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I don't know what domain you're looking for so I just gave one answer. You can adjust accordingly in you want a general solution: <a href="http://www.codecogs.com/eqnedit.php?latex=cosx=2cos^2(\frac{x}{2})-1\\ \therefore 1@plus;cosx=cos(\frac{x}{2})\\ 1@plus;2cos^2\frac{x}{2}-1=cos\frac{x}{2}\\...

Re: 2012 HSC MX2 Marathon This is a question from our 4U test this afternoon (I hope I remembered it correctly...): <a href="http://www.codecogs.com/eqnedit.php?latex=\textup{Prove by induction that }\\ (1-a_{1})(1-a_{2})...(1-a_{n})>1-(a_{1}@plus;a_{2}@plus;...@plus;a_{n})\\ \textup{for...

Best advice anyone ever gave me was "Don't make mistakes". By following this advice alone, I topped ext. maths in prelim at a selective school. On a more serious note, in terms of every night, do however much is needed for you to FULLY understand the concept. For example, when we were doing...

Re: 2012 HSC MX2 Marathon sec(90-x)=1/cos(90-x)=1/sinx=cosecx. Similarly, tan(90-x)=cotx

I'd laugh if they give us the same stimulus...

Look at the value inside g(x) as x and the output as y. Therefore it is a linear (straight line) that passes through the points (3,2) and (4,4). The equation of this line is y=2x-4. Therefore g(x)=2x-4. As g(x) is defined only for x is greater than or equal to zero, y is at least -4. Therefore...

Have got English and Maths Ext 2 on Monday. To make it even more fun, in english we are being given an extract from Frank and a still from Bladerunner and we have to analyse those...

1) i)Cuts the x axis when y=0. Therefore at -2,0,2. iii) Using product rule+chain rule <a href="http://www.codecogs.com/eqnedit.php?latex=y=x\sqrt{4-x^2} \\\frac{dy}{dx}=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}\\ \frac{dy}{dx}=\frac{4-2x^2}{\sqrt{4-x^2}}\\...

Even simpler then that, who do you sit with it class? If you sit next to someone else and you don't get something/they don't, you can help each other and that will instantly help you get to know others.

Re: 2012 HSC MX2 Marathon P=NP if P=0 or N=1. So as long as neither of those cases apply, P does not equal NP. Do I get $1,000,000 now? :)

Uh the line y=x+6 only passes through A so it can't connect A and B. The line y=5x-6 does. Is that what you mean?

You can even go directly: "The visual metaphor of the blah blah blah reveals insert theme" etc.

<a href="http://www.codecogs.com/eqnedit.php?latex=z=x@plus;iy\\ Re(\frac{1-z}{1@plus;z})=Re(\frac{(1-x)-iy}{(1@plus;x)@plus;iy})\\ Re(\frac{1-z}{1@plus;z})=Re(\frac{(1-x)(1@plus;x)-iy(1@plus;x)-iy(1-x)-iy(-iy)}{(1@plus;x)^2@plus;y^2})\\...

<a href="http://www.codecogs.com/eqnedit.php?latex=logx^2@plus;log8x=3\\ 2logx@plus;log8@plus;logx=3\\ 3logx=3-log8\\ logx=1-log8^{\frac{1}{3}} \\logx=log10-log2\\ logx=log5\\ x=5" target="_blank"><img src="http://latex.codecogs.com/gif.latex?logx^2+log8x=3\\ 2logx+log8+logx=3\\ 3logx=3-log8\\...

Basically just mutliply the vector z1a by i to rotate it 90 degrees: <a href="http://www.codecogs.com/eqnedit.php?latex=(i)(z_{1}-a)=(z_{2}-a)\\ iz_{1}-ai=z_{2}-a \\z_{2}=a(1-i)@plus;iz_{1}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?(i)(z_{1}-a)=(z_{2}-a)\\ iz_{1}-ai=z_{2}-a...

I haven't even looked at my prelim stuff so far. Honestly though, if you need to use them every now and again, can't you just borrow them back off your brother?

That's how I did it. Do you have any other questions similar to that one (I love those sorts of questions)?

Yeah go for it.

This is how I did it. <a href="http://www.codecogs.com/eqnedit.php?latex=1-y=x\\ 1-3y@plus;3y^2-y^3@plus;y^3=19 \\3y^2-3y-18=0 \\3(y-3)(y@plus;2)=0\\ y=-2,3\\ \therefore x^2@plus;y^2=(-2)^2@plus;3^3=13" target="_blank"><img src="http://latex.codecogs.com/gif.latex?1-y=x\\ 1-3y+3y^2-y^3+y^3=19...

For this question, the easiest way to do it would be to solve both of those equations simultaneously as the numbers work out nicely giving you (3,-2) as the values for x and y (or vice versa). Then simply square both of those and your done.