and the 180. For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "For every 180 degrees, you have pi radians. 180 degrees over pi radians, these are essentially the equivalent thing. Essentially, you're just multiplying this quantity by 1, but you're changing the units. The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "The radians cancel out, and then the pi's cancel out, and you're left with 180 over 3 degrees. 180 over 3 is 60, and we could either write out the word degrees, or you can write degrees just like that. Now let's think about 45 degrees. So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So what about 45 degrees? And I'll write it like that just so you can figure it out as they're. Figure it out with that notation as well. How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "How many radians will this be equal to? Well, once again, we're going to want to think about how many radians do we have per degree. So we're going to multiply this times, well, we know we have pi radians for every 180 degrees, or we could even write it this way, pi radians for every 180 degrees. And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "And here, this might be a little less intuitive, the degrees cancel out, and that's why I'd like to usually write out the word, and you're left with 45 pi over 180 radians. Actually, let me write this with the words written out. Maybe that's more intuitive when I'm thinking about it in terms of using the notation. So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians."}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "So 45 degrees times, we have pi radians for every 180 degrees. So we are left with, when you multiply, 45 times pi over 180, the degrees have canceled out, and you're just left with radians, which is equal to what? 45 is half of 90, which is half of 180, so this is 1 4th. This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees?"}, {"video_title": "Radian and degree conversion practice Trigonometry Khan Academy.mp3", "Sentence": "This is equal to pi over 4 radians. Let's do one more over here. So let's say that we had negative pi over 2 radians. What's that going to be in degrees? Well, once again, we have to figure out how many degrees are each of these radians. We know that there are 180 degrees for every pi radians, so we're going to get the radians cancel out, the pi's cancel out, and so you have negative 180 over 2. This is negative 90 degrees, or we could write it as negative 90 degrees."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "The relationship between the elapsed time t in decades, let me highlight that, because that's not a typical unit, but in decades since CO2 levels were first measured and the total amount of CO2 in the atmosphere, so the amount of CO2 a sub-decade of t, in parts per million is modeled by the following function. So the amount of CO2 as a function of how many decades have passed is going to be this. So t is in decades in this model right over here. Complete the following sentence about the yearly rate of change, the yearly rate of change in the amount of CO2 in the atmosphere. Round your answer two decimal places. Every year, the amount of CO2 in the atmosphere increases by a factor of, if they said every decade, well, this would be pretty straightforward. Every decade, you increase t by one, and so you're gonna multiply by 1.06 again."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "Complete the following sentence about the yearly rate of change, the yearly rate of change in the amount of CO2 in the atmosphere. Round your answer two decimal places. Every year, the amount of CO2 in the atmosphere increases by a factor of, if they said every decade, well, this would be pretty straightforward. Every decade, you increase t by one, and so you're gonna multiply by 1.06 again. So every decade, you increase by a factor of 1.06. But what about every year? I always find it helpful to make a bit of a table just so we can really digest things properly."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "Every decade, you increase t by one, and so you're gonna multiply by 1.06 again. So every decade, you increase by a factor of 1.06. But what about every year? I always find it helpful to make a bit of a table just so we can really digest things properly. So I'll say t and I'll say a of t. So when t is equal to zero, so at the beginning of our study, well, 1.06 to the zero power is just gonna be one. You have 315 parts per million. So what's a year later?"}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "I always find it helpful to make a bit of a table just so we can really digest things properly. So I'll say t and I'll say a of t. So when t is equal to zero, so at the beginning of our study, well, 1.06 to the zero power is just gonna be one. You have 315 parts per million. So what's a year later? So a year later is going to be a tenth of a decade. Remember, t is in decades. So a year later is 0.1 of a decade."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "So what's a year later? So a year later is going to be a tenth of a decade. Remember, t is in decades. So a year later is 0.1 of a decade. So 0.1 of a decade later, what is going to be the amount of carbon we have? Well, it's going to be 315 times 1.06 to the 0.1 power. And what is that going to be?"}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "So a year later is 0.1 of a decade. So 0.1 of a decade later, what is going to be the amount of carbon we have? Well, it's going to be 315 times 1.06 to the 0.1 power. And what is that going to be? Well, let's see. If we, so 1.06 to the, so to the 0.1 power, I didn't have to actually use a parentheses there, is equal to 1.0058. I'll just stick with that."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "And what is that going to be? Well, let's see. If we, so 1.06 to the, so to the 0.1 power, I didn't have to actually use a parentheses there, is equal to 1.0058. I'll just stick with that. It's 1.0058. So this is the same thing as 3.5 times 1.0058. And I should say approximately equal to."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "I'll just stick with that. It's 1.0058. So this is the same thing as 3.5 times 1.0058. And I should say approximately equal to. I did a little bit of rounding there. So after another year, so now we're at t equals 0.2. We're at two tenths of a decade."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "And I should say approximately equal to. I did a little bit of rounding there. So after another year, so now we're at t equals 0.2. We're at two tenths of a decade. Where are we gonna be? We're gonna be at 3.5 times 1.06 to the 0.2, which is the same thing as 3.5 times 1.06 to the 0.1. And then that raised to the second power."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "We're at two tenths of a decade. Where are we gonna be? We're gonna be at 3.5 times 1.06 to the 0.2, which is the same thing as 3.5 times 1.06 to the 0.1. And then that raised to the second power. So we're gonna multiply by this 1.06 to the one tenth power again. So we're gonna multiply by 1.0058 a second time. Another way to think about it, if we wanna reformulate this model in terms of years, so per year of t, it's going to be 315."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "And then that raised to the second power. So we're gonna multiply by this 1.06 to the one tenth power again. So we're gonna multiply by 1.0058 a second time. Another way to think about it, if we wanna reformulate this model in terms of years, so per year of t, it's going to be 315. And now our common ratio wouldn't be 1.06. It'd be 1.06 to the 0.1 power or 1.0058. And then we would raise that."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "Another way to think about it, if we wanna reformulate this model in terms of years, so per year of t, it's going to be 315. And now our common ratio wouldn't be 1.06. It'd be 1.06 to the 0.1 power or 1.0058. And then we would raise that. Now t would be in years. Now here it is in decades. And I could say approximately since this is rounded a little bit."}, {"video_title": "Interpreting change in exponential models changing units High School Math Khan Academy.mp3", "Sentence": "And then we would raise that. Now t would be in years. Now here it is in decades. And I could say approximately since this is rounded a little bit. And so every year the amount of CO2 in the atmosphere increases by a factor of, I could say 1.06 to the 0.1 power. But if I'm rounding my answer to two decimal places, well we're gonna increase by 1.0058. In fact, they should, they increases by a factor of, they should, I'm guessing they want more than two decimal places."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We're asked to simplify r to the 2 3rds, s to the 3rd, that whole thing squared, times the square root of 20 r to the 4th, s to the 5th. This looks kind of daunting, but I think if we take it step by step it shouldn't be too bad. So first we can look at this first expression right here, where we're taking this product to the second power. We know that instead we can take each of these, each of the terms in the product to the second power, and then take the product. So this is going to be the same thing as r to the 2 3rd squared times s to the 3rd squared. And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1 half power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We know that instead we can take each of these, each of the terms in the product to the second power, and then take the product. So this is going to be the same thing as r to the 2 3rd squared times s to the 3rd squared. And now let's look at this radical over here. We have the square root, but that's the exact same thing as raising something to the 1 half power. So this is equal to, so times this part, let me do this in a different color, this part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as a product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5, that's the 20 part, times r to the 4th, times s to the 5th."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We have the square root, but that's the exact same thing as raising something to the 1 half power. So this is equal to, so times this part, let me do this in a different color, this part right here, that is the same thing as 20. And instead of just writing 20, let me write 20 as a product of a perfect square and a non-perfect square. So 20 is the same thing as 4 times 5, that's the 20 part, times r to the 4th, times s to the 5th. Now let me write s to the 5th also as a product of a perfect square and a non-perfect square. r to the 4th is obviously a perfect square, its square root is r squared. But let's write s to the 5th in a similar way."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So 20 is the same thing as 4 times 5, that's the 20 part, times r to the 4th, times s to the 5th. Now let me write s to the 5th also as a product of a perfect square and a non-perfect square. r to the 4th is obviously a perfect square, its square root is r squared. But let's write s to the 5th in a similar way. So s to the 5th we can rewrite as s to the 4th times s. Right, s to the 4th times s to the 1st, that is s to the 5th. And of course all of this has to be raised to the 1 half power. Now let's simplify this even more."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "But let's write s to the 5th in a similar way. So s to the 5th we can rewrite as s to the 4th times s. Right, s to the 4th times s to the 1st, that is s to the 5th. And of course all of this has to be raised to the 1 half power. Now let's simplify this even more. If we're taking something to the 2 3rds power, and then to the 2nd power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4 3rds power. And just as a bit of review, taking something to the 4 3rds power, you can view it as either finding its cube root, taking it to the 1 3rd power, and then taking its cube root to the 4th power, or you can view it as taking it to the 4th power and then finding the cube root of that."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Now let's simplify this even more. If we're taking something to the 2 3rds power, and then to the 2nd power, we can just multiply the exponents. So this term right here, we can simplify this as r to the 4 3rds power. And just as a bit of review, taking something to the 4 3rds power, you can view it as either finding its cube root, taking it to the 1 3rd power, and then taking its cube root to the 4th power, or you can view it as taking it to the 4th power and then finding the cube root of that. Those are both legitimate ways of something being raised to the 4 3rds power. So you have r to the 4 3rds times s to the 3 times 2, times s to the 6th power. And then we could raise each of these terms right here to the 1 half power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And just as a bit of review, taking something to the 4 3rds power, you can view it as either finding its cube root, taking it to the 1 3rd power, and then taking its cube root to the 4th power, or you can view it as taking it to the 4th power and then finding the cube root of that. Those are both legitimate ways of something being raised to the 4 3rds power. So you have r to the 4 3rds times s to the 3 times 2, times s to the 6th power. And then we could raise each of these terms right here to the 1 half power. So times, let me color code it a little bit. Times, and we actually would need the parentheses once we do that, times 4 to the 1 half, times 5 to the 1 half, that's that term right there, times r to the 4th to the 1 half power, times, I might run out of colors, s to the 4th to the 1 half power, raising each of these terms to that 1 half power. Times s to the 1 half power."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "And then we could raise each of these terms right here to the 1 half power. So times, let me color code it a little bit. Times, and we actually would need the parentheses once we do that, times 4 to the 1 half, times 5 to the 1 half, that's that term right there, times r to the 4th to the 1 half power, times, I might run out of colors, s to the 4th to the 1 half power, raising each of these terms to that 1 half power. Times s to the 1 half power. And there's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect squares here, and we're raising them to the 1 half power, we're taking their square root, so let's simplify those. So this 4 to the 1 half, that's the same thing as 2, we're taking the principle root of 4. 5 to the 1 half, well, we can't take the square root of that, so let's just write that as the square root of 5."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Times s to the 1 half power. And there's a lot of ways we can go with this, but the one thing that might jump out is that there are some perfect squares here, and we're raising them to the 1 half power, we're taking their square root, so let's simplify those. So this 4 to the 1 half, that's the same thing as 2, we're taking the principle root of 4. 5 to the 1 half, well, we can't take the square root of that, so let's just write that as the square root of 5. Square root of 5. r to the 4th to the 1 half, 4 times 1 half, there's two ways you can think about it, 4 times 1 half is 2, so this is r squared, or you could say the square root of r to the 4th is r squared. So this is r squared. Similarly, the square root of s to the 4th, or s to the 4th to the 1 half is also s squared."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "5 to the 1 half, well, we can't take the square root of that, so let's just write that as the square root of 5. Square root of 5. r to the 4th to the 1 half, 4 times 1 half, there's two ways you can think about it, 4 times 1 half is 2, so this is r squared, or you could say the square root of r to the 4th is r squared. So this is r squared. Similarly, the square root of s to the 4th, or s to the 4th to the 1 half is also s squared. And then this s to the 1 half, let's just write that as the square root of s. Square root of s. Just like that. And let's see what else we can do here. So we have, let me write these other terms, we have an r to the 4 thirds, times s to the 6th, times 2 times the square root of 5, times r squared, times s squared, times the square root of s. Now, a couple of things we can do here, we can combine these s terms, let's do that."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "Similarly, the square root of s to the 4th, or s to the 4th to the 1 half is also s squared. And then this s to the 1 half, let's just write that as the square root of s. Square root of s. Just like that. And let's see what else we can do here. So we have, let me write these other terms, we have an r to the 4 thirds, times s to the 6th, times 2 times the square root of 5, times r squared, times s squared, times the square root of s. Now, a couple of things we can do here, we can combine these s terms, let's do that. Actually, let's write the 2 out front first. So let's write the 2 out front first, so you have 2 times, now let's look at these 2 s terms over here. We have s to the 6th times s squared."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we have, let me write these other terms, we have an r to the 4 thirds, times s to the 6th, times 2 times the square root of 5, times r squared, times s squared, times the square root of s. Now, a couple of things we can do here, we can combine these s terms, let's do that. Actually, let's write the 2 out front first. So let's write the 2 out front first, so you have 2 times, now let's look at these 2 s terms over here. We have s to the 6th times s squared. And when someone says to simplify it, there's multiple interpretations for it, but we'll just say, s to the 6th times s squared, that's s to the 8th, right? 6 plus 2 times s to the 8th power, times, now this one's interesting, and we might want to break it up depending on what we consider to be truly simplified. We have r to the 4 third times r squared."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We have s to the 6th times s squared. And when someone says to simplify it, there's multiple interpretations for it, but we'll just say, s to the 6th times s squared, that's s to the 8th, right? 6 plus 2 times s to the 8th power, times, now this one's interesting, and we might want to break it up depending on what we consider to be truly simplified. We have r to the 4 third times r squared. r to the 4 thirds is the same thing as r to the 1 and 1 third, that's what 4 thirds is. So 1 and 1 third, 1 and 1 third, plus 2 is 3 and 1 third. So we could write this times r to the 3 and 1 third."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "We have r to the 4 third times r squared. r to the 4 thirds is the same thing as r to the 1 and 1 third, that's what 4 thirds is. So 1 and 1 third, 1 and 1 third, plus 2 is 3 and 1 third. So we could write this times r to the 3 and 1 third. And that's a little inconsistent over here, I'm adding a fraction over here, with the s I kind of left out the s to the 1 half from the s's here. But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2, we've already dealt with these 2 s's, we've already dealt with these r's, and then you have the square root of 5 times the square root of s. And we could merge them if we want, but I won't do it just yet."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we could write this times r to the 3 and 1 third. And that's a little inconsistent over here, I'm adding a fraction over here, with the s I kind of left out the s to the 1 half from the s's here. But we could play around with it and all of those would be valid expressions. So we've already dealt with the 2, we've already dealt with these 2 s's, we've already dealt with these r's, and then you have the square root of 5 times the square root of s. And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's 2 ways we could do it, we might not like having a fractional exponent here, and then we could break it out, or we might want to take this guy and merge it with the 8th power. Because you know that this is the same thing as s to the 1 half, so let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the 8th times s to the 1 half."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So we've already dealt with the 2, we've already dealt with these 2 s's, we've already dealt with these r's, and then you have the square root of 5 times the square root of s. And we could merge them if we want, but I won't do it just yet. Times the square root of 5 times the square root of s. Now there's 2 ways we could do it, we might not like having a fractional exponent here, and then we could break it out, or we might want to take this guy and merge it with the 8th power. Because you know that this is the same thing as s to the 1 half, so let's do it both ways. So if we wanted to merge all of the exponents, we could write this as 2 times s to the 8th times s to the 1 half. So s to the 8th and s to the 1 half, that would be 2 times s to the 8, I could even write it as a decimal, 8.5 right? 8 plus, you could imagine this is s to the.5 power. So that's 8.5 times r to the 3 and 1 third, I'm kind of mixing notations here, I have just a decimal notation, then I have a fraction notation, mixed number notation, times the square root of 5."}, {"video_title": "Fractional exponent expressions 2 Exponent expressions and equations Algebra I Khan Academy.mp3", "Sentence": "So if we wanted to merge all of the exponents, we could write this as 2 times s to the 8th times s to the 1 half. So s to the 8th and s to the 1 half, that would be 2 times s to the 8, I could even write it as a decimal, 8.5 right? 8 plus, you could imagine this is s to the.5 power. So that's 8.5 times r to the 3 and 1 third, I'm kind of mixing notations here, I have just a decimal notation, then I have a fraction notation, mixed number notation, times the square root of 5. This is one simplification, I kind of have it in the fewest terms possible. The other simplification, if you don't want to have these fractional exponents out here, you could write it as, I'll do this in a different color, you could write this, and these are all equivalent statements, so it's up to debate what simplified really means. So you could write this as 2 times s to the 8th, instead of writing r to the 3 and 1 third, we could write r to the 3rd, r to the 3rd, times the cube root of r, which is the same thing as r to the 1 third."}, {"video_title": "Worked example finding missing monomial side in area model High School Math Khan Academy.mp3", "Sentence": "So we have a rectangle right over here, and let's say that we know that its area is 42x times y to the third. So that is the area of the rectangle. And we also know that the height right over here is 14xy. And what we want to do in this video is figure out what the length is going to be. And as you can imagine, it's going to be in terms of, it's going to be an algebraic term. So I encourage you to pause the video and figure out what is the length going to be if this height is 14xy and the area is 42xy to the third. Well, how do you figure out area?"}, {"video_title": "Worked example finding missing monomial side in area model High School Math Khan Academy.mp3", "Sentence": "And what we want to do in this video is figure out what the length is going to be. And as you can imagine, it's going to be in terms of, it's going to be an algebraic term. So I encourage you to pause the video and figure out what is the length going to be if this height is 14xy and the area is 42xy to the third. Well, how do you figure out area? You take your length, and I'll just use L for length, I'll put L in parentheses. So you take your length and you multiply it times your height, so let's multiply it times 14xy, and that's going to give you your area. So that's going to give us our area of 42xy to the third power."}, {"video_title": "Worked example finding missing monomial side in area model High School Math Khan Academy.mp3", "Sentence": "Well, how do you figure out area? You take your length, and I'll just use L for length, I'll put L in parentheses. So you take your length and you multiply it times your height, so let's multiply it times 14xy, and that's going to give you your area. So that's going to give us our area of 42xy to the third power. So how do we solve for our length? Well, we can just divide both sides by 14xy, so let's do that. So let's divide the left-hand side and the right-hand side by 14xy."}, {"video_title": "Worked example finding missing monomial side in area model High School Math Khan Academy.mp3", "Sentence": "So that's going to give us our area of 42xy to the third power. So how do we solve for our length? Well, we can just divide both sides by 14xy, so let's do that. So let's divide the left-hand side and the right-hand side by 14xy. 14xy. Now, on the left-hand side, I'm multiplying by 14xy and dividing by 14xy, that's the same thing as just multiplying or dividing by one, so that cancels out, so I'm just left with L, or our length, which is the whole point of dividing both sides by 14xy. And on the right-hand side, I can look at the coefficients first."}, {"video_title": "Worked example finding missing monomial side in area model High School Math Khan Academy.mp3", "Sentence": "So let's divide the left-hand side and the right-hand side by 14xy. 14xy. Now, on the left-hand side, I'm multiplying by 14xy and dividing by 14xy, that's the same thing as just multiplying or dividing by one, so that cancels out, so I'm just left with L, or our length, which is the whole point of dividing both sides by 14xy. And on the right-hand side, I can look at the coefficients first. I could say 42 divided by 14, that's going to be three. Three, and then I could say, well, x divided by x, that's just going to be one, and then I have y to the third divided by y. Y to the third divided by y, that is going to be, y to the third divided by y is going to be y squared. And then we're done."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "In the following polynomial, identify the terms along with the coefficient and exponent of each term. So the terms are just the things being added up in this polynomial. So the terms here, let me write the terms here. The first term is 3x squared. The second term, it's being added to negative 8x. You might say, hey, wait, isn't it minus 8x? And you can just view that as it's being added to negative 8x."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The first term is 3x squared. The second term, it's being added to negative 8x. You might say, hey, wait, isn't it minus 8x? And you can just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "And you can just view that as it's being added to negative 8x. So negative 8x is the second term. And then the third term here is 7. It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "It's called a polynomial. Poly, it has many terms. Or you could view each term as a monomial, as a polynomial with only one term in it. So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x, or what's multiplying the x part of the term. So over here, the x part is x squared."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So those are the terms. Now let's think about the coefficients of each of the terms. The coefficient is what's multiplying the power of x, or what's multiplying the x part of the term. So over here, the x part is x squared. That's being multiplied by 3. So 3 is the coefficient on the first term. On the second term, we have negative 8 multiplying x."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So over here, the x part is x squared. That's being multiplied by 3. So 3 is the coefficient on the first term. On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8, it's negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "On the second term, we have negative 8 multiplying x. And we want to be clear, the coefficient isn't just 8, it's negative 8. It's negative 8 that's multiplying x. So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So that's the coefficient right over here. And here you might say, hey wait, nothing is multiplying x here. I just have a 7. There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0. Because we know that x to the 0th power is equal to 1."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "There is no x. 7 isn't being multiplied by x. But you can think of this as 7 being multiplied by x to the 0. Because we know that x to the 0th power is equal to 1. So we would even call this constant, this 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Because we know that x to the 0th power is equal to 1. So we would even call this constant, this 7, this would be the coefficient on 7x to the 0. So you could view this as a coefficient. So this is also a coefficient. So let me make it clear. These three things are coefficients. Now the last part, they want us to identify the exponent of each term."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "So this is also a coefficient. So let me make it clear. These three things are coefficients. Now the last part, they want us to identify the exponent of each term. So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "Now the last part, they want us to identify the exponent of each term. So the exponent of this first term is 2. It's being raised to the second power. The exponent of the second term, remember, negative 8x. x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0."}, {"video_title": "The parts of polynomial expressions Polynomial and rational functions Algebra II Khan Academy.mp3", "Sentence": "The exponent of the second term, remember, negative 8x. x is the same thing as x to the first power. So the exponent here is 1. And then on this last term, we already said, 7 is the same thing as 7x to the 0. So the exponent here on the constant term on 7 is 0. So these things right over here, those are our exponents. And we are done."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And to do that, let me just draw a little circle here. So that's the center of the circle. And then do my best shot, best attempt to freehand draw a reasonable looking circle. Yeah, that's not, I've done worse. I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Yeah, that's not, I've done worse. I've done worse than that. All right. Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Now, if we were to go in degrees, if we were to go one time around the circle like that, how many degrees is that? Well, we know that that would be 360 degrees. Well, if we did the same thing, how many radians is that? If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "If we were to go all the way around the circle? Well, we just have to remember, when we're measuring in terms of radians, we're really talking about the arc that subtends that angle. So if you go all the way around, you're really talking about the arc length of the entire circle, or essentially the circumference of the circle. And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And you're essentially saying how many radiuses this is, or radii, or how many radii is the circumference of the circle? You know a circumference of a circle is two pi, is two pi times the radius. Or you could say that the length of the circumference of the circle is two pi radii. Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Two pi radii. If you wanna know the exact length, you just have to get the length of the radius and multiply it by two pi. That just comes from, really actually the definition of pi, but it comes from what we know as the formula for the circumference of a circle. So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So if we were to go all the way around this, this is also two pi radians. This is also two pi, two pi radians. So that tells us that two pi radians, as an angle measure, is the exact same thing, and I'm gonna write it out as 360, 360 degrees. And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "And then we can take all of this relationship and manipulate it in different ways. If we wanna simplify a little bit, we can divide both sides of this equation by two, in which case you are left with, if you divide both sides by two, you are left with pi radians, pi radians is equal to 180 degrees, is equal to 180 degrees, 180 degrees. So how can we use this relationship now to figure out what 150 degrees is? Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, this relationship we could write in different ways. We could divide both sides by 180 degrees, and we could get pi radians, pi radians over 180 degrees, over 180 degrees is equal to one, which is just another way of saying that there are pi radians for every 180 degrees, or you could say pi over 180 radians per degree. The other option, you could divide both sides of this by pi radians. You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "You could say, you would get, on the left-hand side, you get one, and you would also get, on the right-hand side, you would get 180 degrees for every pi radians, 180 degrees for every pi radians, for every pi radians, or you could interpret this as 180 over pi degrees per radian. So how would we figure out, how would we do what they asked us? So let's convert 150 degrees to radians. Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Let me write the word out. So 150 degrees, 150 degrees. Well, we wanna convert this to radians, so we really care about how many radians there are, how many radians there are per degree. Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Actually, let me do that in that color. So we care about how many radians, radians there are per degree, per, let me do that same green color, per degree. Well, how many radians are there per degree? We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "We already know, there's pi radians for every 180 degrees, or there are pi, pi, let me do that yellow color, there are pi over 180, pi over 180 radians per degree. And so if we multiply, and this all works out, because you have degrees in the numerator, degrees in the denominator, these cancel out, and so you are left with 150 times pi divided by 180 radians. So what do we get? Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "Well, this becomes, let me just rewrite it, 150 times pi, times pi, all of that over 180, all of that over 180, so this is equal to, and we get it in radians, radians. And so if we simplify it, let's see, we can divide the numerator and the denominator both by, looks like 30. So if you divide the numerator by 30, you get five. You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "You divide the denominator by 30, you get six. So you get five pi over six radians, or five six pi radians, depending how you want to do it. Now let's do the same thing for negative 45 degrees. What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians."}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "What do you get for negative 45 degrees if you were to convert that to radians? Well, same exact process. You have negative, and I'll do this one a little quicker, for negative 45 degrees, I'll write down the word, times pi radians, pi radians for every 180 degrees, for every 180 degrees, the degrees cancel out, and you are left with negative 45 pi over 180 radians. So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing?"}, {"video_title": "Example Converting degrees to radians Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to negative 45 pi, pi over 180, over 180 radians, radians. So how can we simplify this? Well, it looks like they're both, at minimum, divisible by nine. Nine times five is 45, this is nine times 20, so actually it's gonna be divisible by more than just, let's see, nine, nine, they're both, well, I will just, actually, they're both divisible by 45, what am I doing? Okay, so if you divide the numerator by 45, you get one. You divide the denominator by 45, 45 goes into 180 four times, four times. You're left with negative pi over four radians."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "You can use it at desmos.com, and I encourage you to use this after this video or even while I'm doing this video. But the goal here is to think about reflection of functions. So let's just start with some examples. Let's say that I had a function f of x, and it is equal to the square root of x. So that's what it looks like, fairly reasonable. Now let's make another function, g of x, and I'll start off by also making that the square root of x. So no surprise there, g of x was graphed right on top of f of x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's say that I had a function f of x, and it is equal to the square root of x. So that's what it looks like, fairly reasonable. Now let's make another function, g of x, and I'll start off by also making that the square root of x. So no surprise there, g of x was graphed right on top of f of x. But what would happen if, instead of it just being the square root of x, what would happen if we put a negative out front, right over there? What do you think is going to happen when I do that? Well, let's just try it out."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So no surprise there, g of x was graphed right on top of f of x. But what would happen if, instead of it just being the square root of x, what would happen if we put a negative out front, right over there? What do you think is going to happen when I do that? Well, let's just try it out. When I put the negative, it looks like it flipped it over the x-axis. It looks like it reflected it over the x-axis. Now, instead of doing it that way, what if we had another function, h of x, and I'll start off by making it identical to f of x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's just try it out. When I put the negative, it looks like it flipped it over the x-axis. It looks like it reflected it over the x-axis. Now, instead of doing it that way, what if we had another function, h of x, and I'll start off by making it identical to f of x. So once again, it's right over there. It traces out f of x. Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, instead of doing it that way, what if we had another function, h of x, and I'll start off by making it identical to f of x. So once again, it's right over there. It traces out f of x. Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign? What if we replaced x with a negative x? What do you think is going to happen there? Well, let's try it out."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Instead of putting the negative out in front of the radical sign, what if we put it under the radical sign? What if we replaced x with a negative x? What do you think is going to happen there? Well, let's try it out. If we replace it, that shifted it over the y-axis. And then, pause this video and think about how would you shift it over both axes? Well, we could do a, well, I'm running out of letters."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, let's try it out. If we replace it, that shifted it over the y-axis. And then, pause this video and think about how would you shift it over both axes? Well, we could do a, well, I'm running out of letters. Maybe I will do a, I don't know, k of x is equal to, so I'm gonna put the negative outside the radical sign, and then I'm gonna take the square root, and I'm gonna put a negative inside the radical sign. And notice, it flipped it over both. It flipped it over both the x-axis and the y-axis to go over here."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Well, we could do a, well, I'm running out of letters. Maybe I will do a, I don't know, k of x is equal to, so I'm gonna put the negative outside the radical sign, and then I'm gonna take the square root, and I'm gonna put a negative inside the radical sign. And notice, it flipped it over both. It flipped it over both the x-axis and the y-axis to go over here. Now, why does this happen? Well, let's just start with the g of x. So, when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It flipped it over both the x-axis and the y-axis to go over here. Now, why does this happen? Well, let's just start with the g of x. So, when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it. So, when x is zero, we got zero. When x is one, instead of one now, you're taking the negative of it, so you're gonna get negative one. When x is four, instead of getting positive two, you're now going to get negative two."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, when you put the negative out in front, when you negate everything that's in the expression that defines a function, whatever value you would have gotten of the function before, you're now going to get the opposite of it. So, when x is zero, we got zero. When x is one, instead of one now, you're taking the negative of it, so you're gonna get negative one. When x is four, instead of getting positive two, you're now going to get negative two. When x is equal to nine, instead of getting positive three, you now get negative three. So, hopefully that makes sense why putting a negative out front of an entire expression is going to flip it over, flip its graph over the x-axis. Now, what about replacing an x with a negative x?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "When x is four, instead of getting positive two, you're now going to get negative two. When x is equal to nine, instead of getting positive three, you now get negative three. So, hopefully that makes sense why putting a negative out front of an entire expression is going to flip it over, flip its graph over the x-axis. Now, what about replacing an x with a negative x? Well, one way to think about it now is whenever you inputted one before, that would now be a negative one that you're trying to evaluate the principal root of. And we know that the principal root function is not defined for negative one. But when x is equal to negative one, our original function wasn't defined there when x is equal to negative one."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Now, what about replacing an x with a negative x? Well, one way to think about it now is whenever you inputted one before, that would now be a negative one that you're trying to evaluate the principal root of. And we know that the principal root function is not defined for negative one. But when x is equal to negative one, our original function wasn't defined there when x is equal to negative one. But if you take the negative of that, well, now you're taking the principal root of one. And so, that's why it is now defined. So, whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x, on the negative value of that x."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "But when x is equal to negative one, our original function wasn't defined there when x is equal to negative one. But if you take the negative of that, well, now you're taking the principal root of one. And so, that's why it is now defined. So, whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x, on the negative value of that x. And so, that's why it flips it over the y-axis. And this is true with many types of functions. We don't have to do this just with a square root function."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, whatever value the function would have taken on at a given value of x, it now takes that value on the corresponding opposite value of x, on the negative value of that x. And so, that's why it flips it over the y-axis. And this is true with many types of functions. We don't have to do this just with a square root function. Let's try another function. Let's say we tried this for e to the x power. So, there you go."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "We don't have to do this just with a square root function. Let's try another function. Let's say we tried this for e to the x power. So, there you go. We have a very classic exponential there. Now, let's say that g of x is equal to negative e to the x. And if what we expect to happen happens, this will flip it over the x-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, there you go. We have a very classic exponential there. Now, let's say that g of x is equal to negative e to the x. And if what we expect to happen happens, this will flip it over the x-axis. So, negative e to the x power. And indeed, that is what happens. And then, how would we flip it over the y-axis?"}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And if what we expect to happen happens, this will flip it over the x-axis. So, negative e to the x power. And indeed, that is what happens. And then, how would we flip it over the y-axis? Well, let's do an h of x. That's going to be equal to e to the, instead of putting an x there, we will put a negative x, negative x. And there you have it."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And then, how would we flip it over the y-axis? Well, let's do an h of x. That's going to be equal to e to the, instead of putting an x there, we will put a negative x, negative x. And there you have it. Notice, it flipped it over the y-axis. Now, both examples that I just did, these are very simple expressions. Let's imagine something that's a little bit more complex."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And there you have it. Notice, it flipped it over the y-axis. Now, both examples that I just did, these are very simple expressions. Let's imagine something that's a little bit more complex. Let's say that f of x, let's give it a nice higher degree polynomial. So, let's say it's x to the third minus two x squared. That's a nice one."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Let's imagine something that's a little bit more complex. Let's say that f of x, let's give it a nice higher degree polynomial. So, let's say it's x to the third minus two x squared. That's a nice one. And actually, let's just add another term here. So, plus two x. Oh, no, I wanna make it minus two x. I wanna accentuate some of those curves. All right, so that's a pretty interesting graph."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "That's a nice one. And actually, let's just add another term here. So, plus two x. Oh, no, I wanna make it minus two x. I wanna accentuate some of those curves. All right, so that's a pretty interesting graph. Now, how would I flip it over the x-axis? Well, the way that I would do that is I could define a g of x. I could do it two ways. I could say g of x is equal to the negative of f of x, and we get that."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "All right, so that's a pretty interesting graph. Now, how would I flip it over the x-axis? Well, the way that I would do that is I could define a g of x. I could do it two ways. I could say g of x is equal to the negative of f of x, and we get that. So, that's essentially just taking this entire expression and multiplying it by negative one. And notice, it's multiplying it, it's flipping it over the x-axis. Another way we could have done it is, instead of that, we could have said the negative of x to the third minus two x squared, and then minus two x, and then we close those parentheses, and we get the same effect."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "I could say g of x is equal to the negative of f of x, and we get that. So, that's essentially just taking this entire expression and multiplying it by negative one. And notice, it's multiplying it, it's flipping it over the x-axis. Another way we could have done it is, instead of that, we could have said the negative of x to the third minus two x squared, and then minus two x, and then we close those parentheses, and we get the same effect. Now, what if we wanted to flip it over the y-axis? Well, then, instead of putting a negative on the entire expression, what we wanna do is replace our x's with a negative x. So, you could do it like this."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "Another way we could have done it is, instead of that, we could have said the negative of x to the third minus two x squared, and then minus two x, and then we close those parentheses, and we get the same effect. Now, what if we wanted to flip it over the y-axis? Well, then, instead of putting a negative on the entire expression, what we wanna do is replace our x's with a negative x. So, you could do it like this. You could say that that's going to be f of negative x, and that has the effect of, everywhere you saw an x before, you replace it with a negative x. And notice, it did exactly what we expect. It flipped it over, over the y-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "So, you could do it like this. You could say that that's going to be f of negative x, and that has the effect of, everywhere you saw an x before, you replace it with a negative x. And notice, it did exactly what we expect. It flipped it over, over the y-axis. Now, the other way we could have done that, just to make it clear, that's the same thing as negative x to the third power minus two times negative x squared minus two times negative x. And of course, we could simplify that expression, but notice, it has the exact same idea. And if we wanted to flip it over both the x and y-axes, well, we've already flipped it over the y-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "It flipped it over, over the y-axis. Now, the other way we could have done that, just to make it clear, that's the same thing as negative x to the third power minus two times negative x squared minus two times negative x. And of course, we could simplify that expression, but notice, it has the exact same idea. And if we wanted to flip it over both the x and y-axes, well, we've already flipped it over the y-axis. To flip it over the x-axis, whoops, I just deleted it. To flip it over the, I'm having issues here. To flip it over the x-axis as well, we would, oh, it gave me a parentheses already."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "And if we wanted to flip it over both the x and y-axes, well, we've already flipped it over the y-axis. To flip it over the x-axis, whoops, I just deleted it. To flip it over the, I'm having issues here. To flip it over the x-axis as well, we would, oh, it gave me a parentheses already. I would just put a negative out front. So, I put a negative out front, and there you have it. This flipped it over both the x and y-axis."}, {"video_title": "Reflecting functions introduction Transformations of functions Algebra 2 Khan Academy.mp3", "Sentence": "To flip it over the x-axis as well, we would, oh, it gave me a parentheses already. I would just put a negative out front. So, I put a negative out front, and there you have it. This flipped it over both the x and y-axis. You can do them in either order, and you will get to this green curve. Now, an easier way of writing that would have been just the negative of f of negative x, and you would have gotten to that same place. So, go to Desmos, play around with it."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "So to do this we can just do the distributive property. We can distribute this entire polynomial, this entire trinomial, times each of these terms. We could have 5a squared plus 7a minus 1 times 10a, and then 5a squared plus 7a minus 1 times negative 3. So let's just do that. So let me just write it out. Let me write it this way. 10a times 5a squared plus 7a minus 1."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "So let's just do that. So let me just write it out. Let me write it this way. 10a times 5a squared plus 7a minus 1. That's that right over here. And then we can have minus 3 times 5a squared plus 7a minus 1. And that is this distribution right over here."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "10a times 5a squared plus 7a minus 1. That's that right over here. And then we can have minus 3 times 5a squared plus 7a minus 1. And that is this distribution right over here. Then we can simplify it. 10a times 5a squared. 10 times 5 is 50. a times a squared is a to the third."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "And that is this distribution right over here. Then we can simplify it. 10a times 5a squared. 10 times 5 is 50. a times a squared is a to the third. 10 times 7 is 70. a times a is a squared. 10a times negative 1 is negative 10a. Then we distribute this negative 3 times all of this."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "10 times 5 is 50. a times a squared is a to the third. 10 times 7 is 70. a times a is a squared. 10a times negative 1 is negative 10a. Then we distribute this negative 3 times all of this. Negative 3 times 5a squared is negative 15a squared. Negative 3 times 7a is negative 21a. Negative 3 times negative 1 is positive 3."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "Then we distribute this negative 3 times all of this. Negative 3 times 5a squared is negative 15a squared. Negative 3 times 7a is negative 21a. Negative 3 times negative 1 is positive 3. And now we can try to merge like terms. This is the only a to the third term here. So this is 58 to the third."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "Negative 3 times negative 1 is positive 3. And now we can try to merge like terms. This is the only a to the third term here. So this is 58 to the third. I'll just rewrite it. Now we have 2a squared terms. We have 70a squared minus 15, or negative 15a squared."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "So this is 58 to the third. I'll just rewrite it. Now we have 2a squared terms. We have 70a squared minus 15, or negative 15a squared. So we can add these two terms. 70 of something minus 15 of that something is going to be 55 of that something. So plus 55a squared."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "We have 70a squared minus 15, or negative 15a squared. So we can add these two terms. 70 of something minus 15 of that something is going to be 55 of that something. So plus 55a squared. And then we also have 2a terms. We have this negative 10a, and then we have this negative 21a. So if we go negative 10 minus 21, that is negative 31a."}, {"video_title": "Example 2 Multiplying a binomial by a polynomial Algebra I Khan Academy.mp3", "Sentence": "So plus 55a squared. And then we also have 2a terms. We have this negative 10a, and then we have this negative 21a. So if we go negative 10 minus 21, that is negative 31a. So that is negative 31a. And then finally, we only have one constant term over here. We have this positive 3."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "We are told the graph of y is equal to log base two of x is shown below, and then say graph y is equal to two log base two of negative x minus three. So pause this video and have a go at it, and the way to think about it is, is that this second equation that we want to graph is really based on this first equation through a series of transformations. So I encourage you to take some graph paper out and sketch how those transformations would affect our original graph to get to where we need to go. All right, now let's do this together. So what we already have graphed, I'll just write it in purple, is y is equal to log base two of x. Now the difference between what I just wrote in purple and where we want to go is, in the first case, we don't multiply anything times our log base two of x, while in our end goal, we multiply by two. In our first situation, we just have log base two of x, while in here, we have log base two of negative x minus three."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "All right, now let's do this together. So what we already have graphed, I'll just write it in purple, is y is equal to log base two of x. Now the difference between what I just wrote in purple and where we want to go is, in the first case, we don't multiply anything times our log base two of x, while in our end goal, we multiply by two. In our first situation, we just have log base two of x, while in here, we have log base two of negative x minus three. In fact, we could even view that as, it's the negative of x plus three. So what we could do is try to keep changing this equation, and that's going to transform its graph until we get to our goal. So maybe the first thing we might want to do is, let's replace our x with a negative x."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "In our first situation, we just have log base two of x, while in here, we have log base two of negative x minus three. In fact, we could even view that as, it's the negative of x plus three. So what we could do is try to keep changing this equation, and that's going to transform its graph until we get to our goal. So maybe the first thing we might want to do is, let's replace our x with a negative x. So let's try to graph y is equal to log base two of negative x. In other videos, we've talked about what transformation would go on there, but we can intuit through it as well. Now, whatever value y would have taken on at a given x value, so for example, when x equals four, log base two of four is two, now that will happen at negative four."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "So maybe the first thing we might want to do is, let's replace our x with a negative x. So let's try to graph y is equal to log base two of negative x. In other videos, we've talked about what transformation would go on there, but we can intuit through it as well. Now, whatever value y would have taken on at a given x value, so for example, when x equals four, log base two of four is two, now that will happen at negative four. So log base two of the negative of negative four, well, that's still log base two of four, so that's still going to be two. And if you were to put in, let's say, whatever was happening at one before, log base two of one is zero, but now that's going to happen at negative one, because you take the negative of negative one, you're gonna get a one over here, so log base two of one is zero. And so similarly, when you had at x equals eight, you got to three, now that's going to happen at x equals negative eight."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "Now, whatever value y would have taken on at a given x value, so for example, when x equals four, log base two of four is two, now that will happen at negative four. So log base two of the negative of negative four, well, that's still log base two of four, so that's still going to be two. And if you were to put in, let's say, whatever was happening at one before, log base two of one is zero, but now that's going to happen at negative one, because you take the negative of negative one, you're gonna get a one over here, so log base two of one is zero. And so similarly, when you had at x equals eight, you got to three, now that's going to happen at x equals negative eight. We are going to be at three. And so the graph is going to look something, something like what I am graphing right over here. All right, fair enough."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "And so similarly, when you had at x equals eight, you got to three, now that's going to happen at x equals negative eight. We are going to be at three. And so the graph is going to look something, something like what I am graphing right over here. All right, fair enough. Now, the next thing we might wanna do is, hey, let's replace this x with an x plus three, because that'll get us at least, in terms of what we're taking the log of, pretty close to our original equation. So now let's think about y is equal to log base two of, and actually I should put parentheses in that previous one just so it's clear. So log base two of, not just the negative of x, but we're going to replace x with x plus three."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "All right, fair enough. Now, the next thing we might wanna do is, hey, let's replace this x with an x plus three, because that'll get us at least, in terms of what we're taking the log of, pretty close to our original equation. So now let's think about y is equal to log base two of, and actually I should put parentheses in that previous one just so it's clear. So log base two of, not just the negative of x, but we're going to replace x with x plus three. Now, what happens if you replace x with an x plus three? Or you could even view x plus three as the same thing as x minus negative three. Well, we've seen in multiple examples that when you replace x with an x plus three, that will shift your entire graph three to the left."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "So log base two of, not just the negative of x, but we're going to replace x with x plus three. Now, what happens if you replace x with an x plus three? Or you could even view x plus three as the same thing as x minus negative three. Well, we've seen in multiple examples that when you replace x with an x plus three, that will shift your entire graph three to the left. So this shifts, shifts three to the left. If it was an x minus three in here, you would shift three to the right. So how do we shift three to the left?"}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "Well, we've seen in multiple examples that when you replace x with an x plus three, that will shift your entire graph three to the left. So this shifts, shifts three to the left. If it was an x minus three in here, you would shift three to the right. So how do we shift three to the left? Well, when the points where we used to hit zero are now going to happen three to the left of that. So we used to hit it at x equals negative one, now it's going to happen at x equals negative four. The point at which y is equal to two, instead of happening at x equals negative four, is now going to happen three to the left of that, which is x equals negative seven."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "So how do we shift three to the left? Well, when the points where we used to hit zero are now going to happen three to the left of that. So we used to hit it at x equals negative one, now it's going to happen at x equals negative four. The point at which y is equal to two, instead of happening at x equals negative four, is now going to happen three to the left of that, which is x equals negative seven. So it's going to be right over there. And so, and the point at which the graph goes down to infinity, that was happening as x approaches zero. Now that's going to happen as x approaches three to the left of that, as x approaches negative three."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "The point at which y is equal to two, instead of happening at x equals negative four, is now going to happen three to the left of that, which is x equals negative seven. So it's going to be right over there. And so, and the point at which the graph goes down to infinity, that was happening as x approaches zero. Now that's going to happen as x approaches three to the left of that, as x approaches negative three. So I could draw a little dotted line right over here to show that as x approaches that, our graph's going to approach zero. So our graph's gonna look something, something like, like this. Like this."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "Now that's going to happen as x approaches three to the left of that, as x approaches negative three. So I could draw a little dotted line right over here to show that as x approaches that, our graph's going to approach zero. So our graph's gonna look something, something like, like this. Like this. And this is all hand-drawn, so it's not perfectly drawn. But we're awfully close. Now to get from where we are to our goal, we just have to multiply the right-hand side by two."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "Like this. And this is all hand-drawn, so it's not perfectly drawn. But we're awfully close. Now to get from where we are to our goal, we just have to multiply the right-hand side by two. So now let's graph y, not two, let's graph y is equal to two log base two of negative of x plus three, which is the exact same goal as we had before, I've just factored out the negative to help with our transformations. So all that means is whatever y value we're taking on at a given x, you're now going to take on twice that y value. So where you were at zero, you're still going to be zero."}, {"video_title": "Graphing logarithmic functions (example 1) Algebra 2 Khan Academy.mp3", "Sentence": "Now to get from where we are to our goal, we just have to multiply the right-hand side by two. So now let's graph y, not two, let's graph y is equal to two log base two of negative of x plus three, which is the exact same goal as we had before, I've just factored out the negative to help with our transformations. So all that means is whatever y value we're taking on at a given x, you're now going to take on twice that y value. So where you were at zero, you're still going to be zero. But where you were at two, you're now going to be equal to four. And so the graph is going to look something, something like what I am drawing right now. And we're done."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "What I have attempted to draw here is a unit circle. And the fact that I'm calling it a unit circle means it has a radius of one. So this length from the center, and I centered it at the origin, this length from the center to any point on the circle is of length one. So what would this coordinate be right over there, right where it intersects along the x-axis? Well, it would be, x would be one, y would be zero. What would this coordinate be up here? Well, we've gone one above the origin, but we haven't moved to the left or the right, so our x value is zero, our y value is one."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So what would this coordinate be right over there, right where it intersects along the x-axis? Well, it would be, x would be one, y would be zero. What would this coordinate be up here? Well, we've gone one above the origin, but we haven't moved to the left or the right, so our x value is zero, our y value is one. What about back here? Well, here, our x value is negative one, we've moved one to the left, and we haven't moved up or down, so our y value is zero. And what about down here?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, we've gone one above the origin, but we haven't moved to the left or the right, so our x value is zero, our y value is one. What about back here? Well, here, our x value is negative one, we've moved one to the left, and we haven't moved up or down, so our y value is zero. And what about down here? Well, we've gone a unit down, or one below the origin, but we haven't moved in the xy direction, so our x is zero, and our y is negative one. Now, with that out of the way, I'm going to draw an angle. And then this, the way I'm gonna draw this angle, I'm gonna define a convention for positive angles."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And what about down here? Well, we've gone a unit down, or one below the origin, but we haven't moved in the xy direction, so our x is zero, and our y is negative one. Now, with that out of the way, I'm going to draw an angle. And then this, the way I'm gonna draw this angle, I'm gonna define a convention for positive angles. I'm gonna say a positive angle, well, the terminal, sorry, the initial side of an angle, we're always gonna do along the positive x-axis. So this is the, you can kind of view it as the starting side of the angle, the initial side of an angle. And then, to draw an angle, a positive angle, the terminal side, we're going to move in a counterclockwise direction."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And then this, the way I'm gonna draw this angle, I'm gonna define a convention for positive angles. I'm gonna say a positive angle, well, the terminal, sorry, the initial side of an angle, we're always gonna do along the positive x-axis. So this is the, you can kind of view it as the starting side of the angle, the initial side of an angle. And then, to draw an angle, a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. Counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And then, to draw an angle, a positive angle, the terminal side, we're going to move in a counterclockwise direction. So positive angle means we're going counterclockwise. Counterclockwise. And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise, clockwise direction. So let me draw a positive angle. So a positive angle might look, might look something like this."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "And this is just the convention I'm going to use, and it's also the convention that is typically used. And so you can imagine a negative angle would move in a clockwise, clockwise direction. So let me draw a positive angle. So a positive angle might look, might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction until I get, until I measure out the angle, and then this is the terminal side. This is a positive angle theta."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So a positive angle might look, might look something like this. This is the initial side. And then from that, I go in a counterclockwise direction until I get, until I measure out the angle, and then this is the terminal side. This is a positive angle theta. And what I wanna do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates A, B. The X value where it intersects is A, the Y value where it intersects is B."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "This is a positive angle theta. And what I wanna do is think about this point of intersection between the terminal side of this angle and my unit circle. And let's just say it has the coordinates A, B. The X value where it intersects is A, the Y value where it intersects is B. And I'm also, the whole point of what I'm doing here is I'm gonna see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I wanna do is I wanna make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here, and let me make it clear that this is a 90 degree angle."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "The X value where it intersects is A, the Y value where it intersects is B. And I'm also, the whole point of what I'm doing here is I'm gonna see how this unit circle might be able to help us extend our traditional definitions of trig functions. And so what I wanna do is I wanna make this theta part of a right triangle. So to make it part of a right triangle, let me drop an altitude right over here, and let me make it clear that this is a 90 degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. So the first question I have to ask you is what is the length of the hypotenuse?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So to make it part of a right triangle, let me drop an altitude right over here, and let me make it clear that this is a 90 degree angle. So this theta is part of this right triangle. So let's see what we can figure out about the sides of this right triangle. So the first question I have to ask you is what is the length of the hypotenuse? What is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of one, so the hypotenuse has length one."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So the first question I have to ask you is what is the length of the hypotenuse? What is the length of the hypotenuse of this right triangle that I have just constructed? Well, this hypotenuse is just a radius of a unit circle. The unit circle has a radius of one, so the hypotenuse has length one. Now, what is the length of this blue side right over here, this blue side? You could view this as the opposite side to the angle. Well, this height is the exact same thing, is the exact same thing as the Y coordinate of this point of intersection."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "The unit circle has a radius of one, so the hypotenuse has length one. Now, what is the length of this blue side right over here, this blue side? You could view this as the opposite side to the angle. Well, this height is the exact same thing, is the exact same thing as the Y coordinate of this point of intersection. So this height right over here is going to be equal to b. The Y coordinate right over here is b. This height is equal to b."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, this height is the exact same thing, is the exact same thing as the Y coordinate of this point of intersection. So this height right over here is going to be equal to b. The Y coordinate right over here is b. This height is equal to b. Now, exact same logic. What is the length of this base going to be, the base just of the right triangle? Well, this is going to be the X coordinate of this point of intersection."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "This height is equal to b. Now, exact same logic. What is the length of this base going to be, the base just of the right triangle? Well, this is going to be the X coordinate of this point of intersection. If you were to drop this down, this is the point X is equal to a, or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine, and what is the cosine, let me use the same green, what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our Soh-Cah-Toa definition."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, this is going to be the X coordinate of this point of intersection. If you were to drop this down, this is the point X is equal to a, or this whole length between the origin and that is of length a. Now that we have set that up, what is the cosine, and what is the cosine, let me use the same green, what is the cosine of my angle going to be in terms of a's and b's and any other numbers that might show up? Well, to think about that, we just need our Soh-Cah-Toa definition. That's the only one we have now. We are actually in the process of extending it. Soh-Cah-Toa definition of trig functions, and the Cah part is what helps us with cosine."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, to think about that, we just need our Soh-Cah-Toa definition. That's the only one we have now. We are actually in the process of extending it. Soh-Cah-Toa definition of trig functions, and the Cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side, for this angle, the adjacent side has length a, so it's going to be equal to a, over what's the length of the hypotenuse?"}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Soh-Cah-Toa definition of trig functions, and the Cah part is what helps us with cosine. It tells us that the cosine of an angle is equal to the length of the adjacent side over the hypotenuse. So what's this going to be? The length of the adjacent side, for this angle, the adjacent side has length a, so it's going to be equal to a, over what's the length of the hypotenuse? Well, that's just one. So the cosine of theta is just equal to a. Let me write this down again."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "The length of the adjacent side, for this angle, the adjacent side has length a, so it's going to be equal to a, over what's the length of the hypotenuse? Well, that's just one. So the cosine of theta is just equal to a. Let me write this down again. So the cosine of theta is just equal to a. It's equal to the x-coordinate of where this terminal side of the angle intersected the unit circle. Now let's think about the sine of theta."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Let me write this down again. So the cosine of theta is just equal to a. It's equal to the x-coordinate of where this terminal side of the angle intersected the unit circle. Now let's think about the sine of theta. Sine of theta, and I'm going to do it in, well, let me see, I'll do it in orange. So what's the sine of theta going to be? Well, we just have to look at the Soh part of our Soh-Cah-Toa definition."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Now let's think about the sine of theta. Sine of theta, and I'm going to do it in, well, let me see, I'll do it in orange. So what's the sine of theta going to be? Well, we just have to look at the Soh part of our Soh-Cah-Toa definition. It tells us that sine is opposite over hypotenuse. Well, the opposite side here has length b, and the hypotenuse has length one. So our sine of theta is equal to b."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, we just have to look at the Soh part of our Soh-Cah-Toa definition. It tells us that sine is opposite over hypotenuse. Well, the opposite side here has length b, and the hypotenuse has length one. So our sine of theta is equal to b. So an interesting thing, this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b, we could also view this as a is the same thing as cosine of theta. A is the same thing as cosine of theta, and b is the same thing as sine of theta. Well, that's interesting."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "So our sine of theta is equal to b. So an interesting thing, this coordinate, this point where our terminal side of our angle intersected the unit circle, that point a, b, we could also view this as a is the same thing as cosine of theta. A is the same thing as cosine of theta, and b is the same thing as sine of theta. Well, that's interesting. That was just, we just used our Soh-Cah-Toa definition. Now, can we in some way use this to extend Soh-Cah-Toa? Because Soh-Cah-Toa has a problem."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Well, that's interesting. That was just, we just used our Soh-Cah-Toa definition. Now, can we in some way use this to extend Soh-Cah-Toa? Because Soh-Cah-Toa has a problem. It works out fine if our angle is greater than zero degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But Soh-Cah-Toa starts to break down as our angle is either zero, or maybe even becomes negative, or as our angle is 90 degrees or more."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Because Soh-Cah-Toa has a problem. It works out fine if our angle is greater than zero degrees, if we're dealing with degrees, and if it's less than 90 degrees. We can always make it part of a right triangle. But Soh-Cah-Toa starts to break down as our angle is either zero, or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90 degree angles in it. It starts to break down. Let me make this clear."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "But Soh-Cah-Toa starts to break down as our angle is either zero, or maybe even becomes negative, or as our angle is 90 degrees or more. You can't have a right triangle with two 90 degree angles in it. It starts to break down. Let me make this clear. So sure, that's, so this is a right triangle, so the angle is pretty large. I can make the angle even larger, and still have a right triangle even larger. But I can never get quite to 90 degrees."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "Let me make this clear. So sure, that's, so this is a right triangle, so the angle is pretty large. I can make the angle even larger, and still have a right triangle even larger. But I can never get quite to 90 degrees. At 90 degrees, it's not clear that I have a right triangle anymore. It all seems to break down, and especially the case what happens when I go beyond 90 degrees. So let's go, let's see if we can use what we set up here."}, {"video_title": "Introduction to the unit circle Trigonometry Khan Academy.mp3", "Sentence": "But I can never get quite to 90 degrees. At 90 degrees, it's not clear that I have a right triangle anymore. It all seems to break down, and especially the case what happens when I go beyond 90 degrees. So let's go, let's see if we can use what we set up here. Let's set up a new definition of our trig functions, which is really an extension of Soh-Cah-Toa, and it's consistent with Soh-Cah-Toa. Instead of defining cosine as, oh, if I have a right triangle, and saying, okay, it's the adjacent over the hypotenuse, sine is the opposite over the hypotenuse, tangent is opposite over adjacent, why don't I just say, for any angle, I can draw it in the unit circle using this convention that I just set up, and let's just say that the cosine of our angle, the cosine of our angle is equal to the x-coordinate where we intersect, is equal to the x-coordinate, coordinate where the terminal side of our angle intersects the unit circle, where angle, I'll write the terminal side, terminal side of angle, side of angle intersects, intersects the unit, the unit circle, and why don't we define sine of theta, sine of theta to be equal to the y-coordinate, where the terminal side of the angle intersects the unit circle. So essentially, for any angle, this point is going to define cosine of theta and sine of theta, and so what would be the reasonable definition for tangent of theta?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "We're asked to graph the function y is equal to 2 sine of negative x on the interval, the closed interval, so it includes the endpoints, negative 2 pi to 2 pi. So to do this, I'm first going to graph the function y is equal to sine of x and then think about how it's changed by the 2 and this negative in front of the x right over here. So let's do y equal to sine of x first. So let me draw our x-axis. Let me draw the y-axis. Pretty straightforward. And we care about it between negative 2 pi and 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me draw our x-axis. Let me draw the y-axis. Pretty straightforward. And we care about it between negative 2 pi and 2 pi. So let's say that this is negative 2 pi, and then this right over here would be negative pi. This, of course, is 0. Then this is positive pi, and then this right over here is 2 pi again."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And we care about it between negative 2 pi and 2 pi. So let's say that this is negative 2 pi, and then this right over here would be negative pi. This, of course, is 0. Then this is positive pi, and then this right over here is 2 pi again. This right over here is 2 pi, and then this could be 1, this could be 2, this could be negative 1, and this could be negative 2. And let me copy this thing so I can use it for later when I adjust this graph. So let me copy."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then this is positive pi, and then this right over here is 2 pi again. This right over here is 2 pi, and then this could be 1, this could be 2, this could be negative 1, and this could be negative 2. And let me copy this thing so I can use it for later when I adjust this graph. So let me copy. All right. So let's think about sine of x. So what happens when sine is 0?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me copy. All right. So let's think about sine of x. So what happens when sine is 0? When sine is 0, or sorry, when x is 0, sine of 0 is 0. And I'll draw a little unit circle here for reference. This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So what happens when sine is 0? When sine is 0, or sorry, when x is 0, sine of 0 is 0. And I'll draw a little unit circle here for reference. This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions. So this is x, this is y. Draw a unit circle. And remember, over here, x is referring to the angle."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "This is what I like to do in my head as I'm trying to figure out the value of the trigonometric functions. So this is x, this is y. Draw a unit circle. And remember, over here, x is referring to the angle. So that's my unit circle, radius 1. So when the angle is 0, sine is going to be the y-coordinate here. So sine of 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And remember, over here, x is referring to the angle. So that's my unit circle, radius 1. So when the angle is 0, sine is going to be the y-coordinate here. So sine of 0 is 0. When, as sine increases, we go up all the way to sine of pi over 2 is 1. So sine of pi over 2 is going to get you to 1. Then you go sine of pi gets you to 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So sine of 0 is 0. When, as sine increases, we go up all the way to sine of pi over 2 is 1. So sine of pi over 2 is going to get you to 1. Then you go sine of pi gets you to 0. Sine of 3 pi over 2 gets you to negative 1. And then sine of 2 pi gets you back to 0. So if I were to graph it, it looks something like this."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then you go sine of pi gets you to 0. Sine of 3 pi over 2 gets you to negative 1. And then sine of 2 pi gets you back to 0. So if I were to graph it, it looks something like this. So this is between 0 and 2 pi. It looks something like that. And we also want to go in the negative direction."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So if I were to graph it, it looks something like this. So this is between 0 and 2 pi. It looks something like that. And we also want to go in the negative direction. So as we go in the negative direction, so sine of negative pi over 2 is negative 1. Negative pi over 2 is negative 1. Then you go back to negative pi, go back to 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And we also want to go in the negative direction. So as we go in the negative direction, so sine of negative pi over 2 is negative 1. Negative pi over 2 is negative 1. Then you go back to negative pi, go back to 0. Negative 3 pi over 2, you're going all the way around like that. That gets you back to sine is equal to 1. So sine is equal to 1."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Then you go back to negative pi, go back to 0. Negative 3 pi over 2, you're going all the way around like that. That gets you back to sine is equal to 1. So sine is equal to 1. And then you go 2 pi, sine is back, is equaling 0. So the curve will look something like this in the negative. So as we go from between 0 and negative 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So sine is equal to 1. And then you go 2 pi, sine is back, is equaling 0. So the curve will look something like this in the negative. So as we go from between 0 and negative 2 pi. This is consistent with everything else that we know about sine. The period of sine of x, well you see here you have a coefficient of 1 here. So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So as we go from between 0 and negative 2 pi. This is consistent with everything else that we know about sine. The period of sine of x, well you see here you have a coefficient of 1 here. So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious. It's just 2 pi. Or you just see here that the period was 2 pi. It took 2 pi length to do one of our smallest repeating pattern right over here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So the period is just going to be 2 pi over the absolute value of 1, which is a little bit obvious. It's just 2 pi. Or you just see here that the period was 2 pi. It took 2 pi length to do one of our smallest repeating pattern right over here. And what is the amplitude? Well, we vary between 1 and negative 1. The total difference between the minimum and the maximum is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It took 2 pi length to do one of our smallest repeating pattern right over here. And what is the amplitude? Well, we vary between 1 and negative 1. The total difference between the minimum and the maximum is 2. Half of that is 1. Or another way of thinking about it, we vary 1 from our middle point. So that was pretty straightforward."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "The total difference between the minimum and the maximum is 2. Half of that is 1. Or another way of thinking about it, we vary 1 from our middle point. So that was pretty straightforward. Let's change it up a little bit. Now let's graph y is equal to 2 sine of x. So let me put my little axes there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So that was pretty straightforward. Let's change it up a little bit. Now let's graph y is equal to 2 sine of x. So let me put my little axes there. I want to do it right under it. And so what is going to happen now that we have y is equal to 2 sine of x? How is the graph going to change?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me put my little axes there. I want to do it right under it. And so what is going to happen now that we have y is equal to 2 sine of x? How is the graph going to change? Well, all we did is we multiplied this function by 2. So whatever values it takes on, it's going to be twice as high now. So 2 times 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "How is the graph going to change? Well, all we did is we multiplied this function by 2. So whatever values it takes on, it's going to be twice as high now. So 2 times 0 is 0. 2 times 1 is now 2. 2 times 1 is 2. 2 times 0 is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So 2 times 0 is 0. 2 times 1 is now 2. 2 times 1 is 2. 2 times 0 is 2. That's at pi over 2. 2 times 0 is 0. 2 times negative 1 is negative 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times 0 is 2. That's at pi over 2. 2 times 0 is 0. 2 times negative 1 is negative 2. 2 times 0 is 0. So it looks something like this between 0 and 2 pi. It looks something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times negative 1 is negative 2. 2 times 0 is 0. So it looks something like this between 0 and 2 pi. It looks something like that. And we could keep going in the negative direction. 2 times negative 1 is negative 2. 2 times 0 is 0."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It looks something like that. And we could keep going in the negative direction. 2 times negative 1 is negative 2. 2 times 0 is 0. 2 times 1 is positive 2. 2 times 0 is 0. So in the negative direction, it looks something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "2 times 0 is 0. 2 times 1 is positive 2. 2 times 0 is 0. So in the negative direction, it looks something like that. My best attempt to draw a relatively smooth curve. Hopefully you get the idea. So it will look something like that."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So in the negative direction, it looks something like that. My best attempt to draw a relatively smooth curve. Hopefully you get the idea. So it will look something like that. So what just happened? Well, the difference between the minimum value and the maximum value just increased by a factor of 2. The total difference is 4."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So it will look something like that. So what just happened? Well, the difference between the minimum value and the maximum value just increased by a factor of 2. The total difference is 4. Half of that difference is now 2. So what is the amplitude here? Well, the amplitude is 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "The total difference is 4. Half of that difference is now 2. So what is the amplitude here? Well, the amplitude is 2. So the amplitude, you could view it as the absolute value of this thing. The absolute value of 2 is now equal to 2. And it's common sense."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the amplitude is 2. So the amplitude, you could view it as the absolute value of this thing. The absolute value of 2 is now equal to 2. And it's common sense. The amplitude here was 1. And now you're swaying from that middle position twice as far because you're multiplying by 2. Now let's go back to sine of x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And it's common sense. The amplitude here was 1. And now you're swaying from that middle position twice as far because you're multiplying by 2. Now let's go back to sine of x. And let's change it in a different way. Let's graph sine of negative x. So let me once again put some graph paper here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Now let's go back to sine of x. And let's change it in a different way. Let's graph sine of negative x. So let me once again put some graph paper here. And now my goal is to graph y is equal to sine of negative x. So for at least the time being, I got rid of that 2 there. And I'm just going straight from sine of x to sine of negative x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me once again put some graph paper here. And now my goal is to graph y is equal to sine of negative x. So for at least the time being, I got rid of that 2 there. And I'm just going straight from sine of x to sine of negative x. So let's think about how the values are going to work out. So when x is 0, this is still going to be sine of 0, which is equal to 0. But then what happens as x increases?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And I'm just going straight from sine of x to sine of negative x. So let's think about how the values are going to work out. So when x is 0, this is still going to be sine of 0, which is equal to 0. But then what happens as x increases? What happens when x is pi over 2? When x is pi over 2, the angle that we're inputting into sine, we're going to have to multiply it by this negative. So when x is pi over 2, we're really taking sine of negative pi over 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "But then what happens as x increases? What happens when x is pi over 2? When x is pi over 2, the angle that we're inputting into sine, we're going to have to multiply it by this negative. So when x is pi over 2, we're really taking sine of negative pi over 2. Well, what's sine of negative pi over 2? Well, we could see that right over here. It's negative 1."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So when x is pi over 2, we're really taking sine of negative pi over 2. Well, what's sine of negative pi over 2? Well, we could see that right over here. It's negative 1. And then when x is equal to pi, well, sine of negative pi we already see is 0. When x is 3 pi over 2, well, it's going to be sine of negative 3 pi over 2, which is 1. And once again, when x is 2 pi, it's going to be sine of negative 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "It's negative 1. And then when x is equal to pi, well, sine of negative pi we already see is 0. When x is 3 pi over 2, well, it's going to be sine of negative 3 pi over 2, which is 1. And once again, when x is 2 pi, it's going to be sine of negative 2 pi. Sine of negative 2 pi is 0. So notice what was happening as I was trying to graph between 0 and 2 pi. I kept referring to the points in the negative direction."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And once again, when x is 2 pi, it's going to be sine of negative 2 pi. Sine of negative 2 pi is 0. So notice what was happening as I was trying to graph between 0 and 2 pi. I kept referring to the points in the negative direction. So you can imagine taking this negative side right over here, between 0 and negative 2 pi, and then flipping it over to get this one right over here. That's what that negative x seemed to do. And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "I kept referring to the points in the negative direction. So you can imagine taking this negative side right over here, between 0 and negative 2 pi, and then flipping it over to get this one right over here. That's what that negative x seemed to do. And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2. Well, it's going to be equal to 1. And then you can flip this over the y-axis. So essentially what we have done is we have flipped it."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And by that same logic, when we go in the negative direction, you say when x is equal to negative pi over 2, well, you have that negative in front of it, so it's going to be sine of pi over 2. Well, it's going to be equal to 1. And then you can flip this over the y-axis. So essentially what we have done is we have flipped it. We have reflected the graph of sine of x over the y-axis. This is the y-axis here, of course. So we have reflected it over the y-axis."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So essentially what we have done is we have flipped it. We have reflected the graph of sine of x over the y-axis. This is the y-axis here, of course. So we have reflected it over the y-axis. So let me make sure I'm so it looks something like this. This is the y-axis. Hopefully you see that reflection right over there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So we have reflected it over the y-axis. So let me make sure I'm so it looks something like this. This is the y-axis. Hopefully you see that reflection right over there. That's what that negative x has done. So now let's think about kind of the combo, having the 2 out the front and the negative x right over there. So let me put our graph, my little axes there one more time."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Hopefully you see that reflection right over there. That's what that negative x has done. So now let's think about kind of the combo, having the 2 out the front and the negative x right over there. So let me put our graph, my little axes there one more time. And now let's try to do what was asked of us. So I'll do it in a new color. I'll do it in blue."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So let me put our graph, my little axes there one more time. And now let's try to do what was asked of us. So I'll do it in a new color. I'll do it in blue. Now let's graph. This is our y-axis. This is a y-axis."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "I'll do it in blue. Now let's graph. This is our y-axis. This is a y-axis. Let's graph y is equal to 2 times sine of negative x. So based on everything we've done, how will this look? What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x?"}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "This is a y-axis. Let's graph y is equal to 2 times sine of negative x. So based on everything we've done, how will this look? What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x? Well, there's two ways you could think about it. You could either take 2 sine of x. So here we multiplied by 2 to get double the amplitude."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "What are the transformations we would do if we're going from original sine of x to y is equal to 2 sine of negative x? Well, there's two ways you could think about it. You could either take 2 sine of x. So here we multiplied by 2 to get double the amplitude. And you could say, well, I'm going to now flip it over to get the negative sine of x. And so if you did that, you would get, so let me make it clear what I'm flipping. So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So here we multiplied by 2 to get double the amplitude. And you could say, well, I'm going to now flip it over to get the negative sine of x. And so if you did that, you would get, so let me make it clear what I'm flipping. So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first. Then it'll go back to 0. Then it'll go positive. And then you get right over there."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So if I took between 0 and negative 2 pi and I were to flip it over, what used to be here, you flip it over, you reflect it over the y-axis, and you now have, so it'll go negative first. Then it'll go back to 0. Then it'll go positive. And then you get right over there. So all I did to go from 2 sine of x to 2 sine of negative x is I just reflected over the y-axis. And then, of course, what is between 0 and negative 2 pi? You just have to look between 0 and 2 pi."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And then you get right over there. So all I did to go from 2 sine of x to 2 sine of negative x is I just reflected over the y-axis. And then, of course, what is between 0 and negative 2 pi? You just have to look between 0 and 2 pi. So now it's going to go up and down. Let me make it a little bit better, draw it a little bit neater. And then down and up."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "You just have to look between 0 and 2 pi. So now it's going to go up and down. Let me make it a little bit better, draw it a little bit neater. And then down and up. So it was a reflection of what was between 0 and 2 pi. So you see that right over here. Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "And then down and up. So it was a reflection of what was between 0 and 2 pi. So you see that right over here. Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x. What's the difference between this graph and this graph? Well, we just have twice the amplitude. We're multiplying this one by 2, and so you get twice the amplitude."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Or if you start with sine of negative x and you go to 2 sine of negative x, notice what happens between sine of negative x and 2 sine of negative x. What's the difference between this graph and this graph? Well, we just have twice the amplitude. We're multiplying this one by 2, and so you get twice the amplitude. So the last thought or question I have for you is how does the period of 2 sine of negative x relate to the period of sine of x? Well, there's two ways to think about it. Actually, I'll let you think about that for a second."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "We're multiplying this one by 2, and so you get twice the amplitude. So the last thought or question I have for you is how does the period of 2 sine of negative x relate to the period of sine of x? Well, there's two ways to think about it. Actually, I'll let you think about that for a second. Well, there's two ways to think about it. You could refer to the graphs right over here, or you could think about the formula, which might be a little bit intuitive right now. If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Actually, I'll let you think about that for a second. Well, there's two ways to think about it. You could refer to the graphs right over here, or you could think about the formula, which might be a little bit intuitive right now. If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here. So the absolute value of negative x, or the absolute value of the negative 1, is just 1. So you get 2 pi. So you have the exact same period as the period of sine of x."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "If you wanted to refer to the kind of classical formula, you would say the period is just going to be 2 pi, and you divide by the absolute value of the coefficient to figure out how much faster are you going to get to 2 pi right over here. So the absolute value of negative x, or the absolute value of the negative 1, is just 1. So you get 2 pi. So you have the exact same period as the period of sine of x. And if you see that, you complete one cycle every 2 pi. Now, what is the difference? Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "So you have the exact same period as the period of sine of x. And if you see that, you complete one cycle every 2 pi. Now, what is the difference? Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it. It doesn't change the period, but it does change how the graph looks. When you start getting increased x's, instead of sine being positive, as it would be in the case of the traditional sine function, as x grows, you're taking the sine of negative x. You're taking the sine of a negative angle, and so that's why you start off having negative values of sine."}, {"video_title": "Example Amplitude and period transformations Trigonometry Khan Academy.mp3", "Sentence": "Well, the period's the same, but remember, this negative x, it's not like you can just completely ignore it. It doesn't change the period, but it does change how the graph looks. When you start getting increased x's, instead of sine being positive, as it would be in the case of the traditional sine function, as x grows, you're taking the sine of negative x. You're taking the sine of a negative angle, and so that's why you start off having negative values of sine. And that's also another way, if you want to think about it, that it's the reflection over the y-axis of just sine of x. These two are reflections, and these two are reflections. This one is 2 times the amplitude as this one."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "So what does that mean? 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And then we can think about what is, well, let me actually write it right over here. I'll put a four here."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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?"}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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? 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?"}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. X times what is going to get us to x squared? Well, x times x is going to get us to x squared."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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? Well, it's just going to be y. So four xy times two x is actually going to give us this first term."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And then you just have the y. Now let's do the same thing with the second term."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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? Well, four times three is 12. X times what is x?"}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And then y times what is y squared? It's going to be y times y is y squared."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. If you multiply these two, you're going to get 12xy squared. Four times three is 12."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. 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."}, {"video_title": "Factoring binomials common factor Mathematics II High School Math Khan Academy.mp3", "Sentence": "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. And so we're done. This right over here is our answer."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, I'm going to show you the last two logarithm properties now. So this one, and I always found this one to be in some ways the most obvious one. But don't feel bad if it's not obvious. Maybe it will take a little bit of introspection. And I encourage you to really experiment with all these logarithm properties, because that's the only way that you'll really learn them. And the point of math isn't just to pass the next exam or to get an A on the next exam. The point of math is to understand math."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Maybe it will take a little bit of introspection. And I encourage you to really experiment with all these logarithm properties, because that's the only way that you'll really learn them. And the point of math isn't just to pass the next exam or to get an A on the next exam. The point of math is to understand math. And so you can actually apply it in life later on and not have to relearn everything every time. So the next logarithm property is if I have A times the logarithm base b of c. If I have A times this whole thing, that that equals logarithm base b of c to the a power. Fascinating."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "The point of math is to understand math. And so you can actually apply it in life later on and not have to relearn everything every time. So the next logarithm property is if I have A times the logarithm base b of c. If I have A times this whole thing, that that equals logarithm base b of c to the a power. Fascinating. So let's see if this works out. So let's say if I have, I don't know, 3 times logarithm base 2 of 8. So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Fascinating. So let's see if this works out. So let's say if I have, I don't know, 3 times logarithm base 2 of 8. So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power. And that's the same thing as, well, we could figure it out. So let's see what this is. 3 times log base, what's log base 2 of 8?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this property tells us that this is going to be the same thing as logarithm base 2 of 8 to the third power. And that's the same thing as, well, we could figure it out. So let's see what this is. 3 times log base, what's log base 2 of 8? The reason why I kind of hesitated a second ago is because every time I want to figure something out, I implicitly want to use log and exponential rules kind of to do it. So I'm trying to avoid that. But anyway, going back."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 times log base, what's log base 2 of 8? The reason why I kind of hesitated a second ago is because every time I want to figure something out, I implicitly want to use log and exponential rules kind of to do it. So I'm trying to avoid that. But anyway, going back. So what is this? 2 to what power is 8? Well, 2 to the third power is 8, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But anyway, going back. So what is this? 2 to what power is 8? Well, 2 to the third power is 8, right? So that's 3. And we have this 3 here. So 3 times 3."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, 2 to the third power is 8, right? So that's 3. And we have this 3 here. So 3 times 3. So this thing right here should equal 9. If this equals 9, then we know that this property works at least for this example. You don't know if it works for all examples."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 3 times 3. So this thing right here should equal 9. If this equals 9, then we know that this property works at least for this example. You don't know if it works for all examples. And for that, maybe you'd want to look at the proof we have in the other videos. But that's kind of a more advanced topic. But the important thing first is just to understand how to use it."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "You don't know if it works for all examples. And for that, maybe you'd want to look at the proof we have in the other videos. But that's kind of a more advanced topic. But the important thing first is just to understand how to use it. So let's see. What is 2 to the ninth power? Well, it's going to be some large number."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But the important thing first is just to understand how to use it. So let's see. What is 2 to the ninth power? Well, it's going to be some large number. Actually, I know what it is. It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256. And so 2 to the ninth should be 512."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, it's going to be some large number. Actually, I know what it is. It's 256, because in the last video, we figured out that 2 to the eighth was equal to 256. And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And so 2 to the ninth should be 512. So 2 to the ninth should be 512. So if 8 to the third is also 512, then we are correct, right, because log base 2 of 512 is going to be equal to 9. What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed. So let's see."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "What's 8 to the third? It's 64 times, right? 8 times 8 squared is 64, so 8 cubed. So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512. Correct."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's see. 4 times 2 is 3. 6 times 8, it looks like it's 512. Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Correct. And there's other ways you could have done it, because you could have said 8 to the third is the same thing as 2 to the ninth. How do we know that? Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, 8 to the third is equal to 2 to the third to the third, right? I just rewrote 8. And we know from our exponent rules that 2 to the third to the third is the same thing as 2 to the ninth. And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And actually, it's this exponent property where you can multiply. When you take something to an exponent and then take that to an exponent, and you can essentially just multiply the exponents, that's the exponent property that actually leads to this logarithm property. But I'm not going to dwell on that too much in this presentation. There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "There's a whole video on proving it a little bit more formally. The next logarithm property I'm going to show you, and then I'll review everything and maybe do some examples. This is probably the single most useful logarithm property if you are a calculator addict. And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I'll show you why. So let's say I have log base B of A is equal to log base C of A divided by log base C of B. Now why is this a useful property if you are a calculator addict? Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, let's say you go to class and there's a quiz. The teacher says, you can use your calculator. And using your calculator, I want you to figure out the log base 17 of 357. And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And you will scramble and look for the log base 17 button on your calculator and not find it. Because there is no log base 17 number, a button on your calculator. You'll probably either have a log button or you'll have an LN button. And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something. It's a number."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And just so you know, the log button on your calculator is probably base 10. And your LN number, your LN button on your calculator is going to be base E. For those of you who aren't familiar with E, don't worry about it. But it's 2.71 something something. It's a number. It's nothing. It's an amazing number, but we'll talk more about that. In a future presentation."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "It's a number. It's nothing. It's an amazing number, but we'll talk more about that. In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "In a future presentation. So there's only two bases you have on your calculator. So if you want to figure out another base logarithm, you use this property. So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if you're given this on an exam, you can very confidently say, oh, well, that is just the same thing as you would have to switch to your yellow color in order to act with confidence. We could do either E or 10. But you could say that's the same thing as log base 10 of 357 divided by log base 10 of 17. So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So you literally could just say, type in 357 in your calculator and press the log button and you're going to get bam, bam, bam, bam. Then you can clear it or if you know how to use a parenthesis on your calculator, you can do that. But then you say, type 17 in your calculator, press the log button, you get bam, bam, bam, bam. And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And then you just divide them and you get your answer. So this is a super useful property for calculator addicts. And once again, I'm not going to go into a lot of depth of how this one to me is the most useful, but it doesn't completely fall out of, obviously, of the exponent properties, but it's hard for me to describe the intuition simply, so you probably want to watch the proof on it if you don't believe why this happens. But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples. So let's just rewrite a bunch of things in simpler forms."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "But anyway, with all of those aside, and this is probably the one you're going to be using the most in everyday life. I still use this in my job, so just so you know, the logarithms are useful. Let's do some examples. So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this. This is the same thing."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's just rewrite a bunch of things in simpler forms. So if I wanted to write the log base 2 of the square root of 32 divided by the square root of 8, how can I rewrite this so it's reasonably not messy? Well, let's think about this. This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is the same thing. This is equal to, I don't know if I move vertically or horizontally, but I'll move vertically. This is the same thing as the log base 2 of 32 over the square root of 8 to the 1 half power, right? And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "And we know from our logarithm properties, the third one we learned, that that is the same thing as 1 half times the logarithm of 32 divided by the square root of 8, right? I just took the exponent and made that the coefficient on the entire thing, and we learned that at the beginning of this video. And now we have a little quotient here, right? Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right?"}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "Logarithm of 32 divided by logarithm of square root of 8. Well, we can use our other logarithm. Let's keep the 1 half out. That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation."}, {"video_title": "Introduction to logarithm properties (part 2) Logarithms Algebra II Khan Academy.mp3", "Sentence": "That's going to equal, oops, parentheses, logarithm, oh, I forgot my base, logarithm base 2 of 32 minus, right, because this is in the quotient, minus the logarithm base 2 of the square root of 8, right? Let's see. Well, here, once again, we have a square root here, so we could say that this is equal to 1 half times log base 2 of 32 minus, this 8 to the 1 half, which is the same thing as 1 half log base 2 of 8. We learned that property at the beginning of this presentation. And then if we want, we can distribute this original 1 half, and this equals 1 half log base 2 of 32 minus 1 half, minus 1 fourth, because we have to distribute that 1 half, minus 1 fourth log base 2 of 8. This is 5 halves minus, this is 3, 3 times 1 fourth minus 3 fourths. Or 10 fourths minus 3 fourths is equal to 7 fourths."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "Use a calculator to find log base e of 67 to the nearest thousandth. So just as a reminder, e is one of these crazy numbers that shows up in nature and finance and all these things and it's approximately equal to 2.71 and it just keeps going on and on and on. So you could view log base e as 67. You might say, what does e mean? E is just a number, just like pi is just a number. This is really the same thing as saying log base 2.71 and the actual number. So you'd have to write all the digits that keep on going forever and never repeat of 67."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "You might say, what does e mean? E is just a number, just like pi is just a number. This is really the same thing as saying log base 2.71 and the actual number. So you'd have to write all the digits that keep on going forever and never repeat of 67. So what power do I have to raise e to to get to 67? So another way of saying that is if this is equal to x, you're saying e to the x is equal to 67. We need to figure out what x is."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "So you'd have to write all the digits that keep on going forever and never repeat of 67. So what power do I have to raise e to to get to 67? So another way of saying that is if this is equal to x, you're saying e to the x is equal to 67. We need to figure out what x is. Now, traditionally you will never see someone write log base e even though e is one of the most common bases to take a logarithm of. So the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "We need to figure out what x is. Now, traditionally you will never see someone write log base e even though e is one of the most common bases to take a logarithm of. So the reason why you wouldn't see log base e written this way is log base e is referred to as the natural logarithm. And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that or seeing that and the more typical way of seeing that is the natural log. And I think this is LN, so I think it's maybe from French or something, log natural of 67. So this is the same thing as log base e of 67."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "And I think that's used because e shows up so many times in nature. So log base e of 67, another way of saying that or seeing that and the more typical way of seeing that is the natural log. And I think this is LN, so I think it's maybe from French or something, log natural of 67. So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is the same thing as log base e of 67. This is saying the exact same thing. To what power do I have to raise e to to get 67? When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth. What power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI-85 out."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "When you see this LN, it literally means log base e. Now, they let us use a calculator and that's good because I don't know off the top of my head what power I have to raise 2.71 and so on and so forth. What power I have to raise that to to get to 67. So we'll get our calculator out. So we get the TI-85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you can literally type in the statement natural log of 67 and then evaluate it. So here this is the button for LN, means natural log, log natural maybe."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we get the TI-85 out. And different calculators will have different ways of doing it. If you have a graphing calculator like this, you can literally type in the statement natural log of 67 and then evaluate it. So here this is the button for LN, means natural log, log natural maybe. LN of 67 and then you press enter and it will give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer. But a graphing calculator can literally type it in the way that you would write it out."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "So here this is the button for LN, means natural log, log natural maybe. LN of 67 and then you press enter and it will give you the answer. If you don't have a graphing calculator, you might have to press 67 and then press natural log to give you the answer. But a graphing calculator can literally type it in the way that you would write it out. And then you would press enter. So 4.20469 and we want to round to the nearest thousandth. So this is a thousandths place right here, this 4."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "But a graphing calculator can literally type it in the way that you would write it out. And then you would press enter. So 4.20469 and we want to round to the nearest thousandth. So this is a thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. This is approximately equal to 4.205."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is a thousandths place right here, this 4. The digit after that is 5 or larger, it's a 6, so we're going to round up. So this is 4.205. This is approximately equal to 4.205. And it actually makes a lot of sense because we know that E is greater than 2 and it is less than 3. And if you think about what 2 to the fourth power gets you to 16 and 3 to the fourth power gets you to 81. 67 is between 16 and 81 and E is between 2 and 3."}, {"video_title": "Natural logarithm with a calculator Logarithms Algebra II Khan Academy.mp3", "Sentence": "This is approximately equal to 4.205. And it actually makes a lot of sense because we know that E is greater than 2 and it is less than 3. And if you think about what 2 to the fourth power gets you to 16 and 3 to the fourth power gets you to 81. 67 is between 16 and 81 and E is between 2 and 3. So at least it feels right that something that's like 2.71 to the little over the fourth power should get you to a number that's pretty close to 3 to the fourth power. And actually that makes sense because it's actually closer to 3. 2.71 is closer to 3 than it is to 2."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "It says divide the polynomials. The form of your answer should either be a straight up polynomial or a polynomial plus the remainder over x minus five, which we have here in the denominator, where p of x is a polynomial and k is an integer. So we've done stuff like this, but like always, I encourage you to pause this video and work on this on your own. And if you were doing this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer. But let's just do it on paper for now. All right, so we're trying to figure out what x minus five divided into two x to the third power. Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "And if you were doing this on Khan Academy, there's a little bit of an input box here where you'd have to type in the answer. But let's just do it on paper for now. All right, so we're trying to figure out what x minus five divided into two x to the third power. Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns. So this is my third degree column. And then I want my second degree column, but there is no second degree term here. There's a first degree term, so I'll write it out here."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Actually, I wanna be careful here because I wanna be very, very organized about my different degree columns. So this is my third degree column. And then I want my second degree column, but there is no second degree term here. There's a first degree term, so I'll write it out here. So minus 47x, and actually, to be even more careful, I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "There's a first degree term, so I'll write it out here. So minus 47x, and actually, to be even more careful, I'll write plus zero x squared. And then I have minus 15. By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene. All right, so now we can work through this. And first, we could say, hey, how many times does x go into the highest degree term here? Well, x goes into two x to the third power, two x squared times."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "By putting that plus zero x squared, that's making sure I'm doing good, I guess, degree place column hygiene. All right, so now we can work through this. And first, we could say, hey, how many times does x go into the highest degree term here? Well, x goes into two x to the third power, two x squared times. And we'd wanna put that in the second degree column. Two x squared. You can see how it would've gotten messy if I put negative 47x here."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Well, x goes into two x to the third power, two x squared times. And we'd wanna put that in the second degree column. Two x squared. You can see how it would've gotten messy if I put negative 47x here. I'd be like, where do I put that two x squared? And you might confuse yourself, which none of us would want to happen. All right, two x squared times negative five is negative 10x squared."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "You can see how it would've gotten messy if I put negative 47x here. I'd be like, where do I put that two x squared? And you might confuse yourself, which none of us would want to happen. All right, two x squared times negative five is negative 10x squared. Two x squared times x is two x to the third power. Now we wanna subtract what we have in red from what we have in blue. So I'll multiply them both by negative one, so that becomes a negative."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "All right, two x squared times negative five is negative 10x squared. Two x squared times x is two x to the third power. Now we wanna subtract what we have in red from what we have in blue. So I'll multiply them both by negative one, so that becomes a negative. And then that one becomes a positive. And that's actually one of the biggest areas for careless errors. If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "So I'll multiply them both by negative one, so that becomes a negative. And then that one becomes a positive. And that's actually one of the biggest areas for careless errors. If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative. It needs to be a positive now. All right, so zero x squared plus 10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "If you have negative here, and you just wanna subtract it because you know you have to subtract, we're like, no, I'm subtracting a negative. It needs to be a positive now. All right, so zero x squared plus 10x squared is 10x squared. And then the two x to the third minus two x to the third is just zero. And then we can bring down that negative 47x. And once again, we look at the highest degree terms. X goes into 10x squared 10x times."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "And then the two x to the third minus two x to the third is just zero. And then we can bring down that negative 47x. And once again, we look at the highest degree terms. X goes into 10x squared 10x times. So plus 10x. 10x times negative five is negative 50x. Negative 50x."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "X goes into 10x squared 10x times. So plus 10x. 10x times negative five is negative 50x. Negative 50x. 10x times x is 10x squared. And once again, we wanna subtract what we have in teal from what we have in red. So we can multiply both of these times negative one."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Negative 50x. 10x times x is 10x squared. And once again, we wanna subtract what we have in teal from what we have in red. So we can multiply both of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus 50x is positive three x."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "So we can multiply both of these times negative one. That becomes a negative. This one becomes a positive. Now, negative 47x plus 50x is positive three x. And then 10x squared minus 10x squared gets canceled out. Bring down that 15. Come on down."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Now, negative 47x plus 50x is positive three x. And then 10x squared minus 10x squared gets canceled out. Bring down that 15. Come on down. I used to watch a lot of Prices Right growing up. Never quite made it to the show. All right, x goes into three x how many times?"}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Come on down. I used to watch a lot of Prices Right growing up. Never quite made it to the show. All right, x goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "All right, x goes into three x how many times? It goes three times. Three times negative five is negative 15. Three times x is three x. We wanna subtract the orange from the teal. And so this becomes a negative. This becomes a positive."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "Three times x is three x. We wanna subtract the orange from the teal. And so this becomes a negative. This becomes a positive. 15, or negative 15 plus 15 is zero. And three x minus three x is zero. So you're just left with zero."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "This becomes a positive. 15, or negative 15 plus 15 is zero. And three x minus three x is zero. So you're just left with zero. So no remainder. So this whole thing you could re-express or simplify as two x squared plus 10x plus three. And once again, if this was on Khan Academy, there would be a little bit of an input box that looks something like this and you would have to type this in."}, {"video_title": "Dividing polynomials by linear expressions missing term Algebra 2 Khan Academy.mp3", "Sentence": "So you're just left with zero. So no remainder. So this whole thing you could re-express or simplify as two x squared plus 10x plus three. And once again, if this was on Khan Academy, there would be a little bit of an input box that looks something like this and you would have to type this in. Now, if you wanted these to be exactly the same expression, you would also need to constrain the domain. You would say, okay, four x does not equal positive five. And the reason why you have to constrain that is the whole reason why we can even divide by x minus five is we're assuming that x minus five is not equal to zero."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Welcome to this presentation on logarithm properties. Now this is going to be a very hands-on presentation. If you don't believe that one of these properties are true and you want them proved, I've made three or four videos that actually prove these properties. But what I'm going to do is I'm going to show you the properties and then show you how they can be used. It's going to be a little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But what I'm going to do is I'm going to show you the properties and then show you how they can be used. It's going to be a little more hands-on. So let's just do a little bit of a review of just what a logarithm is. So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So if I say that a, oh, that's not the right, let's see, I want to change, there you go. Let's say I say that a, let me start over, a to the b is equal to c. a to the b to the power is equal to c. So another way to write this exact same relationship, instead of writing an exponent, is to write it as a logarithm. So we could say that the logarithm base a of c is equal to b. So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So these are essentially saying the same thing, they just have different kind of results. In one, you know a and b and you're kind of getting c. That's what exponentiation does for you. And the second one, you know a and you know that when you raise it to some power, you get c. And then you figure out what b is. So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So the exact same relationship, just dated in a different way. Now I will introduce you to some interesting logarithm properties. And they actually just fall out of this relationship and the regular exponent rules. So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So the first is that the logarithm, let me do a more cheerful color. The logarithm, let's say, of any base, so let's just call the base, let's say b for base, logarithm base b of a plus logarithm base b of c. And this only works if we have the same basis, so that's important to remember. That equals the logarithm of base b of a times c. Now what does this mean and how can we use it? Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve. I don't know, I never know how to say that properly."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Or let's just even try it out with some, I don't know, examples. So this is saying that, I'll switch to another color. Let's make mauve my mauve. I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "I don't know, I never know how to say that properly. Let's make that my example color. So let's say logarithm of base 2 of, I don't know, of 8 plus logarithm base 2 of, I don't know, let's say 32. So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So in theory, this should equal, if we believe this property, this should equal logarithm base 2 of what? Well we say 8 times 32. So 8 times 32 is 240 plus 16, 256. Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's see if that's true just trying out this number. And this really isn't a proof, but it'll give you a little bit of an intuition, I think, for what's going on around here. So we just used our property, this little property that I presented to you, and let's see if it works out. So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So log base 2 of 8. 2 to what power is equal to 8? Well 2 to the third power is equal to 8, right? 2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 to the third power is equal to 8. So this term right here, that equals 3, right? Log base 2 of 8 is equal to 3. 2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right? And 2 to the what power is equal to 256?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "2 to what power is equal to 32? Let's see, 2 to the fourth power is 16, 2 to the fifth power is 32. So this right here is 5, right? And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And 2 to the what power is equal to 256? Well, let's see. Well if you're a computer science major, you'll know that immediately. That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "That a byte can have 256 values in it. So it's 2 to the eighth power. But if you don't know that, you could multiply it out yourself. But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "But this is 8. And I'm not doing it just because I knew that 3 plus 5 is equal to 8. I'm doing this independently. So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is equal to 8. But it does turn out that 3 plus 5 is equal to 8. This may seem like magic to you, or it may seem obvious. And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And for those of you who it might seem a little obvious, you're probably thinking, well, 2 to the third times 2 to the fifth is equal to 2 to the 3 plus 5, right? This is just an exponent rule. What do they call this? The additive exponent? I don't know. I don't know the names of things. And that equals 2 to the eighth."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "The additive exponent? I don't know. I don't know the names of things. And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And that equals 2 to the eighth. And that's exactly what we did here, right? On this side, we had 2 to the third times 2 to the fifth, essentially. And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "And on this side, you have them added to each other. And what makes logarithms interesting is, and why it's a little confusing at first, and you can watch the proofs if you really want a kind of a rigorous, not even my proofs aren't rigorous, but if you want kind of a better explanation of how this works. But this should hopefully give you an intuition for why this property holds, right? Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Because when you multiply two numbers of the same base, two exponential expressions of the same base, you can add their exponents. Similarly, when you have the log of two numbers multiplied by each other, that's equivalent to the log of each of the numbers added to each other. This is the same property. If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "If you don't believe me, watch the proof videos. So let me show you another log property that's pretty much the same one. I almost view them the same. So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So this is log base b of a minus log base b of c is equal to log base b of a divided by c. That says a divided by c. And we can, once again, try it out with some numbers. I use 2 a lot, just because 2 is an easy number to figure out the powers. But let's use a different number. Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Let's say log base 3 of 1 ninth minus log base 3 of 81. So this property tells us that this is the same thing as, well, I'm ending up with a big number. Log base 3 of 1 ninth divided by 81. So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward. So let's see."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So that's the same thing as 1 ninth times 1 over 81. I used two large numbers for my example. But we'll move forward. So let's see. 9 times 8 is 720. Right? 9 times 8 is 720."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So let's see. 9 times 8 is 720. Right? 9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "9 times 8 is 720. So this is 1 over 729. So this is log base 3 over 1 over 729. So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So 3 to what power is equal to 1 ninth? Well, 3 squared is equal to 9, right? 3 squared is equal to 9. So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2. Right?"}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "So we know that if 3 squared is equal to 9, then we know that 3 to the negative 2 is equal to 1 ninth, right? The negative just inverts it. So this is equal to negative 2. Right? And then minus 3 to what power is equal to 81? Let's see. 3 to the third power is 27."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Right? And then minus 3 to what power is equal to 81? Let's see. 3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "3 to the third power is 27. So 3 to the fourth power. So we have minus 2 minus 4 is equal to, well, we could do it a couple of ways. Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Minus 2 minus 4 is equal to minus 6. And now we just have to confirm that 3 to the minus 6 power is equal to 1 over 729. So my question is, 3 to the minus 6 power, is that equal to 1 over 729? Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here, but let's see."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Well, that's the same thing as saying 3 to the 6th power is equal to 729, because that's all the negative exponent does, is inverts it. Let's see. We could multiply that out, but that should be the case. Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close."}, {"video_title": "Introduction to logarithm properties Logarithms Algebra II Khan Academy.mp3", "Sentence": "Because, well, we could look here, but let's see. 3 to the third power. This would be 3 to the third power times 3 to the third power is equal to 27 times 27. That looks pretty close. You can confirm it with a calculator if you don't believe me. Anyway, that's all the time I have in this video. In the next video, I'll introduce you to the last two logarithm properties."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "We're told an electronic circuit has two resistors with resistances R1 and R2 connected in parallel. The circuit's total resistance, R sub T or RT, is given by this formula. Suppose we increase the value of R1 while keeping R2 constant. What does the value of R sub T increase, decrease, or stay the same? So pause this video and see if you can answer this question. All right, now let's work through this together. And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "What does the value of R sub T increase, decrease, or stay the same? So pause this video and see if you can answer this question. All right, now let's work through this together. And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression. There's some quantity, R sub T, that's equal to one over, and then in the denominator, we have one over R1 plus one over R2. So if we increase the value of R1 while keeping R2 constant, what happens? So this is going to increase, and R2 is going to be constant."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "And some of you might be familiar with the idea of an electronic circuit and resistors and what they represent, but you really don't need to understand that in order to understand what's going on in this expression. There's some quantity, R sub T, that's equal to one over, and then in the denominator, we have one over R1 plus one over R2. So if we increase the value of R1 while keeping R2 constant, what happens? So this is going to increase, and R2 is going to be constant. So one way to think about it, we have two variables here, especially in this denominator, but really in this entire expression. But if R2 is going to be constant, we really just have to focus our analysis on R1. If R2 is constant, that means it's just a number."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "So this is going to increase, and R2 is going to be constant. So one way to think about it, we have two variables here, especially in this denominator, but really in this entire expression. But if R2 is going to be constant, we really just have to focus our analysis on R1. If R2 is constant, that means it's just a number. It could be two, it could be five, it could be pi, whatever, but that is not going to change as we increase the value of R1. So let's think about what's happening here. If R sub one increases, then what does that do to one over R1?"}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "If R2 is constant, that means it's just a number. It could be two, it could be five, it could be pi, whatever, but that is not going to change as we increase the value of R1. So let's think about what's happening here. If R sub one increases, then what does that do to one over R1? Well, if you increase the denominator, then you are going to decrease the reciprocal of that. So that means that this whole thing right over here is going to decrease. Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2?"}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "If R sub one increases, then what does that do to one over R1? Well, if you increase the denominator, then you are going to decrease the reciprocal of that. So that means that this whole thing right over here is going to decrease. Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2? Will this entire expression increase or decrease? Well, this part is staying constant. R2 is constant, so one over R2 is constant."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "Now, if one over R1 is decreasing, if one over R1 is decreasing, what's going to happen to one over R1 plus one over R2? Will this entire expression increase or decrease? Well, this part is staying constant. R2 is constant, so one over R2 is constant. Just imagine R2 could be two or three, so this would just be 1 1\u20442 or 1 3rd or whatever it is, while over here, this part of the expression is going down. So if you're taking the sum of two things, one part's going down, the other part's constant, then that means this whole thing is going to be going down. So the entire denominator of this entire thing is going down."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "R2 is constant, so one over R2 is constant. Just imagine R2 could be two or three, so this would just be 1 1\u20442 or 1 3rd or whatever it is, while over here, this part of the expression is going down. So if you're taking the sum of two things, one part's going down, the other part's constant, then that means this whole thing is going to be going down. So the entire denominator of this entire thing is going down. Now, if the entire denominator is going down, if one over R1 plus one over R2, if this whole thing is going down, what's going to happen to the reciprocal of that? One over, one over R1 plus one over R2. Well, if something is going down, the reciprocal of that is going to go up."}, {"video_title": "Interpreting expressions with multiple variables Resistors Modeling Algebra II Khan Academy.mp3", "Sentence": "So the entire denominator of this entire thing is going down. Now, if the entire denominator is going down, if one over R1 plus one over R2, if this whole thing is going down, what's going to happen to the reciprocal of that? One over, one over R1 plus one over R2. Well, if something is going down, the reciprocal of that is going to go up. If you get smaller and smaller denominators, one over that is going to be a larger and larger value. So the value of RT increases if R1 increases and R2 is constant. And for those of you who know about resistance, which is really how well a current can flow through a circuit that will also make intuitive sense, but you don't need to understand resistance to analyze this mathematically."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So immediately you might say, well, this is either going to be a sine function or cosine function. But its midline and its amplitude are not just the plain vanilla sine or cosine function. And we can see that right over here. The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum point, it hits a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "The midline is halfway between the maximum point and the minimum point. The maximum point right over here, it hits a value of y equals 1. At the minimum point, it hits a value of y is equal to negative 5. So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2. So this right over here is the midline. So this thing is clearly, so this is y is equal to negative 2. This is y is equal to negative 2."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So halfway between those, the average of 1 and negative 5, 1 plus negative 5 is negative 4 divided by 2 is negative 2. So this right over here is the midline. So this thing is clearly, so this is y is equal to negative 2. This is y is equal to negative 2. So it's clearly shifted down. So we're going to have, so actually I'll talk in a second about what type of an expression it might be. But now also let's think about its amplitude."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "This is y is equal to negative 2. So it's clearly shifted down. So we're going to have, so actually I'll talk in a second about what type of an expression it might be. But now also let's think about its amplitude. That's how far it might get away from the midline. We see here it went 3 above the midline, going from negative 2 to 1. It went 3 above the midline at the maximum point."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "But now also let's think about its amplitude. That's how far it might get away from the midline. We see here it went 3 above the midline, going from negative 2 to 1. It went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So amplitude of 3."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "It went 3 above the midline at the maximum point. And it can also go 3 below the midline at the minimum point. So this thing clearly has an amplitude of 3. So amplitude of 3. So immediately we can say, well, look, this is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write cosine first."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So amplitude of 3. So immediately we can say, well, look, this is going to have a form something like f of x is equal to the amplitude 3. We haven't figured out yet whether this is going to be a cosine function or a sine function. So I'll write cosine first. Cosine may be some coefficient times x plus the midline. The midline we already figured out was minus 2 or negative 2. So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So I'll write cosine first. Cosine may be some coefficient times x plus the midline. The midline we already figured out was minus 2 or negative 2. So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x. Sine of kx minus 2, plus the midline. So minus 2. So how do we figure out which of these are?"}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So it could take that form, or it could take f of x is equal to 3 times, it could be sine of x, or sine of some coefficient times x. Sine of kx minus 2, plus the midline. So minus 2. So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So how do we figure out which of these are? Well, let's just think about the behavior of this function when x is equal to 0. When x is equal to 0, if this is kx, then the input into the cosine is going to be 0. Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0, so if x is 0, k times 0 is going to be 0. Sine of 0 is 0. So what's this thing doing when x is equal to 0?"}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Cosine of 0 is 1, whether you're talking about degrees or radians, cosine of 0 is 1. While sine of 0, so if x is 0, k times 0 is going to be 0. Sine of 0 is 0. So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So what's this thing doing when x is equal to 0? Well, when x is equal to 0, we are at the midline. If we're at the midline, that means that all of this stuff right over here evaluated to 0. So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So since when x equals 0, all of this stuff evaluated to 0, we can rule out the cosine function. When x equals 0 here, this stuff doesn't evaluate to 0. So we can rule out this one right over there. And so we are left with this. And we just really need to figure out what could this constant actually be? And to think about that, let's look at the period of this function. So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so we are left with this. And we just really need to figure out what could this constant actually be? And to think about that, let's look at the period of this function. So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let's just remind ourselves what the period of sine of x is."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So to go from, and we could, let's see, if we went from this point where we intersect the midline, we go this point to intersect the midline, and we have a positive slope, the next point that we do that is right over here. So our period is 8. So what coefficient could we have here to make the period of this thing be equal to 8? Well, let's just remind ourselves what the period of sine of x is. So the period of sine of x, so I'll write period right over here, is 2 pi. 2 pi, you increase your angle by 2 pi radians or decrease it. You're back at the same point on the unit circle."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, let's just remind ourselves what the period of sine of x is. So the period of sine of x, so I'll write period right over here, is 2 pi. 2 pi, you increase your angle by 2 pi radians or decrease it. You're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now your x, your input, is increasing k times faster. So you're going to get to the same point k times faster."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "You're back at the same point on the unit circle. So what would be the period of sine of kx? Well, now your x, your input, is increasing k times faster. So you're going to get to the same point k times faster. So your period is going to be 1 kth as long. So now your period is going to be 2 pi over k. Notice, you're increasing your argument as x increases. Your argument into the sine function is increasing k times as fast."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So you're going to get to the same point k times faster. So your period is going to be 1 kth as long. So now your period is going to be 2 pi over k. Notice, you're increasing your argument as x increases. Your argument into the sine function is increasing k times as fast. You're multiplying it by k. So your period is going to be shorter. It's going to take you less distance for the whole argument to get to the same point on the unit circle. So let's think about it this way."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "Your argument into the sine function is increasing k times as fast. You're multiplying it by k. So your period is going to be shorter. It's going to take you less distance for the whole argument to get to the same point on the unit circle. So let's think about it this way. So if we wanted to say 2 pi over k is equal to 8, well, what is our k? Well, we could take the reciprocal of both sides. We get k over 2 pi is equal to 1 over 8."}, {"video_title": "Determining the equation of a trig function Graphs of trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's think about it this way. So if we wanted to say 2 pi over k is equal to 8, well, what is our k? Well, we could take the reciprocal of both sides. We get k over 2 pi is equal to 1 over 8. Multiply both sides by 2 pi. And we get k is equal to pi over 4. And we are done."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So we have y is equal to 2 times the quantity x minus 4 squared plus 3. We also have y is equal to negative x squared plus 2x minus 2. So the solution, or it might be 1, or it might be none, it might be two solutions, but the solutions to this system occur for the x values that generate the same y values. So the same x and y that satisfy both of these equations. So in order to find the x values, they need to equal the same y value. So this y has to be that y value. So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So the same x and y that satisfy both of these equations. So in order to find the x values, they need to equal the same y value. So this y has to be that y value. So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there. Is equal to 2 times x minus 4 squared plus 3. And now let's just try to solve for x. So the left-hand side, well, we're going to have to multiply this out."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So the solution is going to occur when this guy right here, negative x squared plus 2x minus 2, is equal to that guy up there. Is equal to 2 times x minus 4 squared plus 3. And now let's just try to solve for x. So the left-hand side, well, we're going to have to multiply this out. So let's do that first. It's negative x squared plus 2x minus 2 is equal to, and on the right-hand side, 2 times x minus 4 squared is x squared minus 8x plus 16 plus 3. This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So the left-hand side, well, we're going to have to multiply this out. So let's do that first. It's negative x squared plus 2x minus 2 is equal to, and on the right-hand side, 2 times x minus 4 squared is x squared minus 8x plus 16 plus 3. This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. And that's, of course, going to be equal to this thing on the left-hand side. Negative x squared plus 2x minus 2. And now let's just get rid of this whole thing from the left-hand side all at once."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "This is going to be equal to 2x squared, I'm just distributing the 2, minus 16x plus 32 plus 3, which is equal to 2x squared minus 16x plus 35. And that's, of course, going to be equal to this thing on the left-hand side. Negative x squared plus 2x minus 2. And now let's just get rid of this whole thing from the left-hand side all at once. By adding x squared to both sides, we could all do it in one step, we're going to add x squared to both sides. Let's subtract 2x from both sides. And let's add 2 to both sides."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "And now let's just get rid of this whole thing from the left-hand side all at once. By adding x squared to both sides, we could all do it in one step, we're going to add x squared to both sides. Let's subtract 2x from both sides. And let's add 2 to both sides. And we will get, on the left-hand side, those cancel out, those cancel out, those cancel out. You're left with 0 is equal to 2x squared plus x squared is 3x squared, negative 16x minus 2x is negative 18x, and then 35 plus 2 is 37. So we just have a plain, vanilla quadratic equation right here."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "And let's add 2 to both sides. And we will get, on the left-hand side, those cancel out, those cancel out, those cancel out. You're left with 0 is equal to 2x squared plus x squared is 3x squared, negative 16x minus 2x is negative 18x, and then 35 plus 2 is 37. So we just have a plain, vanilla quadratic equation right here. And we might as well apply the quadratic formula here to try to solve it. So our solutions are going to be x is equal to negative b. Well, b is negative 18, so negative b is positive 18."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So we just have a plain, vanilla quadratic equation right here. And we might as well apply the quadratic formula here to try to solve it. So our solutions are going to be x is equal to negative b. Well, b is negative 18, so negative b is positive 18. So it's 18 plus or minus the square root of 18 squared minus 4 times 3 times c times 37. All of that over 2 times a, 2 times 3, which is 6. Now let's think about what this is going to be."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Well, b is negative 18, so negative b is positive 18. So it's 18 plus or minus the square root of 18 squared minus 4 times 3 times c times 37. All of that over 2 times a, 2 times 3, which is 6. Now let's think about what this is going to be. Over here we have, so it's 18 plus or minus the square root of, well, let's just use our calculator. I could multiply it out, but I think, so we have 18 squared minus 4 times 3 times 37. Negative 120."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Now let's think about what this is going to be. Over here we have, so it's 18 plus or minus the square root of, well, let's just use our calculator. I could multiply it out, but I think, so we have 18 squared minus 4 times 3 times 37. Negative 120. So it's 18 plus or minus the square root of negative 120. And you might even be able to figure out this is negative. 4 times 3 is 12."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Negative 120. So it's 18 plus or minus the square root of negative 120. And you might even be able to figure out this is negative. 4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. But we definitely end up with a negative number under the radical here."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "4 times 3 is 12. 12 times 37 is going to be a bigger number than 18. Although it's not 100% obvious, but you might be able to just get the intuition there. But we definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "But we definitely end up with a negative number under the radical here. Now, if we're dealing with real numbers, there is no square root of negative 120. So there is no solution to this quadratic equation. There is no solution. And if we wanted to, we could have just looked at the discriminant. The discriminant is this part, b squared minus 4ac. We see the discriminant is negative."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "There is no solution. And if we wanted to, we could have just looked at the discriminant. The discriminant is this part, b squared minus 4ac. We see the discriminant is negative. There's no solution, which means that these two guys, these two equations, never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "We see the discriminant is negative. There's no solution, which means that these two guys, these two equations, never intersect. There is no solution to the system. There are no x values that when you put into both of these equations give you the exact same y value. Now let's think a little bit about why that happened. This one is already in kind of our y-intercept form. And it's an upward opening parabola."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "There are no x values that when you put into both of these equations give you the exact same y value. Now let's think a little bit about why that happened. This one is already in kind of our y-intercept form. And it's an upward opening parabola. So it looks something like this. I'll do my best to draw it. Just a quick and dirty version of it."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "And it's an upward opening parabola. So it looks something like this. I'll do my best to draw it. Just a quick and dirty version of it. Let me draw my axes in a neutral color. So let's say that this right here is my y-axis. That right there is my x-axis, x and y."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Just a quick and dirty version of it. Let me draw my axes in a neutral color. So let's say that this right here is my y-axis. That right there is my x-axis, x and y. This vertex, it's in the vertex form, occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. And it's an upward opening parabola."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "That right there is my x-axis, x and y. This vertex, it's in the vertex form, occurs when x is equal to 4 and y is equal to 3. So x is equal to 4 and y is equal to 3. And it's an upward opening parabola. We have a positive coefficient out here. So this will look something like this. It will look something like this."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "And it's an upward opening parabola. We have a positive coefficient out here. So this will look something like this. It will look something like this. I don't know the exact thing, but that's close enough. Now what will this thing look like? Well, it's a downward opening parabola."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "It will look something like this. I don't know the exact thing, but that's close enough. Now what will this thing look like? Well, it's a downward opening parabola. And we could actually put this in vertex form. Let me do that. Let me put the second equation in vertex form, just so we have it."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Well, it's a downward opening parabola. And we could actually put this in vertex form. Let me do that. Let me put the second equation in vertex form, just so we have it. So we have a good sense. So y is equal to, we could factor out a negative 1, negative x squared minus 2x plus 2. x squared minus 2x plus 2. Actually, let me put the plus 2 further out."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Let me put the second equation in vertex form, just so we have it. So we have a good sense. So y is equal to, we could factor out a negative 1, negative x squared minus 2x plus 2. x squared minus 2x plus 2. Actually, let me put the plus 2 further out. Plus 2, all the way up out there. And then we could say, OK, half of negative 2 is negative 1. You square it."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Actually, let me put the plus 2 further out. Plus 2, all the way up out there. And then we could say, OK, half of negative 2 is negative 1. You square it. So you have a plus 1 and then a minus 1 there. And then this part right over here, we can rewrite as x minus 1 squared. So it becomes negative x minus 1 squared."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "You square it. So you have a plus 1 and then a minus 1 there. And then this part right over here, we can rewrite as x minus 1 squared. So it becomes negative x minus 1 squared. Let me just do it one step at a time. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So it becomes negative x minus 1 squared. Let me just do it one step at a time. I don't want to skip steps. Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "Negative x minus 1 squared minus 1 plus 2. So that's plus 1 out here. Or if we want to distribute the negative, we get y is equal to negative x minus 1 squared minus 1. So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1. The vertex is there. And this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "So here the vertex occurs at x is equal to 1, y is equal to negative 1. x is equal to 1, y is equal to negative 1. The vertex is there. And this is a downward opening parabola. We have a negative coefficient out here on the second degree term, so it's going to look something like this. It's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward."}, {"video_title": "Non-linear systems of equations 3 Algebra II Khan Academy.mp3", "Sentence": "We have a negative coefficient out here on the second degree term, so it's going to look something like this. It's going to look something like this. So as you see, they don't intersect. This vertex is above it and it opens upward. This is its minimum point. And it's above this guy's maximum point. So they will never intersect."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And then over here on the left, we have the sine taken of angle MKJ, cosine of angle MKJ, and tangent of angle MKJ. And angle MKJ is this angle right over here, same thing as theta. So these two angles, these two angles have the same measure. We see that right over there. And what we wanna do is figure out which of these expressions are equivalent to which of these expressions right over here. And so I encourage you to pause the video and try to work this through on your own. So assuming you've had a go at it, so let's try to work this out."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "We see that right over there. And what we wanna do is figure out which of these expressions are equivalent to which of these expressions right over here. And so I encourage you to pause the video and try to work this through on your own. So assuming you've had a go at it, so let's try to work this out. And when you look at this diagram, it looks like the intention here on the left is this evokes the unit circle definition of trig functions because this is really, this is a unit circle right over here. And this evokes kind of the Sokodowa definition because we're just kind of in a plain vanilla right triangle. And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So assuming you've had a go at it, so let's try to work this out. And when you look at this diagram, it looks like the intention here on the left is this evokes the unit circle definition of trig functions because this is really, this is a unit circle right over here. And this evokes kind of the Sokodowa definition because we're just kind of in a plain vanilla right triangle. And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful. So sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And so just to remind ourselves, let's just remind ourselves of Sokodowa because we have a feeling it might be useful. So sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. And tangent is opposite over adjacent. So we can refer to this, and we can also refer to remind ourselves of the unit circle definition of trig functions, that the cosine of an angle is the x-coordinate, and that the sine of where this ray intersects the unit circle, and the sine of this angle is going to be the y-coordinate. And what we'll see through this video is that they're actually, the unit circle definition is just an extension of Sokodowa. So let's look first at x over one."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And tangent is opposite over adjacent. So we can refer to this, and we can also refer to remind ourselves of the unit circle definition of trig functions, that the cosine of an angle is the x-coordinate, and that the sine of where this ray intersects the unit circle, and the sine of this angle is going to be the y-coordinate. And what we'll see through this video is that they're actually, the unit circle definition is just an extension of Sokodowa. So let's look first at x over one. So we have x, x is the x-coordinate. That's also the length of this side right over here. Relative to this angle, theta, that is the adjacent side."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So let's look first at x over one. So we have x, x is the x-coordinate. That's also the length of this side right over here. Relative to this angle, theta, that is the adjacent side. So x is equal to the adjacent side. What is one? Well, this is a unit circle."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Relative to this angle, theta, that is the adjacent side. So x is equal to the adjacent side. What is one? Well, this is a unit circle. One is the length of the radius, which for this right triangle is also the hypotenuse. So if we apply the Sokodowa definition, x over one is adjacent over hypotenuse. Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well, this is a unit circle. One is the length of the radius, which for this right triangle is also the hypotenuse. So if we apply the Sokodowa definition, x over one is adjacent over hypotenuse. Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. So that's going to be, this is equal to cosine of theta, but theta is the same thing as angle MKJ. They have the same measure. So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Adjacent over hypotenuse, adjacent over hypotenuse, that's cosine. So that's going to be, this is equal to cosine of theta, but theta is the same thing as angle MKJ. They have the same measure. So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one. Now let's move over to y over one. Well, y is going to be the length of this side right over here. Y is going to be, let me do this in the blue."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So cosine of angle MKJ is equal to cosine of theta, which is equal to x over one. Now let's move over to y over one. Well, y is going to be the length of this side right over here. Y is going to be, let me do this in the blue. Y is going to be this length relative to angle theta. That is the opposite side. That is the opposite side."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Y is going to be, let me do this in the blue. Y is going to be this length relative to angle theta. That is the opposite side. That is the opposite side. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse, that's sine of theta. Sine of theta."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "That is the opposite side. Now which trig function is opposite over hypotenuse? Opposite over hypotenuse, that's sine of theta. Sine of theta. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure. And now we see that's the same thing as y over one."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Sine of theta. So sine of angle MKJ is the same thing as sine of theta. We see that they have the same measure. And now we see that's the same thing as y over one. Now for both of these, I use the Sohcahtoa definition, but we could have also used the unit circle definition. X over one, that's the same thing. That's the same thing as x."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And now we see that's the same thing as y over one. Now for both of these, I use the Sohcahtoa definition, but we could have also used the unit circle definition. X over one, that's the same thing. That's the same thing as x. And the unit circle definition says, well, this x, this point, the x coordinate of where this, I guess you could say the terminal side of this angle, this ray right over here intersects the unit circle, that by definition, by the unit circle definition, is the cosine of this angle. X is equal to the cosine of this angle. And the unit circle definition, the y coordinate is equal to the sine of this angle."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "That's the same thing as x. And the unit circle definition says, well, this x, this point, the x coordinate of where this, I guess you could say the terminal side of this angle, this ray right over here intersects the unit circle, that by definition, by the unit circle definition, is the cosine of this angle. X is equal to the cosine of this angle. And the unit circle definition, the y coordinate is equal to the sine of this angle. We could have written this as, instead of x comma y, we could have written this as cosine of theta, sine theta, just like that. But let's keep going. Now we have x over y."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And the unit circle definition, the y coordinate is equal to the sine of this angle. We could have written this as, instead of x comma y, we could have written this as cosine of theta, sine theta, just like that. But let's keep going. Now we have x over y. We have adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now we have x over y. We have adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So this is the reciprocal of tangent. So this right over here, if we had to, this is equal to one over tangent of theta. And we later learn about cotangent and all of that, which is essentially this."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Tangent is opposite over adjacent, not adjacent over opposite. So this is the reciprocal of tangent. So this right over here, if we had to, this is equal to one over tangent of theta. And we later learn about cotangent and all of that, which is essentially this. But it's not one of our choices. So we can rule this one out. But then we have y over x."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "And we later learn about cotangent and all of that, which is essentially this. But it's not one of our choices. So we can rule this one out. But then we have y over x. Well, this is looking good. This is, y is opposite, opposite, x is adjacent relative to angle theta, adjacent. So this is the tangent of theta."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "But then we have y over x. Well, this is looking good. This is, y is opposite, opposite, x is adjacent relative to angle theta, adjacent. So this is the tangent of theta. This is equal to tangent of theta. So tangent of angle mkj is the same thing as tangent of theta, which is equal to y over x. Now let's look at j over k. So j over k. Now we're moving over to this triangle."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is the tangent of theta. This is equal to tangent of theta. So tangent of angle mkj is the same thing as tangent of theta, which is equal to y over x. Now let's look at j over k. So j over k. Now we're moving over to this triangle. J over k. So relative to this angle, because this is the angle that we care about, j is the length of the adjacent side, and k is the length of the opposite side, of the opposite side. So this is adjacent over opposite. So this is equal to adjacent over opposite."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now let's look at j over k. So j over k. Now we're moving over to this triangle. J over k. So relative to this angle, because this is the angle that we care about, j is the length of the adjacent side, and k is the length of the opposite side, of the opposite side. So this is adjacent over opposite. So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So once again, this is one, this is a reciprocal of the tangent function, not one of the choices right over here. So we can rule that one out."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So this is equal to adjacent over opposite. Tangent is opposite over adjacent, not adjacent over opposite. So once again, this is one, this is a reciprocal of the tangent function, not one of the choices right over here. So we can rule that one out. Now k over j. Well now this is opposite over adjacent, opposite over adjacent. That is equal to tangent of theta."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "So we can rule that one out. Now k over j. Well now this is opposite over adjacent, opposite over adjacent. That is equal to tangent of theta. This is equal to tangent of theta, or tangent of angle mkj. So this is equal to k over j. Now we have m over j. M over j. Hypotenuse over adjacent side."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "That is equal to tangent of theta. This is equal to tangent of theta, or tangent of angle mkj. So this is equal to k over j. Now we have m over j. M over j. Hypotenuse over adjacent side. This of course, this of course is equal to the hypotenuse. Hypotenuse over adjacent. Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Now we have m over j. M over j. Hypotenuse over adjacent side. This of course, this of course is equal to the hypotenuse. Hypotenuse over adjacent. Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. So this is actually one over the cosine of theta, not one of our choices here, so I'll just rule that one out right over there. Then we have its reciprocal, j over m. That's adjacent over hypotenuse. Adjacent over hypotenuse is cosine."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Well if it was adjacent over hypotenuse, we'd be dealing with cosine, but this is the reciprocal of that. So this is actually one over the cosine of theta, not one of our choices here, so I'll just rule that one out right over there. Then we have its reciprocal, j over m. That's adjacent over hypotenuse. Adjacent over hypotenuse is cosine. So this is equal to cosine of theta, or cosine of angle mkj. So we could write it down. So this is equivalent to j over m. And then one last one, k over m. Well that's opposite over hypotenuse."}, {"video_title": "Matching ratios to trig functions Trigonometry Khan Academy.mp3", "Sentence": "Adjacent over hypotenuse is cosine. So this is equal to cosine of theta, or cosine of angle mkj. So we could write it down. So this is equivalent to j over m. And then one last one, k over m. Well that's opposite over hypotenuse. Opposite over hypotenuse. That's going to be sine of theta. So this right over here is equal to sine of theta, which is the same thing as sine of angle mkj, which is the same thing as all of these expressions."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Real numbers include things like zero and one and 0.3 repeating and pi and e, and I could keep listing real numbers. These are the numbers that you're kind of familiar with. And then we explored something interesting. We explored the notion of, well, what if there was a number that if I squared it, I would get the negative one? And we defined that thing, that if we squared it, we got negative one. We defined that thing as i. And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We explored the notion of, well, what if there was a number that if I squared it, I would get the negative one? And we defined that thing, that if we squared it, we got negative one. We defined that thing as i. And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i and pi times i and e times i. So this might raise another interesting question. What if I combined imaginary and real numbers?"}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so we defined a whole new class of numbers which you could really view as multiples of the imaginary unit. So imaginary numbers would be i and negative i and pi times i and e times i. So this might raise another interesting question. What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real and imaginary numbers? For example, let's say that I had the number, let's say I call it z, and z tends to be what we, the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to the real number five plus the imaginary number three times i."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "What if I combined imaginary and real numbers? What if I had numbers that were essentially sums or differences of real and imaginary numbers? For example, let's say that I had the number, let's say I call it z, and z tends to be what we, the most used variable when we're talking about what I'm about to talk about, complex numbers. Let's say that z is equal to the real number five plus the imaginary number three times i. So this thing right over here, we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. It won't make any sense."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let's say that z is equal to the real number five plus the imaginary number three times i. So this thing right over here, we have a real number plus an imaginary number. You might be tempted to add these two things, but you can't. It won't make any sense. These are kind of going in different, well, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "It won't make any sense. These are kind of going in different, well, we'll think about it visually in a second, but you can't simplify this anymore. You can't add this real number to this imaginary number. A number like this, let me make it clear, that's real and this is imaginary. Imaginary. A number like this we call a complex number. A complex number."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "A number like this, let me make it clear, that's real and this is imaginary. Imaginary. A number like this we call a complex number. A complex number. It has a real part and an imaginary part. And sometimes you'll see notation like this where someone will say, well, what's the real part? What's the real part of our complex number z?"}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "A complex number. It has a real part and an imaginary part. And sometimes you'll see notation like this where someone will say, well, what's the real part? What's the real part of our complex number z? Well, that would be the five right over there. And then they might say, well, what's the imaginary part? What's the imaginary part of our complex number z?"}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "What's the real part of our complex number z? Well, that would be the five right over there. And then they might say, well, what's the imaginary part? What's the imaginary part of our complex number z? And then typically the way that this function is defined, they really wanna know, well, what multiple of i is this imaginary part right over here? And in this case, it is going to be three. And we can visualize this."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "What's the imaginary part of our complex number z? And then typically the way that this function is defined, they really wanna know, well, what multiple of i is this imaginary part right over here? And in this case, it is going to be three. And we can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we, on the vertical axis, we plot the imaginary part. So that's the imaginary part."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And we can visualize this. We can visualize this in two dimensions. Instead of having the traditional two-dimensional Cartesian plane with real numbers on the horizontal and the vertical axis, what we do to plot complex numbers is we, on the vertical axis, we plot the imaginary part. So that's the imaginary part. And then on the horizontal axis, we plot the real part. We plot the real part just like that. We plot the real part."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So that's the imaginary part. And then on the horizontal axis, we plot the real part. We plot the real part just like that. We plot the real part. So for example, z right over here, which is five plus three i, the real part is five. So we would go one, two, three, four, five. That's five right over there."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We plot the real part. So for example, z right over here, which is five plus three i, the real part is five. So we would go one, two, three, four, five. That's five right over there. The imaginary part is three. One, two, three. And so on the complex plane, we would visualize that number right over here."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "That's five right over there. The imaginary part is three. One, two, three. And so on the complex plane, we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And so on the complex plane, we would visualize that number right over here. This right over here is how we would visualize z on the complex plane. It's five, positive five in the real direction, positive three in the imaginary direction. We could plot other complex numbers. Let's say we had the complex number a, which is equal to, let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "We could plot other complex numbers. Let's say we had the complex number a, which is equal to, let's say it's negative two plus i. Where would I plot that? Well, the real part is negative two. Negative two. And the imaginary part is going to be, you could imagine this is plus one i. So we go one up, it's going to be right over there."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Well, the real part is negative two. Negative two. And the imaginary part is going to be, you could imagine this is plus one i. So we go one up, it's going to be right over there. So that right over there is our complex number, our complex number a, would be at that point of the complex, complex, let me write that, that point of the complex plane. And let me just do one more. Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "So we go one up, it's going to be right over there. So that right over there is our complex number, our complex number a, would be at that point of the complex, complex, let me write that, that point of the complex plane. And let me just do one more. Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i. Where would we plot that? Well, one, two, three, four. And then, let's see, minus one, two, three, or negative three gets us right over there."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "Let's say you had the complex number b, which is going to be, let's say it is, let's say it's four minus three i. Where would we plot that? Well, one, two, three, four. And then, let's see, minus one, two, three, or negative three gets us right over there. So that right over there would be the complex number b."}, {"video_title": "Introduction to complex numbers Imaginary and complex numbers Precalculus Khan Academy.mp3", "Sentence": "And then, let's see, minus one, two, three, or negative three gets us right over there. So that right over there would be the complex number b."}]