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I can speak with the utmost grammar, but still talk like a 10 year old. For example, most people who dislike English argue on here like a pack of geese, their words might have some meaning and emotion put into them. But nevertheless...

Tell me, how would the best mathematicians speak if they spoke like monkeys? With mathematical symbols? Sorry no mathematician ever uses solely mathematical symbols and basic words, explanations of proofs need to be made. Scientists need to publish their discoveries, would you listen to a...

In relation to Rawf's post. 1) In response to how you doubt that all composers dont mean what English teachers say they do. You will be surprised at the real depth some deep texts take. But its important to be able to form your own opinion. Not every forest is supposed to be an obscure...

im going to go study Chem now. Im going to die probably

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Go and study Business heh.

Aerrooonautticallll engineeeeeeeeeeerrrr

There is a reason why x=-2 is not a vertical asymptote, its because we can simplify our fraction a bit: y=\frac{x+2}{x^2-4} \\ y=\frac{x+2}{(x+2)(x-2)} \\ \\ y=\frac{1}{x-2} So it cant be an asymptote, because we have a cancellation going on. BUT this doesnt mean that x=-2 is a valid point...

Potential for 99.95, it depends on externals Alot of it now is dependent on externals.

HINT: (concept) ak+bl=ck+dl Therefore a=c b=d EDIT: For the quadratic question it should be b^2 - 4ac lol, silly mistake (and alot of people missed it too heh)

Heylo

ಠ_ಠ

lol the VCE is easier than the HSC, much easier

$i) Solve the equation using completing the square method$ \\ \\ ax^2+bx+c=0 Next question $Given that the equation$ \\ (x-\alpha)(x-\beta)=x^2+bx+c \\ \\ $Find$ \\ $ii)$ \ \alpha \beta \\ $iii) \alpha + \beta \\ \\ \text{Find both in terms of b and c}

For Economics how many statistics do I need in order to get 18+/20 for essays (yearly). Presuming I write and answer brilliantly. Also while Im talking about Economics, does it normally always scale better than any other conventional subject (not a language or extension), or has it only...

\lim_{x\to -\infty}\frac{1}{1+x^2} \\ =\lim_{x \to -\infty} \frac{1/x^2}{1/x^2+1} We know how to do this, what we are doing here is we arent finding the limit, we are simply transforming the question in a way where we can use our limit rule: \lim_{x \to \infty} \frac{1}{x}=0 This is what we...

ftfy

Re: Math help (Because Carrot made me do this) Yes that is exactly right, since we have no vertical asymptote however, there are no points to test that are around the vertical asymptote (because there are none). We have already 'subbed' in infinity (you found the limit). Remember earlier how...

If Im 2nd for 3U and 2U Maths, and both me and 1st place get say 100% Raw, do I miss out on state rank completely, despite my internal mark moderated to about 98-99?

Are your subjects at the very very extremeties of scaling? Ive heard Extension Classical Greek has scaled more than 4U maths once