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A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 15 km/h. Another boat has been heading due east at 20 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?
 2 years ago
 2 years ago
Invizen Group Title
A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 15 km/h. Another boat has been heading due east at 20 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together?
 2 years ago
 2 years ago

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KhaleelA Group TitleBest ResponseYou've already chosen the best response.0
This question is so confusing. the only thing i could come up with was x is 20 (since it travels for an hour) and x' = 20km/h and y'= 15km/h. dw:1353476410429:dw > z^2 = x^2 +y^2 We implicitly derive that equation so it becomes 2zz'=2xx'+2yy' Thats all I have until now, can anyone add to that?
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.0
dw:1353538349121:dw The top equation represents the distance from the dock of the boat heading east at t minutes, the right one represents the distance from the dock of the boat heading south. You can find the distance using Pythagoras' formula This must be a minimum, and as you will notice the distance is a quadratic equation so you should be able to find the minimum :)
 2 years ago
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