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how much is serious enough

Hey cem this may be so vague and so naive that it's not even a reasonable question but what are the markers knowledge of everything ^.^ .. what I mean by that is, if they read someone's essay and that person has for instance taken information from the internet and changed the words around and...

His school rank is shit.. But honestly, it's naive for anyone to put a mark on what you are doing, since you are coming first in almost all of your subjects, it is almost 100%, entirely dependant on your performance in the final exam. You could get low 70's or you could get in the 90's. ^.^

You didn't just post your essay did you?

You still didn't specify why you are actually posting this as spiral said.

>implying E3 is bad.

hahaha I lol'd

Hmmm.. this is a hard one..

Oh boy! :D

Isn't it just that the electromagnets make a magnetic field to the left whilst the current in the coil is going from left to right? so wouldn't it be (B) because of motor effect? I duno probably wrong.

Now guys ^.^, please don't think I'm a faggot trying to "prove my math skills" in a sea of genius's haha, cause I realise I'm only doing ext1 :(, I am just trying my luck at lots of questions and building my Mathematics vocabulary (very slowly ^.^) (and attempting to get better at Latex.. pfft)...

You deserve a medal for your contributions bro, Id'e like to thank you personally for the time you take to do this :3

Oh man how embarrassing, sorry for that :(

Some schools allow you to print for free.. some schools.. not mine :(

Never seen THAT before :O that's so cool, I'de just do ye ol'e \small $,Let u = sinx$\\\\ $When x = pi/4, u = 1/\sqrt{2}$\\ $When x = 0, u = 0\\\\ \therefore \frac{du}{dx} = cosx\\\\ \therefore du = dxcosx\\\\ \therefore \int_{0}^{\frac{1}{\sqrt{2}}} u^{2}du\\\\\\\\ Blah (Don't mind me, I'm...

\small \therefore \frac{2^{n+1}}{n+1} = \binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+...+\frac{1}{n+1}\binom{n}{n}....(1)\\\\ \therefore \frac{1}{n+1} = \binom{n}{0}-\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}-...-\frac{1}{n+1}\binom{n}{n}....(2)\\\\ $Subtract (2) from...

It's targeted at more mathematics 2u and above students because people who don't do this do not know 2sinxcosx = sin2x. So no, they don't expect you to derive it, it just makes things easier and shows you know what your doing.

y = p^{2}+q^{2} \\ y = (p+q)^2 -2pq...(1) \\\\ $however$\\\\p^2 + q^2 -6pq = 0\\ \and$ x^2 = (p+q)^2 \\\\ (p+q)^2 -8pq=0\\ x^2-8pq = 0\\ x^2 = 8pq\\ x^2/4 = 2pq ....(2)\\\\ \therefore $substitute (2) into (1)$\\\\ y = x^2 -\frac{x^2}{4} I am tired and so it could be wrong, and plus the working...

Wtf, how can they have the same code

I did mine today in 3 - 5 as well.