Solve sqrt{27}+{left(-1/2right)}^-2-3tan(60^circ)+left(pi-2right)^0

$$\sqrt{ 27 } + { \left(- \frac{ 1 }{ 2 } \right) }^{ -2 } -3 \tan ( 60 ^ { \circ } ) + { \left( \pi -2 \right) }^{ 0 }$$

Evaluate

$$5$$

Solution Steps

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}.

$$3\sqrt{3}+\left(-\frac{1}{2}\right)^{-2}-3\tan(60)+\left(\pi -2\right)^{0}$$

Calculate -\frac{1}{2} to the power of -2 and get 4.

$$3\sqrt{3}+4-3\tan(60)+\left(\pi -2\right)^{0}$$

Get the value of \tan(60) from trigonometric values table.

$$3\sqrt{3}+4-3\sqrt{3}+\left(\pi -2\right)^{0}$$

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

$$4+\left(\pi -2\right)^{0}$$

Calculate \pi -2 to the power of 0 and get 1.

$$4+1$$

Add 4 and 1 to get 5.

$$5$$