Simplify:
$ \frac{1}{6^{-2}}\left[\left(\frac{2}{3}\right)^{2}\right]^{-1} \times\left(\frac{4}{9}\right) \times 7^{0} $

Given:

\( \frac{1}{6^{-2}}\left[\left(\frac{2}{3}\right)^{2}\right]^{-1} \times\left(\frac{4}{9}\right) \times 7^{0} \)

To do:

We have to simplify the given expression.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$\frac{1}{6^{-2}}[(\frac{2}{3})^{2}]^{-1} \times(\frac{4}{9}) \times 7^{0}=6^2[\frac{4}{9}]^{-1}\times(\frac{4}{9}) \times 1$

$=36\times\frac{9}{4}\times\frac{4}{9}$

$=36$

Hence, $\frac{1}{6^{-2}}[(\frac{2}{3})^{2}]^{-1} \times(\frac{4}{9}) \times 7^{0}=36$.

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Updated on: 10-Oct-2022

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