Recent content by munch0r

Re: Multiple choice answers Yep 14 was A, 15 was C

i just did 8!/9! number of perms with them sitting opposite / number of total perms

is the answer 1/9? i'm currently doubting it but meh

f(x) = x^3 - 3x +4 f'(x) = 3(x^2 - 1) if the curve is decreasing, f'(x) is negative f'(x) is only negative when -1 < x < 1 lol beaten -__-

as x approaches infinity (sinx)/x = 0 but for the question sin(pi/2n) / (pi/2n) as n approaches infinity pi/2n approaches 0 so think of pi/2n as x (sinx)/x = 1 where x is approaching 0

where exactly is sin x/x = 1 as x--> infinity in any textbook. i can only find sin x/x = 0 as x --> infinity according to the "squeeze theorem"

LHS = | x + yi - i | RHS = y Therefore | x + i(y-1) | = y sqrt(x^2 + (y-1)^2) = | x + i(y-1) | x^2 + (y-1)^2 = y^2 2y = x^2 + 1 y = 1/2 . (x^2 + 1)

yes, verified it using slices, got the same answer...

tan-1(4) - tan-1(3/5)= pi/4 tan (tan-1(4) - tan-1(3/5)) = tan (pi/4) RHS = 1 LHS (using double angle formula) = (4 - 3/5) / (1 + 12/5) = 1 LHS = RHS :)

usually the only thing you have to do is, take tan of both sides and use double angle formula for LHS....

volume of a shell = 2 pi x y dy so for y = x/2 x = 2y so 2pi (2y^2)dy and for y = sqrtx x = y^2 so 2pi (y^3) dy

answer is 8pi/3 u^3 if you want to do it via shells rotated around the x axis you have to do shells for y=x/2 minus shells for y = sqrtx V of a shell = 2pi (2y^2 - y^3) dy Integral from 2 to 0 of 2y^2 - y^3 = 2pi [ 2y^3/3 - y^4/4] = 2pi(16/3 - 4) = 8pi/3

oops made mistake :/

do you have an actual answer for this, so i can verify if i am right? :)

Re: HSC Starts tomorrow? I wish the hsc was tomorrow, I've run out of test papers