# Math Help - rearranging equations

1. ## 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?

2. 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.

3. Originally Posted by youmuggles
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!