Search results

I'm not too sure I have screwed up in the same way that you have. I go to a small school, with little competition in the way of students who would improve my internal marks if I had a bad day in the HSC (although we did have a 99.90 kid and about another half dozen in the 99s). I did well...

99.95 :) Super relieved after seriously screwing up a couple of exams.

mmm. I think my solution is actually shorter than that, however it is more hacky and less 'clean'.

if your proof is anything like mind, a post online isn't realistic. Even if i was computer savvy it would require too many diagrams and explanations. The approximate process is the following: 1. do what realisenothing did and show that they cannot be arbitrarily close 2. construct a diagram with...

realisenothing: i think you have to be a bit more rigorous in your proof that there cannot be 8. you are correct that if the two individuals who get shot are arbitrarily close then we have that contradiction because you are basically considering a 'revolution' around a single point - however in...

obviously we cant go above 8 because there must be an absolute shortest distance between any two girls in the field (i don't really know what happens when there is equality in distances but it doesn't really matter for this idea). for this absolute shortest distance, both girls shoot at each...

Overall the paper was straighforward up until the last question. Q1-15 were generic textbook stuff (bar 1 question which required a bit of thinking but was only worth 2 marks and wasn't hard when you worked out what you were doing - i.e. finding coordinates of C ). Last question was mostly...

sort of similar. The 'basel' element was only in the final 2 marks. in fact, parts i - v all culminated in the result in part v. You could do part vi independently (using only the result from (v))

from memory: 16b) i) 1 mark algebraic proof --> super easy ii) recurrence formula for 3 marks using (i) identity... something like prove I(2n+1) = (2n/(2n+1)) x I(2n-1) iii) prove I(2n+1) = (2^n * n!)/(1x3x5x...x(2n-1)) (2/3marks, can't remember) iv) i couldn't even get close to typing this one...