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For Competitive Exams

Simplify : (3a)-2

$\dfrac{1}{9a^{-2}}$
$\dfrac{1}{3a^{2}}$
$\dfrac{1}{9a^{2}}$
$\dfrac{-1}{9a^{2}}$
Explanation:

(3a)-2=$\dfrac{1}{(3a)^{2}}$ [Using Rule a-m =$\dfrac{1}{a^{m}}$]

=$\dfrac{1}{3^{2}a^{2}}$ [Using Rule (am)n=amn]

=$\dfrac{1}{9a^{2}}$ [Evaluate 32]

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