Recent content by TimTheTutor

I agree. It should be but when I downloaded his solutions yesterday they said that answer not the incorrect one mentioned above. I assume he corrected his answers in response to your post? Anyway his answer for Q5(c)(iii) I think is wrong. Aren't we integrating with respect to h from -b to b...

I agree with the above answer given. Just apply the sum of a GP formula separately 2 each of the two GP's then add sums 2gether. ie. G1: a = 2*3^1 = 6 , r = 3 G2: a = 3*2^1 = 6, r = 2

Here's how to do Q2 Let V = projected initial velocity, @ = projected initial angle Therefore Vy (vertical velocity component) = Vsin@ Therefore Vx (horizontal velocity component) = Vcos@ Derive the following two standard results to begin with: (1) Max Height Reached (b) =...

Your answer is correct but your method leaves a bit to be desired. Just as with limits with polynomials divide numerator/denominator through by the term with the highest power and then apply the basic rule that lim (x->infinity) [1/x] = 0 [5(10^n)+3]/(10^n) -------------------------...

Here is a method that does use calculus (integration & constant) (a) dV/dt = K (K<0 since the volume constantly decreases over time) Integrating with respect to t V = Kt + C x^3 = Kt + C When t=0 x=10 => C=1000 x^3 = 1000 + Kt When t=70 x=5 125 = 1000 + 70K 70K = -875 K = -12.5 m/s...

x = t2 - 5t - 4 (t>=0) v = 2t - 5 a = 2 Change in direction will only occur when v=0 and a<>0 0 = 2t - 5 t = 2.5 => x = -2.25 When t=0 x=4 and when t=4 x=0 [--->----] [---<---------------<----] |---------0--------------4--------------------->x -2.25...

iambored - u r indeed right that sin(4pi/3)= -sqrt(3)/2. Therefore the bitmapped answer is wrong! Wogboy's solution is correct.

Mmm.. rather than use "tricks" such as 3 = e^(ln3) I would go for the straightforward option using indice laws then solving equation e^(2x) = 3e^(-x) e^(2x) = 3 / e^x e^(2x + x) = 3 e^(3x) = 3 3x = ln 3 x = (ln3)/3