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?

1 1 1,841

The width can be less than or equal to 9.8ft since the area is no more than 175ft²

Step-by-step explanation:

Taking the shape of the conference table to be rectangular in nature,

Area of a rectangle = Length * Width

Given the area = 175ft²

Let the width of the table be x and;

Length of the table be L

Since length of the table is 18 ft more than the width, x, this means;

L = 18+x

To get the with we plug in the given parameters in the formula to have;

175 = L*x

175 = (18+x) x

175 = x²+18x

x²+8x-175 = 0

Using general formula

x = - b±√b²-4ac/2a

Where a = 1 b = 8 c = - 175

x = - 8±√8²-4 (1) (-175) / 2

x = - 8±√64+700/2

x = - 8±√764/2

x = - 8±27.6/2

x = - 8+27.6/2 (neglecting the negative)

x = 19.6/2

x = 9.8ft

The width can be less than or equal to 9.8ft since the area is no more than 175ft²