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I didn't get the question out... because my school gave a replacement question. Anyone have the q?

Hate to fluff it for you, but we got that on our first induction hw sheet. :p n(n-1)(n+1)(3n+2)/24 + (n+1)(1+2+3...+n) using assumption = n(n-1)(n+1)(3n+2)/24 + n(n+1)^2/2 = n(n+1)[(n-1)(3n+2) +12(n+1)]/24 = n(n+1)[3n^2 +11n +10]/24 = n(n+1)(n+2)(3n+5)/24, as required.

5/14 80/243

That's right. I managed to get it wrong by saying 2 times 2 is 3 instead of 4. How hideous.

m111: Great! Now there's just the oh-so-little matter of getting accepted.

x<0, x>= 1 damn I am bored. edit: integ: sec^2x - secxtanx

All you needed to say was there is a double root at point of contact where the curves are tangential. (edit: of course showing that the eqn happened to be solving the 2 curves simultaneously. ishq: yes)

*shakes head*

to the people above: If you made no mistakes, you'd get 120/120 obviously! lol

No school worth their salt. Teachers know about this web site too you know...

They should get estimates.

lol the restriction took me a while to get out though... ARRGH... 2 times 2 is 4, not 3 =/ Seems like a typical CSSA paper. Q8 is easiest, Q1 is hardest, etc.

Food chain :p Quite a few out there I'm sure can do the CSSA 4U with 50 min to spare and a 95% mark...

"Star" Commerce and Math student with a target UAI of 95.5. *shakes head* Arts degrees should be compulsory. Refer to case study: Albert Speer.

I could say something mean but I won't. :p

Save your advertising for people who care.

The q with the functions.t = f(t)

It's 80. =p

xy + yz + zx = 5 (using the results they gave you) then 1/x + 1/y + 1/x = (xy + yz + zx)/xyz = 5/xyz or 5/xy (by substituting xy = z^2 in an earlier form of the expression) so z = 1, xy = 1, answer is 5.

I put the wrong answer for that one =/ Shocking. The polynomials q was a normal 4u question though.