Transposition of Formula question........

**johnnymercer74** · December 8th 2009, 03:34 AM

Hi, OK well congratulations on all being such a clever and not to mention helpful lot....OK hopefully that's buttered you up nicely ;-)

OK so here's the problem, some crazy fool has decided to send me to do a foundation degree for my job in the forces, the only problem is I've not done maths since I left college in the 90's.

I'm doing the pre course package at the minute and while I'm OK at transposing basic formulae, I get in to real problems when the solution I'm looking for occurs more than once in an expression. I've got a few examples, which hopefully if answered fairly concisely should give me something to go on when answering similar questions in the exercise.....OK here goes.....

1)P-mg=mv(squared)/r for m

2)Z= the square root of [x/(x+y) for x]

Sorry, I can't make them look like they do on paper due to my woeful IT skills, don't think there is much hope for me but if any of you can decipher them I'd love to see how you do them.

Thanks

john

**Amer** · December 8th 2009, 05:15 AM

Originally Posted by johnnymercer74

1)P-mg=mv(squared)/r for m

2)Z= the square root of [x/(x+y) for x]

$p -mg = \frac{mv^2 }{r }$

$p = \frac{mv^2 }{r} + mg$ add "mg"to both sides

$p = m\left(\frac{v^2}{r} + g \right)$ factoring m

$m =\frac{p}{\frac{v^2 + gr}{r}}$ divide both sides by the $\frac{v^2}{r} + g$

$m = \frac{pr}{v^2 + gr}$

2)
$z = \sqrt{ \frac{x}{x+y}}$

$z^2 = \frac{x}{x+y}$

$xz^2 + yz^2 - x = 0$

$x-xz^2 = yz^2$

$x(1-z^2) = yz^2$

$x = \frac{yz^2}{1-z^2}$

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