Search results

oh.. it's 96 for 3U and 97 for 4U,

*nods* in other words 'mathematics extension 1' and 'mathematics extension 2' will each be counted as a 2 unit course.

the teacher would then start to strongly recommend some students not to continue in the course

prelim marks don't count

Just do it in order, you ARE planning to finish the paper aren't you?

it will bite

error made on second last line

*prays so they're not gonna replace it with an extended response section*

I've seen solutions to the fitzpatrick book, my year 12 teacher has them. Think it's a collection of handwritten booklets

48% is alright, and some teachers do set crazy assessment tasks (ie turtle level). statistics show that ppl doing 12 units consistently performed slightly better than those who studied 10 units (students were divided into groups according to their school certificate marks and compared).

Slow down on your reasoning steps, science teachers tend get confused when you jump. And.. you should consult the english department for advice on this situation.

There's almost no positrons in a CRT unless you introduce it. canal rays are positive particles moving from anode to cathode, they are not electrons/positrons

classic one: we know the apparent brightness and the distance of stars to us, the inverse square law allows us to tell how bright the stars really are.

Bashing does work.. think there's a smart way which I forgot. gradient of tangent = b/[asin(t)] using that to solve for Q you should get: Q=(a/e, u) where u = b[ tan(t) - cosec(t)*(sec(t) - 1/e)] gradient(PQ)*gradient(SQ) ={ [btan(t) - u]/[asec(t) - a/e] } * { u/[ a/e - ae] }...

:D

you see the problem is, the UNSW student card's quality is much better than the Usyd one

The probabilty of solving a question in Fitzpatrick > probabilty of solving an arnold question. ofcourse there are deviations

commerce/science UNSW probably will make it engineering+commerce if they decide to allow the program

98 and 97 :S and you? I am so happy that someone still remembers me, just came back from a long trip :D

what about b(n) = {1/[a(n) - L]} + L where L is the limit of a(n) then it doesn't have to converge to 0. (every convergent series is then linked to an unique divergent series)