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November 28, 2014

November 28, 2014

Posted by **nesri** on Tuesday, October 5, 2010 at 8:06am.

- calculus -
**Reiny**, Tuesday, October 5, 2010 at 9:36ammake a diagram, you will have a right-angled triangle

let the horizontal distance covered by the shark since t=0 be x m , and

let the distance between the shark and the lifeguard be y m

x^ + 90^2 = y^2

2x(dx/dt) + 0 = 2y(dy/dt)

when y = 150

x^2 + 90^2 = 150^2

x = √14400 = 120

dy/dt = 2x(dx/dt)/(2y)

= 2(120)(30)/300

= 24 m/sec

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