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sinx + cosx = 0 sinx = -cosx divide by cosx tanx = -1 now solve

They are right in both cases. It all comes down to interpretation. In the first example, the rock never goes upwards, it only ever falls down, thus they take down to be the positive direction. Thus g is also positive, since g goes downwards. However itd work out even if you take g as negative...

Personally, I found q10 in 2004 easier than 2003, the rest I found pretty much on par.

If you get gnuplot, you can define as much composite functions as you want and graph them all together.

The easiest method would be something taught in 4 unit, where you sketch the graphs of those two functions separately, and then add the ordinates. Simlarly for the ones where you divide or multiply. There are a number of things you need to consider: all that Dreamerish*~ said, including the...

Oh right. You've never come across ln in functions? If youre in yr 11 then that'd explain it.

ln is the natural logarithm, also written as loge since it is of base e. Other then that, its pretty much the same as any other log, at least as far as the HSC is concerned. Consult http://en.wikipedia.org/wiki/Natural_logarithm for more info. Now what is fns?

Pre-enrolled in 3 maths and 6 computing subjects. Probably wont be here next year though, but i guess its good to have a backup just in case.

http://community.boredofstudies.org/showthread.php?t=86488 Read that. As ive said before, its impossible to give such estimates for any specific year before that cohort has actually sat the hsc.

You split the integral: Integrate v from 0 to pi/3 and take the absolute value Integrate v from pi/3 to pi Then add the two Ill calculate and edit my post with my answer edit: I get 4sqrt(3) + 2pi/3 Answers to 2003 can be found here...

Its been a problem for some time, the admin are aware and are working on it.

Have you tried Geff Jeha's solutions: http://www.boredofstudies.org/courses/maths/3u/2003_Mathematics_Extension_1_HSC_Sol_Jeff_Geha.pdf ?

which parts specifically?

y = ((tanx + 1)/sec x) - ((cotx+ 1)/cosecx) = cosx(tanx + 1) - sinx(cotx + 1) = cosx(sinx/cosx + 1) - sinx(cosx/sinx + 1) = sinx + cosx - cosx - sinx = 0 Which is a constant

y = 50sin30t - 4.9t^2 Make y = 0 for range 50sin30 - 4.9t = 0 t = 50sin30/4.9 x = 50tcos30 sub that value of t into x and youll get x = 221m

Theres no way to tell, it depends on what you are ranked in your school, how your school performs, what mark you actually get in the hsc exam etc. If you are ranked first in your school, then a hsc exam mark of 95 will give you a final mark of 95.

3C2 = 3 There are 3 possible combinations: (2, 2, number other than 2) (2, number other than 2, 2) (number other than 2, 2, 2) Hence you need the 3C2 Doing it the 2 unit way, the answer is: (1/6.1/6.5/6) + (1/6.5/6.1/6) + (5/6.1/6.1/6) But this just equals 3(1/6.1/6.5/6)...

Hahaha yes i remember that. http://community.boredofstudies.org/showthread.php?t=48400

http://www.boredofstudies.org/courses/maths/3u/1109069740_2004_Mathematics_Extension_1_HSC_Jan_Hansen.pdf http://www.boredofstudies.org/courses/maths/3u/2003_Mathematics_Extension_1_HSC_Sol_Jeff_Geha.pdf

Thats wrong. The board of studies pre-determines what raw marks correspond to what band, so theoretically everyone can get a band 6 if they all get above the band 6 cutoff that the board determined. And to answer the question asked, no one can tell you what this cutoff is. That is also not...