Integral of 2sin(2x) is -cos(2x)
from 0 to pi/2 (sub pi/2 minus sub 0 for x) and then solve
= -cos(2*pi/2)-(-cos0)
= -cos pi + cos 0
= 1 + 1
= 2
I think I got it, but just in case...tell me the whole thing again. I wasn't listening.
B. Is the right answer
First you have to take the common elements then use an identity/formula to get the rest
x^3 - 3x^2 + x-3
x^2 (x-3) +1 (x-3)
(x^2 +1) (x-3)
(x-1)(x+1)(x-3) {using a^2-b^2 on x^2-1^2}
Sorry, not sure how to describe it, but the lower the temperature, the higher the hot chocolate sales.
I mean yeah I'd like to befriends but isn't this a studying app?
