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

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap