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Re: 2012 HSC MX1 Marathon One question that's quite irrelevant, sorry, how do you make the latex "next line" thing so neat? E.g. \\ for next line, but it looks really uneven for some reason. How do you do a double spacing? Also, if I type something that is not latex and then latex...

Re: 2012 HSC MX1 Marathon Actually, I was wondering, why did you multiply with the second 4!? 10C5 determines the combinations for the first table, 4! for the second table. Obviously your calculations are correct as 10!/(5!5!) x 4!4! = 10!/5^2 but I don't understand how it works.

Re: 2012 HSC MX1 Marathon Thankyou! Managed to work out 1) and 2) during the time but for 3), I had no idea how to approach the question.

Re: 2012 HSC MX1 Marathon Actually I needed help with this probability question: There\ are\ 5\ boys\ and\ 5\ girls\ in\ a\ gymnastics\ squad.\ In\ how\ many\ ways\ can\ they\ be\ formed\ into\\ a)\ one\ ring,\ with\ no\ two\ girls\ next\ to\ one\ another;\\ b)\ two\ cocentric\ rings,\ with\ a\...

Was Lv 132 Bowmaster... what a waste of my time haha.

I would probably avoid him but did you add the P plate one as a joke? I REALLY doubt that he would come to believe that. EDIT: Sorry to ruin the joke unless it is a real fact but I doubt anyone in NSW really believes even half of these.

But if you do MX2, its awesome :P

I was wondering why that seemed so similar but then realised your from Mackellar. Anyway sorry I can't really help too much.

Right, thinking logically is the key isn't it :P Thanks for both yours and Spiral's help ^^.

What the... ?

The answer does go and add 1 + 3^1/2 and 1 - 3^1/2. Where did they get the 1 - 3^1/2 from then? They say, since 1 + 3^1/2 is a root, then so is 1 - 3^1/2.

Oh, I figured out that you can just copy + paste the link they provide you with. Oh well :p

So the question is write down an equation of the lowest possible degree with i) complex coefficients, II) rational coefficients and having the following among its roots. It gives you <a href="http://www.codecogs.com/eqnedit.php?latex=\sqrt{3} @plus; 1 \and \2-i" target="_blank"><img...

Thankyou :] I was going to get some sleep anyway. I genuinely didn't know that quadratics have conjugate surds as their roots.

yeah... we have 2 hours + extra if needed a week.

Thanks, that's a pretty nice method. One thing though, wouldn't it be a bit confusing once you get some more complex numbers. After that, the long division bit will get a bit weird. Correct me if I'm wrong though :P

Oh, of course. Thanks for that, 1 last question, I promise. In the next part, the solutions (for the rational bit), start off by saying if 1 + root 3 is a root, than so is 1 - root 3. Surely they aren't using the conjugate root theorem right? Since it isn't even a complex number.

One more question please :D Edited above.

andrew sucks

Also, some schools might have the minimum amount of MX2 classes, (2 a fortnight, 1 hour lessons) or maybe a higher amount of another particular class, leading to 'more free periods' etc etc.