Solve (2/3)^-1+(pi-3)^0+(-2)^-2+|(-2)^3|

$$( \frac { 2 } { 3 } ) ^ { - 1 } + ( \pi - 3 ) ^ { 0 } + ( - 2 ) ^ { - 2 } + | ( - 2 ) ^ { 3 } |$$

Evaluate

$$\frac{43}{4}=10.75$$

Solution Steps

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

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

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

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

Add \frac{3}{2} and 1 to get \frac{5}{2}.

$$\frac{5}{2}+\left(-2\right)^{-2}+\left(|-2|\right)^{3}$$

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

$$\frac{5}{2}+\frac{1}{4}+\left(|-2|\right)^{3}$$

Add \frac{5}{2} and \frac{1}{4} to get \frac{11}{4}.

$$\frac{11}{4}+\left(|-2|\right)^{3}$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -2 is 2.

$$\frac{11}{4}+2^{3}$$

Calculate 2 to the power of 3 and get 8.

$$\frac{11}{4}+8$$

Add \frac{11}{4} and 8 to get \frac{43}{4}.

$$\frac{43}{4}$$

Factor

$$\frac{43}{2 ^ {2}} = 10\frac{3}{4} = 10.75$$