Find the value of ( $3^{0}$ + $4^{- 1}$ ) × $2^{2}$

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Solution

In ( $3^{0}$ + $4^{- 1}$ ) × $2^{2}$ , $4^{- 1}$ = $\frac{1}{4}$ ( $a^{- 1}$ = $\frac{1}{a}$ ), $3^{0}$ = 1( $a^{0}$ = 1)
So ( $3^{0}$ + $4^{- 1}$ ) × $2^{2}$ ,
$4^{- 1}$ = $\frac{1}{4}$ ,
(1+ $\frac{1}{4})$ × $2^{2}$
=5

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