2. The length of a rectangular floor is 5 feet longer than its width w. The area of the floor is 535 ft2. (a) Write a quadratic equation in terms of w that represents the situation.
(b) What are the dimensions of the floor? Round to the nearest tenth.
Show your work.

Thanks!

Solution given:

let width of the floor be w.

length of floor =5ft+w

Area of floor=535ft²

we have

Area of floor =length*width

substituting value

535=(5+w)*w

opening bracket

535=5w+w²

w²+5w-535=0

Comparing above equation with ax²+bx+c=0,we get

a=1

b=5

c=-535

In terms of W in quadratic equation is:

W=

Substituting value

w=

Solving like terms

=Taking positive

w=

w=20.76

taking negative

-25.76(neglected)

So

again

length =20.76+5=25.76ft

quadratic equation in terms of w that represents the situation is w=

The dimension of the floor is:

length =20.76+5=25.76ft

width of floor=20.76ft

Let width be x

Solving further we get

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

c. 1,233 miles