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Flauer [41]
1 year ago
10

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?. .
Mathematics
2 answers:
Neko [114]1 year ago
7 0

Answer:

The correct answer is B.

Step-by-step explanation:


Aleksandr [31]1 year ago
3 0
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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