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16. the area a of the rectangle is described with the inequality 40 ≤ a…

Question

  1. the area a of the rectangle is described with the inequality 40 ≤ a ≤ 120. if the width is 8 and the length is k, write and solve a compound inequality for k.

Explanation:

Step1: Write the area formula

The area of a rectangle is \(A = \text{length}\times\text{width}\). Given width \(= 8\) and length \(=k\), so \(A = 8k\).

Step2: Substitute \(A\) in the inequality

Since \(40\leq A\leq120\), substitute \(A = 8k\) into the inequality: \(40\leq8k\leq120\).

Step3: Solve the compound inequality

Divide each part of the inequality by \(8\). For \(40\leq8k\), we have \(\frac{40}{8}\leq k\), i.e., \(5\leq k\). For \(8k\leq120\), we have \(k\leq\frac{120}{8}\), i.e., \(k\leq15\).

Answer:

The compound inequality for \(k\) is \(5\leq k\leq15\)