Solve (pi-3.14)^0-|sqrt{12}sin60^circ-4|+(1/2)^-1

$$( \pi - 3.14 ) ^ { 0 } - | \sqrt { 12 } \sin 60 ^ { \circ } - 4 | + ( \frac { 1 } { 2 } ) ^ { - 1 }$$

Evaluate

$$2$$

Solution Steps

Calculate \pi -3.14 to the power of 0 and get 1.

$$1-|\sqrt{12}\sin(60)-4|+\left(\frac{1}{2}\right)^{-1}$$

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}.

$$1-|2\sqrt{3}\sin(60)-4|+\left(\frac{1}{2}\right)^{-1}$$

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

$$1-|2\sqrt{3}\times \frac{\sqrt{3}}{2}-4|+\left(\frac{1}{2}\right)^{-1}$$

Cancel out 2 and 2.

$$1-|\sqrt{3}\sqrt{3}-4|+\left(\frac{1}{2}\right)^{-1}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$1-|3-4|+\left(\frac{1}{2}\right)^{-1}$$

Subtract 4 from 3 to get -1.

$$1-|-1|+\left(\frac{1}{2}\right)^{-1}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -1 is 1.

$$1-1+\left(\frac{1}{2}\right)^{-1}$$

Subtract 1 from 1 to get 0.

$$0+\left(\frac{1}{2}\right)^{-1}$$

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

$$0+2$$

Add 0 and 2 to get 2.

$$2$$