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Atar: 99.95

My school: 0 all rounder in 2008 5 this year!

I have no idea what the writing task is about. I'm just preparing for the 'ordinary' questions lol Mine is at 2pm. 5 hours after your interview start time Anyway, good luck with the interview!

ESL 93 Phys, Chem, MX2 97 MX1 98 9th in the state in Chem 99.95 ATAR (because I got Usyd Combined Med Interview Offer)

hi5!! Did you get the email? I didn't get it yet.

Thanks everyone!

Yea, it's an interview offer

You mean, the one who called me? She just asked me if she can speak to me and she offerred the interview, telling me the date and time and that further info will be sent to me through email, although I haven't received it yet.

I just got it! Very excited now :haha:

I forgot lol If I remembered, I would have said it in the first comment.

They didn't

Yea, that's unfair. I've heard that HSC was actually regarded as superior to VCE in the past (don't know when), but they somehow changed it so that ENTER=ATAR lol

I would think so. 1.No 3 hours exams for the VCE. 2.The data sheets that are provided for the VCE are far more extensive. 3.LOL, No 3-8 markers for science (Thus damn easy)

I did this, but in Year 11, not Year 12. For the 1st assessment tak for chem, I came 15th (research task). Then I aced others and came 1st.

LOL, I see. So you guys consider 90 as sufficient mark for one to become a tutor. Personally though, I don't know if it's only me, but I'd never want to pay those with 90/100 for any kind of tutoring service. The knowledge gap between 90 and 97 is quite tremendous. (Of course there are...

cyclohexdiene with numbering in front

\left | z_1-z_2 \right |=\left | z_1+z_2\right | \\$Geometrically, this means that diagonals are equal in the parallelogram formed by $z_1, z_2,$ origin, and $z_1+z_2 \\\therefore $ the parallelogram is a rectangle.$ \\\therefore$ we can let $z_2=kiz_1 \\\frac{z_2}{z_1}=ki \\$which is purely...

lol, um, post the full question?

\\$Since $z_1=kiz_2 \\\frac{z_2}{z_1}=\frac{1}{ki} \\=\frac{-i}{k}$ which is purely imaginary$

\sqrt{1+k^2} =\left | ki-1 \right | \\$is a fact that we know using $\left | a+bi \right |=\sqrt{a^2+b^2}