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  1. scardizzle

    Another Question "-_-

    |2x - 1| < / = |x -3| This is another novincial thing I've forgotten how to do. i took the case where both sides are positive which gives me x >/= -2 then i took the case where both sides are negative which gives me the same answer but what about when one side is negative and the other side...
  2. scardizzle

    Drawing Inverse Functions

    Anyone got any tricks for drawing inverse functions? I'm sure there was one i learnt but i' ve forgotten. Any tips or tricks are appreciated.
  3. scardizzle

    Just did a practice trial, raised numerous questions!!

    oh my bad, misread your post(nice use of OP :P) fine let a = v^2 +3 or something and just follow those steps
  4. scardizzle

    Just did a practice trial, raised numerous questions!!

    1. if A(x1,y1) B(x2,y2) are divided in the interval m:n then multiply the A with n and B with m you pretty much switch it around 2. Lets say we are given the equation a = v^2 + 3 and we want to find an equation for velocity There are 2 ways to do this that i know of: i) a = dv/dt so...
  5. scardizzle

    Binomial Theorem

    Thanks, i get it now
  6. scardizzle

    In need of Sydney Grammer Papers

    ah you have to draw your own line, of course! thanks for the help
  7. scardizzle

    Binomial Theorem

    Hey guys, having trouble with this question: In the expansion (1+x+kx^2)^9 in ascending powers of x, the coefficient of x^2 is zero find the value of k. there are so many coeff of x^2 im sure there is another way then expanding everything out. Thanks in advance
  8. scardizzle

    In need of Sydney Grammer Papers

    Since the paper is so conveniently posted here ... can anyone explain to me q7 b) part (i) ? please forgive me if it is a stupid question but i just cant see it :(
  9. scardizzle

    Parametrics HELP

    okay here's my attempt: differentiating 4ay = x^2 gives us y'= x/2a therefore gradient at x = x1 is x1/2a now to work out the gradient of SQ since we know S is (0,a) we must find out Q Eq'n of the tangent at P is: 2a(y-y1)=x1(x-x1) for the point Q y must = 0 since it intersect the...
  10. scardizzle

    [message deleted]

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  11. scardizzle

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  12. scardizzle

    Probability

    i get 1/5 my working: possible outcomes with B: RB WB BB BR BW each has a 1/6 probability of occuring therefore the probability of selecting BB is 1/5 if one must be B
  13. scardizzle

    Arthur please stop posting on my account t_T

    Arthur please stop posting on my account t_T
  14. scardizzle

    Another perm and comb question

    how have you accounted for the ascending order part of the question? Also since this is a card question order doesn't matter so why have you divided by 40P5
  15. scardizzle

    Year 11 and 12 students, HOW DO YOU DO IT?

    That's not very nice, I'd advice to get a straw so you can suck it up
  16. scardizzle

    Another perm and comb question

    yeh i asked bakes about this question the question is unclear what it means is having the cards in a straight e.g. 12345,23456 etc. therefore solution is 4C1x4C1x4C1x4C1x4C1 x no. of straights you can make just write it out it, doesnt take too long then divide this by 40C5
  17. scardizzle

    Warning: to Windows 7 users

    Torrent it? and use the windows 7 loader that's floating around if you cbf to sign up
  18. scardizzle

    The best maths book for 2unit+3unit [poll]

    Agreed The font makes it worse, the letters are so thin which makes it annoying to read I reckon if they made it a thicker font and popped in a couple of pictures here and there it would be better
  19. scardizzle

    Arrangement in a circle

    this means 3 out of the 3! permutations are the same since you can rotate the circle
  20. scardizzle

    Arrangement in a circle

    Consider a straight line of men alternating with women MFMFMF if this were made into a circle it would the perscribed arrangement so men can be arranged 3! ways and women can be arranged 3! since it is a circle we do not need therefore total arrangement is = to 3! x 3! = 36 not sure if this...
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