# Thread: Maths tutor available (AMOC/NMSS tutor)

1. ## 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.  Reply With Quote

2. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

3. ## 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.  Reply With Quote

4. ## Re: Maths tutor available (AMOC/NMSS tutor) 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)  Reply With Quote

5. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

6. ## 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.  Reply With Quote

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

8. ## Re: Maths tutor available (AMOC/NMSS tutor) 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.  Reply With Quote

9. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

10. ## Re: Maths tutor available (AMOC/NMSS tutor)

Ahh Templar, we all want a job ... *sigh*  Reply With Quote

11. ## Re: Maths tutor available (AMOC/NMSS tutor)

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

12. ## Re: Maths tutor available (AMOC/NMSS tutor) 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). 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   Reply With Quote

13. ## 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  Reply With Quote

14. ## Re: Maths tutor available (AMOC/NMSS tutor) 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?   Reply With Quote

15. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

16. ## Re: Maths tutor available (AMOC/NMSS tutor)

Templar is A++. Highly recommended!  Reply With Quote

17. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

18. ## 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.  Reply With Quote

19. ## Re: Maths tutor available (AMOC/NMSS tutor)

Bump......  Reply With Quote

20. ## Re: Maths tutor available (AMOC/NMSS tutor) 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  Reply With Quote

21. ## 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  Reply With Quote

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