The length of a rectangle is 4 inches more than its width. the area of the rectangle is equal to 4 inches less than 5 times the perimeter. find the length and width of the rectangle.

Solution:
Let the width of the rectsngle (b) be x
and the length of the rectangle (l) =x+4
Area of the rectangle (A)=?
Perimeter of the rectangle (P)=?
Now,
P= 2 (l+b)
= 2 (x+4+x)
= 2 (2x+4)
= 4x+8
And,
A= l×b
= x(x+4)
= x^2+4x
According to the question,
A=5P-4
or, x^2+4x =5 (4x+8)-4
or, x^2 +4x =20x+40-4
or, x^2 = 20x-4x+36
or, x^2 = 16x+36
or, x^2-16x-36 = 0
or, x^2-(18-2)x-36 = 0
or, x^2-18x+2x-36 = 0
or, x (x-18)+2 (x-18) = 0
or, (x-18) (x+2) = 0

Either, Or,
x-18 = 0 x+2 = 0
so, x = 18 so, x = -2

Taking the value of x = 18 ( Because the length is always positive)
Length = x+4 = 18+4 = 22 cm
Width = x = 18 cm

The answer is d 18ft

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.