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The first question is actually quite interesting in that there is several ways that you can think about it. It's also connected to the circle geometry theorem that the angle at the centre of a circle is twice the angle at the circumference.

The answer to this is long and I don't want to type it. It also would not be an enriching and fulfiliing experience for you if I did so. :) That said, I will point you in the right direction (ie. tell you the answer without writing it out :P). Solve the two equations simultaneously (Hint...

Why wouldn't it be? You aren't required to long divide. You are required to factorise polynomials. How you choose to perform that factorisation is up to you!

Housemouse: I remember that question if it is the one I'm thinking of. z = w + 1 / w - 1 where z = cis2@ I think. It would be possible to obtain the solution by realising the denominator but this would be a much, much slower way to do the question... You need to use trigonometric...

I thought I'd start them off on one that didn't have a remainder since it is something that a lot of students actually do struggle with I've noticed. I think the main problem is how much they can remember in their head at any given moment which is sort of what it is. Calculations, addition and...

I'm of the opinion that more capable HMX2 students should practice in being able to put any polynomial function in the form A(x).Q(x) if A(x) is linear simply by inspection. Take the polynomial, p(x) = x3 - 4x2 + x + 6. [Sorry no idea how to do powers on the forum but you get the idea... :)]...

I know everything for I am omnipotent. So, yes, that would include answers to all those questions. :) But I'm sure there are also many capable mathematicians still lurking around who have finished their HSC who could also answer those questions without having seen them before. And...

1. You can't really call that a loophole since it involves you doing the questions, which is the point of a reward scheme in the first place! 2. Oh your poor souls. Hope you enjoy the homeworks I wrote. :P Oh memories.

You've integrated incorrectly - it's not a log, it's a backwards chain rule. You may also find the substitution u = 1 + x^2 easier to work with.

I don't want to post up my entire solution which is bulky at best. I'd like to point out that this question uses a similar idea to the 2003 HSC q8, as well as Bill Pender's Harder 3u inservice from several years ago. With that said, let's do the question. :) i)a) Start with the fact that...

Ok I have worked out where my mistake was. The basic idea of how I was working is fine. You arbitrarily select an original point, and there are then 5C2 = 10 ways to select 2 remaining points. However, THREE of these 10 ways will give you a triangle that does not overlap (ie. the points...

I'm a little busy atm but I found this question too interesting to pass up. However I only took a minute to think about it so I am definitely not sure that what I am about to type is correct. (ii) ANSWER TO PART 2 DENOTED AS X (you will see why in a minute). (iii) For you to have two...

Generally, when you have an integral with a numerator and denominator that look similar you want to transform one. In this case: (x²+1)/(x²+4) = (x²+4)/(x²+4) - (3)/(x²+4) I think you can see where this is going now.

This is indeed a grave oversight on my behalf! :)

No we are situated on Oxford Street in Epping, a few minutes walk from the train station. :)

It appears that as of now many of these posts have already been deleted so this thread may not make a lot of sense to those of you who have just joined us. :)

Hi, My name is Andrew and I work for Intuition Education. This thread is to apologise for the other threads that were made in something like 10 subforums. We were unaware that these threads were being made and do not condone them. Clearly, one of our friend's has seen fit to advertise for us...

Very true :0 Amateur mistake! Disregard what I said :)

Turn it into a circle geometry question. Construct a circle passing through BCB'C'. Label angles C'BB' and CBB' as x (since they are equal as bisecetion is given). Label angles BCC' and B'CC' as y (since they are equal as bisecetion is given). Then it remains to prove that 2x = 2y (or...