Question 14 Marks
Evaluate: $\left(2^2\right)^0+2^{-4} \div 2^{-6}+\left(\frac{1}{2}\right)^{-3}$
Answer
$\begin{aligned} & \left(2^2\right)^0+2^{-4} \div 2^{-6}+\left(\frac{1}{2}\right)^{-3} \\ & =(4)^0+\left(\frac{1}{2}\right)^4 \div\left(\frac{1}{2}\right)^6+\left(\frac{2}{1}\right)^3\left(\because a^0=1\right) \\ & =1+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) \div\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{2}{1} \times \frac{2}{1} \times \frac{2}{1}\right) \\ & =1+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\right)+8 \\ & =1+4+8=13\end{aligned}$
View full question & answer→$\begin{aligned} & \left(2^2\right)^0+2^{-4} \div 2^{-6}+\left(\frac{1}{2}\right)^{-3} \\ & =(4)^0+\left(\frac{1}{2}\right)^4 \div\left(\frac{1}{2}\right)^6+\left(\frac{2}{1}\right)^3\left(\because a^0=1\right) \\ & =1+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right) \div\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)+\left(\frac{2}{1} \times \frac{2}{1} \times \frac{2}{1}\right) \\ & =1+\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times 2 \times 2 \times 2 \times 2 \times 2 \times 2\right)+8 \\ & =1+4+8=13\end{aligned}$