Recent content by paddy@2011

hey conman, buchanan is right, your site is good. what's your connection to sydney boys?

here is a worked solution in the attachment.

P(x) = 5X^4 - 11X^3 + 16X^2 - 11X + 5 = 0 Divide both sides by x^2 and we get 5x^2 - 11x + 16 - 11/x + 5/x^2 = 0 5(x^2 + 1/x^2) -11(x+1/x) +16 =0 [Now if u = x + 1/x, then x^2 + 1/x^2 = u^2 - 2] so the above equation becomes 5(u^2 - 2) -11u + 16 = 0 5u^2 - 11u + 6 = 0 (5u - 6)...

sin(x)*cos(x)^3-cos(x)*sin(x)^3 =sin(x)cos(x)[cos(x)^2-sin(x)^2] =1/2*sin(2x)*cos(2x) =1/4*sin(4x) cheers

that is bull shit about courses being finished by the trials. pardonne-moi le francais. teachers don't have to be finished by then. trials are designed to give you practice with formal examinations with the correct exam time lengths etc... you should be finished by the end of term three...