video_title stringlengths 25 104 | Sentence stringlengths 91 1.69k |
|---|---|
Inconsistent systems of equations Algebra II Khan Academy.mp3 | And so you do that. So you say, well, let me solve both of these for b. So if you want to solve this first equation for b, you just subtract 2a from both sides. If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | If you subtract 2a from both sides of this first equation, you get b is equal to negative 2a plus 3. Now let's solve this second equation for b. So the first thing you might want to do is subtract 6a from both sides. So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | So you would get 3b is equal to negative 6a plus 15. And then you can divide both sides by 3. You get b is equal to negative 2a plus 5. So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | So the second equation, let me revert back to that other shade of green, is b is equal to negative 2a plus 5. And we haven't even graphed it yet, but it looks like something interesting is going on. They both have the exact same slope when you solve for b, but they seem to have different, I guess you could call them b-intercepts. Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis. And then this could be my a-axis. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | Let's graph it to actually see what's going on. So let me draw some axes over here. So let's call that my b-axis. And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | And then this could be my a-axis. And this first equation has a b-intercept of positive 3. So let's see, 1, 2, 3, 4, 5. The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this. I'm trying my best to draw it straight. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | The first one has a b-intercept of positive 3, and then has a slope of negative 2. So you go down, or you go to the right one, you go down 2. So the line looks something like this. I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | I'm trying my best to draw it straight. So it looks something like that. And now let's draw this green one. This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there. And you immediately see now that the bird was right. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | This green one, our b-intercept is 5, so it's right over here. But we have the exact same slope, a slope of negative 2. So it looks something like that right over there. And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect. And so the bird is right. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | And you immediately see now that the bird was right. There is no solution, because these two constraints can be represented by lines that don't intersect. So the lines don't intersect. And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | And so the bird is right. There's no solution. There's no x and y that can make this statement equal true, or that can make 0 equal 6. There is no possible. There is no overlap between these two things. And so something gets into your brain. You realize that our bagel is trying to stump you. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | There is no possible. There is no overlap between these two things. And so something gets into your brain. You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations. Which happens to be the word that is sometimes used to refer to a system that has no solutions, where the lines do not intersect. |
Inconsistent systems of equations Algebra II Khan Academy.mp3 | You realize that our bagel is trying to stump you. And you say, our bagel, you have given me inconsistent information. This is an inconsistent system of equations. Which happens to be the word that is sometimes used to refer to a system that has no solutions, where the lines do not intersect. And therefore, this information is incorrect. We cannot assume that the apple or banana, either you are lying, which is possible, or you accounted for it wrong, or maybe the prices of apples and bananas actually changed between the two visits to the market. At which point, the bird whispered into the king's ear and says, oh, this character isn't so bad at this algebra stuff. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So this has a negative 1, you just see a minus here, but it's really the same thing as having a negative 1 times this quantity, and here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left hand side of our equation we have our negative 9, and then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to, let's distribute the 3, 3 times 4x is 12x, and then 3 times 6 is 18. Now what we want to do, let's combine our constant terms if we can, we have a negative 9 and a 6 here, on this side we've combined all of our like terms, we can't combine a 12x and an 18. So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So let's combine this, let's combine the negative 9 and the 6, our two constant terms on the left hand side of the equation. So we're going to have this negative 9x, so we're going to have negative 9x plus, let's see we have a negative 9 and then plus 6, so negative 9 plus 6 is negative 3. So we're going to have a negative 9x and then we have a negative 3, so minus 3 right here, that's the negative 9 plus the 6, and that is equal to 12x plus 18. Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | Now we want to group all of the x terms on one side of the equation and all of the constant terms, the negative 3 and the positive 18 on the other side. I like to always have my x terms on the left hand side if I can, you don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right, and the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So let me subtract 12x from the right, and subtract 12x from the left. Now on the left hand side I have negative 9x minus 12x, so negative 9 minus 12, that's negative 21x minus 3 is equal to 12x minus 12x, well that's just nothing, that's 0, so I could write a 0 here but I don't have to write anything, that was the whole point of subtracting the 12x from the left hand side. And that is going to be equal to, so on the right hand side we just are left with an 18, we are just left with that 18 here, these guys cancelled out. Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | Now let's get rid of this negative 3 from the left hand side, so on the left hand side we only have x terms, on the right hand side we only have constant terms. So the best way to cancel out a negative 3 is to add 3, so it cancels out to 0, so we're going to add 3 to the left, let's add 3 to the right. And we get the left hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x, the negative 3 and the plus 3 or the positive 3 cancel out, that was the whole point, equals, what's 18 plus 3? 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | 18 plus 3 is 21, so now we have negative 21x is equal to 21 and we want to solve for x. So if you have something times x and you just want it to be an x, let's divide by that something, and in this case that something is negative 21, so let's divide both sides of this equation by negative 21. Divide both sides by negative 21, the left hand side, negative 21 divided by negative 21, you're just left with an x, that was the whole point behind dividing by negative 21. And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | And we get x is equal to, what's 21 divided by negative 21? Well that's just negative 1, right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So that is our answer. Now let's verify that this actually works for that original equation, so let's substitute negative 1 into that original equation. So we have negative 9, I'll do it over here, I'll do it in a different color than we've been using, we have negative 9 minus, that 1 wasn't there originally, it's there implicitly, minus 9 times negative 1. 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | 9 times, I'll put negative 1 in parentheses, minus 6 is equal to, well actually let me just solve for the left hand side when I substitute a negative 1 there. So the left hand side becomes negative 9 minus, 9 times negative 1 is negative 9 minus 6, and so this is negative 9 minus, in parentheses, negative 9 minus 6 is negative 15. So this is equal to negative 15, and so we get negative 9, let me make sure I did that, negative 9 minus 6, yep, negative 15. So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation. I'll do it in green. |
Solving equations with the distributive property Linear equations Algebra I Khan Academy.mp3 | So I have negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left hand side of the equation when we substitute x is equal to negative 1, we get that it equals 6. So let's see what happens when we substitute negative 1 to the right hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1, plus 6, so that is 3 times negative 4 plus 6, negative 4 plus 6 is 2, so it's 3 times 2, which is also 6. So when x is equal to negative 1, you substitute here, the left hand side becomes 6, and the right hand side becomes 6. Definitely works out. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | So just as a quick review of what we did in the last video, we said that this is the same thing as 3 times the principal root of 500. And I'm going to do it a little bit different than I did in the last video, just to make it interesting. This is 3 times the principal root of 500 times the principal root of x to the third. And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | And 500, we can rewrite it, because 500 is not a perfect square. We can rewrite 500 as 100 times 5, or even better, we can rewrite that as 10 squared times 5. 10 squared is the same thing as 100. So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | So we can rewrite this first part over here as 3 times the principal root of 10 squared times 5 times the principal root of x squared times x. That's the same thing as x to the third. Now, the one thing I'm going to do here, actually I won't talk about it just yet, of how we're going to do it differently than we did in the last video. This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | This radical right here can be rewritten as, so this is going to be 3 times the square root, or the principal root I should say, of 10 squared times the square root of 5. If we take the square root of the product of two things, it's the same thing as taking the square root of each of them and taking the product. And so this over here is going to be times the square root of, or the principal root of x squared times the principal root of x. And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | And the principal root of 10 squared is 10. And then what I said in the last video is that the principal root of x squared is going to be the absolute value of x, just in case x itself is a negative number. And so then if you simplify all of this, you get 3 times 10, which is 30, times, and I'm just going to switch the order here, times the absolute value of x. And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | And then you have the square root of 5, or the principal root of 5, times the principal root of x. And this is just going to be equal to the principal root of 5x. Taking the square root of something and multiplying that times the square root of something else is the same thing as just taking the square root of 5x. So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | So all of this simplified down to 30 times the absolute value of x times the principal root of 5x. And this is what we got in the last video. And the interesting thing here is if we assume we're only dealing with real numbers, the domain of x right over here, the x's that will make this expression defined in the real numbers, then x has to be greater than or equal to 0. So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | So let me, so maybe I could write it this way. The domain here is that x is, x, any real number greater than or equal to 0. And the reason why I say that is if you put a negative number in here, if you put a negative number in here and you cube it, you're going to get another negative number. And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | And then it doesn't make, at least in the real numbers, you won't get an actual value. You'll get a square root of a negative number here. So if you make this, if you assume this right here, we're dealing with the real numbers. We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number. |
Simplifying square roots comment response Algebra I Khan Academy.mp3 | We're not dealing with any complex numbers. When you open up the complex numbers, then you can expand the domain more broadly. But if you're dealing with real numbers, you can say that x is going to be greater than or equal to 0. And then the absolute value of x is just going to be x because it's not going to be a negative number. And so if we're assuming that the domain of x is, or if this expression is going to be a valuable, or it's going to have a positive number, then this can be written as 30x times the square root of 5x. If you had the situation where we were dealing with complex numbers, then you would. So numbers that were, and if you don't know what a complex number is or an imaginary number, don't worry too much about it. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | 45x squared minus 125. So whenever I see something like this, I have a second degree term here, I have a subtraction sign, my temptation is to look at this as a difference of squares. We've already seen this multiple times. We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | We've already seen that if we have something of the form a squared minus b squared, that this can be factored as a plus b times a minus b. So let's look over here. Well, over here it's not obvious that this right over here is a perfect square. Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | Neither is it obvious that this right over here is a perfect square. So it's not clear to me that this is a difference of squares. But what is interesting is that both 45 and 125 have some factors in common. And the one that jumps out at me is 5. So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | And the one that jumps out at me is 5. So let's see if we can factor out a 5. And by doing that, whether we can get something that's a little bit closer to this pattern right over here. So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | So if we factor out a 5, this becomes 5 times 45x squared divided by 5 is going to be 9x squared. And then 125 divided by 5 is 25. Now this is interesting. 9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | 9x squared, that's a perfect square. If we call this a squared, then that tells us that a would be equal to 3x. 3x, the whole thing squared is 9x squared. Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | Similarly, I can never say similarly correctly, 25 is clearly just 5 squared. So in this case, if we're looking at this template, b would be equal to 5. So now this is a difference of squares. And we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | And we can factor it completely. So we can't forget our 5 out front that we factored out. So it's going to be 5 times a plus b. So let me write this. So it's going to be 5 times a plus b times a minus b. So it's 5 times a plus b times a minus b. So let me write the b's down. |
Example 2 Factoring a difference of squares with leading coefficient other than 1 Khan Academy.mp3 | So let me write this. So it's going to be 5 times a plus b times a minus b. So it's 5 times a plus b times a minus b. So let me write the b's down. Plus b and minus b. And we're done. 5 times 3x plus 5 times 3x minus 5 is 45x squared minus 125 factored out. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So let's think about what they're saying. They're saying two cups of flour. So two cups of flour for every three cups of oatmeal. For every three cups of oatmeal. And so they're saying how much flour is needed for a big batch of cookies that uses nine cups of oatmeal. So now we're going to a situation where we are using nine cups of oatmeal. Let me write it this way. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | For every three cups of oatmeal. And so they're saying how much flour is needed for a big batch of cookies that uses nine cups of oatmeal. So now we're going to a situation where we are using nine cups of oatmeal. Let me write it this way. Nine cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | Let me write it this way. Nine cups of oatmeal. And I'll show you a couple of different ways to think about it. And whatever works for you, that works. So first of all, one way to think about it, so we're wondering, we're going to say, look, we know if we have three cups of oatmeal, we should use two cups of flour. But what we don't know is if we have nine cups of oatmeal, how many cups of flour do we use? That's what they're asking us. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | And whatever works for you, that works. So first of all, one way to think about it, so we're wondering, we're going to say, look, we know if we have three cups of oatmeal, we should use two cups of flour. But what we don't know is if we have nine cups of oatmeal, how many cups of flour do we use? That's what they're asking us. But if we're going from three cups of oatmeal to nine cups of oatmeal, how much more oatmeal are we using? We're using three times more oatmeal. We're multiplying by three. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | That's what they're asking us. But if we're going from three cups of oatmeal to nine cups of oatmeal, how much more oatmeal are we using? We're using three times more oatmeal. We're multiplying by three. Three cups of oatmeal to nine cups of oatmeal, we're using three times the oatmeal. If we want to use flour in the same proportion, we have to use three times the flour. So then we're also going to have to multiply the flour times three. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | We're multiplying by three. Three cups of oatmeal to nine cups of oatmeal, we're using three times the oatmeal. If we want to use flour in the same proportion, we have to use three times the flour. So then we're also going to have to multiply the flour times three. So we're going to have to use six cups of flour. We're going to have to use 6 cups of, ignore that question mark, 6 cups of flour. Another way you could think about it, and that answers their question. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So then we're also going to have to multiply the flour times three. So we're going to have to use six cups of flour. We're going to have to use 6 cups of, ignore that question mark, 6 cups of flour. Another way you could think about it, and that answers their question. That's how much flour we need for a big batch of cookies that uses 9 cups of oatmeal. The other thing is you could set up a proportion. You could say 2 cups of flour over 3 cups of oatmeal is equal to question mark. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | Another way you could think about it, and that answers their question. That's how much flour we need for a big batch of cookies that uses 9 cups of oatmeal. The other thing is you could set up a proportion. You could say 2 cups of flour over 3 cups of oatmeal is equal to question mark. And I'll say, instead of writing question mark, I'll put a variable in there. I'll say is equal to a question, actually let me put a question mark there just so you really understand, is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | You could say 2 cups of flour over 3 cups of oatmeal is equal to question mark. And I'll say, instead of writing question mark, I'll put a variable in there. I'll say is equal to a question, actually let me put a question mark there just so you really understand, is equal to a question mark in a box number cups of flour over 9 cups of oatmeal. And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross multiplying, but that cross multiplying is still using a little bit of algebra. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | And so I like this first way we did it because it's really just common sense. If we're tripling the oatmeal, then we're going to have to triple the flour to make the recipe in the same proportion. Another way, once you set up an equation like this, is actually to do a little bit of algebra. Some people might call it cross multiplying, but that cross multiplying is still using a little bit of algebra. And I'll show you why they're really the same thing. In cross multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross multiplication, you'll say that 2 times 9 must be equal to question mark times 3. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | Some people might call it cross multiplying, but that cross multiplying is still using a little bit of algebra. And I'll show you why they're really the same thing. In cross multiplication, whenever you have a proportion set up like this, people will multiply the diagonals. So when you use cross multiplication, you'll say that 2 times 9 must be equal to question mark times 3. Whatever's in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So when you use cross multiplication, you'll say that 2 times 9 must be equal to question mark times 3. Whatever's in this question mark, the number of cups of flour times 3. Or we get 18 is equal to whatever our question mark was times 3. So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal to 18? You might be able to do that in your head. That is 6. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So the number of cups of flour we need to use times 3 needs to be equal to 18. What times 3 is equal to 18? You might be able to do that in your head. That is 6. Or you could divide both sides by 3 and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | That is 6. Or you could divide both sides by 3 and you will get 6. So we get question mark in a box needs to be equal to 6 cups of flour. Same answer we got through kind of common sense. Now, you might be wondering, hey, this cross multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this, a proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there, that that needs to be equal to the numerator here times the denominator there? |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | Same answer we got through kind of common sense. Now, you might be wondering, hey, this cross multiplying doesn't make any intuitive sense. Why does that work? If I have something set up like this, a proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there, that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x, just to simplify the writing a little bit. So we have 2 over 3 is equal to, instead of that question mark, I'll write x over 9. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | If I have something set up like this, a proportion set up, why does it work that if I take the denominator here and multiply it by the numerator there, that that needs to be equal to the numerator here times the denominator there? And that comes from straight up algebra. And to do that, I'm just going to rewrite this part as x, just to simplify the writing a little bit. So we have 2 over 3 is equal to, instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you'd have to do the same thing to it. And what we want to do is we want to simplify this so all we have on the right-hand side is an x. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So we have 2 over 3 is equal to, instead of that question mark, I'll write x over 9. And in algebra, all you're saying is that this quantity over here is equal to this quantity over here. So if you do anything to what's on the left, if you want it to still be equal, if the thing on the right still needs to be equal, you'd have to do the same thing to it. And what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x, so that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | And what we want to do is we want to simplify this so all we have on the right-hand side is an x. So what can we multiply this by so that we're just left with an x, so that we've solved for x? Well, if we multiply this times 9, the 9's are going to cancel out. So let's multiply the right by 9. But of course, if we multiply the right by 9, we have to still multiply the left by 9. Otherwise, they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | So let's multiply the right by 9. But of course, if we multiply the right by 9, we have to still multiply the left by 9. Otherwise, they still wouldn't be equal. If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2 thirds, or 9 over 1 times 2 thirds. Or this is equal to 18 over 3. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | If they were equal before being multiplied by 9, for them to still be equal, you have to multiply 9 times both sides. On the right-hand side, the 9's cancel out, so you're just left with an x. On the left-hand side, you have 9 times 2 thirds, or 9 over 1 times 2 thirds. Or this is equal to 18 over 3. And we know that 18 over 3 is the same thing as 6. So these are all legitimate ways to do it. I want you to understand that what I'm doing right here is algebra. |
Proportion word problem (example 1) 7th grade Khan Academy.mp3 | Or this is equal to 18 over 3. And we know that 18 over 3 is the same thing as 6. So these are all legitimate ways to do it. I want you to understand that what I'm doing right here is algebra. That's actually the reasoning why cross multiplication works. But for a really simple problem like this, you could really just use common sense. If you're increasing the cups of oatmeal by a factor of 3, then increase the cups of flour by a factor of 3. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So to figure this out, I've actually already copy and pasted this problem onto my scratch pad. I have it right over here. 1 half times 2a minus 6b plus 8. Actually, let me just rewrite it. So I'm gonna take, and let me color code it too, just for fun. So this is going to be 1 half times, give myself some space, 1 half times 2a minus 6b. So 2a minus 6b minus 6, let me write it this way, minus 6b, and then we have plus 8. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | Actually, let me just rewrite it. So I'm gonna take, and let me color code it too, just for fun. So this is going to be 1 half times, give myself some space, 1 half times 2a minus 6b. So 2a minus 6b minus 6, let me write it this way, minus 6b, and then we have plus 8. Plus, and I will do 8 in this color. Plus 8. And so I just need to distribute the 1 half. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So 2a minus 6b minus 6, let me write it this way, minus 6b, and then we have plus 8. Plus, and I will do 8 in this color. Plus 8. And so I just need to distribute the 1 half. If I'm multiplying 1 half times this entire expression, that means I multiply 1 half times each of these terms. So I'm gonna multiply 1 half times this, 1 half times this, and 1 half times that. So 1 half times 2a, so it's gonna be 1 half times 2a times, let me do that same color so you see where the 2a came from. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And so I just need to distribute the 1 half. If I'm multiplying 1 half times this entire expression, that means I multiply 1 half times each of these terms. So I'm gonna multiply 1 half times this, 1 half times this, and 1 half times that. So 1 half times 2a, so it's gonna be 1 half times 2a times, let me do that same color so you see where the 2a came from. 1 half times 2a minus 1 half times 6b minus 1 half times 6b times 6b plus 1 half times 8. 1 half times 8. And so what's this going to be? |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So 1 half times 2a, so it's gonna be 1 half times 2a times, let me do that same color so you see where the 2a came from. 1 half times 2a minus 1 half times 6b minus 1 half times 6b times 6b plus 1 half times 8. 1 half times 8. And so what's this going to be? Well, let's see, I have 1 half times 2a. 1 half times 2 is just 1, so you're just gonna be left with a. And then you have minus 1 half times 6b. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And so what's this going to be? Well, let's see, I have 1 half times 2a. 1 half times 2 is just 1, so you're just gonna be left with a. And then you have minus 1 half times 6b. Well, we can just think about what 1 half times 6 is going to be. 1 half times 6 is going to be 3, and then you still are multiplying by this b. So it's gonna be 3b, and then we have plus 1 half times 8. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And then you have minus 1 half times 6b. Well, we can just think about what 1 half times 6 is going to be. 1 half times 6 is going to be 3, and then you still are multiplying by this b. So it's gonna be 3b, and then we have plus 1 half times 8. 1 half of 8 is 4, or as you can see, 8 halves is equal to 4 wholes. All right, so this is going to be 4. So it's a minus 3b plus 4. a minus 3b plus 4. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So it's gonna be 3b, and then we have plus 1 half times 8. 1 half of 8 is 4, or as you can see, 8 halves is equal to 4 wholes. All right, so this is going to be 4. So it's a minus 3b plus 4. a minus 3b plus 4. So let's type that in. It's gonna be a minus 3b plus 4. And notice, it's literally half of each of these terms. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So it's a minus 3b plus 4. a minus 3b plus 4. So let's type that in. It's gonna be a minus 3b plus 4. And notice, it's literally half of each of these terms. Half of 2a is a, half of 6b is 3b, so we have minus 6b, so it's gonna be minus 3b, and then plus 8, instead of that, half of that, plus 4. So let's check our answer. And we got it right. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And notice, it's literally half of each of these terms. Half of 2a is a, half of 6b is 3b, so we have minus 6b, so it's gonna be minus 3b, and then plus 8, instead of that, half of that, plus 4. So let's check our answer. And we got it right. Let's do another one of these. So let's say, so they say apply the distributive property to factor out the greatest common factor. And here we have 60m minus 40. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And we got it right. Let's do another one of these. So let's say, so they say apply the distributive property to factor out the greatest common factor. And here we have 60m minus 40. So let me get my scratch pad out again. So, I'm running out of space that way. So we have, all right, like this. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And here we have 60m minus 40. So let me get my scratch pad out again. So, I'm running out of space that way. So we have, all right, like this. We have 60m minus 40. Minus 40. So what is the greatest common factor of 60m and 40? |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So we have, all right, like this. We have 60m minus 40. Minus 40. So what is the greatest common factor of 60m and 40? Well, 10 might jump out at us. We might say, okay, look, you know, 60 is 10, so we could say this is the same thing as 10 times 6, and actually, and then of course you have the m there, so you could view this 10 times 6m, and then you could view, you could view this as 10 times 4. But 10 still isn't the greatest common factor, and you say, well, Sal, how do you know that? |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So what is the greatest common factor of 60m and 40? Well, 10 might jump out at us. We might say, okay, look, you know, 60 is 10, so we could say this is the same thing as 10 times 6, and actually, and then of course you have the m there, so you could view this 10 times 6m, and then you could view, you could view this as 10 times 4. But 10 still isn't the greatest common factor, and you say, well, Sal, how do you know that? Well, because 4 and 6 still share a factor in common. They still share 2. So if you're actually factoring out the greatest common factor, what's left should not share a factor with each other. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | But 10 still isn't the greatest common factor, and you say, well, Sal, how do you know that? Well, because 4 and 6 still share a factor in common. They still share 2. So if you're actually factoring out the greatest common factor, what's left should not share a factor with each other. So let me think even harder about what a greatest common factor of 60 and 40 is. Well, 2 times 10 is 20. So you could actually factor out a 20. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So if you're actually factoring out the greatest common factor, what's left should not share a factor with each other. So let me think even harder about what a greatest common factor of 60 and 40 is. Well, 2 times 10 is 20. So you could actually factor out a 20. So you have 20 and 30m, sorry, 20 and 3m, and 40 can be factored out into 20 and 20 and 2. And now 3m and 2, 3m and 2 share no common factor, so you know that you have fully factored these two things out. Now if you think that this is something, kind of a strange art that I just did, one way to think about greatest common factors, you say, okay, 60, you can literally do a prime factorization. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So you could actually factor out a 20. So you have 20 and 30m, sorry, 20 and 3m, and 40 can be factored out into 20 and 20 and 2. And now 3m and 2, 3m and 2 share no common factor, so you know that you have fully factored these two things out. Now if you think that this is something, kind of a strange art that I just did, one way to think about greatest common factors, you say, okay, 60, you can literally do a prime factorization. You say 60 is 2 times 30, which is 2 times 15, which is 3 times 5. So that's 60's prime factorization, 2 times 2 times 3 times 5. And then 40's prime factorization is 2 times 20. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | Now if you think that this is something, kind of a strange art that I just did, one way to think about greatest common factors, you say, okay, 60, you can literally do a prime factorization. You say 60 is 2 times 30, which is 2 times 15, which is 3 times 5. So that's 60's prime factorization, 2 times 2 times 3 times 5. And then 40's prime factorization is 2 times 20. 20 is 2 times 10. 10 is 2 times 5. So that right over here, this is 40's prime factorization. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | And then 40's prime factorization is 2 times 20. 20 is 2 times 10. 10 is 2 times 5. So that right over here, this is 40's prime factorization. And to get out the greatest common factor, you want to get out as many common prime factors. So you have here, you have two 2's and a 5. Here you have two 2's and a 5. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So that right over here, this is 40's prime factorization. And to get out the greatest common factor, you want to get out as many common prime factors. So you have here, you have two 2's and a 5. Here you have two 2's and a 5. You can't go to three 2's and a 5 because there aren't three 2's and a 5 over here. So you have two 2's and a 5 here, two 2's and a 5 here. So 2 times 2 times 5 is going to be the greatest common factor. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | Here you have two 2's and a 5. You can't go to three 2's and a 5 because there aren't three 2's and a 5 over here. So you have two 2's and a 5 here, two 2's and a 5 here. So 2 times 2 times 5 is going to be the greatest common factor. So 2 times 2 times 5, that's 4 times 5. 4 times 5, that is 20. That's one way of kind of very systematically figuring out a greatest common factor. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | So 2 times 2 times 5 is going to be the greatest common factor. So 2 times 2 times 5, that's 4 times 5. 4 times 5, that is 20. That's one way of kind of very systematically figuring out a greatest common factor. But anyway, now that we know that 20 is the greatest common factor, let's factor it out. So this is going to be equal to 20 times, so 60m divided by 20, you're just going to be left with 3m. Just going to be left with 3m. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | That's one way of kind of very systematically figuring out a greatest common factor. But anyway, now that we know that 20 is the greatest common factor, let's factor it out. So this is going to be equal to 20 times, so 60m divided by 20, you're just going to be left with 3m. Just going to be left with 3m. And then minus, minus, 40 divided by 20, you're just left with a 2. Minus 2. Minus 2. |
How to use the distributive property with variables 6th grade Khan Academy.mp3 | Just going to be left with 3m. And then minus, minus, 40 divided by 20, you're just left with a 2. Minus 2. Minus 2. So let's type that in. So this is going to be 20 times, 20 times 3m, 3m minus 2. And once again, we feel good that we, literally, we did take out the greatest common factor because 3m and 2, especially 3 and 2, are now relatively prime. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | And it deals with right triangles. So a right triangle is a triangle that has a 90 degree angle in it. So the way I drew it right here, this is our 90 degree angle. If you've never seen a 90 degree angle before, the way to think about it is if this side goes straight left to right, this side goes straight up and down. These sides are perpendicular, or the angle between them is 90 degrees, or it is a right angle. And the Pythagorean Theorem tells us that if we're dealing with a right triangle, let me write that down, if we're dealing with a right triangle, not a wrong triangle, if we're dealing with a right triangle, which is a triangle that has a right angle or a 90 degree angle in it, then the relationship between their sides is this. So if this side is A, this side is B, and this side is C, and remember, the C that we're dealing with right here is the side opposite the 90 degree angle. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | If you've never seen a 90 degree angle before, the way to think about it is if this side goes straight left to right, this side goes straight up and down. These sides are perpendicular, or the angle between them is 90 degrees, or it is a right angle. And the Pythagorean Theorem tells us that if we're dealing with a right triangle, let me write that down, if we're dealing with a right triangle, not a wrong triangle, if we're dealing with a right triangle, which is a triangle that has a right angle or a 90 degree angle in it, then the relationship between their sides is this. So if this side is A, this side is B, and this side is C, and remember, the C that we're dealing with right here is the side opposite the 90 degree angle. It's important to keep track of which side is which. The Pythagorean Theorem tells us that if and only if this is a right triangle, then A squared plus B squared is going to be equal to C squared. And we can use this information. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So if this side is A, this side is B, and this side is C, and remember, the C that we're dealing with right here is the side opposite the 90 degree angle. It's important to keep track of which side is which. The Pythagorean Theorem tells us that if and only if this is a right triangle, then A squared plus B squared is going to be equal to C squared. And we can use this information. If we know two of these, we can then use this theorem, this formula, to solve for the third. I'll give you one more piece of terminology here. This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here, it's C in this example, this is called a hypotenuse. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | And we can use this information. If we know two of these, we can then use this theorem, this formula, to solve for the third. I'll give you one more piece of terminology here. This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here, it's C in this example, this is called a hypotenuse. Very fancy word for a very simple idea. The longest side of a right triangle, the side that is opposite the 90 degree angle, is called the hypotenuse. Now that we know the Pythagorean Theorem, let's actually use it. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | This long side, the side that is the longest side of our right triangle, the side that is opposite of our right angle, this right here, it's C in this example, this is called a hypotenuse. Very fancy word for a very simple idea. The longest side of a right triangle, the side that is opposite the 90 degree angle, is called the hypotenuse. Now that we know the Pythagorean Theorem, let's actually use it. It's one thing to know something, but it's a lot more fun to use it. So let's say I have the following right triangle. Let me draw a little bit neater than that. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Now that we know the Pythagorean Theorem, let's actually use it. It's one thing to know something, but it's a lot more fun to use it. So let's say I have the following right triangle. Let me draw a little bit neater than that. The following right triangle. This side over here has length 9. This side over here has length 7. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Let me draw a little bit neater than that. The following right triangle. This side over here has length 9. This side over here has length 7. And my question is, what is this side over here? Maybe we can call that C. Well, C in this case, once again, it is a hypotenuse. It is the longest side. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | This side over here has length 7. And my question is, what is this side over here? Maybe we can call that C. Well, C in this case, once again, it is a hypotenuse. It is the longest side. So we know that the sum of the squares of the other side is going to be equal to C squared. So by the Pythagorean Theorem, 9 squared plus 7 squared is going to be equal to C squared. 9 squared is 81 plus 7 squared is 49. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | It is the longest side. So we know that the sum of the squares of the other side is going to be equal to C squared. So by the Pythagorean Theorem, 9 squared plus 7 squared is going to be equal to C squared. 9 squared is 81 plus 7 squared is 49. 80 plus 40 is 120. Then we're going to have the 1 plus the 9, that's another 10. So this is going to be equal to 130. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 9 squared is 81 plus 7 squared is 49. 80 plus 40 is 120. Then we're going to have the 1 plus the 9, that's another 10. So this is going to be equal to 130. So this is going to be equal to 130. And that is equal to C squared. So what's C going to be equal to? |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So this is going to be equal to 130. So this is going to be equal to 130. And that is equal to C squared. So what's C going to be equal to? Let me rewrite it over here. C squared is equal to 130. Or we could say that C is equal to the square root of 130. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So what's C going to be equal to? Let me rewrite it over here. C squared is equal to 130. Or we could say that C is equal to the square root of 130. And notice, I'm only taking the principal root here. Because C has to be positive. We're dealing with a distance. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.