Results 1 to 7 of 7

Math Help - Solve for E

  1. #1
    Junior Member
    Joined
    Feb 2008
    From
    NWT, Canada
    Posts
    51

    Solve for K

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

    My steps....

    VS=T(E-V)

    TE-TV-VS=0

    -TV-VS=-TE

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

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

    My question is, is it possible to simplify it further?
    Last edited by iamanoobatmath; September 5th 2008 at 03:34 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    You can factor out the -V giving E = \frac{V(S+T)}{T}.

    Bobak
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    First you can multiply top and bottom by -1 to get

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

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

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

    ... apart from that, good job.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2008
    From
    NWT, Canada
    Posts
    51

    Smile

    Thanks for the help
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2008
    From
    NWT, Canada
    Posts
    51

    Unhappy

    I've gone from
    T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}
    to
    \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
    k=\pm\sqrt{\frac{h(T^2g-4\pi^2h)}{2\pi}}
    Last edited by iamanoobatmath; September 5th 2008 at 04:32 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    Quote Originally Posted by iamanoobatmath View Post
    I've gone from
    T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}
    to
    \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

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

    Bobak
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Quote Originally Posted by iamanoobatmath View Post
    S=T*\frac{(E-V)}{V}

    My steps....

    VS=T(E-V)

    TE-TV-VS=0

    -TV-VS=-TE

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

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

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

    VS=T(E-V)

    VS=TE-TV (multiplying out the T)

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

    TE=V(S+T) (factoring by V)

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


    But it doesn't matter, what you did was perfectly accurate.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. how do i solve this IVP: y'=y^2 -4
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2010, 11:14 AM
  3. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 02:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum