Make r the subject $\displaystyle P=\pi r+2r+2a$ I could only get $\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$
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Originally Posted by Mukilab Make r the subject $\displaystyle P=\pi r+2r+2a$ I could only get $\displaystyle \frac{P-2a}{\pi}=r+\frac{2r}{\pi}$ Take 2a as normal. $\displaystyle \pi r + 2r = P-2a$ You can then factor out an r on the LHS $\displaystyle r(\pi+2) = P-2a$ Can you find $\displaystyle r$ now?
Originally Posted by e^(i*pi) Take 2a as normal. $\displaystyle \pi r + 2r = P-2a$ You can then factor out an r on the LHS $\displaystyle r(\pi+2) = P-2a$ Can you find $\displaystyle r$ now? $\displaystyle r=\frac{P-2a}{\pi +2}?$
Originally Posted by Mukilab $\displaystyle r=\frac{P-2a}{\pi +2}?$ Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question
Originally Posted by e^(i*pi) Indeed. $\displaystyle \pi +2$ is a constant so the domain is all the real numbers but you don't need to know that for this question But thank you for the supply of extra information ^^
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