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my pleasure

the shorter side is l split the trapezium up into 2 triangles and a rectangle so the longer side consists of the bases of the 2 triangles plus l the bases are l cos @ from trigonometry from the same triangles the perpendicular height is lsin@ so area = 1/2. . (2l + 2lcos@). lsin@...

where exactly are you stuck? to find the area of trapezium you need the length of both parallel sides one is given, the other is found using trig on the two triangles on the sides the max part is dont my differentiating area with respect to theta

Here are 4 circle theorems relating to concurrence. I think its very useful to go through these proofs as harder 3 unit geometry.

the answer to part c is correct, but our teacher always advised us not to assume that formula for v^2, but prove it from d/dx(v^/2) = -n^2x alternatively differentiate the expression for x with respect to t to get velocity, then the max value is the max value of that trig curve

from what ive been told, both rounds of interview have ample spaces to give interviews to all those that could possible make the course. there is no disadvantage in having a second round interview furthermore, if any schools consistently bump up their students uai, the faculty is aware of it...

i think t formula may be useful once you get that expression just try trig bashing

let (a,0) be O find gradient of PO, PQ gradient(PO) * gradient (PQ) = -1

no if you get a predicted UAI +99 you are not guaranteed an interview. You must get a umat of at least 152 even if you get predicted 100 and no, every mark in umat isn’t really worth 10 in the hsc. If you don’t get a good uai you have little chance of getting in despite your umat score

dont be so modest 3 time international olympian

in that case its 1/t

2. A particle moves in a straight line so that its accl. at any time is d²x/dt² = -9x. Find its persion, amp. and displacement at time t if initially the particle is 2cm from the orogin and has velocity 2root(3) cm/s ans: p: 2pi/3 amp: 4root(3)/3 x= 4root(3) cos (3t- pi/6) d/dx(v^2/2)...

although considering the huge range of the raw marks, and the fact that the grammar cohort has a long 'tail,' im guessing there would a large range of hsc marks too. i reckon you wouldve been up there at grammar.

I have no idea.

SHM At extremeties, x = amplitude v = 0 a = max at origin, x = 0 v = max a = 0 For the form a = -n^2 x, let a = d/dx(v^2/2) integrate for v, and use any information they give and the table above to find C if a particle moves about a point p other than the origin, replace x by...

int cosx/sinx = log|sinx| + C

d/dx(v^2/2) = -k/x² v^2/2 = k/x + c when x = 6400, v = 0 so c = -k/6400 substitute and solve for v

ahh the walrus. from what ive heard he isnt a great teacher anyway, you should try to be top 75 if you want to do 4u pender doesn't allow you to do it if you are significantly below that rank

from i you know that the solutions of polynomial are cos(-pi/9), cos(7pi/9) and cos (13pi/9) so from polynomial, some of roots = coeff of x^2/1 = 0 so cos(-pi/9) + cos(7pi/9) + cos (13pi/9) = 0 as cosx is even, cos(-pi/9) = cos(pi/9) also, remember that cosA = -cos(pi - A) [can be proved...

use the fact that sum of roots = 0