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For example, in this case, the solution that @tywebb has posted would be much harder for an HSC marker to criticise if it stated what a cross-product is and included working showing the reasoning. Perhaps something like: The cross product of any two vectors, \overrightarrow{u} \times...

It has the same implication as using anything that is not in the syllabus... if it is correct and valid, it should be credited, but no one can promise how an individual teacher may respond in marking an assessment task. You need to give careful consideration to what reasoning and evidence is...

If there is water present, and the product is soluble, it will be aqueous.

The cross product is NOT part of the MX2 syllabus. Its exclusion from both the MX1 and MX2 syllabi is a bit silly, I agree, but the fact is that it is not in either syllabus.

... If more students were given the resources to train as doctors or nurses we would have more doctors and nurses, if more students were given the highest quality of study note in math we would have more students capable of receiving degrees in finance or engineering, the existence of these...

Zakary, I'm not sure how you are trying to come across. I sincerely hope it is not as you are coming across, at least from my perspective, because that would reflect very poorly on you, in my opinion. Replying to some of your points: ... there is no point in attending a university if you...

That's pretty judgemental. Some competition can become unhealthy, sure, and high school students can be immature and petty at times - but being "net bad" for society and "bringing the general society down"? Isn't this a bit over the top? Are "some nerds" really the group that are "never real...

Solving assignment questions is not particularly ethical...

The point that I was making was not whether the method being used works - as it clearly does - but whether a student's response might be judged as showing all necessary working, especially in an assessment task where individual marking can be considerably more variable than HSC marking. The...

This approach effectively turns the question into a problem in coordinate geometry... if A(1, -1) and C(3, 4) are the ends of a diagonal of square ABCD find coordinates for B and D. If asked as a MCQ, it is certainly a quick approach. However, for a written question, and given the application...

It will be in the unit of study guides, at the individual department websites. For a unit worth 6 credit points, it should average to 6 h per week of formal contact. It may not be the same each week, too. It could be 3 x 1 h lectures / week for 13 weeks (39 hours) plus 10 weeks with 1 x 3 h...

y = \cos{2x} looks like a typical cosine graph but with a period of \pi. The line y = mx - 1 must have a y-intercept at (0, -1), and so as m decreases from infinity, there must be only a single solution until the line becomes a tangent. This will happen for some x-value near but smaller than...

I would work with the equations as given. There will be one solution when m is large, so there will be two sets of possible values, m > k and m < -k where the value m= \pm k correspond to the line being a tangent to the curve at some value \pm x_0 and the line also crossing the curve at some...

I advise strongly against making unsupported assertions about indeterminate forms. Unjustified statement (dangerous in marking terms): \lim_{x\to 0^+} x\ln{x} = 0 Reasonable working: \lim_{x\to 0^+} x\ln{x} = 0^- \text{ as $\ln{x} \to -\infty$ and $x \to 0^+$ and polynomial functions dominate...

Recognising the GP in the second polynomial allows the link between the two parts to be seen more easily: \begin{align*} z^2 + z^4 + z^6 &= 0 \\ \frac{z^2\left[\left(z^2\right)^3 - 1\right]}{z^2 - 1} &= 0 \qquad \text{on summing the GP with $a = z^2$, $r = z^2$, and $n = 3$} \\...

Noting that the polynomial is even, the roots being x=\cos{\frac{(2k + 1)\pi}{12}}\ \text{for}\ k\in\{0, 1, 2, 3, 4, 5\} are more conveniently taken as x=\cos{\frac{(2k + 1)\pi}{12}}\ \text{for}\ k\in\{-3, -2, -1, 0, 1, 2\} and hence are x=\pm\cos{\frac{\pi}{12}},\ \pm\cos{\frac{\pi}{4}} =...

Another approach is to say that AB = i.AD = DC. Thus, AC = AD + DC = AD + i.AD = (1+ i).AD. Now, AC = AO + OC = OC - OA = 3 + 4i - (1 - i) = 2 + 5i. With these two results, it follows that: \begin{align*} \vec{AD} &= \frac{1}{1 + i}\vec{AC} \\ &= \frac{2 + 5i}{1 + i} = \frac{(2 + 5i)(1 -...

Note, also, that the method generalises, to: |\alpha - \beta| - |z - \alpha| \le |z - \beta| \le |\alpha - \beta| + |z - \alpha| is the result without any substitutions. Applying it to part (i), we use: \alpha = 2 + i \ \ \text{and} \ \ \beta = 0 \quad \implies \quad |\alpha - \beta| = |2 + i...

I agree with @liamkk112's advice. Double integrals, with one integral sign after another, have specific meaning that is beyond HSC Maths, so I advise avoiding notation like you have written and just solve the DE by two separate integration steps.

Yes, you can. For part (ii), take \alpha = 2 + i and \beta = 3i. We are given that |z - \alpha| = 1 and can calculate that |\alpha - \beta| = |(2 + i) - 3i| = |2 - 2i| = \sqrt{2^2 + (-2)^2} = 2\sqrt{2}. We seek |z - \beta|. Applying the triangle inequality gives us that: \begin{align*} |z -...