Question

part a
values for the area a of the rectangle shown are ( 12 leq a leq 60 ).

a compound inequality for this situation is ( 12 leq ) choose... ( leq 60 ).

part b
the solution of the compound inequality in part a is ( leq leq )

Explanation:

Step1: Find the area formula of the rectangle

The area formula of a rectangle is \(A = length\times width\). Here, the length is \(8x\) and the width is \(\frac{3}{4}\), so \(A=8x\times\frac{3}{4}\). Simplify \(8x\times\frac{3}{4}\) to get \(6x\).

Step2: Solve the compound inequality

We have the compound inequality \(12\leq6x\leq60\).

• Divide each part of the 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: \(\frac{6x}{6}\leq\frac{60}{6}\), which simplifies to \(x\leq10\).

Answer:

Part A: \(12\leq6x\leq60\)
Part B: \(2\leq x\leq10\)