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Hi @dan964, Thank you, thank you, thank you so so so much for all the work done on THSC, it is an absolutely fantastic resource! Last year, there were indications that THSC might not be updated as regularly going forward... do you have any information that you can share on how you see the...

Are these alignments to AQF levels based on some published reference or your own thoughts / estimates? The AQF has kept away from stating an AQF level that matches any Senior Secondary Certificate of Education (like the HSC), and its alignments are based on completed courses rather than...

ZJN is correct that there is a substantial difference in the level of the content in Std2 v. Advanced. The lack of any calculus in Std2 makes it a poor preparation for studies in most sciences and in other disciplines where maths is an essential tool, even beyond the STEM grouping. For a...

Cool... I was just showing some other approaches :)

Or... \begin{align*} -2i\left(e^{-\frac{\pi}{2}i}\right) &= -2 \times 1\left(e^{\frac{\pi}{2}i}\right) \times \left(e^{-\frac{\pi}{2}i}\right) \qquad \text{as $|i| = 1$ and $\arg{i} = \frac{\pi}{2}$} \\ &= -2\left(e^0\right) \\ &= -2 \times 1 \\ &= -2 \end{align*} Or... \begin{align*}...

Declaring that, in your mind, not having a degree equates to being a nobody is an extremely effective way to portray your perspective on people. I sincerely hope that my future is filled with people with more mature perspectives. "Similar, but not entirely different" Such eloquent... drivel -...

FYI, this a variation on the 2006 MX1 HSC question 5(d).

Not being interested in someone because their sexual interests and future behaviours do not align with your values is perfectly reasonable. As an obvious example, person A wanting a monogamous relationship is unlikely to want as a partner person B who wants to set new records for number of...

\begin{align*} A_{\text{total}} &= A_{\Delta OKJ} + A_{\Delta OMP} \\ &= \frac{1}{2} \times JK \times JO \times \sin{\angle{OJK}} + \frac{1}{2} \times MO \times MP \times \sin{\angle{OMP}} \\ &= \frac{1}{2}sx\sin{\alpha} + \frac{1}{2}(l - x) \times MP \times \sin{\alpha} \qquad \text{as...

But, assuming you have two hands and two eyes, using two pads allows two papers to be done at once... :P

In any triangle, the longest side is opposite the largest angle. Or... draw the circle with the red line as radius, centred where all the coloured lines meet. This circle and the one you have drawn touch with a common tangent at the end of your red line, and nowhere else... yet, your coloured...

For the last one, the max and min values of arg z occur when the ray for each is a tangent to the circle, and so perpendicular to the radius at the point of contact. Use right-angled triangle trigonometry on the resulting triangles.

For 12G, you could say that P(x) = x^3 + x^2 + 6x - 3 = \left(x^2 + 4\right)Q(x) + R(x) where R(x) = Ax + B and solve for A and B using P(2i) = R(2i) = 2Ai + B \qquad \text{and} \qquad P(-2i) = R(-2i) = -2Ai + B as you know P(2i) and P(-2i) from the earlier parts.

For the first one... From any point outside a circle, the closest and furthest points must lie on the line through the external point and the centre of the circle. Draw a circle, centred at O, and an external point P. Join OP, which crosses the circle at C. PO produced meets the circle at F...

Another approach is to take the forces as p\overrightarrow{i} and q\overrightarrow{j} when they are at right angles, so that: \left|p\overrightarrow{i} + q\overrightarrow{j}\right| = 2\sqrt{7} \qquad \implies \qquad p^2 + q^2 = 28 \qquad \text{. . . (1) And, when at 30 degrees, take one force...

Note: The result that I derived in blue, that \color{blue} 1 + \tan{\theta}\tan{(k + 2)\theta} = \frac{\tan{(k +2)\theta} - \tan{\theta}}{\tan{(k + 1)\theta}} can actually be derived more easily by: \color{blue} \begin{align*} \tan{(k + 1)\theta} &= \tan{\left[(k + 2)\theta - \theta\right]}...

Here is a complete proof, though there may be simplifications available... Theorem: For all integers n \ge 1 \sum_{r=1}^n \tan{r\theta}\tan{(r+1)\theta} = \tan{(n + 1)\theta}\cot{\theta} - (n + 1) Proof: By induction on n: A: Put n = 1: \begin{align*} \text{LHS} &=...

Your assertion / belief that my understanding of the education system is poor does not make it so. Your experience with "this kind of dance", on the other hand, should mean your arguments are well-tested, solidly grounded, and eloquently expressed. It is curious, then, that they contain...

I have already stated that you have made assumptions about me that are inaccurate. I am not a child - in fact, following the definition of a human child as an individual who has yet to begin puberty, almost no one here is a child. Even under the United Nations Convention on the Rights of the...

Sorry, but you need to know basic solubility rules