Question

1. the area a of a rectangle with dimensions 1.5 and 4x is described with the inequality ( 36 leq a leq 72 ).

part a
select the compound inequality for the area written in terms of x.
a ( 36 leq 4x + 1.5 leq 72 )
b ( 36 leq 5.5x leq 72 )
c ( 36 leq 6x leq 72 )
d ( 36 leq 8x + 3 leq 72 )
part b
the solution to the compound inequality is (square) 6 (square) 42 (square) 12 (square) 78 (leq x leq) (square) 6. (square) 30. (square) 12. (square) 70.
are all of your solutions viable?
(square) yes (square) no

Explanation:

Step1: Find area formula for rectangle

The area \(A\) of a rectangle is length × width, so \(A = 1.5 \times 4x = 6x\).

Step2: Substitute into given inequality

Replace \(A\) with \(6x\) in \(36 \leq A \leq 72\), giving \(36 \leq 6x \leq 72\).

Step3: Solve left inequality for \(x\)

Divide all parts by 6: \(\frac{36}{6} \leq x\), so \(6 \leq x\).

Step4: Solve right inequality for \(x\)

Divide all parts by 6: \(x \leq \frac{72}{6}\), so \(x \leq 12\).

Step5: Check viability of solution

Since \(x\) represents a dimension-related value, \(6 \leq x \leq 12\) gives positive lengths, so all solutions are viable.

Answer:

Part A

C. \(36 \leq 6x \leq 72\)

Part B

\(6 \leq x \leq 12\)
yes