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re: HSC Chemistry Marathon Archive The oxidation state of C in CH4 is: x + 4 = 0 -> x = -4. The oxidation state of C in CO2 is: x + (-4) = 0 -> x = +4. The oxidation state of C increased from -4 to +4, so Carbon has been oxidised.

re: HSC Chemistry Marathon Archive Wikipedia says its element category is "unknown, but probably a noble gas", and that its phase being solid is only a prediction. I'd still go with C.

Re: HSC 2015 3U Marathon $We can find $\cos{\theta}$ using the cosine rule, but to do that we need the lengths of all three sides in $\Delta VMC$. We know that $VC=\sqrt{5}a$. \\ \\$\Delta VAB$ is an equilateral triangle as $VA=VB=AB=2a$. Therefore $\angle{VBM}=60^{\circ}$. Since $VM$ bisects...

Sydney Grammar 2U Trial 2015.

re: HSC Chemistry Marathon Archive I think it isn't, as you want the chloride to form a precipitate after everything else has been removed. If you add Silver nitrate first, it will react with the chloride to form Silver chloride, a white precipitate. Not sure what adding Sulfuric acid does...

$Not sure if this is the intended method. You can work out the values of every angle in each triangle, so using the sine rule, find an expression for the length of the diagonal $AC$ in terms of $x$. Then find the area of each triangles using $A=\frac{1}{2}ab\sin{C}$. You should get something...

It is 2ln(x)+C.

Re: HSC 2015 3U Marathon

For the first one: Use m=tan(x) to find the gradient of line l. Note that the gradient of line l is negative. Now, since lines l and k are perpendicular, what relationship exists between their gradients? Use this to find the gradient of line k. For the second one: The correct answer is 2ln(4).

Read through this: http://www.boredofstudies.org/courses/maths/2u/1119956014_2004_Mathematics_Notes_Emily_Meers.pdf

Do something similar for the second part.

Re: HSC 2015 2U Marathon If 2f(x)+3f(-x)=15-4x, find f(x).

Re: HSC 2015 3U Marathon Using the Binomial Theorem, prove that 7n+2 is divisible by 3, where n is a positive integer.

You have to use the numbers 2 and 5 to get 0.4 and 50. (Let log(x)=log7(x)) For (i), log(0.4) = log(2/5) = log(2) - log(5) = 0.36 - 0.83 = -0.47. For (ii) log(50) = log(25 * 2) = log(52) + log(2) = 2log(5) + log(2) = 2(0.83) + 0.36 = 2.02.

Re: MX2 2015 Integration Marathon They are in this thread: Official BOS Trial 2015 Thread (right before the introduction). Here is a direct link: BOS Trials.zip.

Yes.

Depends what your scale is. You could do 1cm:2microns.

Re: HSC 2015 2U Marathon $Let $A_n$ be the amount in the bank account at the end of the $n$th year.$ \begin{align*}A_1&=A(1+r_1)(1-r_2)+A \\A_2&=\left [A(1+r_1)(1-r_2)+A \right ] \left [(1+r_1)(1-r_2) \right ] + A \\&=A(1+r_1)^2(1-r_2)^2+A(1+r_1)(1-r_2)+A \\A_3&=\left...

re: HSC Chemistry Marathon Archive Yes. I think it is part of this dot point: "Identify data and choose resources to gather information about the colour changes of a range of indicators"