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I meant for people to oppose so the discussion is actually going, are people referring to the question anymore? No. Has it created discussion? Yes. So it has served its purpose and doesn't need to be brought into light again, so please instead of pointing out silly observations, maybe...

Ok, so just going over a paragraph over and over in my head, with no concern over the meaning of anything that you are reading and 'roting'. Sure you can just spit the words out on another page but you don't gain any intuition/skill or anything at all from it. Last time I remember, I was told...

But it's not learning.

History can be a very limited exception, you do need rote learning but critical thinking is definitely required. Moderate =/= Mediocre, I would say that in general Moderate would be 80 and yeah Mediocre would be around 70. And that's fine, if people want to cheat the education system and skip...

Re: HSC 2013 4U Marathon $i) $d_{r+1}=u_{r+2}-u_{r+1} d_r=u_{r+1}-u_r \\ 2d_r=2u_{r+1}-2u_r \\ 2d_r+u_{r+1}=3u_{r+1}-2u_r = u_{r+2} \\ \therefore 2d_r=u_{r+2}-u_{r+1}= d_{r+1} $ii) $ d_{n-1}=2d_{n-2}=2(2d_{n-3})=2(2(2d_{n-4}))= 2(2...(d_0)..) = 2^{n-1}d_0

Rote learning is NOT learning. Here is a message to everyone who misunderstands the damn point of this thread. The point is, rote learning leading to a good mark should not exist in the first place.

Woah, I thought 3 hours was long heh, I haven't gone tutoring before so my views are probably not that important to you. I think 3-4 (maybe 5) hours straight per week would be optimum for me, but yeah my views should be taken with a grain of salt.

What time period is considered short or long? ============== In my opinion about 90% of the learning done is by the student. It is the student's choice to listen throughout the whole lesson, if they can't keep up the whole time and fail to ask questions then that is their fault and no amount...

Re: HSC 2013 4U Marathon Yes, edited now.

Re: HSC 2013 4U Marathon I haven't learnt conics yet but this was on the second page (and no one answered all of it yet)

Re: HSC 2013 4U Marathon $The plane curve defined by the relation$ \\ \\ x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}} \\ \\ $is called an astroid curve$ Note, replace the 1 with a. The image in the graph sets a=1 $Prove that the length of the tangent to the astroid curve at the point$ \...

Re: HSC 2013 4U Marathon Just the main points will do, and like a one sentence explanation between each 'major' line of working (i.e. skip the non important algebra)

Ok since latex doesnt work. I will walk you through the steps. Part i) Yes you are correct y= (x^2)/4 + 1 ii) Now we are asked to find a minimum argument, this can be observed that the minimum argument is the point where it is a tangent to the parabola from the origin. So our objective is to...

I made a crucial mistake somewhere, I apologize -> rewriting my solution now. (and for part i you are right)

Re: HSC 2013 4U Marathon \text{Given that} \ \ x^m y^n = (x+y)^{m+n} \\ \\ \text{Show that} \ \ \ \frac{dy}{dx}=\frac{y}{x} ========== Is latex broken or something?

Re: HSC 2013 4U Marathon Ah I see there, well you just posted the solution for Step 1, if I get rid of the strength of the assumption and loosen it up is it correct? EDIT: And I am guessing that: P(x)=\sum_{k=0}^{n} f^{(k)}(a) x^k ??

Technically you can get any atar with any subjects, but do 4U maths to increase your chances at a better ATAR (only do so if you are competent and willing to learn mathematics)

Re: HSC 2013 4U Marathon This is my attempt: $Step 1: n=1$ \\ \\ $By definition$ f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a} \\ $There must exist a polynomial P, of degree 1, in the form$ \ \ P(x-a) \ \ $such that$ \\ P(x-a)=f(a) \ \ \ $IFF$ \ \ \ x \to a $i.e. a polynomial must exist such...

Re: HSC 2013 4U Marathon Last question - last part yeah, I thought the intuition you needed would be great for a marathon question (short and sweet).

Re: HSC 2013 4U Marathon $Given that$\\ \\ \frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+ \frac{1}{7^2}+\frac{1}{9^2}+... $Find (displaying full working) the value of$ \\ \\ 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...