Question

Question 3(i)
Find the value of:

$(3^0+4^{-1})\times 2^2$

Answer 1

$(3^0+4^{-1})\times 2^2=(1+\frac{1}{4})\times 2^2[\because a^{-m}=\frac{1}{a^m}]=\left(\frac{4+1}{4}\right)\times 2^2=\frac{5}{4}\times 2^2=\frac{5}{2^2}\times 2^2=5\times 2^{2-2}[\because a^m÷a^n=a^{m-n}]=5\times 2^0=5\times 1=5[\because a^0=1]$