Question 160302
two ships leave port, one sailing east and the other south. Some time later they are 17 miles apart, with the eastbound ship 7 miles further from the port than the southbound ship. how far is each from port?
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If you look at this, you can see that the hypotenuse of a right triangle is the
distance between the ships. The two legs are the distance from the ships to port
A rough diagram will make it easy to understand.
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Let x = distance of the southbound ship from port
then
(x+7) = distance of the eastbound ship from port
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Hypotenuse = 17 mi
:
x^2 + (x+7)^2 = 17^2
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FOIL (x+7)(x+7)
x^2 + (x^2 + 14x + 49) = 289
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Arrange as a quadratic equation which we can solve:
x^2 + x^2 + 14x + 49 - 289 = 0
:
2x^2 + 14x - 240 = 0
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Simplify, divide by 2
x^2 + 7x - 120 = 0
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Factor
(x - 8)(x + 15) = 0
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Positive solution
x = 8 mi is southbound ship
and
8 + 7 = 15 mi is the eastbound ship
:
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Check solution in pythag
8^2 + 15^2 = 17^2
64 + 225 = 298
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Did this make sense? Any questions?
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