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Ok, here's my situation. I've already attended all classes up till now in B Planning @ UNSW but I may have to defer for 1 year (for reasons out of my control)... Will there by any academic penalties for deferring this late? P.S I'm not really concerned about the financial penalties at...

Family and World Order I find family somewhat interesting but it's jam packed with content :mad: World Order needs more out-of-school research than any other topic; a bit hard to keep up

Cleopatra VII Roman Society from Augustus to Titus Fall of the Roman Republic Augustus and the Julio-Claudians The test was pretty straightforward but the family question in Roman Society kinda threw me off...I knew the stuff, just didn't any sources and I had to resort to name-dropping in...

this is one of the most naive comments i've ever heard. no matter what happens, the system works so that you get what you deserve

I found the questions to be on par with most past papers but I think the problem is the LENGTH. About 10 more minutes would've made a lot of difference to me I'm looking at about 55/84 on this one.

Depends on what the judges think of the exam.

He's right on the money :) I expected MUCH more from a q10.

Yup. I got 1:2root3 as well...

The test had some aspects that was harder than previous years, but it was definitely easier than last year's paper.I'm capped at around 114/120 but I took a slower approach than usual, and answered the rest with some confidence so I'm hoping for about 107-110/120, which would get aligned pretty...

NO. For finding an unknown angle, b and c are the sides that "forms" the angle. a is the one that is "opposite" to the angle.

It's not too late. Take a look at past papers and look for trends. Then take a complete gamble with it if you're really that screwed. The topics with most marks allocated to: -Plane geometry -Integration -Coordinate geomatry -Differentiation/Applications of Differentiation -Sequences...

very few trial exams have q10's that 'up to standard' of HSC q10s.

The universal one is: h/3 [ f(A) + f(B) + 4*(first one after a) + 2*(the second) + 4* + 2* +...], where h is B-A/n. This should get you a corect answer every time. But even more importantly, you only have to use it once! My opinions on the Simpson's Rule: This is very very poorly taught by...