Doomtown is 300 miles due west of sagebrush and joshua is due west of doomtown. at 9 a. m. mr. archer leaves sagebrush for joshua. at 1 p. m. mr. sassoon leaves doomtown for joshua. if mr. archer travels at an average speed 20 mph faster than mr. sassoon and they each reach joshua at 4 p. m., how fast is each traveling?

This is an example of distance problem in algebra.

Ley x be the rate of mr. sassoon

X + 20 be the rate of mr. archer

Y = distance from doomtown to joshua

So the distance travelled by mr. sassoon:

Y = 4x

The distance travelled by mr. archer:

(300 + y) = (x + 20)*7

Substitute y = 4x to the second eqation

300 + 4x = (x +20) *7

Solving for x

X = 53.33 mph is the rate of mr sassoon

X +20 = 73.33 mph the rate of mr. archer

answer: b.

step-by-step explanation:

the y coordinate for point b is 0, and answer b is the only one with that y coordinate.