Fajardolemus Fajardolemus
  • 12-07-2018
  • Mathematics
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How do I resolve this?
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Respuesta :

jdoe0001 jdoe0001
  • 12-07-2018

[tex] \bf \textit{area of a circle}\\\\
A=\pi r^2~~
\begin{cases}
r=radius\\
-----\\
A=144\pi
\end{cases}\implies 144\pi =\pi r^2\implies \cfrac{144\pi }{\pi }=r^2
\\\\\\
144=r^2\implies \sqrt{144}=r\implies \boxed{12=r}
\\\\\\
\textit{circumference of a circle}\\\\
C=2\pi r\qquad \qquad C=2\pi \boxed{12}\implies C=24\pi [/tex]

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