Question

Find the circumference of a circle whose area is /.

A.
B.
C.
D.

Answer 1

\(\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus
\\ \quad \\
60\pi =\pi r^2\impliedby \textit{solve for "r"}\)

what does that give you for "r"? or radius

Answer 2

60=rsqure

Answer 3

hmm rsquare?

Answer 4

idk what r is

Answer 5

\(\bf \textit{area of a circle}=\pi r^2\qquad area=60\pi \qquad thus
\\ \quad \\
60\pi =\pi r^2\implies \cfrac{60\cancel{\pi }}{\cancel{\pi }}=r^2\implies \sqrt{60}=\sqrt{r^2}\implies \sqrt{60}=r\)

Answer 6

that was close though

Answer 7

idk

Answer 8

7.777

Answer 9

so now that we know what "r" is
let us use it in the circumference formula then

\(\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then
\\ \quad \\
circumference=2\pi \left( \sqrt{60} \right)\)

Answer 10

and then you'd want to simplify that 60, see if you can squeeze something out of the radical

Answer 11

30

Answer 12

30 pie

Answer 13

one may note that \(\huge \pi \ne pie\) but yours is tastier though

Answer 14

so is it 30pie

Answer 15

so is it 30 pie?

Answer 16

\(\bf circumference=2\pi r\qquad \sqrt{60}=r\qquad then
\\ \quad \\
circumference=2\pi \left( \sqrt{60} \right)
\\ \quad \\
{\color{brown}{ 60\to 2\cdot 2\cdot 15\to 2^2\cdot 15 }}\qquad thus
\\ \quad \\
2\pi \left( \sqrt{60} \right)\implies 2\pi \left( \sqrt{{\color{brown}{ 2^2\cdot 15}}} \right)\implies 2\pi \sqrt{2^2}\sqrt{15}\)

Answer 17

so its 2square root 5