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Simplify the following:$(\frac{22}{7})^{-4} \times(\frac{7}{44})^{-4}$.
Given :
The given expression is $(\frac{22}{7})^{-4} \times(\frac{7}{44})^{-4}$.
To do :
We have to simplify the given expression.
Solution :
$(\frac{22}{7})^{-4} \times(\frac{7}{44})^{-4}$
$\Rightarrow (\frac{22}{7} \times \frac{7}{44})^{-4}$ $[a^m \times b^m=(a\times b)^m]$
$\Rightarrow (\frac{1}{2})^{-4}$
$\Rightarrow (2)^4$ $[a^{-m}=(\frac{1}{a})^m]$
$\Rightarrow 2 \times 2\times 2 \times 2 = 16$
Therefore, the value of $(\frac{22}{7})^{-4} \times(\frac{7}{44})^{-4}$ is 16.
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