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    HSC & Preliminary Tuition (99.75 ATAR)

    Hi all, I am advertising tuition services in the subjects: Mathematics Advanced/Ext. 1/Ext. 2, Chemistry & Physics. My 2013 HSC marks were: English Advanced - 93 Mathematics Ext. 1 - 98 Mathematics Ext. 2 - 93 Chemistry - 93 Physics - 94 This gave me an ATAR of 99.75 I am...
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    HIGH SCHOOL TUTORING (YR7-12) (99.75 & 99.85 ATAR tutors)

    "Achieve your dream ATAR, by studying smart not hard." HOURS OF OPERATION: Monday: 9:00am to 5:00pm Tuesday: 9:00am to 5:00pm Wednesday: 9:00am to 5:00pm Thursday: 9:00am to 5:00pm Friday: 9:00am to 5:00pm Saturday: 10:00am to 4:00pm Sunday: 10:00am to 4:00pm We...
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    UNSW Co-op Scholarship (Finance & Banking)

    Apologies, was on phone :)
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    UNSW Co-op Scholarship (Finance & Banking)

    Yup, I am only interviewing for the Finance & banking scholarship. Mine is on 26th Nov, 9:30 am Wbu?
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    UNSW Co-op Scholarship (Finance & Banking)

    Thanks for the response, If you don't mind me asking, which course are you undertaking for your coop?
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    UNSW Co-op Scholarship (Finance & Banking)

    Hi everyone, This is just a general post on the UNSW Co-op Scholarship interviews (specifically, Finance & Banking): Could any current co-op scholars or any successful candidates who have been eligible for an interview, shed some light on what the interview will be like (as detailed as...
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    HSC 2013 MX2 Marathon (archive)

    Re: HSC 2013 4U Marathon Hey guys, a short question. Pretty simple but i really liked it :) $ If $ \alpha , \beta $ are two complex numbers where $ \alpha\neq\beta $ and $ |\alpha|=1, $ show that $ \left |\frac {\alpha\bar{\beta}-1}{\alpha-\beta} \right |=1
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    Complex numbers help?

    I think it was one of those promotion pens? :) Just a normal ball-point pen i think, Why?
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    Complex numbers help?

    What they said but here's a diagram which may ALSO help you understand it a bit better :) NOTE: This solution is "diagram dependent", so it only proves one case. However it will work wherever you draw Z1 and Z2, yielding a result of either pi/2 or -pi/2, which means it is always purely imaginary.
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    Yh, Sorry just saw you're message about the Dot Points books, which ones did you have? Could you...

    Yh, Sorry just saw you're message about the Dot Points books, which ones did you have? Could you link me to the thread?
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    Preliminary Electricity: Potential in a Circuit

    Kirchoff's Law of Voltage Drop :)
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    Hey, does it have any writing in it? How's the condition, etc.? Can you go down a bit further...

    Hey, does it have any writing in it? How's the condition, etc.? Can you go down a bit further still, it's still a bit pricy >< a lot of people are selling on BOS for 10 or 15 and i'm buying 2 at once ahaha :)
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    Simpson's Rule (Integration)

    \\ $ 5. (I THINK this question is done similarly to question 4.) \\\\ $ V = \int ^{h}_{0} A(x)dx \\\\ \begin{align*} A &\approx \frac{h}{3}[y_{0} + y_{4} + 2y_{2} + 4(y_{1} + y_{3})] \\ &= \frac{20}{3}[64 + 88 + 2(146) + 4(98 + 160)] \\ &= 9840 m^2 \end{align*} \\\\ $ Therefore $ V = 9840 m^3
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    Max and Min Problems (Differentiation)

    \\ \pi r^2 h = 16 \pi \Rightarrow r^2 = \frac{16}{h} \\\\ SA=2 \pi r^2 + 2 \pi r h = \frac{32 \pi}{h} + 8 \pi \sqrt{h} \\\\ \frac{d}{dh} SA = - \frac{32 \pi}{h^2} +\frac{4 \pi}{\sqrt{h}} = 0 $ for max/min$. \\\\ \frac{4 \pi}{\sqrt{h}} - \frac{32 \pi}{h^2} = 0 \\\\ 4 \pi h^2 - 32 \pi \sqrt{h} = 0...
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