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 t han the width, x, which interval can be the possible widths?
1 answer:
The answer is 0 < x <span>≤ 7 </span> First, we know that width = x Which means that length = x +18 So, the possible equation for the Table's area is X (X + 18) ≤ 175 X^2 + 18x - 175 <span>≤ </span>0 Next, we need to calculate is by using complete square method x^2 + 18x + 81 <span>≤ 175 + 81 (x + 9)^2 </span><span>≤ 256 |x + 9| </span><span>≤ sqrt(256) |x + 9| </span><span>≤ +-16 -16 </span>≤ x + 9 <span>≤ 16 </span>-16 - 9 ≤ x <span>≤ 16 - 9 </span>-25 ≤ x <span>≤ 7 Since the width couldn't be negative, we can change -25 with 0, so it become </span> 0 < x ≤ 7
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