Solve -(1/2)^-3+sqrt{(-7)^2-(3sqrt{5})^0-4cos30^circ+6/sqrt{3}}

$$- ( \frac { 1 } { 2 } ) ^ { - 3 } + \sqrt { ( - 7 ) ^ { 2 } - ( 3 \sqrt { 5 } ) ^ { 0 } - 4 \cos 30 ^ { \circ } + \frac { 6 } { \sqrt { 3 } } }$$

Evaluate

$$4\left(\sqrt{3}-2\right)\approx -1.07179677$$

Solution Steps

Calculate \frac{1}{2} to the power of -3 and get 8.

$$-8+\sqrt{\left(-7\right)^{2}-\left(3\sqrt{5}\right)^{0}-4\cos(30)+\frac{6}{\sqrt{3}}}$$

Calculate -7 to the power of 2 and get 49.

$$-8+\sqrt{49-\left(3\sqrt{5}\right)^{0}-4\cos(30)+\frac{6}{\sqrt{3}}}$$

Calculate 3\sqrt{5} to the power of 0 and get 1.

$$-8+\sqrt{49-1-4\cos(30)+\frac{6}{\sqrt{3}}}$$

Subtract 1 from 49 to get 48.

$$-8+\sqrt{48-4\cos(30)+\frac{6}{\sqrt{3}}}$$

Get the value of \cos(30) from trigonometric values table.

$$-8+\sqrt{48-4\times \frac{\sqrt{3}}{2}+\frac{6}{\sqrt{3}}}$$

Cancel out 2, the greatest common factor in 4 and 2.

$$-8+\sqrt{48-2\sqrt{3}+\frac{6}{\sqrt{3}}}$$

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$-8+\sqrt{48-2\sqrt{3}+\frac{6\sqrt{3}}{\left(\sqrt{3}\right)^{2}}}$$

The square of \sqrt{3} is 3.

$$-8+\sqrt{48-2\sqrt{3}+\frac{6\sqrt{3}}{3}}$$

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

$$-8+\sqrt{48-2\sqrt{3}+2\sqrt{3}}$$

Combine -2\sqrt{3} and 2\sqrt{3} to get 0.

$$-8+\sqrt{48}$$

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}.

$$-8+4\sqrt{3}$$