SOLUTION: 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.
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-> SOLUTION: 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.
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You can put this solution on YOUR website! 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
;
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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You can put this solution on YOUR website!
As you can sketch they form a right triangle ---> Ship1 goes East and Ship2 goes South with 17 miles apart (hypotenuse).
Then, is miles fromt the port. (adjacent), that depends of course which angle of inclination. For argument sake we put adjacent.
Also, is miles farther from the port than (opposite)
By Pythagorean Theorem: --------------> working eqn divide whole eqn by 2 perfect square ,
use , distance of from the port , distance of from the port
check via working eqn:
Thank you,
Jojo
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