Question

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

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Answer 1

In ($3^0$ + $4^{-1})\times 2^2$, {$4^{-1}$ = $\frac{1}{4}$($a^{-1}$ = $\frac{1}{a}$).$3^0$ = 1($a^0$ = 1)}

So ($3^0$ + $4^{-1}$) $\times$ $2^2$ = (1+$\frac{1}{4}$) $\times$ $2^2$

$=\frac{5}{4}\times 4$

$=5$