Subject of the formula

• Jan 6th 2010, 11:40 AM
Mukilab
Subject of the formula
Make r the subject

$P=\pi r+2r+2a$

I could only get

$\frac{P-2a}{\pi}=r+\frac{2r}{\pi}$
• Jan 6th 2010, 11:44 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab
Make r the subject

$P=\pi r+2r+2a$

I could only get

$\frac{P-2a}{\pi}=r+\frac{2r}{\pi}$

Take 2a as normal.

$\pi r + 2r = P-2a$

You can then factor out an r on the LHS

$r(\pi+2) = P-2a$

Can you find $r$ now?
• Jan 6th 2010, 11:50 AM
Mukilab
Quote:

Originally Posted by e^(i*pi)
Take 2a as normal.

$\pi r + 2r = P-2a$

You can then factor out an r on the LHS

$r(\pi+2) = P-2a$

Can you find $r$ now?

$r=\frac{P-2a}{\pi +2}?$
• Jan 6th 2010, 11:51 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab
$r=\frac{P-2a}{\pi +2}?$

Indeed. $\pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question (Giggle)
• Jan 6th 2010, 11:53 AM
Mukilab
Quote:

Originally Posted by e^(i*pi)
Indeed. $\pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question (Giggle)

But thank you for the supply of extra information ^^