Question

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

Answer 1

Now,

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

$=(1+\frac{1}{3})\times 3^2$

$=\frac{4}{3}\times 3^2$

$=4\times 3$

$=12$.