calculus help please
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dy/dx=2y^2 and if y=1 when x=1, then when x=2, y=? (you already posted this, but i still don't get it) dy/dx = 2y^2 Integrating...y=2/3 y^3 + C put 1,1 into the equation, and solve for C c=1/3 so the equation is y=(2y^3)/31/3
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I have got help from Damon on this question but all I need is help on integrating the answer he gave me into my original equation: y=15/20.25(x+1.5)^2+0 Here's Damon's help: (x+1.5)(x+1.5) = x^2 + 1.5 x + 1.5 x + (1.5*1.5) = x^2
asked by Alex on December 4, 2014 
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have got help from Damon on this question but all I need is help on integrating the answer he gave me into my original equation: y=15/20.25(x+1.5)^2+0 Here's Damon's help: (x+1.5)(x+1.5) = x^2 + 1.5 x + 1.5 x + (1.5*1.5) = x^2 +
asked by Alex on December 4, 2014