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Math Help - Please Check my Answer TIA

  1. #1
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    Please Check my Answer TIA

    If:

    V=(6-\frac{r^2}{3})*h

    and

    \frac{h}{2-r}= \frac{5}{2}

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

    My answer: 1st, I solved for R in the equation:

    \frac{h}{2-r}= \frac{5}{2} and figured it to be: \frac{4h^2-40h+100}{75}= \frac{r^2}{3}

    I then, plugged it into the equation V=(6-\frac{\pi^2}{3})*h
    since I now know R. I am positive my answer is correct but the computer is rejecting it.

    Here is my answer:
    V=\frac{4h^3+40h^2}{75} +350
    Last edited by mvho; June 19th 2009 at 05:47 PM.
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  2. #2
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    V is already a function of h...


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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mvho View Post
    If V=(6-\frac{\pi^2}{3})*h and \frac{h}{2-r} = \frac{5}{2}

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

    My Answer: I solved for r first and plugged in R in the original equation where v=(6-etc...). I solved for R in the equation: h/2-r=5/2 which I found as: 4h^2-40h+100/75=r^2.



    My algebra is not the best so if you catch any mistakes let me know!

    Answer (I think): (4*h^3/75) + (40*h^2)+350h

    TIA
    It seems like you've already solved the problem.

    You've already got V in terms of h. If V was in terms of r and h, then to write it as a function of h by substituting for r.

    This is a strange question altogether because I've never seen quadratic volume.
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  4. #4
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    To be honest, I am quite lost as well. Our TA just told us to solve for R first and that was the "clue".
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mvho View Post
    To be honest, I am quite lost as well. Our TA just told us to solve for R first and that was the "clue".

    Now that you've corrected the problem, I can help.

    \frac{h}{2-r}=\frac{5}{2}
    cross multiply
    2h=5(2-r)
    multiply out parentheses
    2h=10-5r
    subtract and divide
    \frac{2h-10}{-5}=r

    NOW you have r in terms of h.

    So substitute

    V(h)=[6-\frac{(\frac{2h-10}{-5})^2}{3}]*h

    now simplify

    V(h)=[6-\frac{(\frac{4h^2-40h+100}{25})}{3}]*h=6h-\frac{h(4h^2-40h+100)}{75}

    If you need to simplify further I'm sure you can do it. You may not be inputting correctly into your computer.
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  6. #6
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    Quote Originally Posted by mvho View Post
    If:

    V=(6-\frac{{\color{red}r^2}}{3})*h

    and

    \frac{h}{2-r}= \frac{5}{2}

    write V as a function of h . Simplify the function so that you can express it as a polynomial in h .
    Oh, now that's different.

    \begin{aligned}<br />
\frac{h}{2-r} &= \frac{5}{2} \\<br />
10 - 5r &= 2h \\<br />
-5r &= 2h - 10 \\<br />
r &= \frac{10 - 2h}{5}<br />
\end{aligned}

    \begin{aligned}<br />
V &= \left(6-\frac{\left(\frac{10 - 2h}{5}\right)^2}{3}\right)h \\<br />
&= \left(6-\frac{\left(\frac{100 - 40h + 4h^2}{25}\right)}{3}\right)h \\<br />
&= \left(6-\frac{4h^2 - 40h + 100}{75}\right)h \\<br />
&= \left(\frac{450 - (4h^2 - 40h + 100)}{75}\right)h \\<br />
  \end{aligned}
    \begin{aligned}<br />
&= \left(\frac{-4h^2 + 40h + 350}{75}\right)h \\<br />
&= \frac{-h(4h^2 - 40h - 350)}{75}<br />
\end{aligned}

    I'm not getting what you're getting, mvho...


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  7. #7
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yeongil View Post

    I'm not getting what you're getting, mvho...


    01
    We've all got the same thing.
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  8. #8
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    ok Thanks guys! Ill have to review this again. The hardest part about calculus is algebra.
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