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Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Or we could say that C is equal to the square root of 130. And notice, I'm only taking the principal root here. Because C has to be positive. We're dealing with a distance. So we can't take the negative square root. So we'll only take the principal square root right here. And if we want to simplify this a little bit, we know how to simplify our radicals. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | We're dealing with a distance. So we can't take the negative square root. So we'll only take the principal square root right here. And if we want to simplify this a little bit, we know how to simplify our radicals. 130 is 2 times 65, which is 5 times 13. Well, these are all prime numbers, so that's about as simple as I can get. C is equal to the square root of 130. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | And if we want to simplify this a little bit, we know how to simplify our radicals. 130 is 2 times 65, which is 5 times 13. Well, these are all prime numbers, so that's about as simple as I can get. C is equal to the square root of 130. Let's do another one of these. Maybe I want to keep this Pythagorean theorem right there, just so we always remember what we're referring to. So let's say I have a triangle that looks like this. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | C is equal to the square root of 130. Let's do another one of these. Maybe I want to keep this Pythagorean theorem right there, just so we always remember what we're referring to. So let's say I have a triangle that looks like this. Let's say it looks like that. And this is the right angle up here. That's my right angle. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So let's say I have a triangle that looks like this. Let's say it looks like that. And this is the right angle up here. That's my right angle. Let's say that this side, I'm going to call it A. This side, I'm going to call 20, or it's going to have length 21. And this side right here is going to be of length 35. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | That's my right angle. Let's say that this side, I'm going to call it A. This side, I'm going to call 20, or it's going to have length 21. And this side right here is going to be of length 35. So your instinct to solve for A might say, 21 squared plus 35 squared is going to be equal to A squared. But notice, in this situation, 35 is the hypotenuse. 35 is our C. It's the longest side of our right triangle. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | And this side right here is going to be of length 35. So your instinct to solve for A might say, 21 squared plus 35 squared is going to be equal to A squared. But notice, in this situation, 35 is the hypotenuse. 35 is our C. It's the longest side of our right triangle. So what the Pythagorean theorem tells us is that A squared plus the other non-longest side, the other non-hypotenuse squared, so A squared plus 21 squared is going to be equal to 35 squared. You always have to remember, the C squared right here, the C that we're talking about is always going to be the longest side of your right triangle, the side that is opposite of our right angle. This is the side that's opposite of the right angle. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 35 is our C. It's the longest side of our right triangle. So what the Pythagorean theorem tells us is that A squared plus the other non-longest side, the other non-hypotenuse squared, so A squared plus 21 squared is going to be equal to 35 squared. You always have to remember, the C squared right here, the C that we're talking about is always going to be the longest side of your right triangle, the side that is opposite of our right angle. This is the side that's opposite of the right angle. So A squared plus 21 squared is equal to 35 squared. And what do we have here? So 21 squared, I'm tempted to use a calculator, but I won't. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | This is the side that's opposite of the right angle. So A squared plus 21 squared is equal to 35 squared. And what do we have here? So 21 squared, I'm tempted to use a calculator, but I won't. So 21 times 21, 1 times 21 is 21, 2 times 21 is 42, it is 441. 35 squared, once again I'm tempted to use a calculator, but I won't. 35 times 35, 5 times 5 is 25, carry the 2. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So 21 squared, I'm tempted to use a calculator, but I won't. So 21 times 21, 1 times 21 is 21, 2 times 21 is 42, it is 441. 35 squared, once again I'm tempted to use a calculator, but I won't. 35 times 35, 5 times 5 is 25, carry the 2. 5 times 3 is 15, plus 2 is 17, put a 0 here, get rid of that thing. 3 times 5 is 15, 3 times 3 is 9, plus 1 is 10. So it is 11, let me do it in order. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 35 times 35, 5 times 5 is 25, carry the 2. 5 times 3 is 15, plus 2 is 17, put a 0 here, get rid of that thing. 3 times 5 is 15, 3 times 3 is 9, plus 1 is 10. So it is 11, let me do it in order. 5 plus 5, 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1, 1225. This tells us that A squared plus 440, 441 is going to be equal to 35 squared, which is 1225. Now we could subtract 441 from both sides of this equation. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So it is 11, let me do it in order. 5 plus 5, 0 is 5, 7 plus 5 is 12, 1 plus 1 is 2, bring down the 1, 1225. This tells us that A squared plus 440, 441 is going to be equal to 35 squared, which is 1225. Now we could subtract 441 from both sides of this equation. 441, the left hand side just becomes A squared. The right hand side, what do we get? 5 minus 1 is 4, we want to, let me write this a little bit neater here. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Now we could subtract 441 from both sides of this equation. 441, the left hand side just becomes A squared. The right hand side, what do we get? 5 minus 1 is 4, we want to, let me write this a little bit neater here. Let me write this a little bit neater. Minus 441, so the left hand side once again they cancel out, A squared is equal to. And then on the right hand side, what do we have to do? |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 5 minus 1 is 4, we want to, let me write this a little bit neater here. Let me write this a little bit neater. Minus 441, so the left hand side once again they cancel out, A squared is equal to. And then on the right hand side, what do we have to do? That's larger than that, but 2 is not larger than 4, so we're going to have to borrow. So that becomes a 12, or regroup depending on how you want to view it. That becomes a 1. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | And then on the right hand side, what do we have to do? That's larger than that, but 2 is not larger than 4, so we're going to have to borrow. So that becomes a 12, or regroup depending on how you want to view it. That becomes a 1. 1 is not greater than 4, so we're going to have to borrow again. Get rid of that, and then this becomes an 11. 5 minus 1 is 4, 12 minus 4 is 8, 11 minus 4 is 7. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | That becomes a 1. 1 is not greater than 4, so we're going to have to borrow again. Get rid of that, and then this becomes an 11. 5 minus 1 is 4, 12 minus 4 is 8, 11 minus 4 is 7. So A squared is equal to 784, and we could write then that A is equal to the square root of 784. And once again, I'm very tempted to use a calculator, but let's not. Let's not use it. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 5 minus 1 is 4, 12 minus 4 is 8, 11 minus 4 is 7. So A squared is equal to 784, and we could write then that A is equal to the square root of 784. And once again, I'm very tempted to use a calculator, but let's not. Let's not use it. So this is 2 times 392, and then 390 times 2 is 78. And then this is 2 times 196. That's right, 190 times 2 is 2 times 196. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Let's not use it. So this is 2 times 392, and then 390 times 2 is 78. And then this is 2 times 196. That's right, 190 times 2 is 2 times 196. 196 is 2 times 98. Let's keep going down here. 98 is 2 times 49. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | That's right, 190 times 2 is 2 times 196. 196 is 2 times 98. Let's keep going down here. 98 is 2 times 49. And of course we know what that is. Notice we have 2 times 2 times 2 times 2, so this is 2 to the 4th power, so 16 times 49. So A is equal to the square root of 16 times 49. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 98 is 2 times 49. And of course we know what that is. Notice we have 2 times 2 times 2 times 2, so this is 2 to the 4th power, so 16 times 49. So A is equal to the square root of 16 times 49. I picked those numbers because they're both perfect squares. So this is equal to the square root of 16 is 4 times the square root of 49 is 7. It's equal to 28. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So A is equal to the square root of 16 times 49. I picked those numbers because they're both perfect squares. So this is equal to the square root of 16 is 4 times the square root of 49 is 7. It's equal to 28. So this side right here is going to be equal to 28 by the Pythagorean theorem. Let's do one more of these. It can never get enough practice. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | It's equal to 28. So this side right here is going to be equal to 28 by the Pythagorean theorem. Let's do one more of these. It can never get enough practice. So let's say I have another triangle. I'll draw this one big. There you go. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | It can never get enough practice. So let's say I have another triangle. I'll draw this one big. There you go. That's my triangle. That is the right angle. This side is 24. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | There you go. That's my triangle. That is the right angle. This side is 24. This side is 12. We'll call this side right here B. Now, once again, always identify the hypotenuse. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | This side is 24. This side is 12. We'll call this side right here B. Now, once again, always identify the hypotenuse. That's the longest side, the side opposite the 90-degree angle. You might say, hey, I don't know that's the longest side. I don't know what B is yet. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | Now, once again, always identify the hypotenuse. That's the longest side, the side opposite the 90-degree angle. You might say, hey, I don't know that's the longest side. I don't know what B is yet. How do I know this is longest? And there, in that situation, you just say, well, it's the side opposite the 90-degree angle. So if that's the hypotenuse, that right there is the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | I don't know what B is yet. How do I know this is longest? And there, in that situation, you just say, well, it's the side opposite the 90-degree angle. So if that's the hypotenuse, that right there is the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. Pythagorean theorem, B squared plus 12 squared is equal to 24 squared. Or we could subtract 12 squared from both sides. We say B squared is equal to 24 squared minus 12 squared, which we know is 144. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So if that's the hypotenuse, that right there is the hypotenuse, then this squared plus that squared is going to be equal to 24 squared. Pythagorean theorem, B squared plus 12 squared is equal to 24 squared. Or we could subtract 12 squared from both sides. We say B squared is equal to 24 squared minus 12 squared, which we know is 144. And that B is equal to the square root of 24 squared minus 12 squared. Now, I'm tempted to use the calculator, and I'll give into the temptation. So let's do it. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | We say B squared is equal to 24 squared minus 12 squared, which we know is 144. And that B is equal to the square root of 24 squared minus 12 squared. Now, I'm tempted to use the calculator, and I'll give into the temptation. So let's do it. The last one was so painful, I'm still recovering. So 24 squared minus 12 squared is equal to 24.78. So this actually turns into 24 squared minus 12 squared is equal to 432. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So let's do it. The last one was so painful, I'm still recovering. So 24 squared minus 12 squared is equal to 24.78. So this actually turns into 24 squared minus 12 squared is equal to 432. So B is equal to the square root of 432. And let's factor this again. We saw what the answer is, but maybe we can write it in kind of a simplified radical form. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So this actually turns into 24 squared minus 12 squared is equal to 432. So B is equal to the square root of 432. And let's factor this again. We saw what the answer is, but maybe we can write it in kind of a simplified radical form. So this is 2 times 216. 216, I believe, is a perfect square. So let me take the square root of 216. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | We saw what the answer is, but maybe we can write it in kind of a simplified radical form. So this is 2 times 216. 216, I believe, is a perfect square. So let me take the square root of 216. Nope, not a perfect square. So 216, let's just keep going. 216 is 2 times 108. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So let me take the square root of 216. Nope, not a perfect square. So 216, let's just keep going. 216 is 2 times 108. 108 is, we could say, 4 times what? 25 plus another 2, 4 times 27, which is 9 times 3. So what do we have here? |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 216 is 2 times 108. 108 is, we could say, 4 times what? 25 plus another 2, 4 times 27, which is 9 times 3. So what do we have here? We have 2 times 2 times 4. So what we have right here is a 16. 16 times 9 times 3. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | So what do we have here? We have 2 times 2 times 4. So what we have right here is a 16. 16 times 9 times 3. Is that right? I'm using a different calculator. 16 times 9 times 3 is equal to 432. |
Pythagorean theorem Right triangles and trigonometry Geometry Khan Academy.mp3 | 16 times 9 times 3. Is that right? I'm using a different calculator. 16 times 9 times 3 is equal to 432. So this is going to be equal to, B is equal to, the square root of 16 times 9 times 3, which is equal to the square root of 16, which is 4 times the square root of 9, times the square root of 3, which is equal to 12 roots of 3. 12, so B is 12 times the square root of 3. I hope you found that useful. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | Which of the following statements is true? So before even reading these statements, let's just think about what's going on. So let me draw my axes here. Let's draw my axes. So this is going to be my vertical axis. That could be one of the variables. And then this is my horizontal axis. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | Let's draw my axes. So this is going to be my vertical axis. That could be one of the variables. And then this is my horizontal axis. That's one of the other variables. And maybe for the sake of convention, this could be x and this could be y. But they're whatever our two variables are. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | And then this is my horizontal axis. That's one of the other variables. And maybe for the sake of convention, this could be x and this could be y. But they're whatever our two variables are. So it's a system of two linear equations. So if we're graphing them, each of the linear equations in two variables can be represented by a line. Now, there's only three scenarios here. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | But they're whatever our two variables are. So it's a system of two linear equations. So if we're graphing them, each of the linear equations in two variables can be represented by a line. Now, there's only three scenarios here. One scenario is where the lines don't intersect at all. So the only way that you're going to have two lines in two dimensions that don't intersect is if they have the same slope and they have different y-intercepts. So that's one scenario. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | Now, there's only three scenarios here. One scenario is where the lines don't intersect at all. So the only way that you're going to have two lines in two dimensions that don't intersect is if they have the same slope and they have different y-intercepts. So that's one scenario. But that's not the scenario that's being described here. They say you have found more than one solution that satisfies the system. Here there are no solutions. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | So that's one scenario. But that's not the scenario that's being described here. They say you have found more than one solution that satisfies the system. Here there are no solutions. So that's not the scenario that we're talking about. There's another scenario where they intersect in exactly one place. So they intersect in exactly one place. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | Here there are no solutions. So that's not the scenario that we're talking about. There's another scenario where they intersect in exactly one place. So they intersect in exactly one place. There's one point, one xy coordinate right over there that satisfies both of these constraints. But this also is not the scenario they're talking about. They're telling us that you have found more than one solution that satisfies the system. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | So they intersect in exactly one place. There's one point, one xy coordinate right over there that satisfies both of these constraints. But this also is not the scenario they're talking about. They're telling us that you have found more than one solution that satisfies the system. So this isn't the scenario either. So the only other scenario that we can have, we don't have parallel lines that don't intersect. We don't have lines that only intersect in one place. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | They're telling us that you have found more than one solution that satisfies the system. So this isn't the scenario either. So the only other scenario that we can have, we don't have parallel lines that don't intersect. We don't have lines that only intersect in one place. The only other scenario is that we're dealing with a situation where both linear equations are essentially the same constraint. They both are essentially representing the same xy relationship. That's the only way that I can have two lines. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | We don't have lines that only intersect in one place. The only other scenario is that we're dealing with a situation where both linear equations are essentially the same constraint. They both are essentially representing the same xy relationship. That's the only way that I can have two lines. And this only applies to linear relationship and lines. But the only way that two lines can intersect more than one place is if they intersect everywhere. So in this situation we know that we must have an infinite number of solutions. |
Solving systems of two linear equations example Algebra I Khan Academy.mp3 | That's the only way that I can have two lines. And this only applies to linear relationship and lines. But the only way that two lines can intersect more than one place is if they intersect everywhere. So in this situation we know that we must have an infinite number of solutions. So which of these choices say that? This one right here. There are infinitely many more solutions to the system. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | The area of a rectangle is equal to base times height. In another video, we saw that if we're looking at the area of a parallelogram, and we also know the length of a base, and we know its height, that the area is still going to be base times height. Now, it's not as obvious when you look at the parallelogram, but in that video, we did a little manipulation of the area. We said, hey, let's take this little section right over here. So, we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that as long as the base and the height is the same as this rectangle here, I'm able to construct the same rectangle by moving that area over. And that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | We said, hey, let's take this little section right over here. So, we took that little section right over there, and then we move it over to the right-hand side, and just like that, you see that as long as the base and the height is the same as this rectangle here, I'm able to construct the same rectangle by moving that area over. And that's why the area of this parallelogram is base times height. I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. So, hopefully that convinces you, that convinces you that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So, let's look at some triangles here. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | I didn't add or take away area, I just shifted area from the left-hand side to the right-hand side to show you that the area of that parallelogram was the same as this area of the rectangle. It's still going to be base times height. So, hopefully that convinces you, that convinces you that the area of a parallelogram is base times height, because we're now going to use that to get the intuition for the area of a triangle. So, let's look at some triangles here. So, that is a triangle, and we're given the base and the height. Now, we're gonna try to think about, well, what's this area, what's the area of this triangle going to be? And you can imagine it's going to be dependent on base and height. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | So, let's look at some triangles here. So, that is a triangle, and we're given the base and the height. Now, we're gonna try to think about, well, what's this area, what's the area of this triangle going to be? And you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So, let me copy, and then let me paste it. And what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, rotate it around like that, and then add it to the original area, and you see something very interesting is happening. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | And you can imagine it's going to be dependent on base and height. Well, to think about that, let me copy and paste this triangle. So, let me copy, and then let me paste it. And what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram, I have now constructed a parallelogram that has twice the area of our original triangle, because I have two of our original triangles right over here, you saw me do it. I copied and pasted it, and then I flipped it over, and I constructed the parallelogram. Now, why is this interesting? |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | And what I'm gonna do is, so now I have two of the triangles, so this is now going to be twice the area, and I'm gonna rotate it around, I'm gonna rotate it around like that, rotate it around like that, and then add it to the original area, and you see something very interesting is happening. I have now constructed a parallelogram, I have now constructed a parallelogram that has twice the area of our original triangle, because I have two of our original triangles right over here, you saw me do it. I copied and pasted it, and then I flipped it over, and I constructed the parallelogram. Now, why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base, base times this height, you also have the height written on the H upside down over here, it's going to be base times height. That's going to be for the parallelogram, for the entire, let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | Now, why is this interesting? Well, the area of the entire parallelogram, the area of the entire parallelogram is going to be the length of this base, base times this height, you also have the height written on the H upside down over here, it's going to be base times height. That's going to be for the parallelogram, for the entire, let me draw a parallelogram right over here. That's going to be the area of the entire parallelogram. So, what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | That's going to be the area of the entire parallelogram. So, what would be the area of our original triangle? What would be the area of our original triangle? Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So, our original triangle is just going to have half the area. So, this area right over here is going to be 1 1 2, the area of the parallelogram. 1 1 2, base, let me do those same colors. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | Well, we already saw that this area of the parallelogram, it's twice the area of our original triangle. So, our original triangle is just going to have half the area. So, this area right over here is going to be 1 1 2, the area of the parallelogram. 1 1 2, base, let me do those same colors. 1 1 2, base times height. 1 1 2, base times height. And you might say, okay, maybe it worked for this triangle, but I want to see it work for more triangles. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | 1 1 2, base, let me do those same colors. 1 1 2, base times height. 1 1 2, base times height. And you might say, okay, maybe it worked for this triangle, but I want to see it work for more triangles. And so, to help you there, I've added another triangle right over here. You could view this as an obtuse triangle. This angle right over here is greater than 90 degrees. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | And you might say, okay, maybe it worked for this triangle, but I want to see it work for more triangles. And so, to help you there, I've added another triangle right over here. You could view this as an obtuse triangle. This angle right over here is greater than 90 degrees. But I'm going to do the same trick. We have the base, and then we have the height. Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, or if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | This angle right over here is greater than 90 degrees. But I'm going to do the same trick. We have the base, and then we have the height. Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, or if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height. It's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, well, this is the height. How far off the ground is it? |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | Here, you can think of, if you start at this point right over here, and if you drop a ball, the length that the ball goes, or if you had a string here, to kind of get to the ground level, you could view this as the ground level right over there, that that's going to be the height. It's not sitting in the triangle like we saw last time, but it's still the height of the triangle. If this was a building of some kind, you'd say, well, this is the height. How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine. It's going to be 1 1 2 base times height. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | How far off the ground is it? Well, what's the area of this going to be? Well, you can imagine. It's going to be 1 1 2 base times height. How do we feel good about that? Well, let's do the same magic here. So, let me copy and paste this. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | It's going to be 1 1 2 base times height. How do we feel good about that? Well, let's do the same magic here. So, let me copy and paste this. I'm going to copy and then paste it. Whoops, that didn't work. Let me copy and then paste it. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | So, let me copy and paste this. I'm going to copy and then paste it. Whoops, that didn't work. Let me copy and then paste it. And so, I have two of these triangles now, but I'm going to flip this one over so that I can construct a parallelogram. I'm going to flip it over and move it over here. I'm going to have to rotate it a little bit more. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | Let me copy and then paste it. And so, I have two of these triangles now, but I'm going to flip this one over so that I can construct a parallelogram. I'm going to flip it over and move it over here. I'm going to have to rotate it a little bit more. So, I think you get the general idea. So, now I've constructed a parallelogram that has twice the area of our original triangle. It has twice the area of our original triangle. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | I'm going to have to rotate it a little bit more. So, I think you get the general idea. So, now I've constructed a parallelogram that has twice the area of our original triangle. It has twice the area of our original triangle. And so, if I talked about the area of the entire parallelogram, it would be base times the height of the parallelogram. Base times the height of the parallelogram. But if we're only talking about the area of, if we're only talking about this area right over here, which is our original triangle, it's going to be half the area of the parallelogram. |
Area of triangles intuition Algebra I High School Math Khan Academy.mp3 | It has twice the area of our original triangle. And so, if I talked about the area of the entire parallelogram, it would be base times the height of the parallelogram. Base times the height of the parallelogram. But if we're only talking about the area of, if we're only talking about this area right over here, which is our original triangle, it's going to be half the area of the parallelogram. So, it's going to be 1 1 2 of that. So, our area of our original triangle is 1 1 2 base times height. So, hopefully that makes you feel pretty good about this formula that you will see in geometry, that area of a triangle is 1 1 2 base times height, while the area of a rectangle or a parallelogram is going to be base times height. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | So you can see the triangles here, that's the delta symbol, it literally means change in. Or another way, and you might see this formula and it tends to be really complicated, but just remember it's just these two things over here. Sometimes slope will be specified with the variable m, and they'll say that m is the same thing, and this is really the same thing as change in y. They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | They'll write y2 minus y1 over x2 minus x1. So the notation tends to be kind of complicated, but all this means is you take the y value of your end point and subtract from it the y value of your starting point. That'll essentially give you your change in y. And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | And it says take the x value of your end point and subtract from that the x value of your starting point, and that'll give you change in x. So whatever of these work for you, let's actually figure out the slope of the line that goes through these two points. So we're starting at, and actually we could do it both ways. We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | We could start at this point and go to that point and calculate the slope, or we could start at this point and go to that point and calculate the slope. So let's do it both ways. So let's say that our starting point is the point 4, 2, and let's say that our end point is negative 3, 16. So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | So what is the change in x over here? What is the change in x in this scenario? So we're going from 4 to negative 3. If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Actually, let me write it this way. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | If something goes from 4 to negative 3, what was its change? Well, you have to go down 4 to get to 0, and then you have to go down another 3 to get to negative 3. So our change in x here is negative 7. Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Actually, let me write it this way. Our change in x is equal to negative 3 minus 4, which is equal to negative 7. If I'm going from 4 to negative 3, I went down by 7. Our change in x is negative 7. Let's do the same thing for the change in y. Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Our change in x is negative 7. Let's do the same thing for the change in y. Notice, I implicitly used this formula over here. Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Our change in x was this value, our end point, our end x value, minus our starting x value. Let's do the same thing for our change in y. Our change in y, if we're starting at 2 and we go to 16, that means we moved up 14. Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here? The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Or another way you could say it, you could take your ending y value and subtract from that your starting y value, and you get 14. So what is the slope over here? The slope is just change in y over change in x. The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | The slope over here is change in y over change in x, which is our change in y is 14, and our change in x is negative 7. If we want to simplify this, 14 divided by negative 7 is negative 2. What I want to show you is that we could have done it the other way around. We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | We could have made this the starting point and this the end point. What we would have gotten is the negative values of each of these, but then they would have cancelled out and we would still get negative 2. Let's try it out. Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Let's say that our start point was negative 3 comma 16, and let's say that our end point is the 4 comma 2. In this situation, what is our change in x? If I start at negative 3 and I go to 4, that means I went up 7. Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7. What is our change in y? Our change in y over here, or we could say our rise. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Or if you want to just calculate that, you would do 4 minus negative 3. Needless to say, we just went up 7. What is our change in y? Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Our change in y over here, or we could say our rise. If we start at 16 and we end at 2, that means we went down 14. Or you could just say 2 minus 16 is negative 14. We went down by 14. This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | We went down by 14. This was our run. If you say rise over run, which is the same thing as change in y over change in x, our rise is negative 14 and our run here is 7. Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2. Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Notice, these are just the negatives of these values from when we swapped them. Once again, this is equal to negative 2. Let's just visualize this. Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points. This is my x-axis. That is my y-axis. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Let me do a quick graph here just to show you what a downward slope would look like. Let me draw our two points. This is my x-axis. That is my y-axis. This point over here, 4, 2. Let me graph it. We are going to go all the way up to 16. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | That is my y-axis. This point over here, 4, 2. Let me graph it. We are going to go all the way up to 16. Let me save some space here. We have 1, 2, 3, 4. It is 4, 1, 2. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | We are going to go all the way up to 16. Let me save some space here. We have 1, 2, 3, 4. It is 4, 1, 2. 4, 2 is right over here. Then we have the point negative 3, 16. Let me draw that over here. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | It is 4, 1, 2. 4, 2 is right over here. Then we have the point negative 3, 16. Let me draw that over here. We have negative 1, 2, 3. We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | Let me draw that over here. We have negative 1, 2, 3. We have to go up to 16. This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here. This is negative 3, 16. The line that goes between them is going to look something like this. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | This is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. It goes right over here. This is negative 3, 16. The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line. That line will keep going. The line will keep going. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | The line that goes between them is going to look something like this. I will try my best to draw a relatively straight line. That line will keep going. The line will keep going. That is my best attempt. Notice it is downward sloping. As you increase an x value, the line goes down. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | The line will keep going. That is my best attempt. Notice it is downward sloping. As you increase an x value, the line goes down. It is going from the top left to the bottom right. As x gets bigger, y gets smaller. That is what a downward sloping line looks like. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | As you increase an x value, the line goes down. It is going from the top left to the bottom right. As x gets bigger, y gets smaller. That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | That is what a downward sloping line looks like. Just to visualize our change in x's and our change in y's that we dealt with here. When we started at 4, 2 and ended at negative 3, 16, that was analogous to starting here and ending over there. We said our change in x was negative 7. We had to move back. Our run, we had to move in the left direction by 7. That is why it was negative 7. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | We said our change in x was negative 7. We had to move back. Our run, we had to move in the left direction by 7. That is why it was negative 7. Then we had to move in the y direction positive 14. That is why our rise was positive. It was 14 over negative 7 or negative 2. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | That is why it was negative 7. Then we had to move in the y direction positive 14. That is why our rise was positive. It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point. In that situation, our run was positive 7. |
Slope from two ordered pairs example 1 Algebra I Khan Academy.mp3 | It was 14 over negative 7 or negative 2. When we did it the other way, we started at this point and ended at this point. We started at negative 3, 16 and ended at that point. In that situation, our run was positive 7. Now we had to go down in the y direction since we switched the starting and the end point. Now we had to go down negative 14. Our run is now positive 7 and our rise is now negative 14. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And we have 836 is equal to 10 over what? Or 8 to 36, or the ratio of 8 over 36, is equal to the ratio of 10 to what? And there's a bunch of different ways to solve this. And I'll explore really all of them. Or not all of them, or a good selection of them. So one way to think about it is these two need to be equivalent ratios, or really equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | And I'll explore really all of them. Or not all of them, or a good selection of them. So one way to think about it is these two need to be equivalent ratios, or really equivalent fractions. So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well, you could multiply 8 times 10 over 8 will definitely give you 10. So we're multiplying by 10 over 8 over here. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So whatever happened to the numerator also has to happen to the denominator. So what do we have to multiply 8 by to get 10? Well, you could multiply 8 times 10 over 8 will definitely give you 10. So we're multiplying by 10 over 8 over here. Or another way to write 10 over 8, 10 over 8 is the same thing as 5 over 4. So we're multiplying by 5 over 4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. |
Solving a proportion with an unknown variable (example) 7th grade Khan Academy.mp3 | So we're multiplying by 10 over 8 over here. Or another way to write 10 over 8, 10 over 8 is the same thing as 5 over 4. So we're multiplying by 5 over 4 to get to 10, from 8 to 10. Well, if we did that to the numerator, in order to have an equivalent fraction, you have to do the same thing to the denominator. You have to multiply it. You have to multiply it times 5 over 4. And so we could say this n, this thing that we just solved for, this n is going to be equal to 36 times 5 divided by 4. |
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