Question

State true or false.

${\left(\frac{27}{125}\right)}^{\frac{2}{3}}\times {\left(\frac{9}{25}\right)}^{-\frac{3}{2}}$ is equal to $1\frac{2}{3}$.

• A. True
• B. False

Answer 1

The correct option is A True

${\left(\frac{27}{125}\right)}^{\frac{2}{3}}\times {\left(\frac{9}{25}\right)}^{-\frac{3}{2}}$

$={\left(\frac{3^3}{5^3}\right)}^{\frac{2}{3}}\times {\left(\frac{3^2}{5^2}\right)}^{-\frac{3}{2}}$

$={\left(\frac{3}{5}\right)}^{\frac{2\times 3}{3}}\times {\left(\frac{3}{5}\right)}^{-\frac{3\times 2}{2}}$

$={\left(\frac{3}{5}\right)}^2\times {\left(\frac{3}{5}\right)}^{-3}$

$={\left(\frac{3}{5}\right)}^{-1}$

$=\frac{5}{3}$

$=1\frac{2}{3}$

The given statement is true.