Solve [(2/7)^-3*sqrt{64/49}+1]div(1/5)^-4=

$$[ ( \frac { 2 } { 7 } ) ^ { - 3 } \cdot \sqrt { \frac { 64 } { 49 } } + 1 ] \div ( \frac { 1 } { 5 } ) ^ { - 4 } =$$

Evaluate

$$\frac{2}{25}=0.08$$

Solution Steps

Calculate \frac{2}{7} to the power of -3 and get \frac{343}{8}.

$$\frac{\frac{343}{8}\sqrt{\frac{64}{49}}+1}{\left(\frac{1}{5}\right)^{-4}}$$

Rewrite the square root of the division \frac{64}{49} as the division of square roots \frac{\sqrt{64}}{\sqrt{49}}. Take the square root of both numerator and denominator.

$$\frac{\frac{343}{8}\times \frac{8}{7}+1}{\left(\frac{1}{5}\right)^{-4}}$$

Multiply \frac{343}{8} and \frac{8}{7} to get 49.

$$\frac{49+1}{\left(\frac{1}{5}\right)^{-4}}$$

Add 49 and 1 to get 50.

$$\frac{50}{\left(\frac{1}{5}\right)^{-4}}$$

Calculate \frac{1}{5} to the power of -4 and get 625.

$$\frac{50}{625}$$

Reduce the fraction \frac{50}{625} to lowest terms by extracting and canceling out 25.

$$\frac{2}{25}$$

Factor

$$\frac{2}{5 ^ {2}} = 0.08$$