Solve (2^5)^-1div(-1/4)_i^3*(16^-1)^2

$$( 2 ^ { 5 } ) ^ { - 1 } \div ( - \frac { 1 } { 4 } ) _ { i } ^ { 3 } \times ( 16 ^ { - 1 } ) ^ { 2 }$$

Evaluate

$$-\frac{1}{8}=-0.125$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 5 and -1 to get -5.

$$\frac{2^{-5}}{-\frac{1}{4}}$$

Calculate 2 to the power of -5 and get \frac{1}{32}.

$$\frac{\frac{1}{32}}{-\frac{1}{4}}$$

Divide \frac{1}{32} by -\frac{1}{4} by multiplying \frac{1}{32} by the reciprocal of -\frac{1}{4}.

$$\frac{1}{32}\left(-4\right)$$

Multiply \frac{1}{32} and -4 to get -\frac{1}{8}.

$$-\frac{1}{8}$$

Factor

$$-\frac{1}{8} = -0.125$$