While traveling to and from a certain destination, you realized increasing your speed by 30 mph saved 3 hours on your return. If the total distance of the roundtrip was 360 miles, find the speed driven while returning.

43.95mph

Step-by-step explanation:

let x=returning speed

x-30=foward speed

Travel time=distance/speed

180/(x-30)-180/x=3

LCD: x(x-30)

180x-180(x-30)=3x(x-3)

180x-180x+5400=3x^2-9x

3x^2-9x-5400=0

Use quadratic formula to solve

(-b+-√b^2-4ac) / 2a.

a = 3, b= -9, c = - 5400

x=[-(-9)±sqrt((-9)^2-4*3*-5400)]/2*3

x = 9±sqrt(81+64800)]/2*3

x = 9±sqrt(64881)]/6

x = 9±254.717/6

x = 263.717/6 or -254.717/6

x = 43.95 or - 42.45( will be rejected because it is less than 0)

x = 43.95

The returning speed will be 43.95mph.

i would have to say it is c.