Solve sqrt{(128/25)^2+(96/25)^2}

$$\sqrt { ( \frac { 128 } { 25 } ) ^ { 2 } + ( \frac { 96 } { 25 } ) ^ { 2 } }$$

Evaluate

$$\frac{32}{5}=6.4$$

Solution Steps

Calculate \frac{128}{25} to the power of 2 and get \frac{16384}{625}.

$$\sqrt{\frac{16384}{625}+\left(\frac{96}{25}\right)^{2}}$$

Calculate \frac{96}{25} to the power of 2 and get \frac{9216}{625}.

$$\sqrt{\frac{16384}{625}+\frac{9216}{625}}$$

Since \frac{16384}{625} and \frac{9216}{625} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{16384+9216}{625}}$$

Add 16384 and 9216 to get 25600.

$$\sqrt{\frac{25600}{625}}$$

Reduce the fraction \frac{25600}{625} to lowest terms by extracting and canceling out 25.

$$\sqrt{\frac{1024}{25}}$$

Rewrite the square root of the division \frac{1024}{25} as the division of square roots \frac{\sqrt{1024}}{\sqrt{25}}. Take the square root of both numerator and denominator.

$$\frac{32}{5}$$

Factor

$$\frac{2 ^ {5}}{5} = 6\frac{2}{5} = 6.4$$