Question

part a. values for the area a of the rectangle shown are 12 ≤ a ≤ 36. part b. a compound inequality for this situation is 12 ≤ 6x ≤ 36. the solution of the compound inequality in part a is 2 ≤ x ≤ 6

Explanation:

Step1: Recall rectangle - area formula

The area formula of a rectangle is $A = l\times w$. Here, $l = 3$ and $w = 2x$, so $A=3\times2x = 6x$.

Step2: Solve the compound - inequality

We are given the compound - inequality $12\leq A\leq36$. Substitute $A = 6x$ into the inequality, we get $12\leq6x\leq36$.
Divide each part of the compound - inequality by 6. For the left - hand side, $\frac{12}{6}\leq\frac{6x}{6}$, which simplifies to $2\leq x$. For the right - hand side, $\frac{6x}{6}\leq\frac{36}{6}$, which simplifies to $x\leq6$.

Answer:

$2\leq x\leq6$