Solve (-1/2)^-2-8x(-2)^-2+(-1)^2019-(0.5)

$$( - \frac { 1 } { 2 } ) ^ { - 2 } - 8 x ( - 2 ) ^ { - 2 } + ( - 1 ) ^ { 2019 } - ( 0.5 )$$

Evaluate

$$2.5-2x$$

Solution Steps

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

$$4-8x\left(-2\right)^{-2}+\left(-1\right)^{2019}-0.5$$

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

$$4-8x\times \frac{1}{4}+\left(-1\right)^{2019}-0.5$$

Multiply 8 and \frac{1}{4} to get 2.

$$4-2x+\left(-1\right)^{2019}-0.5$$

Calculate -1 to the power of 2019 and get -1.

$$4-2x-1-0.5$$

Subtract 1 from 4 to get 3.

$$3-2x-0.5$$

Subtract 0.5 from 3 to get 2.5.

$$2.5-2x$$

Factor

$$\frac{5-4x}{2}$$