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allsm [11]
3 years ago
8

A=pir^2 solve for pi

Mathematics
1 answer:
VashaNatasha [74]3 years ago
6 0

Given problem;

   A = \pi r²

   Solve for π;

To solve for π implies that we make it the subject of the expression.

So;

       A = π r²

      Now multiply both sides by \frac{1}{r^{2} }

 So;

                A x \frac{1}{r^{2} }   = \pi x r² x \frac{1}{r^{2} }

              r² cancels out from the right side and leaves only π;

                  π = \frac{A}{r^{2} }

So    \pi  = \frac{A}{r^{2} }

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\bold{Heya!}

Your answer to this is:

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

<h2>→ <u>EXPLANATION :-</u></h2>

<u />

<u />\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: (1)

\sf{Apply \: rule:} \: (a) = a

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\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0} \: ^. \: 1

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\sf{\frac{2^1^0^0}{1 \: + \: x^2^0^0} ^.^ \: 1 \: = \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

\sf{= \frac{1}{2} \: + \frac{2}{1 \: + \: x^2} \: + \: \frac{1}{1 \: + \: x^4} \: + \: \frac{2^1^0^0}{1 \: + \: x^2^0^0}

Hopefully This Helps ! ~

#LearnWithBrainly

\underline{Answer :}

<em>Jaceysan ~</em>

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