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February 1, 2015

February 1, 2015

Posted by **Tiffany** on Monday, April 2, 2012 at 10:33pm.

- calculus -
**Reiny**, Monday, April 2, 2012 at 11:41pmLet the coordinates of the ship be P(x,y)

so AP =2 and BP=5

√(x^2+y^2) = 2

x^2+y^2 = 4 , (#1)

√( (x^2 + (y-6)^2 ) = 5

x^2 + y^2 - 12y + 36 = 25

x^2 + y^2 - 12y = -11 , (#2)

#1 - #2 :

12y = 15

y = 15/12 = 5/4 = 1.25

then in #1:

x^2 + 225/144 = 4

x^2 = 39/16

x = ± √39/4

The ship could be at (√39/4 , 5/4) or (-√39/4 , 5/4)

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