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November 26, 2014

November 26, 2014

Posted by **Thien Nguyen** on Friday, April 26, 2013 at 1:04am.

- math -
**Steve**, Friday, April 26, 2013 at 10:55amdraw a diagram. The distance from the observer to the shuttle when the height is y, is

d^2 = 6^2 + y^2

2d dd/dt = 2y dy/dt

when y=8, d=10 so

20 dd/dt = 2(8)(500)

dd/dt = 400 km/hr

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