Arectangular parking lot has a length 20 h more
than its width. its area is 2400h². what are
the dimensions of the lot?

Let the width = X

The length would be X + 20

Area = Length x width:

2400 = x * (x+20)

Simplify:

2400 = x^2 + 20x

Subtract 2400 from both sides:

x^2 + 20x - 2400 = 0

Factor:

(x-40) (x+60) = 0

Solve for both x's:

x = 40 and x = -60

The dimensions of a parking lot have to be positive values so you would use x = 40.

The width = x = 40

The length = x +20 = 40+20 = 60

The dimension of the parking lot is 40H x 60H

answer:

c 60% on edg