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Re: HSC 2015 3U Marathon What methods of randomly selecting the two points are there? I thought it would be enough to know both points lie on the circle.

Re: HSC 2015 3U Marathon Here's another probability one: Two points are picked at random on the unit circle x^2+y^2=1. What is the probability that the chord joining the two points has length at least 1? Idk if it is 3U difficulty or not. If it isn't I'll post it to the X2 Perms/Combs Marathon.

Re: HSC 2015 3U Marathon The answers say 1560.

Re: 2015 permutation X2 marathon The solution provided with the question: $Let $x$ be the number of matches in which women defeated men. The number of matches between two women is $\frac{n(n-1)}{2}$, the number of matches between two men is $\frac{2n(2n-1)}{2}$, and the number of matches...

Answer is 0.66c apparently. Found this site: https://www.physicsforums.com/threads/relativistic-speeds-within-a-relativistic-frame-of-reference.760124/

Enzymes called cellulase are used. Read here: http://www.easychem.com.au/production-of-materials/biomass-research/cellulose-as-a-raw-material

Yep. Alternatively, crude oil -> fractional distillation -> cracking -> ethene and so on.

re: HSC Chemistry Marathon Archive It is a neutralisation reaction, so we have: acid + base -> salt + water. Since citric acid is triprotic it ionises like so: C6H8O7 -> C6H5O73- + 3H+. For sodium hydroxide: NaOH -> Na+ + OH-. So for the salt the reaction is: C6H5O73- + 3Na+ -> Na3C6H5O7...

Re: HSC 2015 3U Marathon A mathematics contest consists of four problems. Each of the six member team from Central High School is assigned to work on exactly one of the four problems. If each of the four problems is worked on by at least one of the team, in how many different ways can the...

Re: HSC 2015 3U Marathon From a box with red and blue balls we randomly choose two balls. Assume that the box has at least two balls and at least one of them is blue. The probability that the chosen balls are both red is five times the probability that the chosen balls are both blue and the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread If you haven't learn the auxiliary method for combining two sine and cosine functions, you can use addition of ordinates. For each x-coordinate, add the y-coordinates of y=sinx and y=cosx. So at x=0, sin(0)=0 and cos(0)=1 ->...

Make a thread here: http://community.boredofstudies.org/110/atar-hsc-marks-guidance-counsel-estimates/ Include your subjects, ranks and school rank.

Re: HSC 2015 3U Marathon $Firstly, note that $\cos{x}=\sqrt{\frac{1+\cos{2x}}{2}}$ and $\sin{x}=\sqrt{\frac{1-\cos{2x}}{2}}$, using $\cos{2x}=1-2\sin^2{x}=2\cos^2{x}-1$...

Re: HSC 2015 3U Marathon 2. (a) \begin{align*}y&=x\to y^2=x^2 \\x^2+(y-k)^2&=1 \\\textup{Solving simultaneously:}& \\y^2+(y-k^2)&=1 \\y^2+y^2-2ky+k^2-1&=0 \\2y^2-2ky+(k^2-1)&=0 \\\textup{Since the line }y&=x \textup{ is a tangent the discriminant is equal to zero} \\\Delta&=b^2-4ac...

re: HSC Chemistry Marathon Archive The HInd is the indicator. Note that Ind is short for Indicator. For this question you have to realise that the indicator is similar to a buffer system. The resulting solution contains Ind- and HInd as it is in equilibrium. I figured that HInd would be...

re: HSC Chemistry Marathon Archive {$HInd$}_{(aq)}~+~$H$_2$O$_{(l)}~ \rightleftharpoons ~$Ind$^-_{(aq)}~+~$H$_3$O$^+_{(aq)} When the solution is added to an acid, the high hydronium concentration causes the equilibrium to shift to the left, as per Le Chatelier's Principle. This increases the...

Angle BAD is 90 degrees. Angles BAF, BAD and GAD add up to 180 degrees as they are on a straight line. Using this, angle BAF is 90-θ. Using the angle sum of a triangle, angle FBA is θ.

Since it is on the x-axis in this graph, f'(x)=0, not f(x). Also, just before B, f'(x)<0 so the gradient is negative, and right after B, f'(x)>0 so the gradient is positive. So f'(x) isn't the same on either side.

Re: HSC 2015 3U Marathon Yea pretty much. Initially tried what you did but then I saw the easier way.

Re: HSC 2015 3U Marathon . Just so that we're clear, the area you're finding is enclosed in red. I didn't integrate to get the answer, so I'm not sure about your solution. How did you integrate \sqrt{1-x^2}?