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October 20, 2014

October 20, 2014

Posted by **Kelly** on Saturday, July 17, 2010 at 4:13pm.

Equation:35x^2+20y^2=1

Given that dx/dt=4, find dy/dt

when (x,y)=(1/7sqrt2, 1/5sqrt2)

- Calculus -
**Damon**, Saturday, July 17, 2010 at 4:58pm70 x dx/dt + 40 y dy/dt = 0

280 x + 40 y dy/dy = 0

280/7sqrt2 + 40/5sqrt2 dy/dt = 0

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