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Of course I do, I'm just not sure I understand the context.

Whatever do you mean by that?

Haha, that was such a fantastic ep!

Caveat: this dog bites.

Re: 2012 HSC MX2 Marathon |z-2|^{2}=3 \\\\ (z-2)\overline{(z-2)}=3 \\\\ (z-2)(\overline{z}-2)=3 \\\\ z\overline{z}-2z-2\overline{z}+4=3 \\\\ z\overline{z}-2z-2\overline{z}+1=0 \\\\ 2z\overline{z}-4z-4\overline{z}+2=0 \\\\ 3z\overline{z}-3z-3\overline{z}+3=z\overline{z}+z+\overline{z}+1 \\\\...

Re: 2012 HSC MX2 Marathon I suspect that it's sufficient to solve Z=Z^3 and discard -1 as a solution. My reasoning is that there clearly can't be an imaginary part present in Z. With this in mind, we can proceed noting that |Z| = Z, Z > 0 and |Z| = -Z, Z < 0. My method yields Z = 0 or/ 1...

Me.

Re: 2012 HSC MX2 Marathon Possibly a simpler way to do this would be to use the sum of roots to find the final root, and the product of roots to find d.

Re: 2012 HSC MX2 Marathon Prove that: 3|z-1|^2 = |z+1|^2 if and only if |z-2|^2 = 3

That's a very inefficient way of doing the first part. The two quarter-circles cancel out and you're left with finding the area of a trapezium. No integration necessary.

Typically you do, but the above is clearly cyclic.

Re: 2012 HSC MX1 Marathon How do you plan on integrating 1/x over an interval that contains 0?

Re: 2012 HSC MX1 Marathon First proof, fourth line?

It's really a tough one - astute investors have been holding onto it for years, waiting for a gem like last week. Given that the TOL can be scrapped at any moment, I'd say it's more of a punt than I'd normally be comfortable with. Having said that, fortune favours the brave?

JB HiFi is an awful investment at this time. I would punt on something like Lynas (which just got a temporary operating license for processing in Malaysia).

Seems very...enforceable.

Screw you, 35kg is crazy heavy for me.

Re: UNSW Food - Expensive or Cheap? I think it's pretty cheap. You can get a decent feed for under $10.

Awful.

Don't even buy the textbook.