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March 29, 2015

March 29, 2015

Posted by **Sandra g** on Thursday, March 15, 2012 at 9:30pm.

- Calculus -
**Steve**, Friday, March 16, 2012 at 11:44amthe distance between the ships at time t is

d^2 = (150-35t)^2 + (25t)^2

at t=4,

d^2 = (150-140)^2 + (100)^2 = 100 + 10000 = 10100, so d=100.5

2d dd/dt = 2(150-35t)(-35) + 2(25t)(25)

at t=4,

201 dd/dt = -700 + 5000

dd/dt = 21.39km/h

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