Mr. beecher and mrs. carter are teachers at the same school. they leave their houses at the same time in the morning to get to school. mr. beecher lives 8 miles away from school and rides his bicycle to work. every minute, he gets 1 6 of a mile closer to school. mrs. carter lives 20 miles away from school and drives her car to work. every minute, she gets 1 2 of a mile closer to school. after how many minutes will mr. beecher and mrs. carter first be the same distance away from school?

Let the school be at point x

Mr. Beecher lives 8 miles away from school

Therefore, his starting point is x – 8

Every minute, he gets 1/6 of a mile closer to school

Therefore, at time, t(in minutes), Mr. Beecher’s distance from school, Db = x – 8 + (1/6)t

Db = x – 8 + (1/6)t            … (Equation I)

Mrs. Carter lives 20 miles away from school

Therefore, her starting point is x – 20

Every minute, she gets 1/2 of a mile closer to school

Therefore, at time, t(in minutes), Mrs. Carter’s distance from school, Dc = x – 20 + (1/2)t

Dc = x – 20 + (1/2)t            … (Equation II)

After how many minutes will Mr. Beecher and Mrs. Carter first be the same distance away from school?

That’s when Db = Dc, given that t is same value in both equations

 Db = x – 8 + (1/6)t            … (Equation I)

Dc = x – 20 + (1/2)t            … (Equation II)

Db = Dc. Therefore,

x – 8 + (1/6)t = x – 20 + (1/2)t

Subtract x from both sides of the equation

x – x – 8 + (1/6)t = x – x – 20 + (1/2)t

– 8 + (1/6)t = – 20 + (1/2)t

(1/6)t – 8 = (1/2)t – 20

Add 20 to both sides of the equation

(1/6)t – 8 + 20 = (1/2)t – 20 + 20

(1/6)t + 16 = (1/2)t

 (1/2)t = (1/6)t + 16

Subtract (1/6)t from both sides of the equation

(1/2)t – (1/6)t = (1/6)t – (1/6)t + 16

1/3t = 16

Multiply both sides of the equation by 3

1/3t x 3 + 16 x 3

t = 48 minutes

Therefore, after 48 minutes, Mr. Beecher and Mrs. Carter will first be the same distance away from school.

452,489 to the nearest thousand is 452,000

the anwser is d) 75°

step-by-step explanation: