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OpenStudy (anonymous):
rearrange this formular to make r the subject. A=4πr²
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OpenStudy (saifoo.khan):
sqrt(A/4π)
OpenStudy (anonymous):
sqrt? I don't understand
OpenStudy (saifoo.khan):
\[\sqrt{A/4π} \]
OpenStudy (anonymous):
so, r=√A/4π?
OpenStudy (saifoo.khan):
yes.
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myininaya (myininaya):
\[A=4 \pi r^2\] \[r^2=A/(4 \pi)\] take square root of both sides
OpenStudy (anonymous):
how did you work that out? It's for an upcoming exam
OpenStudy (saifoo.khan):
look, 4π is being multiplied on the right... so it will b divided with A. A/4π = r^2
OpenStudy (saifoo.khan):
then a Sqrt to cancel the square.
OpenStudy (saifoo.khan):
finally u r done with r.
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OpenStudy (anonymous):
thanks c:
OpenStudy (saifoo.khan):
C:
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