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\frac{1}{\sqrt5+\sqrt2}+\frac{3}{3\sqrt2-\sqrt5}\\\\$Let's rationalise the denominator for both...

Probably should have done this beginning of last week but since i would like to know more people i'm gonna ask who here does PHYS1901 and/or MATH1901 and/or MATH1902? :D

bump

oh damn not fast enough XD

LHS=(\cot\mathit{t}+ \textup{cosec}\:\mathit{t})^2\\\\\indent\indent\:\:\:=\cot^2\mathit{t}+2\cot\mathit{t}\:\textup{cosec}\:\mathit{t}+ \textup{cosec}\:^2\mathit{t}\\\\\indent\indent\:\:\:=\frac{\cos^2t}{\sin^2t}+\frac{2\cos t}{\sin t\times\sin...

i'm in maths :)

I am :D which subject group you going to?

OMG they're out! XD*checks*

visual techniques are used to convey the ideas from an image or visual thing they want you to imagine/see. I never did 'The Loaded Dog' for HSC so i can't give you an example:/

\log_a2=\log_b16\\\\\indent\frac{\log_{10}2}{\log_{10}a}=\frac{\log_{10}16}{\log_{10}b}\\\\$We times $\log_{10} b$ up to LHS and times the $\log_{10}2$ down to the RHS so we can simplify the...

Ok well zero can also be written as 0+0i, right? This basically means that real part is equal to zero and imaginary part is equal to zero as well.

$So you are given that $\left | z \right |^2-2iz+2t(1+i)=0$ and we let $z=x+iy:\\\\x^2+y^2-2i(x+iy)+2t(1+i)=0\\\\$Collecting real and imaginary parts we get:$\\\\x^2+y^2+2y+2t+i(2t-2x)=0\\\\$So if we equate real parts to zero then we have:$\\\\x^2+y^2+2y+2t=0\\\\x^2+y^2+2y=-2t\\\\$You can see...

Re: 2012 HSC MX1 Marathon Oh woops I just realised haha my bad...fixed it anyway :D

Re: 2012 HSC MX1 Marathon $Ok so our aim here is to find the constant term of both ${\left(x+\frac{4}{x}\right)}^{2n}$ and ${\left(\frac{x}{a}+\frac{b}{x}\right)}^{2n}$ so we use $T_{r+1}$ for both. Firstly do this...

me too :)

HAHAHA i just happened to stumble across this thread! HIIIIIIIIIIII:D yeh basically what ben150891 said. vet science ain't gonna be easy and would require all your 5 days/week. The same goes for any science degree, I know i'm gonna be doing 5 days/wk as well for adv science so yeh. Go to link...

$So you are given: $z + \frac{1}{z}=k,$ where k is real$\\\\$First, let z=x+iy so:$\\\\x+iy+\frac{1}{x+iy}= k\\\\x+iy+\frac{1}{x+iy}\times \frac{x-iy}{x-iy}= k\\\\x+iy+\frac{x-iy}{x^2+y^2}= k\\\\$Since we know k is REAL then we can group all the real parts and the imaginary parts and make the...

No not really. Usually some of the high achievers have one exam they stuff up but as long as they do well in other exams they can get a good atar. Like Aysce said, learn from your mistakes. Always ask questions on how you can improve and keep working hard so you can do better:)

ok thanks

I tried registering online but it keeps saying incorrect SID or Last Name :/