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Ok, a bridging course makes more sense to me, especially if the school will agree to testing at the start of the year and not waiting until then to decide if a test will be administered. As for the "elephant in the room", I agree strongly that school rankings / ratings are distorting decision...

Yes, I did. My idea was for the student to demonstrate their ability to the school. But, if a coaching college were involved / certifying the student's ability, I think the resistance from some schools would be extreme. There are teachers who reflexively reject anything that they know comes...

Suppose this was done... how would the school react to requests to be admitted to Advanced based on this coaching college course?

Maybe they can offer to do some Advanced work over the holidays and be tested at the start of the year, to prove they can catch up / cope?

Working as a tutor is not easy. I find myself trying to balance the student's expectations, their parents' expectations, my experience and perceptions of areas of difficulty in general and for the individual, their style (including spoken and unspoken feedback), the need to build trust with the...

Worth investigating, then. I got a subsidised place that reduced fees but that applied to fees both under Grad Dip and Masters enrolment. Good luck!

I started my Masters as a Grad Dip and upgraded / transferred, so it certainly is possible. I don't recall if the fee per unit was different, though.

Transfer of credit to Masters should not be a problem, allowing the upgrade will depend far more on how you have performed in the Grad Cert units rather than your academic performance in undergrad.

Transfer the Grad Cert credits to what?

Suppose we have a group of 4 people, conveniently named A, B, C, and D. If everyone shakes hands with everyone else, the number of handshakes undertaken by A is 3 (A with B, A with C, and A with D)... but the total number of handshakes is 6 (A with B, A with C, A with D, B with C, B with D, and...

One approach that I would suggest is to look for activities / experiments / investigations that both can do but encourage the older child to delve more deeply into the topic, and based on what they find interesting, encourage them to explore into extensions. Giving them ownership over parts of...

The wording was "calculate the equilibrium constant" so I think they will expect an answer in scientific notation. Further, 10-8.99 is (as a number) an exact value and my answer is a 2sf approximation to it. I am not inclined, on first thinking, to call 10-8.99 an answer to 2sf, though I might...

I understand... I was wondering whether it meant to use part (iv), or about noting that part(iv) was derived from (ii), and so using it is building from part (ii)

My answer to Q36 would be: H2SO3 (aq) + 2 H2O (l) <---in equilibrium---> 2 H3O+ (aq) + SO32- (aq) \begin{align*} K &= \cfrac{\left[\text{H}_3\text{O}^+\right]^2\left[\text{SO}_3^{2-}\right]}{\left[\text{H}_2\text{SO}_3\right]} \\ &=...

Would you clarify what you have in mind, in relation to "and/or"? Are you thinking one 2 h session per week? 1 h per week for each of them (or together, when appropriate)? Alternating between the children, on some timetable?

The two results that you have derived: and Can be found without complex numbers: \cos{2\theta} = 1 - 2\sin^2\theta \quad \implies \quad \sin^2\theta = \cfrac{1 - \cos{2\theta}}{2} and then set first \theta = \cfrac{2\pi}{5} and then \theta = \cfrac{\pi}{5}

If part (iv) has established that \cos{\cfrac{4\pi}{5}} = -\cfrac{1 + \sqrt{5}}{4} then applying the result \cos{(\pi - \theta)} = -\cos\theta shows that \cos{\cfrac{\pi}{5}} = -\cos{\left(\pi - \cfrac{\pi}{5}\right)} = -\cos{\cfrac{4\pi}{5}} = \cfrac{1 + \sqrt{5}}{4} from which we can...

Bear in mind, though, if z + iw is purely real but w and z are complex numbers, it does not follow that w = 0 and that the number is z.

NESA offers materials back to 2001 with reports from exam centres, etc Here is the 1998 examiner's report for all the Maths exams, and here is an archive for 1995-2000 HSC papers plus reports.

\omega^4 = \cfrac{\omega^5}{\omega} = \cfrac{1}{\omega} = \cfrac{\overline{\omega}}{\omega\overline{\omega}} = \cfrac{\overline{\omega}}{|\omega|^2} = \cfrac{\overline{\omega}}{1} = \overline{\omega} \quad \implies \quad \omega + \omega^4 = \omega + \overline{\omega} = 2\Re{(\omega)}