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Yep, pretty much.

I think it's just 8C5, which is (8!/5!3!), because there's 8 spots, and you're choosing 5 to put the stars (vowels) into.

Thaaaaanks, you too! Good luck everyone, we're gonna smash this.

The vowels will be in the order: A E E E U Now, first, we arrange the 7 consonants: 7!/2!2! (7 consonants being arranged, N and C are repeated twice) Between each consonant, and before and after the consonants, we will have 8 slots we can place the vowels into (this is called...

All the working out looks correct (I've read over it a few times), and you get to the final answer, so yeah, you should get two marks for that.

Re: HSC 2015 4U Marathon oh, lol, that was a typo, should've been the other way round. Fixed, thanks!

Re: HSC 2015 4U Marathon x^4 $ could be used instead of $ \frac{5}{3} x^4 $, since $ \frac{x^5}{5} < \frac{x^5}{3} $ for $ x > 0 $.$ $Also, the terms should be integrated from 0 to x, so that constants do not matter.$

Re: HSC 2015 3U Marathon I realised after posting the question that I didn't actually have the answers, lol But yeah, I got that too, so we'll just say it's right

Re: HSC 2015 3U Marathon $Construct OD, OE and OF.$ $In $ \triangle $OEF, OB is perpendicular to EF, because the tangent is perpendicular to the radius of a circle at its point of contact. Therefore, EF is the base of the triangle, and OB is the height.$ \begin{align*} A_{\triangle...

Re: HSC 2015 3U Marathon $I'll skip straight to the assumption, since the base case is pretty straight forward.$ $Assume true for $ n = k $: $ P_k = 2^k [1 *3 * ... * (2k-1)] $Prove true for $ n = k + 1 $: $ P_{k+1} = 2^{k+1} [1 * 3 * ... * (2k - 1)(2k+1)] P_k =...

I thought I went okay, but man, after reading some of the comments on here! I think I got 18 for MC (forgot to count reserve assets, rip), which I'm alright with, and the short answers and essays were okay, so I hope I managed to write some decent answers for those. On the bright side, it's...

Yeah, now that you mention it, I'm really not sure about D. It is pretty much the opposite effect, increasing export revenue in the short term (while a lower ToT decreases it) and hindering international competitiveness in the long term (while a lower ToT improves it). Does anyone know?

Since the price of exports has reduced, export revenue will be less (in the short term). Consequently, aggregate demand will be reduced (AD = C + I + G + X - M), and so to counter this, the government can increase expenditure (G). (therefore, the answer is C) Increasing interest rates and...

trying to practice essay writing, so looking forward to finishing the exam though! "Additionally, Fire Permits are not required for fires for the purpose of heating or cooking, provided that: the fire is in a permanently constructed fireplace; at a site surrounded by ground that is cleared of...

I'm extremely keen for it! And it's only illegal if we get caught... (who's going to notice heaps of smoke anyway?) Soooo, how is everyone feeling about economics tomorrow?

Re: 2015 HSC Economics Marathon Quota refers to the quantity of a good/service allowed to be imported into a country. Therefore, increasing the quota of meat (as an example) means increasing the quantity of meat allowed to be imported into the country; hence, increasing foreign competition...

Re: HSC 2015 3U Marathon $Differentiating: $ y = 4x^2 + \frac{1}{x} \frac{dy}{dx} = 8x - \frac{1}{x^2} When \frac{dy}{dx} = 0 \\ 8x^3 = 1 \\ x = \frac{1}{2} $The second derivative is positive when $ x = \frac{1}{2} $Therefore, the minimum value occurs when $ x = \frac{1}{2} $, i.e. $ y =...

Re: HSC 2015 3U Marathon I just used some Pythagoras and circle geometry. AB will be a chord of the circle, and since CD is the perpendicular bisector of AB, it passes through O. Letting the radius equal r, and using Pythagoras in triangle DOB, we get: r^2 = (r - x)^2 + y^2 which simplifies...

Re: HSC 2015 3U Marathon The questions can be assigned either as [1,1,1,3] or [1,1,2,2]: [1,1,1,3]: 4 * (6C3 * 3 * 2 * 1) = 480 [1,1,2,2]: 4C2 * (6C2 * 4C2 * 2) = 1080 Therefore, total ways = 1560

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We're dealing with the lines y = x and y = n sin x. Now, for there to be 69 positive solutions (70 including x = 0), the line y = x must be greater than the line y = n sin x after the 70th solution, which occurs during the...