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do you remember the question? re-draw?

Lucky

Because x^2 + y^2 = r^2 is centred at the origin. And to translate the centre h units to the right, we replace x with (x-h) and k units up, we replace y with (y-k), hence why the circle centred (h,k) has equation (x-h)^2 + (y-k)^2 = r^2 As for why we replace x with (x-h) for a horizontal...

Yeah for 2U

$$ \sqrt{1+5\sqrt{1+6\sqrt{1+7\sqrt{1+\ldots}}}} = \ ?

$\noindent Let $u = \sqrt{x+\sqrt{x+\sqrt{x+...}}} \implies u^2 - u - x = 0 \implies u = \frac{1+\sqrt{1+4x}}{2}$, taking only the positive answer since $u$ is a positive square root. \\\\ $\int_{2}^{12}\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ d$x = \frac{1}{2} \int_{2}^{12} 1+ \sqrt{1+4x}$ d$x =...

Coordinate geometry is definitely not the easiest, nor is it free marks. There are enough tools in coordinate geometry to make a challenging question (compared to superannuation). Superannuation is the same thing with different numbers most of the time, I've only seen a handful of exceptions...

Probability Superannuation is the same every time

Re: HSC 2017 MX2 Marathon For higher n, there becomes multiple ways to construct the star, each with their own angle sum. ie. the points of the star may create an n-gon, or another star, as shown for n = 7, 11. Note in each case, the first star creates an n-gon by joining every second vertex...

10,000 people 7,500 German 4,500 French 1,000 Neither If N people speak both, (7500 - N) + (4500 - N) + N + 1000 = 10000 N = 3000 people

cos(x) is an even function so cos(x) = cos(-x). If you have 7pi/3 and 11pi/3, you must also have -7pi/3 and -11pi/3.

Re: HSC 2017 MX2 Marathon $\noindent $ \begin{align*}S_{(n,k)} &= k + (k+10k) + (k +10k+100k)+ \ldots + (k+10k+\ldots+10^{n-1}k) \\ &= \frac{k(10^1-1)}{9}+\frac{k(10^2-1)}{9}+\frac{k(10^3-1)}{9}+\ldots+\frac{k(10^n-1)}{9} \\ 9S_{(n,k)} &=...

$\noindent By inspection it is $y = x + 4$ but we can also use the perpendicular distance formula using $P(x,y)$, $x - y + 5 = 0$ and $x - y + 3 = 0$. \\ \begin{align*}\quad \frac{|x-y+3|}{\sqrt{2}} &= \frac{|x-y+5|}{\sqrt{2}} \\ |x-y+3| &= |x-y+5| \\ x - y + 3 &= -(x-y+5) \\ y &= x+4 \end{align*}

What did you write in the exam? Are you sure the answer is sqrt(3/2)? I put -sqrt(3/2) because Q was defined to be in the third quadrant and Q(2q, 3/q) so q could not be defined for positive values. And L = 16q^2 + 32/q^2, so a negative q does not yield a negative length L.

What did everyone get for the last question and the verticle height of M (wheel) question? Also I recall the volume question having a weird answer.

Are they allowed to accept two different answers?

Yeah the wheel question was the train track question, I made a trapezium connecting O, M', H and M where M' was the new position of M and M'H was the vertical height of M'. Then used trigonometry to find M'H

The train track question was good, so was the one about the point M(4, -1) lying outside the circle. Q14 was harder than Q16 I think. Was the financial question n = 12 years or n = 13 years, because she can't make a 13th withdrawal of $M without the amount remaining becoming negative?

$\noindent $\sec{\theta} = 3 \implies \cos{\theta} = \frac{1}{3} \implies \theta = 2\pi n \pm \cos^{-1}{\left(\frac{1}{3}\right)},$ $n \in \mathbb{Z}$

$\noindent \textbf{Q1.} Should be written $\frac{f(5)-f(1)}{f(2)}$, hence why they said $f(2) \neq 0$. Now if $b^2f(a) = a^2f(b)$ then $\frac{f(a)}{f(b)} = \frac{a^2}{b^2}$. \\ \begin{align*} \quad a = 5 \text{ and } b =2 \implies \frac{f(5)}{f(2)} &= \frac{25}{4} \\ a = 1 \text{ and } b = 2...