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Mathematics 13 Online
OpenStudy (anonymous):

Two cars leave an intersection at the same time. One travels north and the other travels east. When the eastbound car is 1 mile further than the northbound car from the intersection, the distance between them is 2 miles more than the northbound car's distance from the intersection. How far is the northbound car from the intersection?

OpenStudy (anonymous):

use the pythagoras theorem :\[c^2 = x^2 + y^2\] where c = 2 miles x = 1 mile so plug this in the theorem and you'll get the following: \[(2)^2 - 1 = y^2\] \[y = \sqrt(4-1) \rightarrow y = \sqrt(3)\] I think so, correct me if I'm wrong ^_^

OpenStudy (anonymous):

I think you made a mistake :)

OpenStudy (anonymous):

I concur.

OpenStudy (anonymous):

If the distance of the northbound car from the intersection point is x, then the distance of the eastbound car from the intersection is y+1, and the distance between the two cars is 2+y.

OpenStudy (anonymous):

So, \[(2+y)^2=y^2+(y+1)^2\]

OpenStudy (anonymous):

Solve the quadratic equation, you get y=-1 or y=3. Take only the positive value of y. Therefore the northbound car is 3 miles away from the intersection.

OpenStudy (anonymous):

I should go prepare for my exam, or I am definitely getting a very bad grade.

OpenStudy (anonymous):

why is it 2 + y?

OpenStudy (anonymous):

oh, right, nvm lol, I again misread the question, sorry ^_^"

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