If f(x) = $\dfrac{1}{x^{2}}$ and g(x) = $e^{-9x}$
then (fog ) (x ) is
$\dfrac{1}{e^{18x}}$
$\dfrac{1}{e^{-18x}}$
$\dfrac{1}{e^{-18x^{2}}}$
$\dfrac{1}{e^{18x^{2}}}$
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