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September 16, 2014

September 16, 2014

Posted by **Anonymous** on Saturday, April 3, 2010 at 8:21am.

- Calculus -
**bobpursley**, Saturday, April 3, 2010 at 9:16amdraw the figure, at noon, then at 4PM.

At 4 pm, I have a right triangle of A,B of sides (base 10+19*4; altitude 15*4)

d^2=(10+va*t)^2 + (vb*t)^2

2d dd/dt=2(10*va*t)va+2(vb*t)vb

so when t=4, solve for dd/dt

- Calculus -
**Anonymous**, Tuesday, October 5, 2010 at 9:11pm9898

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