Question

Quantos é (3^-2)-(2^3)+(7^0)?

Answer 1

($3^{-2}$)-($2^{3}$)+($7^{0}$)=
($\frac{1}{3} ^{2}$)-(8)+(1)=
$\frac{1}{9}$ -8+1=
$\frac{1}{9} - \frac{8}{1}$=
$\frac{1-72}{9}$=
$\frac{71}{9}$