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nexis

  • 2 years ago

In 4a) http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/exam-3/materials-for-exam-3/MIT18_01SCF10_exam3sol.pdf How do we find the formula for hypotenuse of the triangle? (h-hx/r) (Just for x) (Sorry, somehow this question got closed before)

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  1. Coolsector
    • 2 years ago
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    you ask how they got y = h- hx/r ?

  2. nexis
    • 2 years ago
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    yes

  3. tjones89
    • 2 years ago
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  4. Coolsector
    • 2 years ago
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    we have two points : ( 0,h) and (r,0)

  5. Coolsector
    • 2 years ago
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    hence the slope : (h-0) /(0-r) = -h/r and using y = y1 + m(x-x1) we get : y = h -hx/r

  6. nexis
    • 2 years ago
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    sorry, why are we using h for y1?

  7. nexis
    • 2 years ago
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    i can see how we get the slope

  8. Coolsector
    • 2 years ago
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    we can take one of the points : (0,h) or (r,0) to be (x1,y1) in both cases we will get the same answer

  9. Coolsector
    • 2 years ago
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    lets take the (r,0) so y1 = 0 : y = 0 + -h/r(x-r) y = h -hx/r

  10. nexis
    • 2 years ago
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    indeed O_O

  11. Coolsector
    • 2 years ago
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    we find slope then we can choose any given point on the line to construct our line equation :)

  12. nexis
    • 2 years ago
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    yes

  13. nexis
    • 2 years ago
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    great, got it, thanks :)

  14. Coolsector
    • 2 years ago
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    yw

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