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find Particular antiderivative satisfy the given condition R'(x) = 600 -0.6x; R(0) = 0

Mathematics
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Integrate. R(x)=600x-0.3x^2+C. R(0)=C, so C=0.
got you.. that's what I thought but I was not confident enough to determine if it is right
Ah, well, there you go.

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you are fast let me post another question
so can I show you one I did to make sure I really got it
Sure. can you fan me again as well? :D
dR/dt = 100/t^2 R(1) = 400 so I came upt with R= 100^t +500
I did it earlier..I can do it again
it won't allow me to fan you again.
You should be able to do it once per question, I think. Anyhow, you did it correctly.
Ok.. I will post another question then I wil fan you

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