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Thread: Solve for E

  1. #1
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    Solve for K

    $\displaystyle S=T*\frac{(E-V)}{V}$

    My steps....

    $\displaystyle VS=T(E-V)$

    $\displaystyle TE-TV-VS=0$

    $\displaystyle -TV-VS=-TE$

    $\displaystyle \frac{(-TV-VS)}{-T}=\frac{-TE}{-T}$

    $\displaystyle \frac{(-TV-VS)}{-T}=E$

    My question is, is it possible to simplify it further?
    Last edited by iamanoobatmath; Sep 5th 2008 at 03:34 PM.
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  2. #2
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    You can factor out the -V giving $\displaystyle E = \frac{V(S+T)}{T}$.

    Bobak
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  3. #3
    MHF Contributor Matt Westwood's Avatar
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    First you can multiply top and bottom by -1 to get

    $\displaystyle E = \frac{TV+VS}{T}$

    then you can extract V as a factor on the top:

    $\displaystyle E = \frac{V (T+S)}{T}$

    ... apart from that, good job.
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  4. #4
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    Smile

    Thanks for the help
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  5. #5
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    Unhappy

    I've gone from
    $\displaystyle T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}$
    to
    $\displaystyle \frac{T^2gh}{4\pi^2}-h^2=k^2$

    I'm supposed to solve for k, but I have absolutely no idea how to get rid of the squares.

    edit: thanks for the help bobak I figured it out now. I'll just edit this post so I dont' bump the thread
    $\displaystyle k=\pm\sqrt{\frac{h(T^2g-4\pi^2h)}{2\pi}}$
    Last edited by iamanoobatmath; Sep 5th 2008 at 04:32 PM.
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  6. #6
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    Quote Originally Posted by iamanoobatmath View Post
    I've gone from
    $\displaystyle T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}$
    to
    $\displaystyle \frac{T^2gh}{4\pi^2}-h^2=k^2$

    I'm supposed to solve for k, but I have absolutely no idea how to get rid of the squares.
    take the square root. so

    $\displaystyle k = \pm \sqrt{ \frac{T^2gh}{4\pi^2}-h^2} $

    Bobak
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  7. #7
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by iamanoobatmath View Post
    $\displaystyle S=T*\frac{(E-V)}{V}$

    My steps....

    $\displaystyle VS=T(E-V)$

    $\displaystyle TE-TV-VS=0$

    $\displaystyle -TV-VS=-TE$

    $\displaystyle \frac{(-TV-VS)}{-T}=\frac{-TE}{-T}$

    $\displaystyle \frac{(-TV-VS)}{-T}=E$

    My question is, is it possible to simplify it further?
    You can do this easier by:

    $\displaystyle VS=T(E-V)$

    $\displaystyle VS=TE-TV$ (multiplying out the T)

    $\displaystyle TE=VS+VT$ (adding VT (which equals TV) to both sides and changing sides around to put TE on the left)

    $\displaystyle TE=V(S+T)$ (factoring by V)

    $\displaystyle E = \frac{V(S+T)}{T}$


    But it doesn't matter, what you did was perfectly accurate.
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