SOLUTION: 2 cars leave intersection, one travel east; other north. When the car traveling east had gone 15 miles the diatance between the cars was 5 miles more than the distance traveled by
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Question 232505: 2 cars leave intersection, one travel east; other north. When the car traveling east had gone 15 miles the diatance between the cars was 5 miles more than the distance traveled by the car heading north, how far had the north bound car traveled. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 2 cars leave intersection, one travel east; other north.
When the car traveling east had gone 15 miles the distance between the cars was
5 miles more than the distance traveled by the car heading north, how far had
the north bound car traveled.
:
this is a right triangle problems a^2 + b^2 = c^2
:
Let a = 15; the distance by the Eastbound car
Let b = distance traveled by the Northbound car
:
The distance between the cars is the hypotenuse
"distance between the cars was 5 miles more than the distance traveled by
the car heading north,", therefore
c = b+5
:
The equation:
15^2 + b^2 = (b+5)^2
:
FOIL the right side
225 + b^2 = b^2 + 10b + 25
:
Subtract b^2 from both sides
225 = 10b + 25
:
Subtract 25 from both sides
225 - 25 = 10b
200 = 10b
b =
b = 20 miles the distance of he northbound car
;
:
See if this is true
a=15
b=20
c=25
:
15^2 + 20^2 = 25^2
225 + 400 = 625!
:
How about this? Did it make sense to you?
