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?. .
2 answers:
Answer:
The correct answer is B.
Step-by-step explanation:
Width = x
Length = x+18
Assuming the table is rectangular:
Area = x(x + 18)
Therefore:
x(x + 18) <span>≤ 175
x^2 + 18x </span><span>≤ 175
Using completing the square method:
x^2 + 18x + 81 </span><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
</span><span>
But x > 0 (there are no negative measurements):
</span><span>
Therefore, the interval 0 < x </span><span>≤ 7 represents the possible widths.</span><span>
</span>
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