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Re: HSC 2012 Marathon :) I cleared it up with seanieg89 earlier after my post on the e series proof. I questioned him about this: http://en.wikipedia.org/wiki/Binomial_theorem#Applications I you look at the section Applications, you will see a part called 'Series for e'. Which they then prove...

Well when you find the area underneath the curve when you integrate it and apply the Fundemental Theorem of Calculus: i.e \int_{a}^{b} f(x) \ dx=F(b)-F(a) \\ F'(x)=f(x) When you do this, the +C's cancel out when you subtract them. And because the C's cancel out, people dont put them in for...

4! 2! 2! 2! 2! = Arrangements where all 4 types of clothing are next to their respective type (i.e. AABBCCDD) 4C3 (5! 2! 2! 2!) = Arrangements where 3 types of clothing are next to their respective type (i.e. CDDAABBC or DDCAACBB ...) It is multiplied by 4C3 due to the fact that I choose 3 out...

Re: HSC 2012 Marathon :) Oops, I misread the question aswell :/ I thought it was your standard old word arrangement questions so I just went right ahead and did it..

Re: HSC 2012 Marathon :) Oh nice, Ive learnt a new technique for limits. (Of course substitution is something that can be used in almost anything, but it didn't really occur to me to use it for limits) Thanks

Re: HSC 2012 Marathon :) I am pretty sure I am right, I too may be completely wrong, but I thought, because there are 4 'a's 2 'r's and 2 't's. Because we cant distinguish between the same letters, and there are 4! 2! and 2! ways of sorting each one, I thought I could divide the total number by...

Re: HSC 2012 Marathon :) \frac{10!}{4! 2! 2!} Also Spiral, a Binomial proof one would be nice, (or anything to do with proofs) but anything will do, as long as its not Probability.

$Call the points$ \ \ A(-3k,1) \ \ B(k-1, k-5) \ \ C(k-4, k-5) \\ \\ m_{AB}=m_{BC} \ \ $Collinear points where m refers to gradient$ \\ \\ \frac{k-3-1}{k-1+3k}=\frac{k-5-k+3}{k-4-k+1} \\ \\ \frac{k-4}{4k-1}=\frac{2}{3} \\ \\ 3k-12=8k-2 \\ k=-2 I just evaluate the gradients of AB and BC...

Re: HSC 2012 Marathon :) Nice work, you have just proved the area of a circle. (Im not sure if you would have to prove that a semi-circle is half a full circle first though before saying it) Either way, nicely done, also for part b, its just there so people dont get lost when they try to...

Once you do HSC 2U Math everything will become very very clear. But for now, here is the gradient concept. A gradient is essentially the rate of change of something. Rise over Run is simply the amount of Rise there is per unit of Run, per se. The higher the gradient (i.e. the steeper upwards)...

Re: HSC 2012 Marathon :) $a) Show that$ \ \ \frac{d}{dx} x\sqrt{r^2-x^2}=\sqrt{r^2-x^2}-\frac{x^2}{\sqrt{r^2-x^2}} \\ \\ $b) Show that$ \frac{x^2}{\sqrt{r^2-x^2}}=\frac{x^2-r^2}{\sqrt{r^2-x^2}}+\frac{r^2}{\sqrt{r^2-x^2}} \\ \\ $Hence show that$ \ \ 2 \times \int_{-r}^{r} \sqrt{r^2-x^2} \...

Whats the answer in the book say?

Looking at the OP's subjects, and the response by many about having to be first across the board, if they had a lower than rank 50 school. If I were to get a 99.95, what rankings would I need if I did Math Extension 2: /8 Math Extension 1: /20 English Advanced: /30 English Extension 1: /12...

x=vt\cos{\theta} \\ y=vt\sin{\theta}-\frac{1}{2}gt^2 \\ \\ $when$ \ \ t=\frac{v\sin{\theta}}{g} \ \ \ y=h \ \rightarrow x=c \\ \\ \therefore c=\frac{v^2 \sin{\theta} \cos{\theta}}{g} \\ h=\frac{v^2 \sin^2{\theta}}{2g} I substituted the time when x=c and y=h (the highest point of the...

dangnabbit. I think you have got it then. Im still happy I got the concept of the head pirate not giving to himself by myself. (unless we are both wrong)

I got 201. Since when there are 201 pirates remaining after everyone killed the head pirate since there were 500 pirates since he couldnt satisfy everyone. Now there are 201 pirates left, and the head pirate gives to everyone 1 piece of gold except himself, so there are 100 people with gold...

Is the answer to Pirates 201? I dont want to spam the thread, so Im just asking here why, if its correct I will give my explanation in the thread. (I am assuming that when there are 201 pirates remaining, he gives 100 to everyone he can BUT himself, this ensures his survival, because thats his...

I keep trying, even with small numbers, I the pirates keep dying. One argument that I had was (after deriving it by first experimenting with 50 Pirates and 10 GOld). At 200 pirates, 100 of them recieve gold, and 100 do not, and Pirate #3 would then think, "if I can eliminate #1 and #2, then I...

Ok, Ill take a stab at it, for the first 500 pirates, there will be an overwhelming majority against the distribution, because they didnt get any money. As in, even if lead pirate gave everyone he could 1 piece of gold, then there still will be 400 left unsatisfied. Therefore that night, the...