Solve sqrt{18}-2sin45^circ+(pi-3)^0-(1/2)^-2

$$\sqrt { 18 } - 2 \sin 45 ^ { \circ } + ( \pi - 3 ) ^ { 0 } - ( \frac { 1 } { 2 } ) ^ { - 2 }$$

Evaluate

$$2\sqrt{2}-3\approx -0.171572875$$

Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$3\sqrt{2}-2\sin(45)+\left(\pi -3\right)^{0}-\left(\frac{1}{2}\right)^{-2}$$

Get the value of \sin(45) from trigonometric values table.

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

Cancel out 2 and 2.

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

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

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

Calculate \pi -3 to the power of 0 and get 1.

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

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

$$2\sqrt{2}+1-4$$

Subtract 4 from 1 to get -3.

$$2\sqrt{2}-3$$