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Sulfuric acid is used in both cases. Concentrated sulfuric acid is a dehydrating catalyst, and dilute sulfuric acid is a hydrating catalyst.

http://www.uac.edu.au/documents/atar/Scaling-report-2017-TableA3.pdf This is for 2017's HSC but these numbers are usually stable (note that the new HSC might possibly change things) Look at the scaled mean (out of 50). Generally speaking - the higher the scaled mean, the higher the scaling for...

Re: HSC 2018 MX2 Marathon (1+i)^m = (1-i)^m Converting to polar form, m\sqrt{2}\left(\cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4} \right)=m\sqrt{2}\left(\cos -\frac{m\pi}{4} + i \sin -\frac{m\pi}{4} \right) Since m \neq 0 , \cos \frac{m\pi}{4} + i \sin \frac{m\pi}{4}=\cos -\frac{m\pi}{4}...

Re: HSC 2018 MX2 Marathon Is the answer m=4? (I intend to post a writeup later)

1. The parabola x^2 - 3x + 2 is negative when 1 < x < 2 . So when you take the absolute value, the only change in the graph is that that area will be flipped across the x axis to become positive. Same area, just the direction is changed. Therefore, to obtain the area between | x^2 - 3x + 2|...

Reliability is probably the easiest. You should have a wide range of soil samples differing in their compactness to reduce the impact of outlier results. You need to control all these variables. The soil should all be the same, the holes should be the same, the plastic cups should be the same...

Re: HSC 2018 MX2 Marathon Oh whoops... I read that as \sin(x^n).

Good luck in Maths. You can always post here for help if you're stuck on a question or you don't understand something. You might like to try Cambridge or maybe Fitzpatrick for tougher questions. My teacher mentioned today that Maths In Focus is not very challenging.

https://drive.google.com/open?id=19U4R1-0dlv8kjcYKa-jYI9zCCsKqnUwe

You can get the upper bound for the first integral by solving y = \sqrt{2x} for y=4. Then evaluate the integral and subtract it from a rectangle to get your answer. Alternatively, you can get the answer directly by switching x and y , rearranging to make y the subject and integrating from...

\begin{aligned}\left|\frac{a-ib}{a^2+b^2}\right| &= \left|\frac{a}{a^2+b^2} - i\frac{b}{a^2+b^2}\right|\\ &=\sqrt{\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2}\end{aligned}

Let z = a + ib where a^2+b^2 \neq 0 \begin{aligned} \left|\frac{1}{z}\right| &= \left|\frac{a-ib}{a^2+b^2}\right| \\ &= \sqrt{\frac{a^2}{(a^2+b^2)^2} + \frac{b^2}{(a^2+b^2)^2}} \\ &= \sqrt{\frac{a^2+b^2}{(a^2+b^2)^2}} \\&= \sqrt{\frac{1}{a^2+b^2}} = \frac{1}{\sqrt{a^2+b^2}} =...

CaCO3 and sand have different melting points. So you could possibly use that (I think), but it certainly wouldn't be doable in a classroom lab.

Re: HSC 2018 MX2 Marathon Yep. If you used z = x + iy , you ended up with the locus x^3 + xy^2 = x Which is the correct answer. It's tempting to divide both sides by x , but if you performed the division you were implicitly assuming that x \neq 0, which lost solutions...

Then test the first derivative around the turning points.

Re: HSC 2018 MX2 Marathon It definitely seems to be something that most people forget. For this type of question it tends to not matter (in terms of marks), but being able to spot when a variable's domain has been restricted has saved me a few marks in exams. (This might help with the locus...

Re: HSC 2018 MX2 Marathon For anyone interested, the Patel question I was talking about is: \text{Prove that}\,\, (1-\cos\theta + 2i\sin\theta)^{-1} = \frac{1-2i\cot\left(\frac{\theta}{2}\right)}{5+3\cos\theta}

Re: HSC 2018 MX2 Marathon An exercise in Patel (I believe 4H, Q8) has a similar, but harder question. \frac{2}{1+z} = \frac{2}{(\cos \theta + 1) + i \sin \theta} \times \frac{(\cos \theta + 1) - i \sin \theta}{(\cos \theta + 1) - i \sin \theta} =\frac{2\cos\theta +...

Re: HSC 2018 MX2 Marathon As I said above, that's not the correct answer. The locus is not x^2 + y^2 = 1. There are certain values of z (specifically x) that need to be considered. (By the way, (0,0) is not on the unit circle so it's not necessary to explicitly exclude it if the locus...

Re: HSC 2018 MX2 Marathon x^2 + y^2 = 1 is not the correct answer. (Be careful of any restrictions you place on the locus)