Solve =sqrt{400/1000+625/1200}

$$= \sqrt { \frac { 400 } { 1000 } + \frac { 625 } { 1200 } }$$

Evaluate

$$\frac{\sqrt{3315}}{60}\approx 0.959600611$$

Solution Steps

Reduce the fraction \frac{400}{1000} to lowest terms by extracting and canceling out 200.

$$\sqrt{\frac{2}{5}+\frac{625}{1200}}$$

Reduce the fraction \frac{625}{1200} to lowest terms by extracting and canceling out 25.

$$\sqrt{\frac{2}{5}+\frac{25}{48}}$$

Least common multiple of 5 and 48 is 240. Convert \frac{2}{5} and \frac{25}{48} to fractions with denominator 240.

$$\sqrt{\frac{96}{240}+\frac{125}{240}}$$

Since \frac{96}{240} and \frac{125}{240} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{96+125}{240}}$$

Add 96 and 125 to get 221.

$$\sqrt{\frac{221}{240}}$$

Rewrite the square root of the division \sqrt{\frac{221}{240}} as the division of square roots \frac{\sqrt{221}}{\sqrt{240}}.

$$\frac{\sqrt{221}}{\sqrt{240}}$$

Factor 240=4^{2}\times 15. Rewrite the square root of the product \sqrt{4^{2}\times 15} as the product of square roots \sqrt{4^{2}}\sqrt{15}. Take the square root of 4^{2}.

$$\frac{\sqrt{221}}{4\sqrt{15}}$$

Rationalize the denominator of \frac{\sqrt{221}}{4\sqrt{15}} by multiplying numerator and denominator by \sqrt{15}.

$$\frac{\sqrt{221}\sqrt{15}}{4\left(\sqrt{15}\right)^{2}}$$

The square of \sqrt{15} is 15.

$$\frac{\sqrt{221}\sqrt{15}}{4\times 15}$$

To multiply \sqrt{221} and \sqrt{15}, multiply the numbers under the square root.

$$\frac{\sqrt{3315}}{4\times 15}$$

Multiply 4 and 15 to get 60.

$$\frac{\sqrt{3315}}{60}$$