1. ## 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?

2. You can factor out the -V giving $\displaystyle E = \frac{V(S+T)}{T}$.

Bobak

3. 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.

4. Thanks for the help

5. 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
$\displaystyle k=\pm\sqrt{\frac{h(T^2g-4\pi^2h)}{2\pi}}$

6. Originally Posted by iamanoobatmath
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

7. Originally Posted by iamanoobatmath
$\displaystyle S=T*\frac{(E-V)}{V}$

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.