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Question 438277: two cars left and intersection at the same time. One traveled North, the other traveled 14m farther but to the east. How far apart were they then, if the distance between them was 4mi more than the distance traveled east?
please help me solve this!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! two cars left and intersection at the same time.
One traveled North, the other traveled 14m farther but to the east.
How far apart were they then, if the distance between them was 4mi more than the distance traveled east?
:
This is a right triangle problem, distance between the cars is the hypotenuse (h)
:
let d = distance traveled north
It says,"the other traveled 14m farther but to the east." therefore:
(d+14) = distance traveled east
It says,"the distance between them was 4mi more than the distance traveled east? "
Therefore
h = (d+14)+ 4
h = (d+18)
:
So we have:
d^2 + (d+14)^2 = (d+18)^2
FOIL
d^2 + d^2 + 28d + 196 = d^2 + 36d + 324
Combine like terms on the left
d^2 + d^2 - d^2 + 28d - 36d + 196 - 324 = 0
A quadratic equation
d^2 - 8d - 128 = 0
Factors to
(d-16)(d+8) = 0
Positive solution
d = 16 min north
then
16 + 14 = 30 mi east
and
16 + 18 = 34 mi is the distance between the cars
:
:
See if this checks out:
16^2 + 30^2 = 34^2
256 + 900 = 1156
:
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