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March 31, 2015

March 31, 2015

Posted by **Sasha** on Wednesday, December 15, 2010 at 11:28pm.

- calculus -
**Reiny**, Thursday, December 16, 2010 at 12:34am3x^2 + 4y dy/dx + 3xdy/dx + 3y = 0

dy/dx(4y+3x) = -3x^2 - 3y

dy/dx = (-3x^2 - 3y)/(4y+3x)

at (1,1) dy/dx = (-3-3)/(4+3) = -6/7

y-1 = (-6/7)(x-1

times 7

7y - 7 = -6(x-1)

7y - 7 = -6x + 6

6x + 7y = 13

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