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Exactly what i did. Q1-4 only took around 30 minutes, Q 5 things started going down hill, but things got a bit better with the projectile question.. binomials raped me hard (except induction i think i got that) hoping for 60+/84

That would actually make sense... there were some articles on SMH about the lack of teaching prowess math teachers have too. regardless. it's cruel of them to give 2009 the hardest papers in recent years. 2u and 3u i talk from experience general and 4u i hear from my friends doing said courses

um, r was 0.9 not 1.9. you can't have a negative limiting sum. the answer is 12

yeah that was my point ahaha, 90+ as in... way more than 90 ;) you only get the marks if you have no answer written down. since you wrote the answer for 3 function values they will mark that. so you probably wont get the full marks.

please tell me you're joking. read what you just posted.

my bad, forgot to write that in, had to go eat ;) thanks for that

This is correct. and in 9)b i) P to R is just 5 * 1000 = 5000 R to S is 3 * 2600 = 7800 Total =12800 Q and R are ON the shore line, not above ii) P to S, find hypotenuse which is Root 34 root 34 * 2600 = iii) not hard, just find PQ and QS using pythagoras' theorem iv) minimise it by...

to be cocky is merely to be very self confident. also, in case you didn't notice, rank 100 is still, 100th. think about James Ruse for a second. Those kids get 100% in almost everything and have an average UAI of 90+ i'd get your rank 1 math guy to do a couple of james ruse trials. think...

Um, what he said was correct, 6 sub intervals relates to h= b-a/2 and when he said 7 function values he was referring to the (f + l + 4(odd) + 2(even)) in that you use all 7 values. I concur. (Albeit my 3rd unit of math isn't counting in my top 10) but I'm just wondering why you did the 2...

This makes sense, as long as you make it clear that "E" and "F" are the midpoints you should be fine, you must have a right angle for "AC" though, in order to make it relate to parts c) i) to c) iv) so i suppose you don't HAVE to have right angles in the smaller triangles

I can't believe so many people couldn't do this question! what do you think c) i) to c) iv) were for? simply construct a line from the top corner to the hypotenuse, and you've got two of the triangles like the one right above it, then construct line bisectors in both and voila. Here:

Yeah, I think Cataclysm's got it going on, ends up being: 2536 to the nearest year