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anonymous
 5 years ago
Factor
5x^3 + 10x^2 + 15x
anonymous
 5 years ago
Factor 5x^3 + 10x^2 + 15x

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is prime. It cannot be factored. If you graph this you will see the graph never crosses the xaxis. to see this you might try using quadratic formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre an idiot, you can factor out x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha very true. sorry i was looking at factoring the remaining part after the x had been factored out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you good at surface area integral? 2pi y dS

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0somewhat i haven't used that too much

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, i have a question posted on the left . its about a torus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Surface area of a torus which has radius r which is generated by the circle centered at (R,0) with radius r. And assume R>r. So the equation of the circle is (xR)^2 +y^2 = r^2. FOr some reason my integral does not work. i have integral 2pi y dS, where y = sqrt [r^2  (xR)^2], and ds = sqrt ( 1 + dy/dx ^2). dy/dx = 2(xR)/ (2sqrt(r^2  (xR)^2)). anyways, integration limits , the lower limit is Rr, and upper R+r

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer i should get is 4pi^2 r*R

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my integral is int sqrt [r^2  (xR)^2] sqrt [ 1 + [2(xR)/ (2sqrt(r^2  (xR)^2))]^2 .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the setup looks correct as do the functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it prob is a computational error, thats a pretty big integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that last part sqrt[] all squared so it cancels out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm nevermind i thought i could reduce it for you. as far as the integration you'll have to find someone else to help you or use some graphing software. sorry
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