Simplify $\dfrac{(0.5)^{-3} \cdot (0.25)^{2}}{(0.125)^{-1}}$.

Question

Aicharya · Accepted Answer

Write each as a power of 2: $0.5 = 2^{-1}$, $0.25 = 2^{-2}$, $0.125 = 2^{-3}$. Then $(2^{-1})^{-3} \cdot (2^{-2})^{2} \div (2^{-3})^{-1} = 2^{3} \cdot 2^{-4} \div 2^{3} = 2^{-4} = \dfrac{1}{16}$.

Simplify $\dfrac{(0.5)^{-3} \cdot (0.25)^{2}}{(0.125)^{-1}}$.

Solution