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Thread: Maths tutor available (AMOC/NMSS tutor)

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    Lightbulb Maths tutor available (AMOC/NMSS tutor)

    Need last minute tutoring for HSC? Preparing ahead for HSC next year?

    I am currently a third year science student majoring in (pure) maths. I am willing to tutor all levels of high school maths. I completed the HSC in 2004 and achieved 94 in both ext 1 and 2 maths. In addition I've also achieved, as a student, bronze in the 2004 Australian Maths Olympiad and invitation to National Maths Summer School in 2004 and 2005.

    As a tutor, I have been working as a mentor in the AMOC Correspondence Program since 2005, having guided numerous students in the program to gold and silver in the AMO. I have lectured to high school students on maths at AMOC schools, and have been tutoring at National Maths Summer School since 2007.

    In addition, I have ample experience in private tutoring, having tutored no less than 10 students in the past 12 months, with a significant improvement in results for almost all students.

    I am willing to tutor around the Inner West and around Camperdown (I am willing to travel a bit), and I charge $35/hour. Please PM me if you're interested or need more details.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    If you think you're good at maths then do this question

    Find the point on the straight line 2x + 3y = 6 which is closest to the origin.
    McNulty: I've gotta ask you: if every time Snot Boogie would grab the money and run away... why'd you even let him in the game?
    Prodnose: What?
    McNulty: Well, if every time, Snot Boogie stole the money, why'd you let him play?
    Prodnose: Got to. It's America, man.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by A High Way Man
    If you think you're good at maths then do this question

    Find the point on the straight line 2x + 3y = 6 which is closest to the origin.
    I personally don't see what's the point of posting that, but anyway:

    This point will be connected to the origin by a line perpendicular to the given line. So all you need to do is find that perpendicular line and solve for (x,y), which turns out to be (12/13, 18/13)

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    hey yinan,

    who are the tutors at NMSS 08? and the EG tutor?

    btw ppl looking for a tutor, this guy's pretty good at maths.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by Templar
    Bump......
    Yinan, you still look exactly the same as you did 4 years ago lol.
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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by dark.deceptionz
    who are the tutors at NMSS 08? and the EG tutor?
    Biography should get to you soon. As for the EG tutor, your guess is as good as mine.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Ahh Templar, we all want a job ... *sigh*
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    Re: Maths tutor available (AMOC/NMSS tutor)

    I've got a job, just it finishes by March.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by A High Way Man
    If you think you're good at maths then do this question

    Find the point on the straight line 2x + 3y = 6 which is closest to the origin.
    If you mean intiger point:

    2x + 3y = 6
    2x + 3y = 6 (mod 3 )
    2x = 6 (mod 3)
    2 and 3 are coprimes
    x = 3 (mod 3), y = 0

    or


    2x + 3y = 6 (mod 2 )
    3y = 6 (mod 2)
    2 and 3 are coprimes
    y = 2 (mod 2)
    x = 0

    The closest intiger point is (0,2)

    If you mean the closest real point:

    Consider the two equations

    2a + 3b -6 = 0
    2c + 3d -6 =0

    2(a-c) + 3(b-d) = 0
    But this is equivilent to the dot product of the vectors (2 3) and (a-c b-d)

    as the dot product equals zero, the vector (2 3) is perpendicular to the vector (a-c b-d) which runs along the given line.

    Therefore the line y = 1.5x pass through the origon and is perpendicular to
    2x + 3y = 6

    Solving these two equations simultaneously gives the point (12/13 , 18/13).

    Quote Originally Posted by Templar
    This point will be connected to the origin by a line perpendicular to the given line. So all you need to do is find that perpendicular line and solve for (x,y), which turns out to be (12/13, 18/13).
    Correct
    Last edited by danielbarter; 23 Dec 2007 at 10:32 PM.
    Tutoring All levels of Year 11 / 12 Mathematics. Contact me at danielbarter@gmail.com

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    Re: Maths tutor available (AMOC/NMSS tutor)

    templar's smart indeed...

    you thinking of doing chem honour next year templar? (hiding recruiting intention)


    free bump for ya
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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by xiao1985
    (hiding recruiting intention)
    What makes you think you will succeed when Lou Rendina, Meredith Jordan and various other staff in the department have failed?

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Templar is A++. Highly recommended!
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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    What is the thickness of a thin film (n = 1.58) when it is put over one slit of a double slit arrangement with light wavelength = 550nm incident to it. On an adjacent viewing screen, the 7th order bright fringe appears in the center.
    McNulty: I've gotta ask you: if every time Snot Boogie would grab the money and run away... why'd you even let him in the game?
    Prodnose: What?
    McNulty: Well, if every time, Snot Boogie stole the money, why'd you let him play?
    Prodnose: Got to. It's America, man.

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Bump......

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Quote Originally Posted by A High Way Man
    What is the thickness of a thin film (n = 1.58) when it is put over one slit of a double slit arrangement with light wavelength = 550nm incident to it. On an adjacent viewing screen, the 7th order bright fringe appears in the center.
    2.00000576nm

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    Re: Maths tutor available (AMOC/NMSS tutor)

    Well if you're so good, determine all three digit numbers N having the property that N is divisible by 11 and N/11 is equal to the sum of the squares of the digits of N?
    Also, solve (cosx)^n - (sinx)^n = 1, where n is a natural number

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