Solve sqrt{18}+-3^0-6cos(45)+-2^-2

$$\sqrt{ 18 } + { -3 }^{ 0 } -6 \cos ( 45 ) + { -2 }^{ -2 }$$

Evaluate

$$\frac{5}{4}=1.25$$

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}.

$$3\sqrt{2}+\left(-3\right)^{0}-6\cos(45)+\left(-2\right)^{-2}$$

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

$$3\sqrt{2}+1-6\cos(45)+\left(-2\right)^{-2}$$

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

$$3\sqrt{2}+1-6\times \frac{\sqrt{2}}{2}+\left(-2\right)^{-2}$$

Cancel out 2, the greatest common factor in 6 and 2.

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

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

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

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

$$1+\frac{1}{4}$$

Add 1 and \frac{1}{4} to get \frac{5}{4}.

$$\frac{5}{4}$$