Solve frac{{200}+4sqrt{128}}{frac{{32}+3}{3}}

$$\frac { { 200 } + 4 \sqrt { 128 } } { \frac { { 32 } + 3 } { 3 } }$$

Evaluate

$$\frac{96\sqrt{2}}{35}+\frac{120}{7}\approx 21.021842914$$

Solution Steps

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$\frac{200+4\times 8\sqrt{2}}{\frac{32+3}{3}}$$

Multiply 4 and 8 to get 32.

$$\frac{200+32\sqrt{2}}{\frac{32+3}{3}}$$

Add 32 and 3 to get 35.

$$\frac{200+32\sqrt{2}}{\frac{35}{3}}$$

Divide 200+32\sqrt{2} by \frac{35}{3} by multiplying 200+32\sqrt{2} by the reciprocal of \frac{35}{3}.

$$\frac{\left(200+32\sqrt{2}\right)\times 3}{35}$$

Use the distributive property to multiply 200+32\sqrt{2} by 3.

$$\frac{600+96\sqrt{2}}{35}$$

Factor

$$\frac{24 {(4 \sqrt{2} + 25)}}{35} = 21.021842913937633$$