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nope. Just first 3 basic chapters (I think).

Solve tan^2 x - 1 = 0 for 0 <= x <= 2pi

ah cheers, I had it right. 1/2 * (ln(17) - ln(5)) = 0.61 units^2 I went one step further and made it 1/2 *(ln(12)) ' which is indeed very very wrong. thanks.

nonono not 4.2 its just 4

Find the area bounded by the curve y=x/ (x^2 + 1), the x-axis and the lines x = 2 and x = 4. 2 dec place. what I ended up with is 1/2(ln(17) - ln(5)) which is wrong

yes, correct

11 and original poster is hot.

17 the &^uk? thats nearly two HSC ATARs :/ lol i wonder if its possible to do 10 units accelerated 1 year then 10 more the next yr and get 2 atars, that would screw up the system and you as well :P

what's the max? and what is the highest anyone has ever done?

so basically you can get a band 1 for english but get band 6s for you other 8 units of subjects and still get an all rounders?

yeah thought so. How is that lazy? its more effort to actually create a thread and then wait and keep checking back to see if someone responds -_-

is it band 6 in your 10 best units, or band 6 in ALL your subjects?

it just means to program by using separate modules for different functions / tasks. So in VB for example: you can write a sub. e.g sub Startmovement1 Timer1.enabled = True End sub now this sub can be used anywhere within the program etc, and it will still perform the same function, i.e...

no it wouldn't, 4 o'clock would be 120 degrees

60 degrees. (180 / 6) = 30 degrees, its 2 o clock - 30*2 note: divide by 6 because there are 6 hours within the 180 degrees. i.e 1,2,3,4,5,6

ah yes yes, sorry LMAo, I'm all over the shop today. yes find the indefinite integral (primitive function)

lol wut? the answer is (2x + 5)^8 / 16 + C

I think the answer for the book is wrong on this question: Integrate through indefinite integral: (2x + 5)^7

okay think I got it now. yes I did get (pi - 2) units^2 so for the quarter of a circle, you don't use integration as it won't give you the exact answer of pi. (requires the user to look at a different angle when solving this question)

x^2 +4x + 4 = 4x - x^2 4x = 4x - 4 x = x - 1 the fuck? __________________________ (x+2)(x+2) = x^2 + 4x + 4 therefore x^2 + 4x + 4 = 4x - x^2 2x^2 + 4 = 0 therefore 2x^2 = -4 x^2 = -2 x = sqroot -2 = no real roots the fuck?