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Re: 回复: Re: Answers Sorry, I still don't quite agree with you. I think you might be mixed up with x-> 0 of sin(x)/x = x Because as x->0, sin(x) -> 0. Then again maybe I'm just stupid? :-P

Ehh... I cbf doing the integrals right now... but for the volumes 3c each cylinder has height y, radius x NOTE: A cylinder is 2pi(r) [The circumference of the circle] * h [The height of the cylinder] therefore each is 2pi(x)(y) 2pi e-INT-1 (x)(y) dx 2pi e-INT-1 (x)(lnx/x) dx 2 pi e-INT-1...

Which ones?

6biv -0.2v^2 is opposing gravity, so it would be air resistence.

5di was kind of annoying... Basically, BCP and BAP are similar coz they have equal sides/angles whatever. And for a 4 sided polygon each interior angle is (n-2)180/5 degrees = 3pi/5 Therefore Angle BAP = 3pi/5 but bap and ead are pi/5 therefore angle cad = pi/5. Also, Side AC = AP+PC And...

Re: 回复: Re: Answers For the rain drop, using v^2 = 49 (1 - e^[-2x/5]) as x -> infinity e^[-2x/5] -> 0 So v^2 -> 49 v -> 7 EDIT: I also got 0 for the last question. Because - As n -> Infinity sin(pi/2n) -> 0 therefore 2(nsin(pi/2n)^2 -> 0 Think about it logically... Why would a...

How did everyone go? How did everyone go in that? It was MUCH easier than I expected... Adding up all the marks I was certain about I got 90/120... Which would be awesome :-)

:bomb: Is anyone else wishing they had just stuck with 3unit? Anyway, good luck to everyone!

We had a guy leave the exam like an hour in... He's been bottom of the class since yr 11 though :-P

Re: 回复: Re: 2007 HSC Maths Ext2 UnOfficial Discussion Thread xtan(x) would involve recurrance wouldn't it? correct me if I'm wrong anywhere here... IBP- let u = tan (x) v' = x u' = sec^2 (x) v = 1/2 x^2 1/2 (x^2 * tan (x)) - 1/2 Int ( x^2 * sec^2 (x) ) Using IBP again...

Re: Exam Thoughts That was so much easier than I expected. It took like an hour to finish the exam. I was considering leaving early, but I'm glad I didn't, as everytime I went over the paper I found an extra point I could add, found silly mistakes etc.

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