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*Sketches graph on Geogebra and investigates* *Sees asymptote* mind=blown

Because for a number: y= a^x \\ \\ x=\log_a{y} \\ \therefore a>0 That is the technical aspect of it. Therefore if a is raised to a power of an undefined number (such as pronumeral variable x or infinity), then a must be greater than zero

Pick up Extension History, then drop General Math imo.

Over the weekend I will have enough time to latex up all the 2U formulas into one new thread. And I will do 3U as well probably

If you dont want to drop Math, then pick 2U Mathematics, no point in doing General Math.

If you are going to do math, dont do general Math, seriously. Just pick 2U math, it really isnt that hard, and if you generally suck really badly at math, then drop it. Dont do General Math on purpose it will be a mistake you will regret, just pick Mathematics and see how you go, if you do...

Definitely, they do recommend Extension 1 Math and Advanced English, but 2U is sufficient enough to get in.

Hmm, how did you do it? I got 14400 by doing 2!2!7!-2!2!2!6! The logic is that, the first part of my answer (the one with 2 of the 2!), is the number of arrangements possible, that group girls together in pairs, disregarding the possibility that the 2 pairs of girls could be placed next to...

UNSW http://www.unsw.edu.au/future-students/domestic-undergraduate/assumed-knowledge http://www.unsw.edu.au/sites/default/files/documents/2012%20ATAR%20Cut-Offs%20Table.pdf USYD - Doesnt have a compact table where I can copy paste a link. But you can search your desired course from this...

\frac{2n\pi}{5}+\frac{-\pi}{15} \\ \frac{6n\pi}{15}+\frac{-\pi}{15} \\ \\ \frac{6n\pi-\pi}{15} \\ \\ \frac{(6n-1)\pi}{15}

1. \ \log_b{x}=y \\ b^{\log_b{x}}=b^y \\ x=b^y \\ \\ 2. \ x=b^5 \\ x^2=(b^5)^2 \\ x^2=b^{10} \\ \log_b{x^2}=10 They are just 'proofs' Bringing the exponent down is sufficient

Re: HSC 2012 Marathon :) Do they have to be integers? If not, then a possible answer is \sqrt{2013} \ \ \ \ and \ \ \ \ \sqrt{2} Also, my I dont see how my answer has a mistake. I may have forgotten the brackets in \ln{1+\frac{1}{x}}=3\ln{0.5} That 0.5 is 1/2, but I wasnt bothered to put...

Re: HSC 2012 Marathon :) Hmm, I cant find a mistake there

Dont memorise any formulas given to you in any book you might have. Because 1. They give it to you most of the time and 2. If you are asked to derive it, its most likely a different scenario (i.e. the origin not being where the projectile is fired) Remember that at \dot{y}=0 when the particle...

Re: HSC 2012 Marathon :) Thanks man. 1. \ \ \ddot{x}=18x^3+27x^2+9x \\\left(\frac{1}{2}v^2\right)=18x^3+27x^2+9x \\ \frac{1}{2}v^2=18\frac{x^4}{4}+27\frac{x^3}{3}+9\frac{x^2}{2}+C \\ \\ x=-2 \Rightarrow v=-6 \\ \frac{1}{2}(-6)^2=18\frac{(-2)^4}{4}+27\frac{(-2)^3}{3}+9\frac{(-2)^2}{2}+C \\ \\...

Re: HSC 2012 Marathon :) Heh, what is the question asking? I have not heard of product-sum or sum-product ratios, and when I google it, I get sinx cos y= 1/2(sin(x+y)+sin(x-y)) Is that what I have to prove?

Re: HSC 2012 Marathon :) Also I dont understand your question, do you want me to find the volume, rotated by x-axis via normal method, AND simpsons rule, and explaining why they are the same volume? I did that and I got \frac{39\pi \sqrt{3}}{4} They are the same, because Simpsons Rule is...

Re: HSC 2012 Marathon :) Hmm, the volume is not defined? Because the parabola is fully above the x-axis, and there is no enclosed part to rotate about. Or is that the trick? EDIT: I have no idea why the inverse function turned out like that, Ive rechecked my working out, and my algebra seems...

Re: HSC 2012 Marathon :) \text{Is the following statement true?} \\ \\ \int \frac{1}{\sqrt{1-x^2}} = \sin^{-1}{x}+C \\ \int \frac{-1}{\sqrt{1-x^2}}=\cos^{-1}{x}+C \\ \\ Since \ \ \ \int \frac{-1}{\sqrt{1-x^2}}=-\sin^{-1}{x}+C \\ \\ \therefore \cos^{-1}{x}=-sin^{-1}{x} \ \ \ \ \ \ \ \ ? \\ \\...

Re: HSC 2012 Marathon :) y=e^{4x}+5e^{2x}+1 \\ e^{4x}+5e^{2x}+1-y=0 \\ \\ e^{2x}=\frac{-5 \pm \sqrt{5^2-4(1-y)}}{2} \\ \\ e^{2x}=\frac{-5 \pm \sqrt{4y+21}}{2} \\ \\ x=\frac{1}{2}\ln{\frac{-5 \pm \sqrt{4y+21}}{2}} \\ \\ \ln{g(x)}>0 \\ \therefore x=\frac{1}{2}\ln{\frac{-5+\sqrt{4y+21}}{2}} \\ \\...