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  1. Affinity

    Anybody from Asia?

    楼主还真幽默
  2. Affinity

    How to do this projectile question

    oh I made a mistake there.. 26.25 - 15 is 11.25 not 11.5 my bad.. Anyway, the main purpose of that post was to show that one can use the familiar idea of energy from physics.
  3. Affinity

    More applications of calculus...

    let R = sqrt(a^2 + b^2) x = R [ (a/R)sin(nt) + (b/R)cos(nt)] recognise that a,b,R form a right angled triangle, so a/R = sin(w) and b = cos(w) for some angle w = arcsin(a/R) so can rewrite x = R(sin(w)sin(nt) + cos(w)cos(nt)) = R(cos(nt - w)) should be easier now I hope beh upstairs beat me...
  4. Affinity

    Engineers of UNSW, need your help

    so erm.. which particular subjects do you have so far
  5. Affinity

    How to do this projectile question

    In maths if you aren't given g then you write g as g.
  6. Affinity

    How to do this projectile question

    From the question we know that at some particular time T , y'(T) = 0, y(T) = 26.5 and x(T) = 30 1. show that [(y')^2 / 2] + gy is constant 2. from that we can deduce that y'(0) = sqrt(2*11.25*g) = sqrt(22.5g) 3. ofcourse y'(t) = y'(0) - gt, so in particular, 0=y'(T) = y'(0) - gT so T = y'(0)/g...
  7. Affinity

    is it possible part 2

    look up devil's staircase for something just as interesting. it is also probably possible to construct a continuous function which is differentiable almost nowhere.
  8. Affinity

    Probability questions =S

    35/36 (you just need 1 even number) (6 choose 2) (1/6)^2 (1/2)^4 (pick where the 6 goes then multiply the probabilities) 1/6 * (5 choose 3)(1/2)^3 (1/3)^2
  9. Affinity

    Is it easy to transfer from a double bachelor degree to a single one?

    generally yes and yes... you should call the university administration though.
  10. Affinity

    Syllabus development

    collective myopia and politics drives any process.
  11. Affinity

    conics again

    method 1: find the equation of a tangent and the perpendicular, solve, etc. (this is the usual way) dy/dx = dy/dy / dx/dt = -1/t^2 so the tangent is (y - c/t) = -(x - ct)/t^2 -------> x + yt^2 =2ct and the perpendicular has gradient t^2 through the origin, so its formula is y = t^2 x...
  12. Affinity

    acturial anybody?

    no, you can try for internships after first year though... how many actuaries do you think they need?
  13. Affinity

    a function....

    then you go lim [h-> 0] |f(a+h)-f(h)| = lim [h -> 0] |f(h)| = 0 so f is continuous at a. or even without hte absolute value signs
  14. Affinity

    Need Help with Accounting

    oh you used running averages which is technically more correct :P
  15. Affinity

    Need Help with Accounting

    FOB delivery point means the buyer bears the costs... I thought that would be called FOB or FOB shipping point. In any case it's when the title changes. read the question.. it's not asking about the cost of goods sold but about how much the remaining inventory is worth... you calculated cost...
  16. Affinity

    Need Help with Accounting

    first question: If I understood correctly FOB delivery point means that title changes when it is received and seller pays for freight, so no. second q: FIFO: 8 units are sold and 12 were bought.. so inventory value = cost of last 4 units = 3*106+99 = 417 Average cost: 4 units are left out 12...
  17. Affinity

    Question

    1.) differentiate implicitly to get dy/dx = - x/(6y) then solve dy/dx = 0 with x^2 + 6y^2 = 15 which should give you (x,y) = (3,-1) pr (-3,1) etc. 2.) Q2 is easier than Q1, find y then continue.
  18. Affinity

    Interesting complex number Q

    The correct answer is sin(5.5x)/sin(0.5x)
  19. Affinity

    Triangle theorem?

    they do.. it's called the orthocenter you can prove it by some simple circle geometry. Let 2 altitudes intersect at H, consider the line through H and the 3 vertex. prove that this is perpendicular to the corresponding base (there are many cyclic quadrilaterals in this figure)
  20. Affinity

    Geometry proof

    It is not surprising that proprietary notes contain errors
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