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If you don't do any of the work your teacher may give you an N-award and you may eventually fail the subject, which will prevent you from being able to start the HSC course, as you need 12 preliminary HSC units completed. If you're 100% sure you're dropping it, just do the minimum amount of...

Re: HSC 2018 MX2 Integration Marathon \begin{aligned} &\int \frac{1}{\sqrt{x^2+x}} \,dx \\ &= \int \frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2 - \frac{1}{4}}}\,dx\end{aligned} \begin{aligned}\mathrm{Let}\,\, x + \frac{1}{2} &= \frac{1}{2} \sec \theta \\ x &= \frac{1}{2}(\sec \theta - 1)...

This is a deceptively tricky question... definitely not on a 2U level.

Hint: use the identity \cos x + \cos y = 2 \cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) And note that \cos\left( \frac{\theta - \left(\theta + \frac{\pi}{3}\right)}{2}\right) is a constant.

Apparently everybody I know gets 4-6 hours of sleep. I usually get 7 hours of sleep at home + 1-2 elsewhere.

Aluminium oxide is not amphiprotic. It is amphoteric. It can act as either acid or base, i.e. it can react with either a base or an acid. I believe you could use these neutralisation reactions: \mathrm{Al_2O_3_{(aq)} +6HCl_{(aq)} \to 3H_2O_{(aq)} + 2AlCl_3_{(aq)}} \mathrm{Al_2O_3_{(aq)}...

Just get Cambridge.

That's not enough information for anyone to give you an accurate estimate. For example, a B5 in Chemistry is very different to a B5 in Senior Science.

Probably not because it's one assessment out of several that counts towards two of your eleven units. Just don't let it happen again.

X1 Maths is one unit, Chem is two. If you really do enjoy them equally but want to drop one, then drop Chem because you get one more freed up unit. If you drop X1 Maths then you end up with 11 units. One is going to get wasted, you'll be doing two units of work for one unit towards your ATAR...

Half is memorisation, the other half is skill-based questions.

Re: HSC 2018 MX2 Marathon Woops. Forgot to rationalise the denominator. \frac{y}{\sqrt{2(x + \sqrt{x^2+y^2})}} = \sqrt{\frac{\frac{1}{2}y^2}{x+\sqrt{x^2+y^2}}} \times \sqrt{\frac{x-\sqrt{x^2+y^2}}{x-\sqrt{x^2+y^2}}} = \sqrt{\frac{\frac{1}{2}y^2(x-\sqrt{x^2+y^2})}{x^2-x^2-y^2}} =...

Re: HSC 2018 MX2 Marathon Let (a+ib)^2 = x+iy. (Note that if a = 0 then y = 0 , x \leq 0 and we have (ib)^2 = x \implies a + ib = \pm i\sqrt{-x}. Otherwise,) Expanding, we get (a^2-b^2) +i(2ab) = x + iy . Equating real and imaginary parts gives us \begin{cases} a^2-b^2 &= x...

For aqueous solutions, diluting is the same as adding water. The expression "adds 200 mL distilled water to 20 mL of solution" is the same as saying "dilutes the solution to 220 mL". But you have to remember the existing 20 mL of solution.

1) Because there are 10^{-6} moles of silver sulfate in every litre, to find the number of moles in 100 mL we simply divide by 10. \mathrm{n(Ag_2SO_4)} = 10^{-6} \times 0.1 =10^{-7} For every mole of silver sulfate there are two silver ions. $n(Ag)$ = 2 \times 10^{-7} Multiply by the...

Re: HSC 2018 MX2 Marathon z_1 = x_1 + iy_1 and z_1 satisfies the locus in i), so y = -\frac b a x which gives us z_1 = x_1 - i\frac b a x_1 therefore |z_1| = \sqrt{x_1\, ^2 + \frac {b^2} {a^2} x_1\, ^2 }

They probably want you to recall the experiment (e.g. "Describe how you performed the .... "), which could include the equipment, procedure, risks, etc. "Interpret" is a bit vague but one example could be interpreting data to draw conclusions.

Re: HSC 2018 MX2 Marathon Yep, it's broken for me as well.

1) Straight away this integral must be negative, so the answer is either A or D. f(x) appears to be symmetric around the y-axis, so \int _{-2}^2 f(x)\, dx = 2 \int _{0}^2 f(x)\, dx You are given \int _0^2 f(x)+g(x)\, dx = \int _0^2 f(x)\, dx+\int _0^2g(x)\, dx = \pi - 10...

\begin{aligned} 0 &= 500000-100000(1.05)^n \\ 500000&=100000(1.05)^n \\ 5 &= (1.05)^n \\ \log 5 &= n \log 1.05 \\ n &= \frac{\log 5}{\log 1.05}\end{aligned} Where did the 105000 come from? Also, if you happened to do this, note that \log\left(\frac{a}{b}\right) \neq \frac{\log a}{\log b}