Solve 2sqrt{144/81}+7^1*sqrt[3]{8}

$$2 \sqrt { \frac { 144 } { 81 } } + 7 ^ { 1 } \cdot \sqrt[ 3 ] { 8 }$$

Evaluate

$$\frac{50}{3}\approx 16.666666667$$

Solution Steps

Reduce the fraction \frac{144}{81} to lowest terms by extracting and canceling out 9.

$$2\sqrt{\frac{16}{9}}+7^{1}\sqrt[3]{8}$$

Rewrite the square root of the division \frac{16}{9} as the division of square roots \frac{\sqrt{16}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$2\times \frac{4}{3}+7^{1}\sqrt[3]{8}$$

Multiply 2 and \frac{4}{3} to get \frac{8}{3}.

$$\frac{8}{3}+7^{1}\sqrt[3]{8}$$

Calculate 7 to the power of 1 and get 7.

$$\frac{8}{3}+7\sqrt[3]{8}$$

Calculate \sqrt[3]{8} and get 2.

$$\frac{8}{3}+7\times 2$$

Multiply 7 and 2 to get 14.

$$\frac{8}{3}+14$$

Add \frac{8}{3} and 14 to get \frac{50}{3}.

$$\frac{50}{3}$$

Factor

$$\frac{2 \cdot 5 ^ {2}}{3} = 16\frac{2}{3} = 16.666666666666668$$