Two cars leave an intersection. One car travels north: the other east. When the car traveling north had gone 15 miles, the distance between the cars was 5 miles more than the distance traveled by the car heading east. How far had the eastbound car traveled?

20 miles

Step-by-step explanation:

Given that :

When the car traveling north 'N' had gone 15 miles, the distance between the cars was 5 miles more than the distance traveled by the car heading east

Let the distance moved by the east bound car be e,

therefore, distance between the cars when the northbound car had traveled a distance of 15 miles = e + 5

Using Pythagoras rule:

(Hypotenus)^2 = (adjacent)^2 + (Opposite)^2

(e+5)^2 = 15^2 + e^2

(e+5)(e+5) = 225 + e^2

e^2 + 5e + 5e + 25 = 225 + e^2

e^2 + 10e + 25 = 225 + e^2

e^2 - e^2 + 10e = 225 - 25

10e = 200

e = 200 / 10

e = 20 miles

Check attached picture for solution diagram

step-by-step explanation:

-22

times both sides by 2, times () by 8, add alile terms, move terms, add alike and caculate divide both sides by 3