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NESA had just released the first in course list for the 2021 HSC: https://educationstandards.nsw.edu.au/wps/portal/nesa/about/events/merit-lists/first-in-course/2021

Not surprising. There are cases of people merely saying certain words next to their phone and then shortly afterwards ads related to said words appear on facebook despite having no apparent search history of it (worth trying as an experiment with your phone if you're bored).

Integrate both sides of the inequality m\leq f(x) \leq M from a to b

FYI E(X^2) = \int_0^c x^2 \dfrac{1}{c}\,dx

Parametric equations can be used to describe curves which may not necessarily be easily expressed in Cartesian equations. This is particularly the case once you start going beyond two dimensions, For example, the equation of a helix: x = cos t y = sin t z = t The x-y parameters are that of a...

From the result in part (a) - Replace a with \sqrt{ab} - Replace b with \sqrt{ac} - Replace c with \sqrt{bc}

Suggest having a look at this flowchart https://boredofstudies.org/threads/flowchart-of-how-hsc-marks-and-atar-are-determined.390261/

The HSC results will be released on Thursday 20 January, rather than Monday 24 January. This means it will be released on the same day as the ATARs. First in course ceremony will occur on 19 January. NESA’s media release...

The sum of n consecutive numbers is an arithmetic series with common difference of 1. If a is the first term then the sum is S = 0.5n(2a + n - 1) For S to be an integer divisible by n, we require (2a + n - 1) to be even, which implies that an odd n is required.

I actually think that understatement of deaths more than offsets any supposed overstatement, particularly if you look at the excess deaths count compared to a normal year (for highly covid infected countries that is). Either way, you have to remember that Australia’s low death rate is inclusive...

From the inductive step 1^3+2^3+3^3+...+k^3+(k+1)^3\\\\=\dfrac{k^2(k+1)^2}{4}+(k+1)^3\\\\=\dfrac{(k+1)^2(k^2+4(k+1))}{4}\\\\=\dfrac{(k+1)^2(k+2)^2}{4}

A combined degree like Engo/Comm is usually a good starting point because it exposes you to a diverse range of subject areas, so you get a bit of everything. Within the first year or two you will then get a better idea of what you like and don't like, which will better inform your decision of...

You can do it here https://boredofstudies.org/forums/buy-sell.46/

The key is to realise that what you are writing is the general term of an infinite sequence of numbers. When you write n=\pm 4k, you actually are writing the general term for n = {...,-8,-4,0,4,8,...}. In fact, since k is any integer it doesn't matter whether we write n = 4k or n = -4k, they...

But seriously, it is natural to be concerned given the amount of uncertainty. You have to remember that back in 2020 the world brushed off the seriousness of the original virus outbreak, so it is understandable that they are being cautious and don’t want to be caught out again.

Relevant

But plugging in numbers to see what the exact trig value is lacks proper rigour, as you have effectively found the answer by direct computation rather than by proof. It is the same as claiming you have proven the exact value of something simply because the calculator says they're equal rather...

It's a circular argument. Prove that this trig result holds. Oh look, they happen to equal on a calculator therefore it is proven!

Hi all, It is with regret that I have to inform you that we will not be able to release a solutions document in time before the HSC Chemistry exam. :( Unfortunately, some things have gone wrong and/or taken longer than expected behind the scenes. We will aim to ensure this does not occur again...

T_{k+1} \\= \dfrac{2(k+1)}{2(k+1)+3}T_k\\=\dfrac{2(k+1)}{2(k+1)+3}\times\dfrac{k!(k+1)!}{(2k+3)!}4^{k+1}\\=\dfrac{2(k+1)}{2(k+1)+3}\times\dfrac{2(k+2)}{2(k+2)}\times\dfrac{k!(k+1)!}{(2k+3)!}4^{k+1} ...and the rest follows