Question

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

Answer 1

As we know that,

$n^0=1$

$n^{-1}=\frac{1}{n}$

Therefore,

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

$=(1+\frac{1}{4})\times 4$

$=\left(\frac{4+1}{4}\right)\times 4$

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

$=5$