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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