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.

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nuhulawal20 nuhulawal20 · Answer 2 · 2022-05-25T12:21:37+02:00

Given the area of the conference table, If the length of the table is 18ft more than the width x, the interval that can be the possible widths is 0 ≤ x ≤ 7.

What is a Quadratic Equation?

Quadratic equation is simply an algebraic expression of the second degree in x. Quadratic equation in its standard form is;

ax² + bx + c = 0

Where x is the unknown

Given that;

Area of the table A = 175ft²
Width = x
Length = x + 18ft
Interval that can be the possible widths = x

Since Area = length × width

175ft² ≤ x × ( x + 18ft )

x( x + 18 ) ≤ 175

x² + 18x ≤ 175

Add 81 to each sides to complete the square

x² + 18x + 81 ≤ 175 + 81

(x+9)(x+9) ≤ 256

(x+9)² ≤ 256

Get the square root of both sides

x+9 ≤ √256

x+9 ≤ ±16

-16 - 9 ≤ x ≤ 16 - 9

-25 ≤ x ≤ 7

Since negative measurement is not possible, we just say;

0 ≤ x ≤ 7

Given the area of the conference table, If the length of the table is 18ft more than the width x, the interval that can be the possible widths is 0 ≤ x ≤ 7.

Learn more about quadratic equations here: brainly.com/question/1863222

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