Question

${183}^{0} \times {5}^{3} \div {3}^{2} + \frac{2}{9}$

Answer 1

$183^0*5^3\div 3^2+\frac{2}{9}=1*5^3*\frac{1}{9}+\frac{2}{9}=\frac{125}{9}+\frac{2}{9}=\frac{127}{9}=14\frac{1}{9}$