Solve [(1/2)^-3+1sqrt{3}-2]+tan60^circ-(-2019)

$$[ ( \frac { 1 } { 2 } ) ^ { - 3 } + 1 \sqrt { 3 } - 2 ] + \tan 60 ^ { \circ } - ( - 2019 )$$

Evaluate

$$2\sqrt{3}+2025\approx 2028.464101615$$

Solution Steps

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

$$8+1\sqrt{3}-2+\tan(60)-\left(-2019\right)$$

Subtract 2 from 8 to get 6.

$$6+1\sqrt{3}+\tan(60)-\left(-2019\right)$$

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

$$6+1\sqrt{3}+\sqrt{3}-\left(-2019\right)$$

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

$$6+2\sqrt{3}-\left(-2019\right)$$

The opposite of -2019 is 2019.

$$6+2\sqrt{3}+2019$$

Add 6 and 2019 to get 2025.

$$2025+2\sqrt{3}$$