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? .

Width = x
Length = x+18

Assuming the table is rectangular:
Area = x(x + 18)

Therefore:
x(x + 18) ≤ 175
x^2 + 18x ≤ 175

Using completing the square method:
x^2 + 18x + 81 ≤ 175 + 81
(x + 9)^2 ≤ 256
|x + 9| ≤ sqrt(256)
|x + 9| ≤ +-16
-16 ≤ x + 9 ≤ 16
-16 - 9 ≤ x ≤ 16 - 9
-25 ≤ x ≤ 7

But x > 0 (there are no negative measurements):

Therefore, the interval 0 < x ≤ 7 represents the possible widths.

With a continuous thought process.