Solve sqrt{4}+(-3)^2-2018^0*|-4|+(1/6)^-1

$$\sqrt { 4 } + ( - 3 ) ^ { 2 } - 2018 ^ { 0 } \times | - 4 | + ( \frac { 1 } { 6 } ) ^ { - 1 }$$

Evaluate

$$13$$

Solution Steps

Calculate the square root of 4 and get 2.

$$2+\left(-3\right)^{2}-2018^{0}|-4|+\left(\frac{1}{6}\right)^{-1}$$

Calculate -3 to the power of 2 and get 9.

$$2+9-2018^{0}|-4|+\left(\frac{1}{6}\right)^{-1}$$

Add 2 and 9 to get 11.

$$11-2018^{0}|-4|+\left(\frac{1}{6}\right)^{-1}$$

Calculate 2018 to the power of 0 and get 1.

$$11-1|-4|+\left(\frac{1}{6}\right)^{-1}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -4 is 4.

$$11-1\times 4+\left(\frac{1}{6}\right)^{-1}$$

Multiply 1 and 4 to get 4.

$$11-4+\left(\frac{1}{6}\right)^{-1}$$

Subtract 4 from 11 to get 7.

$$7+\left(\frac{1}{6}\right)^{-1}$$

Calculate \frac{1}{6} to the power of -1 and get 6.

$$7+6$$

Add 7 and 6 to get 13.

$$13$$

Factor

$$13$$