Question

1. the length of a rectangle is $\frac{w + 3}{w^{2}+w - 12}$ and the width is $\frac{w^{2}+7w + 12}{w^{2}-9}$. what is the area of the rectangle in simplest form?

Explanation:

Step1: Recall the area formula

The area $A$ of a rectangle is $A = l\times w$, where $l$ is the length and $w$ is the width. Here, $l=\frac{w + 3}{w^{2}+w - 12}$ and $w=\frac{w^{2}+7w + 12}{w^{2}-9}$.

Step2: Factor the polynomials

Factor the denominators and numerators:
$w^{2}+w - 12=(w + 4)(w-3)$, $w^{2}+7w + 12=(w + 3)(w + 4)$, $w^{2}-9=(w + 3)(w - 3)$.
So, $l=\frac{w + 3}{(w + 4)(w-3)}$ and $w=\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}$.

Step3: Calculate the area

$A=l\times w=\frac{w + 3}{(w + 4)(w-3)}\times\frac{(w + 3)(w + 4)}{(w + 3)(w - 3)}$
Cancel out the common factors $(w + 3)$ and $(w + 4)$:
$A=\frac{w + 3}{(w-3)^{2}}$

Answer:

$\frac{w + 3}{(w-3)^{2}}$