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School rank is ~25. This year we have around 170-180 people. I don't think it changes a lot every year.

We do better at maths and science I think. For maths, around half the cohort usually gets a band 6 (~60 for both 2u, 3u and ~25 for 4u last year) and for science it ranges from a third to a quarter (~25 for bio, phys and ~40 for chem last year).

Yea they are. Doesn't help when your teacher disappears after trials either.

re: HSC Chemistry Marathon Archive Its in the syllabus, so its a possibility I think. You are required to "gather information from secondary sources by identifying practising male and female Australian scientists, the areas in which they are currently working and information about their...

Re: HSC 2015 3U Marathon Is this correct for induction? \begin{align*}\textup{RTP }\frac{\mathrm{d}}{\mathrm{d}x}x^{n}&=nx^{n-1} \\\textup{1. Show true for }n&=1 \\f(x)&=x^1=x \\\textup{LHS}&=\frac{\mathrm{d}}{\mathrm{d}x}x \\ &=\frac{f(x+h)-f(x)}{h}\\&=\frac{x+h-x}{h} \\&=\frac{h}{h} \\&=1...

Re: HSC 2015 3U Marathon \begin{align*}f(x)&=x^n,\,f(x+h)=(x+h)^n \\f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\&=\lim_{h\to0}\frac{(x+h)^n-x^n}{h} \\&=\lim_{h\to0}\frac{{\color{Blue} \binom{n}{0}x^nh^0} + \binom{n}{1}x^{n-1}h^1 + \binom{n}{2}x^{n-2}h^2 +...+\binom{n}{n}x^0h^n{-\color{Blue}...

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Didn't you make these questions yourself Jono? :P

Just in case you were unaware, it is a 3D diagram. Have a read through this: Isosceles Triangle Theorems Picture of the question:

It is when the molecules configuration of atoms is unique when viewed in 3D. For the last and third ones, you have one chlorine atom on each carbon. The following are all the same - they all have one chlorine atom bonded with each carbon atom: ......H H Cl - C - C - H ......H Cl ......H...

Your first and last ones are both 1,2-dichloroethane. There is one chlorine atom per carbon for both.

There are 2. You can either have both chlorine atoms on one carbon or one chlorine atom on each carbon.

Re: HSC 2015 2U Marathon 2f(x)+3f(-x)=15-4x$ (1).\\ Let $x=-x \to 2f(-x)+3f(x)=15+4x$ (2). $\\$Now solve these equations simultaneously$.

You should post here: http://community.boredofstudies.org/482/tutoring-classifieds/

\begin{align*}\frac{A_2}{A_1}&=\frac{\pi r^2-\frac{1}{2}r^2(\theta-\sin{\theta})}{\frac{1}{2}r^2(\theta-\sin{\theta})} \\&=\frac{\frac{2\pi-(\theta-\sin{\theta})}{2}}{\frac{\theta-\sin{\theta}}{2}} \\&=\frac{2\pi-\theta+\sin{\theta}}{\theta-\sin{\theta}} \\\end{align*}

A_1$ is the major segment and $A_2$ is the minor segment. $A_1$ is given by the formula $\frac{1}{2}r^2(\theta-\sin{\theta})$. $A_2$ is the the area of the circle less $A_1$. So $A_2=\pi r^2-\frac{1}{2}r^2(\theta-\sin{\theta})$. Simplifying $\frac{A_2}{A_1}$ will give you the answer.

$Did you end up with $\arcsin{(3x+1)}=\arccos{x} \to \arcsin{(3x+1)}=\arcsin{(\sqrt{1-x^2})} \to 3x+1=\sqrt{1-x^2}? If you squared at this point, you end up with incorrect solutions. Assuming you did, you will get x=0,-\frac{3}{5}. If you sub x=-\frac{3}{5} into 3x+1=\sqrt{1-x^2}, you'll see...

Ah okay, I didn't realise they weren't the original. I've attached what I think are the original ones.

Look here http://www.boredofstudies.org/view.php?course=10.

You might have better luck looking at the ATAR Notes website.