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@skepticalmiss... the CSSA take a strong stance on copyright and so have forced most places hosting their papers to take them down. HSC papers can be accessed only from NESA, who similarly object to anyone else posting their papers online. Likely that is why this old link no longer works.

It's worth noting that this proof would not have established the maximum area at all without the equality part. Leaving out the equality part, the proof establishes that the area of any triangle with semi-perimeter s cannot be more than s2 / root 27, but that doesn't mean that a triangle with...

Bear in mind that the paper is an assessment task 1 from about April, not a yearly paper

There are two answers to 5(a), but the second one is not 117o, it is 21o. If the third angle is x (say) then: \frac{\sin{x}}{8}=\frac{\sin{42^\circ}}{6} \qquad \implies \qquad \sin{x} = \frac{4}{3} \times \sin{42^\circ} \approx 0.892174 ... There are two possible solutions: \text{Answer 1: }...

Another approach: Arrange the IMION in 5! / 2! ways. Choose the 5 positions from 10 for the word PRESS in 10C5 = 10! / (5!)2 ways Total arrangements = 10! / (5!)2 times 5! / 2! = 10!5! / (5!)22! = 10! / 5!2! after cancelling Thus, the answer is indeed (B)

It is NOT true to say that, just because the dot product of two vectors a and b is zero, it follows that a and b are perpendicular. The correct inference is that EITHER a and b are perpendicular OR at least one of the vectors a or b is the zero vector. Formally, once you have shown that a . b...

The net effect in (A) has forward faster than reverse, though decreasing as it approaches equilibrium. The forward rate is increasing in (B) but the reverse rate is higher / faster so the net movement is to the left.

For Q 13, adding aqueous NaOH to a saturated Mg(OH)2 solution will lead to further precipitation of Mg(OH)2 (s) and with a decrease in the concentration of magnesium ions and an increase in the concentration of hydroxide ions such that [Mg2+][OH-]2 remains constant. This assumes that the sodium...

For the second one, shifting to the right means the rate of the forward reaction (A + B ---> C) is faster / higher than that of the reverse reaction. Prior to returning to equilibrium, graph A has rate forward > rate reverse (moving right) while (B) has rate reverse > rate forward.

I am guessing that you are just starting to deal with angles of any magnitude, and thus that trig functions aren't always positive. In each case, the noted information is meant to tell you that the two angles are both acute and complementary... you need the (highlighted) complementary part to...

More likely that the answer is correct and the question is wrong... y = 8 leads to a messier answer, they probably meant it to be y = 9.

It is true that the derivation of Euler's formula by differentiation f(\theta)=e^{i\theta} requires extending calculus beyond the syllabus, but so does projectile motion calculus with vectors which is not justified as valid within the syllabus

\sin{\theta}=2 or similar questions have appeared on MX2 assessment tasks

They were standard integrals on the old syllabus, but that's going back a way - see page 16 of the 2001 paper accessible from https://educationstandards.nsw.edu.au/wps/portal/nesa/resources-archive/hsc-exam-papers-archive/mathematics/mathematics-2001-hsc-exam-pack-archive, for example

Since element B has the highest first IE amongst the four consecutive elements, it must be a noble gas. Thus: element A is a halogen (group 17) and forms an A- anion element C is in group 1 and forms a C+ cation element D is in group 2 and forms a D2+ cation Forming a D3+ cation requires much...

You can also find the boundary values by solving \begin{align*} |3x - 2| &= |x + 4| \\ \\ \textbf{Case 1:} \qquad 3x - 2 = x + 4 \qquad \implies \qquad 2x &= 6 \qquad \implies \qquad x = 3 \\ \\ \textbf{Case 2:} \qquad 3x - 2 = -(x + 4) \qquad \implies \qquad 4x &= -2 \qquad \implies \qquad x =...

Define a height axis, y, along the central axis of the cylinder, taking y = 0 at the top and placing the light source at y = h above the cylinder. When h > 0, the light casts a shadow of radius i cm from the axis. We are told that h is increasing at dh/dt = 3 cm/s, and r decreases as this...

Right-angle triangle trigonometry should allow you to show that \text{LHS} = \sin{\frac{\theta}{2}} + \frac{1}{2}\cos{\frac{\theta}{2}} = \frac{b + h}{2x} Applying the AM-GM inequality leads to \text{LHS} \geq \sqrt{\frac{bh}{x^2}} Now, using the fact that \text{Area of } \Delta =...

The cover is a modified version of that used by the Independent Schools papers, as is the table on page 2 for MCQ papers... but it's not the 2020 Independent paper. The way things like square roots appear makes it look like it's not from a professional source. The repeating of the diagram in...

Some excellent points. From a cognition perspective, there is a significant difference depending on how you are with writing on a computer and creating illustrations, etc. If you are typing up notes that are largely already written, then the cognitive demand of composing prose is relatively...