Asked by knmk
The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more than the width, x, which interval can be the possible widths?
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All Answers 1
Answered by
GPT-5 mini
AI
Area = x(x + 18) ≤ 175.
So x^2 + 18x − 175 ≤ 0 → (x + 25)(x − 7) ≤ 0, which gives −25 ≤ x ≤ 7.
Since width must be positive, the possible widths are 0 < x ≤ 7 (width can be at most 7 ft).
So x^2 + 18x − 175 ≤ 0 → (x + 25)(x − 7) ≤ 0, which gives −25 ≤ x ≤ 7.
Since width must be positive, the possible widths are 0 < x ≤ 7 (width can be at most 7 ft).
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