Solve (sqrt{48}+1/4sqrt{12})divsqrt{27}

$$( \sqrt { 48 } + \frac { 1 } { 4 } \sqrt { 12 } ) \div \sqrt { 27 }$$

Evaluate

$$\frac{3}{2}=1.5$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{4\sqrt{3}+\frac{1}{4}\sqrt{12}}{\sqrt{27}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{4\sqrt{3}+\frac{1}{4}\times 2\sqrt{3}}{\sqrt{27}}$$

Multiply \frac{1}{4} and 2 to get \frac{2}{4}.

$$\frac{4\sqrt{3}+\frac{2}{4}\sqrt{3}}{\sqrt{27}}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$\frac{4\sqrt{3}+\frac{1}{2}\sqrt{3}}{\sqrt{27}}$$

Combine 4\sqrt{3} and \frac{1}{2}\sqrt{3} to get \frac{9}{2}\sqrt{3}.

$$\frac{\frac{9}{2}\sqrt{3}}{\sqrt{27}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{\frac{9}{2}\sqrt{3}}{3\sqrt{3}}$$

Cancel out \sqrt{3} in both numerator and denominator.

$$\frac{\frac{9}{2}}{3}$$

Express \frac{\frac{9}{2}}{3} as a single fraction.

$$\frac{9}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{9}{6}$$

Reduce the fraction \frac{9}{6} to lowest terms by extracting and canceling out 3.

$$\frac{3}{2}$$

Factor

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$