Question

If the area of a rectangle is 24a^2b and the length is 8ab^2, what would be the width of the rectangle, given that width is found by dividing area by length? Simplify the answer. 1. 3a/b 2. b/3a 3. 3/ab 4. 3ab

Answer 1

$\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\ -------------------------------$

$\bf \textit{area of a rectangle}\\\\ A=WL~~ \begin{cases} W=width\\ L=length\\ -----\\ A=24a^2b\\ L=8ab^2 \end{cases}\implies 24a^2b=W(8ab^2)\implies \cfrac{24a^2b}{8ab^2}=W \\\\\\ \cfrac{24}{8}\cdot \cfrac{a^2b}{ab^2}=W\implies 3\cdot \cfrac{a^2a^{-1}}{b^2b^{-1}}=W\implies 3\cdot \cfrac{a^{2-1}}{b^{2-1}}=W\implies \cfrac{3a}{b}=W$