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There is also IB people to consider. Not to mention ATAR isn't linear.

Re: HSC 2013 4U Marathon Google taylor polynomials.

Re: HSC 2013 4U Marathon I didn't read the rest but i can already tell your confused. Unsurprising since this question is definitely out of the scope of the hsc. (note that you said f'(a)=0, which is note true in general). Here's a further hint: $$\text{by definition, f is differentiable...

I want to address two points raised so far, First is that the humanities are scaled 'unfairly'. Subjects are scaled relative to how the students do in other subjects, in particular English since it's the only compulsory subject. Intuitively, one would expect the opposite if you assume students...

I'm unsure of what exactly your asking. Is it possible to form an infinite series of irrational numbers such that the series equals a rational number? Then the answer is yes. A 'nontrivial' construction is as follows: $$\text{let } \sum_{k=0}^{\infty}a_k \text{ be a converging series that...

Thats because your not adding any new information using pythagoras theorem since the theorem already encodes it in a way. Notice that your not using the fact that OT=MT anywhere so it seems likely the answer depends on this condition.

Nothing, night weaver misquoted the theorem. If you extend the line OA to intersect on the other side of the circle which we label N. Then the theorem is: AT^2 = AM*AN. EDIT: bonus points: It is actually proven using Pythagoras' theorem AT^2 = OA^2 - OT^2 = (OA-OT)(OA+OT)=AM \times AN .

Well if cd and dx are parallel and they both intersect at D then they must be the same line.

That the two points are the same.

Personally I would like to see a section to each topic that isn't aimed at doing HSC questions, but actually learning the theory properly. Something more in the style of this. Obviously, not so difficult. But with more leaving things as problems for the student to figure out and open ended...

Re: HSC 2013 4U Marathon That's true. It also gives us that the only possible form of x,y and z are re^{i\theta},re^{i(\theta+2\pi/3)},re^{i(\theta-2\pi/3)} which after a quick substitution prove to be solutions.

Re: HSC 2013 4U Marathon There isn't any. You can prove it by rearranging the conditions to get it in terms of z; $$ z = -(x+y) \text{and, } z = \frac{-xy}{x+y}$$. Then form the difference function $$f(x,y) = \frac{-xy}{x+y} +x+y = \frac{x^2+xy+ y^2}{x+y}$$ which can be shown...

It doesn't give you any extra information so I guess you could say that. The reverse triangle inequality is the identity $$ | |x|-|y| | \leq |x-y| $$. From this when can derive bounds on |z|. |z| - |4+3i| \leq |z - 4-3i| \Rightarrow |z| \leq 8 and |4+3i| - |z| \leq |z - 4-3i| \Rightarrow 2 \leq...

Your algebra is right, you're just misinterpreting what the equations are telling you. What you have proven is that for any z on the circle, its modulus >=-2. It's not saying that for some z on the circle, its modulus is -2 (a nonsensical statement since modulus is always positive). Edit: If...

The short answer is a pendulum moves in simple harmonic motion only when the angle is small, and the spring and mass moves in SHM. The more detailed answer involves talking about what exactly 'doing physics' means. In physics, we try and model reality via mathematical models. However reality is...

Counter-example: $arg(i) \neq \arg(-i), \quad | |i| - |-i| | = 0 = | i + (-i) |$

One way to do this is to use the Inclusion-exclusion principle. If we label the styles by the letters A, B, C and D as Sy123 did we consider the sets of strings such that AA are together, BB are together and so on. Using this we get the formula: $$...

Re: 2012 HSC MX2 Marathon lolcakes, the if F(x) is the primitive of f(x) then -F(a-x) is the primitive of f(a-x)

Just a clarification on the question: What happens if you start off less then one kilometer below the north pole? does the man continue walking south after he passes the north pole or does he just stop since there is no more N/W/E and then goes S for 1km (How would one even determine what...

Just a small technical point but shouldn't the answer be + or - pi/2 ?