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Math Help - rearranging equations

  1. #1
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    rearranging equations

    write V as a function of h. Simplify the function so that you can express it as a polynomial in h.

    V = ( 2-r^2 / 3 ) h and h / (2-r) = 5/2

    I got h = (10-5r)/2

    and then V = (5*r^3)-(10*r^2)-(10*r)+20

    somehow it's not correct?
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  2. #2
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    Looks like you've made a small multiplication error somewhere.

    You are right that h / (2 - r) = 5/2 is the same as (10 - 5r)/2. But you can make things easier for yourself if you simplify that further. Why not just write (5/2)(2-r) = h?

    Now, take V, and replace h with what you found.
    (( 2 - r^2)/3)(5/2)(2-r).

    Now, to make things look simpler, let's pull out that 1/3 and the 5/2 and simplify them.
    (1/3)*(5/2) = 5/6

    So now we have (5/6)(2-r^2)(2-r)

    You can multiply out (2 - r^2) and (2 - r), right? Then either multiply each term of your answer by 5/6 or leave the 5/6 factored out in front.
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  3. #3
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    Quote Originally Posted by youmuggles View Post
    write V as a function of h. Simplify the function so that you can express it as a polynomial in h.

    V = ( 2-r^2 / 3 ) h and h / (2-r) = 5/2

    I got h = (10-5r)/2

    and then V = (5*r^3)-(10*r^2)-(10*r)+20 <<<< somehow the h has vanished

    somehow it's not correct?
    1. Solve this equation for r:

    \frac h{2-r} = \frac52~\implies~r = \frac52 (5-h)

    2. Now replace r in the first equation by this term:

    V(h)= h \cdot \left(\dfrac{2-\left( \frac52 (5-h) \right)^2}{3}  \right)

    3. Simplify!
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