chancemorris123
  • chancemorris123
Find the circumference of a circle whose area is /. A. B. C. D.
Mathematics
chestercat
  • chestercat
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chancemorris123
  • chancemorris123
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jdoe0001
  • jdoe0001
\(\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus \\ \quad \\ 60\pi =\pi r^2\impliedby \textit{solve for "r"}\) what does that give you for "r"? or radius
chancemorris123
  • chancemorris123
60=rsqure

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jdoe0001
  • jdoe0001
hmm rsquare?
chancemorris123
  • chancemorris123
idk what r is
jdoe0001
  • jdoe0001
\(\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus \\ \quad \\ 60\pi =\pi r^2\implies \cfrac{60\cancel{\pi }}{\cancel{\pi }}=r^2\implies \sqrt{60}=\sqrt{r^2}\implies \sqrt{60}=r\)
jdoe0001
  • jdoe0001
that was close though
chancemorris123
  • chancemorris123
idk
chancemorris123
  • chancemorris123
7.777
jdoe0001
  • jdoe0001
so now that we know what "r" is let us use it in the circumference formula then \(\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then \\ \quad \\ circumference=2\pi \left( \sqrt{60} \right)\)
jdoe0001
  • jdoe0001
and then you'd want to simplify that 60, see if you can squeeze something out of the radical
chancemorris123
  • chancemorris123
30
chancemorris123
  • chancemorris123
30 pie
jdoe0001
  • jdoe0001
one may note that \(\huge \pi \ne pie\) but yours is tastier though
chancemorris123
  • chancemorris123
so is it 30pie
chancemorris123
  • chancemorris123
so is it 30 pie?
jdoe0001
  • jdoe0001
\(\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then \\ \quad \\ circumference=2\pi \left( \sqrt{60} \right) \\ \quad \\ {\color{brown}{ 60\to 2\cdot 2\cdot 15\to 2^2\cdot 15 }}\qquad thus \\ \quad \\ 2\pi \left( \sqrt{60} \right)\implies 2\pi \left( \sqrt{{\color{brown}{ 2^2\cdot 15}}} \right)\implies 2\pi \sqrt{2^2}\sqrt{15}\)
chancemorris123
  • chancemorris123
so its 2square root 5

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