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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there .
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why ca n't the slope be defined at this point ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point .
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so if a line on a function is constant ( straight ) then is the derivative going to always be a straight horizontal line as well ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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why do the beginning of the pink line and the end of the orange line not end in open circles ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity .
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should n't the derivative be undefined at those spots , because we only know the limit from one side ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative .
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5 , why does sal put the point on the x axis ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line .
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why is the y value zero ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity .
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why the last jump of the derivate ( ) is a defined point and not undefined ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope .
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how does one know what specific points to draw the derivative through ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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in other words , how is the y-intercept of the derivative determined ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here .
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is there a difference between f ( x ) lines that in end in closed circles vis-a-vis lines that end in open circles ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there .
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sal said the derivative of the point does n't exist , why ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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if derivatives are functions themselves , i 'm assuming it 's possible to take the derivative of a derivative ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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how does one determine a derivative exists ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope .
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for a derivative to exist the function has to be continuous , right ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative .
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in the orange line ( the last section of the derivative graph ) have an open or closed hole ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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can the graph of the derivative be drawn in any number of ways , since different graphs often have the same interpretation or is there a set graph that a derivative of a function has to be ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive .
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why does n't sal shade the circle ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value .
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so the prime graph reflects only the slope ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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if i were told to sketch the graph of an even function f ( x ) which has a derivative at every point , how would it look like ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 .
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how do we know that for a fact that wherever the turning point is , that is the same value of the x-intercept for the derivative ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here .
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can you consider then , that since the derivative of the slope is measuring the change in y with respect to x , that graphing the derivative of a function , that is measuring something like say , speed , would resulting in the graphing of the change in velocity over time when compared to the graph of the function ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line .
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how is the slope , the way it is drawn , positive when sal is talking exactly ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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could someone explain why sal drew the lines on different levels and not one continuous line ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here .
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, what i mean to say is that why are some lines above the x-axis like the yellow and red lines while the blue one is on the x-axis ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here .
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at what values of x is f ( x ) continuous but not differentiable ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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in the example , was there a reason behind sal sketching the line at that specific point on the y-axis ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity .
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so , my question is , how can you take the slope of an undefined slope ( whether because of discontinuity or not ) and thus taking the derivative of a function where that function is undefined ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope .
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the derivative of the parabolic portion of the function is explained/represented as a straight line , whereas in the video , `` figuring out which function is the derivative '' , what appears to be a similar figure for f ( x ) has a derivative line that is not straight ( rather , sigmoidal ) - ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 .
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in first interval there is sine function then why the derivative of that function is straight line ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined .
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yet a unique limit for that point exists , so it should also exist a derivative right ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color .
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is the derivative of that slope the stangent line stretched from one point to another on a slope ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here .
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why is the derivative of y=x computed as x=1 , should n't it be y=-x to make it according to the slope signs ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function .
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if you took that ( red ) segment as an interval alone and evaluated it , would n't the f ' ( x ) of its left endpoint still be defined because the limit would only include evaluation from the positive side ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive .
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would you say that the drawing of the slope/derivative on the second graph must be between [ -1,1 ] because the slope must always be a fraction , therefore between these two numbers ?
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so i 've got this crazy discontinuous function here , which we 'll call f of x . and my goal is to try to draw its derivative right over here . so what i 'm going to need to think about is the slope of the tangent line , or the slope at each point in this curve , and then try my best to draw that slope . so let 's try to tackle it . so right over here at this point , the slope is positive . and actually , it 's a good bit positive . and then as we get larger and larger x 's , the slope is still positive , but it 's less positive -- and all the way up to this point right over here , where it becomes 0 . so let 's see how i could draw that over here . so over here we know that the slope must be equal to 0 -- right over here . remember over here , i 'm going to try to draw y is equal to f prime of x . and i 'm going to assume that this is some type of a parabola . and you 'll learn shortly why i had to make that assumption . but let 's say that , so let 's see , here the slope is quite positive . so let 's say the slope is right over here . and then it gets less and less and less positive . and i 'll assume it does it in a linear fashion . that 's why i had to assume that it 's some type of a parabola . so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 . and then the slope is getting more and more negative . and at this point , it seems like the slope is just as negative as it was positive there . so at this point right over here , the slope is just as negative as it was positive right over there . so it seems like this would be a reasonable view of the slope of the tangent line over this interval . now let 's think about as we get to this point . here the slope seems constant . our slope is a constant positive value . so once again , our slope here is a constant positive line . let me be careful here because at this point , our slope wo n't really be defined , because our slope , you could draw multiple tangent lines at this little pointy point . so let me just draw a circle right over there . but then as we get right over here , the slope seems to be positive . so let 's draw that . the slope seems to be positive , although it 's not as positive as it was there . so the slope looks like it is -- i 'm just trying to eyeball it -- so the slope is a constant positive this entire time . we have a line with a constant positive slope . so it might look something like this . and let me make it clear what interval i am talking about . i want these things to match up . so let me do my best . so this matches up to that . this matches up over here . and we just said we have a constant positive slope . so let 's say it looks something like that over this interval . and then we look at this point right over here . so right at this point , our slope is going to be undefined . there 's no way that you could find the slope over -- or this point of discontinuity . but then when we go over here , even though the value of our function has gone down , we still have a constant positive slope . in fact , the slope of this line looks identical to the slope of this line . let me do that in a different color . the slope of this line looks identical . so we 're going to continue at that same slope . it was undefined at that point , but we 're going to continue at that same slope . and once again , it 's undefined here at this point of discontinuity . so the slope will look something like that . and then we go up here . the value of the function goes up , but now the function is flat . so the slope over that interval is 0 . the slope over this interval , right over here , is 0 . so we could say -- let me make it clear what interval i 'm talking about -- the slope over this interval is 0 . and then finally , in this last section -- let me do this in orange -- the slope becomes negative . but it 's a constant negative . and it seems actually a little bit more negative than these were positive . so i would draw it right over there . so it 's a weird looking function . but the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point . and by doing so , we have essentially drawn the derivative over that interval .
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so it gets less and less and less positive . notice here , for example , the slope is still positive . and so when you look at the derivative , the slope is still a positive value . but as we get larger and larger x 's up to this point , the slope is getting less and less positive , all the way to 0 .
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what would be the derivative of an infinitely changing slope ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats .
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where did the idea come from that these 3 slices would consume the entire remains of the pie ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful .
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could n't you just do 180-30=150 , then divide 150 by 3 which equals 50 , and then multiply 50 times 2 to get 100 ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees .
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what is the measure of the larger angle in degrees ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats .
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where did the idea come from that these 3 slices would consume the entire remains of the pie ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats .
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does your family have to eat the whole pie ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see .
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how many slices did sam cut the pie into ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees .
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what is the complement of an angle ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees .
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and what is the supplement of an angle ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ?
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how many degrees is the smaller slice ?
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there is a half an apple pie left . you want to eat twice what your little brother eats , but you also need to save a slice for your mom . you can cut her a slice that is 30 degrees . what is the measure of your piece of the pie in degrees ? so to tackle this , we just have to remember a few things . we have to remember that the degrees in a circle -- so if we were to go all the way around a circle that that would be 360 degrees . but we only have half a pie here . and actually , let me draw a little bit differently . so if we had a whole pie -- so if we started here , and we had a whole pie , we went all the way around , that would be 360 degrees . but we only have a half pie . so we only have 180 degrees and let me do that in a color you 're more likely to see . that 's hard to see as well . we only have 180 degrees of pie left , but let 's just keep that in mind and think about how we 're going to split it between ourselves , our brother , and our mom . so let 's define some variables . let 's let x be equal to the degree measure of my brother 's pie or your little brother 's pie . so this is degree measure of brother 's pie . and then what would the amount of pie you eat be ? well , it says you eat twice what your little brother eats . so 2x would be equal what you eat , and it 's really the degree measure of what you eat . and then how much does your mom eat ? well , it says you cut her slice that is 30 degrees . so your mom 's going to get a 30 degree slice , going to get a slice of something like that . so the amount that your brother eats , x , plus the amount that you eat , plus the amount that your mom gets , plus 30 degrees , is going to be equal to this half pie . and remember , all of these are degree measures . so it 's going to be equal to 180 degrees . and just to visualize it over here , let me draw it down here where it 's easier to see it . we have half a pie that we 're dealing with . we 're going to save 30 degrees for our mom . so that 's 30 degrees right over there . your brother is going to eat some amount . so that is x . and then you 're going to eat twice that amount . so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees . now , we can simplify this . if we have one of something and then have another two of it , how much do we have now ? well , i now have three x 's . so 1x plus 2x is going to be 3x . so i have 3x plus 30 degrees is going to be equal to , of course , 180 degrees . now , to solve for x , we can subtract 30 degrees from both sides . so minus 30 degrees , minus 30 degrees . and i could have just written -- i could have all assumed that i 'm doing it in degrees and then just done it at the end . i keep writing it so i 'll just keep going with that . and then we are left with 3x equaling 180 degrees minus 30 degrees is equal to 150 degrees . and now we can just divide both sides by 3 , and we 're left with x equaling 150 divided by 3 is 50 degrees . x is equal to 50 degrees . now , we have to be careful . x is not what they 're asking for . they 're asking for the measure of your piece of pie in degrees . x is the degree measure of your brother 's piece of pie . what you eat is 2 times that . so if x is 50 degrees , 2 times that is going to be 100 degrees . so what is the measure of your piece of the pie in degrees ? it is going to be 100 degrees . so if we draw our pie again , if we draw our half pie , you have 30 degrees for your mom . so that 's 30 degrees for your mom . you have 50 degrees for your brother . and then you have twice that for yourself , 100 degrees .
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so that is 2x . x is the measure of this angle , and then 2x is a measure of that angle . so you see that 30 degrees plus x plus 2x , or x plus 2x plus 30 degrees , is going to be equal to 180 degrees .
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what is the measure of abc ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica .
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why are these wars ( greco-persian , peloponnesian ) so important that they are taught to children across europe and in other societies of european origin ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse .
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what was the point of syracuse ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse .
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did syracuse use navel or ground or both to fend off athens ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica .
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when rome took over athens the romans took the statue of athena from the parthenon right ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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so what 's the difference between `` macedon '' and `` macedonia '' ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey .
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which city-states were neutral during the peloponnesian war ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back .
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did the helot uprising continue thrueout the peloponnesian war ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey .
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what happened after the peloponnesian war ?
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies . sparta gets worried that the athenian navy is looking maybe a little bit too strong , they were decisively important in the second persian invasion , and they build this delian league , whose purpose is to go on the offensive against the persians , but more and more so it was starting to look like an athenian empire . and a lot of this you can imagine , it 's like a game of risk . the more you take over , the more resources you have , the more that you can build more ships and have more soldiers so that you can take on even more . so as sparta and athens are starting to become more and more rivals , sparta 's getting insecure about athens ' influence and their resources , and their military power . and we even saw it in the last video , you have an earthquake in sparta , potentially right around the same time that sparta was planning an invasion of athens , leaving the spartans vulnerable . there 's a helot uprising , these spartan slaves . the athenians send hoplites to apparently help the spartans but the spartans are suspicious of it and they send them back . and then it culminates with the skirmish you have between megara and corinth , which were historically two spartan allies , but athens decides to take sides , gets an alliance with megara , and that 's one thing too many for the spartans , and so they declare war , and you have what 's called the first peloponnesian war . but that 's not the peloponnesian war , that 's the first time that you start having very open conflict between the spartans and the athenians . and these conflicts last for , on the order of about 15 years , and they 're ended with what 's called the peace of 30 years . but as we will see , this peace of 30 years only lasts about 15 years . and the whole time , tensions continue to arise . the delian league , or you could call it the athenian power , the athenian empire i should say , is getting more and more powerful . the athenian navy is getting more and more powerful . and once again , things culminate , and now this is the beginning of the actual peloponnesian war , the thing that people are referring to when they talk about the peloponnesian war . in 431 bce , the king of sparta is convinced or is , i guess you could say he is influenced to , even though he himself was somewhat skeptical of the idea , he decides to invade attica . attica 's something you 'll hear a lot about . it is this region right over here , this little out-jutting of land , that athens is on . and this first phase of the peloponnesian war is called the archidamian war , named for the king of sparta who somewhat reluctantly decides to invade attica . and so that 's stage one , archidamian war . and the archidamian war , its essential ingredients are you have sparta with its dominant army going and having repeated attacks onto attica , and the athenians with their dominant navy going and having raids throughout the peloponnese and along the aegean coast . so here you have the athenians , let me do the athenians and the spartans in two different colors , actually . so here you go . let 's do the spartans in this red color , attacking attica with their army , and now let 's have the athenians with this light blue color coming here and attacking various points , various points on the peloponnese and in the aegean , further extending their influence . now that first phase ends , and it goes on for quite a bit , it goes on for about 10 years , ends with the peace of nicias , but as you will see and that was actually intended to be a 50-year peace , but you 'll see it does n't last long at all . the skirmishes continue , and in 415 , the athenians say hey , we want to extend our power . remember , it 's like this game of risk , the more city-states you take over , the more citizens that you 're able to tax , the more soldiers you will have , the more wealth you will have , the more shipbuilding capability you will have . so the athenians get this great idea to sail all the way to syracuse and just to be clear where syracuse is , syracuse is over here on the coast of sicily . now this is a long distance in this ancient world . they get this idea to sail all the way to syracuse to try to take it over , and essentially to extend their empire . so this right over here is in 415 , this long voyage to take over syracuse , and it is disastrous . the fleet that tries to go is able to be destroyed , and they get the folks in syracuse are able to get the assistance of the spartans , and so that greek fleet is destroyed . and this is often viewed as the second phase of the peloponnesian war . and so this is the attack at syracuse . attack at syracuse , the failed attempt of the athenians to get syracuse , to get syracuse . and this is a two-year period of time , because once again , this is no joke to send your navy and to try to get at syracuse . and then that takes us into the third phase , the third phase of the peloponnesian war , let me scroll this over a little bit . the third phase is often called the ionian war . ionian war . ionia is this region that 's now in modern day , off the coast of modern-day turkey . you have many of the city-states that were part of this athenian empire , they 're starting to revolt . you can imagine that the spartans are trying to help those revolts because they 're trying to get the athenians wherever they are . this is often also called the decelean war . decelea , i 'm probably not pronouncing it well , is a little village right over here at the top of attica , and it was a strategic location that went over to the spartans and that 's where the decelean war gets its name . but over the course of this third phase , the spartans get help from the persians . remember this was their enemy , they had unified the spartans and athenians had unified against the persians , but some time has passed . we 're now 70 or 80 years later , and the spartans say hey , we want to win this thing decisively . they get the help of the persians and finally in 405 bce there 's decisive naval battle at aegospotami , and i 'm sorry to all of you greeks out there that might be listening , i know my pronunciation is not perfect or close to perfect , but this is a decisive victory right over here for the spartans . they 're able to destroy the athenian navy , and that causes , that 's the decisive victory , and then in 404 , this is in 405 , in 404 , the athenians surrender and the famous peloponnesian war is over , the spartans are victorious . but it 's not a great victory because as you can imagine , you have towns that have been destroyed , large parts of greece have been weakened , and it leaves the whole area open to attack from others . and as we will see in the next century , in the 4th century , we have phillip of macedon , or macedon depending on how you want to pronounce it , who 's able to use that vulnerable , and the macedonians or the macedonians they are related to the greek people , but he 's able to use that vulnerability that happens over the course of the 5th century to attack the city-states of greece , but he actually is able to unify them .
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as we 've already seen , the 5th century bce starts off with athens and sparta and various greek city-states fighting on the same side against the persian invaders . but as we saw in the last video , as soon as the persians are dealt with , tensions start to rise between athens and sparta and their various allies .
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do we know that from the greek historian sources as well ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia .
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can someone be diagnosed with alzheimer 's , or do you have to guess , like schizophrenia ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia .
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also , are the symptoms universal , or do they vary from patient to patient ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ?
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would chronic traumatic encephalopathy be considered a form of dementia ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ?
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can dementia happen in younger people ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things .
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do all people have the same symptoms or are they different per person ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people .
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are infectious agents the other cause of dementia apart from alzheimer ?
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let 's say you 're going to the grocery store , but you realize you left your keys in your room . you go up to get them and then walk into the room and look around for a second , and realize you completely forgot why you even came here in the first place . we all know how frustrating this can be , right ? but this scenario is totally normal and is something everybody experiences from time to time . and besides causing you to take one more trip up the stairs to your room after you remember again , it does n't throw off your day too much . when problems remembering or problems with your thinking skills in general become so severe and so common that they actually interfere with your daily life , it might be diagnosed as dementia . dementia , though , is not a specific disease . what do i mean by that ? well , it 's more of a general term that describes a range of potential difficulties with reasoning , judgement , and memory . for example somebody with dementia might have troubles with speaking or writing coherently , or understanding what was spoken or written . they also might have trouble recognizing their surroundings , especially when those surroundings should normally be very familiar to them . planning and performing tasks that require multiple steps can also be difficult for patients with dementia . even tasks that you might think are really simple , like getting dressed and eating , these can become a serious challenge for patients with later-stage dementia . and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging . so , dementia , in very general terms , is something we use to describe when someone has troubles learning , remembering , and communicating . but where does alzheimer 's disease fit into this ? well , alzheimer 's disease is a type of dementia . specifically , it 's what 's known as a neuro degenerative disease , and it counts for about 60 to 80 % of all cases of dementia , affecting about five million people . so , if we look at `` neuro , '' we know that this refers to the nerves of the nervous system , or basically your brain , and then , `` degenerative '' or `` degeneration '' means to decline or to deteriorate . so , with alzheimer 's disease , there 's this deterioration of your nerve cells in your brain . and this brain of yours houses about a hundred billion nerve cells , which are also called neurons . and these guys communicate using a hundred trillion connections . that 's a lot . now , this communication and those connections are what control essentially every other organ and every other function in your body , not to mention your thinking abilities . unfortunately , though , the main type of cells that alzheimer 's disease targets and affects are these precious neurons . and depending on where the affected neurons live in your brain , different functions of your brain can be affected . for example , if nerve cells in this area of your cerebrum are affected , you might have trouble solving problems or making plans , because these cells help you do those things . or , if the neurons in this area are affected , you might have problems remembering something or storing new memories . and as alzheimer 's disease progresses , more and more of these neurons die , and your brain tissue actually begins to shrink due to this loss of nerve cells . all right . so , nerve cells in your brain are destroyed in alzheimer 's disease , but how are they destroyed ? well , unfortunately , answers to seemingly fundamental questions like these are n't really fully understood yet . but scientists have pinpointed a couple of possible culprits that usually seem to be involved , and these are called plaques and tangles . plaques are like these weird , abnormal clusters of protein fragments that build up between neighboring neurons . so , this is a normal group of nerve cells going about their business , you know , communicating and what-not . with alzheimer 's disease , these plaques start to form in between these neurons , which is thought to make it a lot harder for them to communicate . now , besides plaques , the other hallmark of alzheimer 's disease are called tangles . unlike plaques , though , these guys are found inside the neurons , and most of these are made up of a protein called tau . and tau proteins are helpful usually , but in alzheimer 's disease , they 're all twisted and abnormal , which ultimately ends up hurting the cells . okay . but we still have n't said why these plaques and tangles form . well , again , that question is still a huge , huge area of research , and is something scientists continue to gain insight on but have yet to completely understand .
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and to make matters worse , many times , patients are n't even aware that they 're experiencing any troubles or any sort of cognitive deficiencies at all . now , dementia is most common in the elderly , especially after age 65 . but it is certainly not a normal part of aging .
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do you have any ideas on keeping an elderly person 's mind active to prevent the worsening of dementia ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here .
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is the hypotenuse the longest side of the right triangle , or the side opposite of the right angle ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit .
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if you square root the whole equation , should n't it be a+b=c ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side .
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can you find the height of the triangle when you know the side lengths ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 .
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other than ( 3,4,5 ) , which sets of whole numbers are solutions to the equation a^2+b^2=c^2 ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into .
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how do you know when to simplify and when not to ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit .
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what is a principle root ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem .
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how do you know which leg is `` a '' and the other is `` b '' ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there .
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so the square in the middle of the triangle multiplies the outer numbers ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse .
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is the hypotenuse the longest side of any triangle or just the longest side of a right triangle ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical .
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is there a negative square root ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared .
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why does the pythagorean theorem not work on other triangles like obtuse or acute ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 .
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is there any way to solve for the missing angle using pythagoras theorem ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it .
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is there a way to make the hypotenuse the same as another side ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical .
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why ca n't we just do the square root of 108 even if it ends up a decimal ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it .
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i learnt that c is the hypotenuse , however , how do you know which side is a and which is b ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared .
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is there a type of pythagorean theorem for other triangles ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical .
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at the end why would you put under a square root/ radical sign ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared .
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what does pythagorean theorem actually means ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 .
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is 3,4,5 and its multiples the only numbers that make the pythagorean theorem work ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side .
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why does it have to be a right triangle ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 .
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why would n't the pythagorean theorem work on non-right angle triangles ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that .
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does the theorem work for non-right triangles ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 .
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why does sal write the square root of the last 3 ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side .
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so only right triangle works ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there .
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if the triangle edge opposite the right angle is called the hypotenuse , do the other two sides have their own names as well ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here .
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is hypotenuse the opposite of the 90 degree angle of a triangle or the opposite of any angle that we want to take ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology .
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is there a way to apply the pythagorean theorem to find the other two sides if you know only one side ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it .
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why is the hypotenuse the longest side ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry .
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how many laws and theorems are there in math ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there .
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if the triangle edge opposite the right angle is called the hypotenuse , do the other two sides have their own names as well ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here .
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if it says the hypotenuse is the side opposite the right angle does that mean that pythagoras ' theorem is only used to figure out the hypotenuse ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 .
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if a^2 + b^2 = c^2 ; does that mean a + b = c ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 .
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if a^2 + b^2 = c^2 , does n't that mean you can make it more simple and do a + b = c ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well .
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how could you solve for the square root of pi ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here .
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could n't we just do a + b = c ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology .
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can a triangle have two right angles ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle .
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90 + 90 is 180 , and the other angle would have 0 which is no angle ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared .
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can the pythagorean theorem be applyed to polygons or other shapes ?
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in this video we 're going to get introduced to the pythagorean theorem , which is fun on its own . but you 'll see as you learn more and more mathematics it 's one of those cornerstone theorems of really all of math . it 's useful in geometry , it 's kind of the backbone of trigonometry . you 're also going to use it to calculate distances between points . so it 's a good thing to really make sure we know well . so enough talk on my end . let me tell you what the pythagorean theorem is . so if we have a triangle , and the triangle has to be a right triangle , which means that one of the three angles in the triangle have to be 90 degrees . and you specify that it 's 90 degrees by drawing that little box right there . so that right there is -- let me do this in a different color -- a 90 degree angle . or , we could call it a right angle . and a triangle that has a right angle in it is called a right triangle . so this is called a right triangle . now , with the pythagorean theorem , if we know two sides of a right triangle we can always figure out the third side . and before i show you how to do that , let me give you one more piece of terminology . the longest side of a right triangle is the side opposite the 90 degree angle -- or opposite the right angle . so in this case it is this side right here . this is the longest side . and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it . and just so we always are good at identifying the hypotenuse , let me draw a couple of more right triangles . so let 's say i have a triangle that looks like that . let me draw it a little bit nicer . so let 's say i have a triangle that looks like that . and i were to tell you that this angle right here is 90 degrees . in this situation this is the hypotenuse , because it is opposite the 90 degree angle . it is the longest side . let me do one more , just so that we 're good at recognizing the hypotenuse . so let 's say that that is my triangle , and this is the 90 degree angle right there . and i think you know how to do this already . you go right what it opens into . that is the hypotenuse . that is the longest side . so once you have identified the hypotenuse -- and let 's say that that has length c. and now we 're going to learn what the pythagorean theorem tells us . so let 's say that c is equal to the length of the hypotenuse . so let 's call this c -- that side is c. let 's call this side right over here a . and let 's call this side over here b . so the pythagorean theorem tells us that a squared -- so the length of one of the shorter sides squared -- plus the length of the other shorter side squared is going to be equal to the length of the hypotenuse squared . now let 's do that with an actual problem , and you 'll see that it 's actually not so bad . so let 's say that i have a triangle that looks like this . let me draw it . let 's say this is my triangle . it looks something like this . and let 's say that they tell us that this is the right angle . that this length right here -- let me do this in different colors -- this length right here is 3 , and that this length right here is 4 . and they want us to figure out that length right there . now the first thing you want to do , before you even apply the pythagorean theorem , is to make sure you have your hypotenuse straight . you make sure you know what you 're solving for . and in this circumstance we 're solving for the hypotenuse . and we know that because this side over here , it is the side opposite the right angle . if we look at the pythagorean theorem , this is c. this is the longest side . so now we 're ready to apply the pythagorean theorem . it tells us that 4 squared -- one of the shorter sides -- plus 3 squared -- the square of another of the shorter sides -- is going to be equal to this longer side squared -- the hypotenuse squared -- is going to be equal to c squared . and then you just solve for c. so 4 squared is the same thing as 4 times 4 . that is 16 . and 3 squared is the same thing as 3 times 3 . so that is 9 . and that is going to be equal to c squared . now what is 16 plus 9 ? it 's 25 . so 25 is equal to c squared . and we could take the positive square root of both sides . i guess , just if you look at it mathematically , it could be negative 5 as well . but we 're dealing with distances , so we only care about the positive roots . so you take the principal root of both sides and you get 5 is equal to c. or , the length of the longest side is equal to 5 . now , you can use the pythagorean theorem , if we give you two of the sides , to figure out the third side no matter what the third side is . so let 's do another one right over here . let 's say that our triangle looks like this . and that is our right angle . let 's say this side over here has length 12 , and let 's say that this side over here has length 6 . and we want to figure out this length right over there . now , like i said , the first thing you want to do is identify the hypotenuse . and that 's going to be the side opposite the right angle . we have the right angle here . you go opposite the right angle . the longest side , the hypotenuse , is right there . so if we think about the pythagorean theorem -- that a squared plus b squared is equal to c squared -- 12 you could view as c. this is the hypotenuse . the c squared is the hypotenuse squared . so you could say 12 is equal to c. and then we could say that these sides , it does n't matter whether you call one of them a or one of them b . so let 's just call this side right here . let 's say a is equal to 6 . and then we say b -- this colored b -- is equal to question mark . and now we can apply the pythagorean theorem . a squared , which is 6 squared , plus the unknown b squared is equal to the hypotenuse squared -- is equal to c squared . is equal to 12 squared . and now we can solve for b . and notice the difference here . now we 're not solving for the hypotenuse . we 're solving for one of the shorter sides . in the last example we solved for the hypotenuse . we solved for c. so that 's why it 's always important to recognize that a squared plus b squared plus c squared , c is the length of the hypotenuse . so let 's just solve for b here . so we get 6 squared is 36 , plus b squared , is equal to 12 squared -- this 12 times 12 -- is 144 . now we can subtract 36 from both sides of this equation . those cancel out . on the left-hand side we 're left with just a b squared is equal to -- now 144 minus 36 is what ? 144 minus 30 is 114 . and then you subtract 6 , is 108 . so this is going to be 108 . so that 's what b squared is , and now we want to take the principal root , or the positive root , of both sides . and you get b is equal to the square root , the principal root , of 108 . now let 's see if we can simplify this a little bit . the square root of 108 . and what we could do is we could take the prime factorization of 108 and see how we can simplify this radical . so 108 is the same thing as 2 times 54 , which is the same thing as 2 times 27 , which is the same thing as 3 times 9 . so we have the square root of 108 is the same thing as the square root of 2 times 2 times -- well actually , i 'm not done . 9 can be factorized into 3 times 3 . so it 's 2 times 2 times 3 times 3 times 3 . and so , we have a couple of perfect squares in here . let me rewrite it a little bit neater . and this is all an exercise in simplifying radicals that you will bump into a lot while doing the pythagorean theorem , so it does n't hurt to do it right here . so this is the same thing as the square root of 2 times 2 times 3 times 3 times the square root of that last 3 right over there . and this is the same thing . and , you know , you would n't have to do all of this on paper . you could do it in your head . what is this ? 2 times 2 is 4 . 4 times 9 , this is 36 . so this is the square root of 36 times the square root of 3 . the principal root of 36 is 6 . so this simplifies to 6 square roots of 3 . so the length of b , you could write it as the square root of 108 , or you could say it 's equal to 6 times the square root of 3 . this is 12 , this is 6 . and the square root of 3 , well this is going to be a 1 point something something . so it 's going to be a little bit larger than 6 .
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and the way to figure out where that right triangle is , and kind of it opens into that longest side . that longest side is called the hypotenuse . and it 's good to know , because we 'll keep referring to it .
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why is the longer line called the hypotenuse ?
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