Hello

Iv been told to rearrang the following equation:

n^{m+1}= (C^{m+1}x (V - U)^{m+1}) / (Q x D)^{m+1}

so that:

n^{m+1}x Q x D = ??????

every time iv done it iv been told i got it wrong.

Can some one please help me.

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- Oct 14th 2012, 07:03 AMkillickhelp to rearrang equation
Hello

Iv been told to rearrang the following equation:

n^{m+1}= (C^{m+1}x (V - U)^{m+1}) / (Q x D)^{m+1}

so that:

n^{m+1}x Q x D = ??????

every time iv done it iv been told i got it wrong.

Can some one please help me. - Oct 14th 2012, 08:06 AMskeeterRe: help to rearrang equation
$\displaystyle n^{m+1} = \frac{C^{m+1}(V-U)^{m+1}}{(QD)^{m+1}}$

$\displaystyle n^{m+1} = \frac{C^{m+1}(V-U)^{m+1}}{(QD)(QD)^m}$

$\displaystyle n^{m+1}QD = \frac{C^{m+1}(V-U)^{m+1}}{(QD)^m}$ - Oct 14th 2012, 09:22 AMkillickRe: help to rearrang equation
Thanks for the help skeeter.

This there a way to get a simalar equation but without the (QD)^m factor? - Oct 14th 2012, 11:21 AMskeeterRe: help to rearrang equation
I get the feeling there is more to this problem than what you've posted. Please state the original problem and the given directions ...

- Oct 14th 2012, 12:48 PMHallsofIvyRe: help to rearrang equation
I suspect you

**really**want n x Q x D= rather than the n+1 power of r.

From $\displaystyle n^{m+1}= \frac{C^{m+1}(V- U)^{m+1}}{(QD)^{m+1}}$, you can write it as $\displaystyle n^{m+1}= \left(\frac{C(V- U)}{QD}\right)^{m+1}$

and take the n+1 root to get $\displaystyle n=\frac{C(V- U)}{QD}$ and then multiply both sides by QD:

$\displaystyle nQD= C(V- U)$ - Oct 17th 2012, 04:47 AMkillickRe: help to rearrang equation
Iv got the equation:

I = n^(m+1) QD K W X

But need to substitute the n^(m+1) QD part using this equation:

n^(m+1) = (C^(m+1) x (V - U)^(m+1)) / (Q x D)^(m+1)

But with out any QD factors left on the right hand side