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YES! I miss 4 unit already.. I'm sitting here studying 3 unit for tomorrow and I'd come past a formula and say "ooh, that would be helpful in 4- oh..." :(

yes but that was a different society. As I think someone has already said, kids are much more rebellious now.

reminds me of that article about a teacher killing a blind student because he couldn't read the koran.. or something like that. Um yeah no. It's a form of violence and violence, for any reason, is not needed anywhere, and definitely not in schools. It's teaching kids that big people can hit...

lol my mistake.. I tend to class everything of this sort under that hideous topic..

okay i worked out that for each of the 5 men, there are 4 ways they can be paired with a woman that is not their wife. now for the other spot for a man, there are 4 possibilties, and there are 4 ways each of the 4 men can be paired with a woman that is not his wife. can anyone convert this to...

are you working out the probability? cause your answers are fractions.. and it asks for arrangments what i understand from the question is it's asking how many ways can you make a pair of two couples from 5 married couples. but the couples can't be married.

well that's what i was doing.. and then divide by 2 since Ma + Wb playing with Mb + Wa is doubled cause it takes into consideration the side of the courts... i think...

lol that gets a fraction.. and not close to the answer either.. i'll try call the hsc dudes when the lines open In the mean time.. anyone who gets it will get a cookie :p

Catholic trial 1988 curse those trials.

nope.. *joins in* :(

In a tennis club, there are five married couples available to play a “mixed doubles” match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In how many ways can a group of four persons be chosen for this match if...

if they ask for general form, i'm pretty sure you make everything = 0 if they don't specify, simplest form.. which would be either in your eg above.

Re: more maths problem a) monotonic decreasing meaning your gradient, ie, f'(x) < 0 b) for stationary points you make f'(x) = 0 3. f(0) = 1 therefore y-intercept at y = 1 f'(x) > 0 for 0<= x <= 3 meaning the gradient of the curve is positive between x = 0 and x = 3 f'(3) = 0, well you...

Re: maths problem (cosx)^2 - cos 2x = 0 (cosx)^2 - [2(cosx)^2 - 1] = 0, using the double the angle rule (cosx)^2 - 2(cosx)^2 +1 = 0 (cosx)^2 = 1 (cosx) = +-1 x = 0, pi, 2pi

no x^2 = 4 x = +- root4 x = +- 2 because as I said before, you sub in either x =2 or x = -2 into the original equation and you'll still get 4 there is however a bit of a contradiction but but it only creates more confusion and it doesn't matter.. just learn it and get the marks :) so...

yes that is true. my mistake. I was trying to point out that it becomes a semi circle because you have to neglect the bottom half since y >= 0. when you have x^2 = 4, then x = +-2 because if you sub either value, you get 4 but if you have x =root4, x must be 2

well for y = root(4 - x^2), since you have a root.. whatever's under the root MUST be > or = 0 if you solve 4 - x^2 >= 0 you get -2 <= x => 2 now, the root of anything always has to be positive, correct? and under the root COULD = 0 so y > or = 0.. so yes it is a circle but your range is y >=0...

it actually takes much less time tracing than anything else and you get a perfect diagram so i suggest tracing.. and I suggest drawing it too. Why would you risk anything in your hsc?

Always, always, ALWAYS do as MUCH working out for curve sketching; x-intercepts, y-intercepts, label every stationary point, inflexion point, EVERYTHING. That is, unless your running out of time, then do the bare minimum, but that includes testing for concavity anyway.

Yep, precisely.