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So if you don't try then it's hard..... anyway I disagree, savant's don't put much effort in and yet they can produce wonders.

I had a bit of fun with this using taylor series. e^x=\sum_{n=1}^{\infty}\frac{x^n}{n!} so basically \frac{e^x-1}{x}=\sum_{n=2}^{\infty}\frac{x^{n-1}}{n!} differentiating both sides we get \frac{e^x(x-1)+1}{x^2}=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{(n+1)!} let x equal one and bam we...

Grid paper, just for convenience. Otherwise I use blank paper. Lined paper is the bane of every exam.

I would usually encourage you but I hardly studied 2 unit in year 11 and got 100% every exam. It makes it seem like you have to study a lot for whatever reason and can be put off by spin-ball questions. 4 unit is essentially spin-ball questions every step of the way, so if you can't handle them...

Yes, and basically for the same reason as carrotsticks. Let's say I'm incredibly rich and have 5 hovercraft's, if I now go and count how many hovercraft's Kim Kardashian has then I can either say zero or I can stand their dumbfounded at an apparent lexical gap. I like the first option more than...

Yes, what sy123 said is what we do when we find the area usually. We just skip the step because its trivial.

otherwise the equation has no solution.

It really comes down to whether you can handle getting bad marks and still doing a course. 4 unit is a really low workload for the marks it can score you. relatively.

I would go with the least amount of units with maths ext 2. I do 9 units, as i did chemistry last year, and I have a great workload.

oh woops lol.

edit: that was super embarrassing

4 pirates and 497 and 498 have 50 gold each. There are a few important things to note: That unless the number of pirates is two, the average amount of gold per pirate will be less than 50. Pirates will always vote against the distribution if they are going to die. This is because they cannot...

if it was a one or two marker then subbing in the points of p and q would be fine. Otherwise you would have to derive the equation. I would use y-y1 = m(x-x1) then solve using by resolving tan and sec in terms of sin and cos then using sin(A-B) and stuff to simplify the gradient.

Re: 2012 HSC MX2 Marathon $find $ \int \frac{\sin{x}-\cos{x}}{2+\sin{2x}} dx I made this one up myself.... I hope its alright....

Can you make one thread instead of three, it makes it easier for people to respond to all your questions without clogging the forums. Also try to post a picture, scan or at least type up the question so we know exactly what you need help with :)

1) The Legend of Zelda: A Link to the Past 2) Metroid Fusion 3) Pokemon Crystal Version 4) Halo 3 5) Red Dead Redemption The first one is the best game of all time as far as I am concerned and I would easily spend thousands of dollars to be able to experience it for the first time again...

You get exposed to different ways of thinking mathematically, my ability to do equations and working in my head shot up immensely at the beginning of year 11.

Okay, ADHD is not an illness as such but a cluster of common traits that synergistically result in poor concentration. I have it and I am on concerta, dex may not be doing it for you as it isn't slow release. Now I can empathise with low productivity but I find it is due to anxiety issues and...

Look, at the point x=0 y=arctan(x) has a gradient of 1 and y=x has a gradient of 1 and they both have a value of y=0. SO AT THE POINT x=0 THEY ARE BASICALLY THE SAME GRAPH! So they are equal at that point and the ratio between the two is 1, comprendai?

You need to get rid of your colossal moobs