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Okay I'm not quite sure what you did, but what I got was MUCH simpler. Firstly, for the gradient of PQ m = {b [ sin(theta+pi)-sintheta ] } / {a [cos(theta+pi) - costheta] } and then expanding out the sin(theta+pi) and cos(theta+pi) it simplifies to m = b sin theta / a cos theta Then subbing...

LOL ermm.. I only got 99.90 :P What about those people that got 99.95? They're crazier than me :)

Re: TUTOR 2u/3u/4u Maths | 99.90 ATAR | 3u 99 4u 97| State Rank for Jap | Western Syd bump :)

Just do it xP I'm in it, and I'm not not even doing like, a degree thats related in ANY way to Mathematics x__x. I think it'll be fun. But if it'll be your only class on Tuesday, maybe not. If you really wanna do it, try changing your timetable a bit? I had to ahaha.

4 cos^2 - 4 sin = 1 4 (1 - sin^2) - 4 sin = 1 4 - 4sin^2 - 4sin = 1 4sin^2 + 4sin -3 = 0 (2sin-1)(2sin+3)=0 therefore, sinx = 1/2, sinx = -3/2 for sin x = 1/2 ,, x = 30, 150 for sinx = -3/2, there are no solutions as -1< sinx <1

f(x) = x^2 g(x) = 1/x^2 First, find the point of intersection. x^2 = 1/x^2 x^4 = 1 Therefore x = 1 since x>0 If we draw a graph we can see that f(x)=x^2 > g(x) = 1/x^2 for x>1 Therefore the area = S {3,1} x^2 - 1/x^2 dx = [ x^3 / 3 + 1/x ] {3,1} = (27 / 3 + 1/3) - (1/3 + 1) = 8 units^2

Derive f(x) = sin2x -2x.cos2x, and us this to find S{pi/4,0} xsin2x dx f'(x) = 2cos2x - 2 (cos2x -2xsin2x) = 4xsin2x therefore, S{pi/4,0} xsin2x dx = 1/4 [ sin2x - 2xcos2x ] {pi/4,0} = 1/4 [ (1 - 0) - (0 - 0) = 1/4

3x^4 - 3x^3 + x^2 - x + 2 ---------------------------------------------- (x+1) | (3x^5 + 0x^4 - 2x^3 + 0x^2 + x - 1) 3x^5 + 3x^4 ---------------- - 3x^4 - 2x^3 - 3x^4 - 3x^3...

Ahh I see. Yeah that's what my teacher taught me. I couldn't quite remember what it was but now I do. :) But he only taught it to us like, after trials. Do other teachers teach this method initially? Or just as a side thing?

hahah well my teacher taught us a different method rather than just multiplying by the square of the denominator.. but i think it only works if one of the sides is = 0 so in the example, (5m+4)/2m < 1/4 and moving the 1/4 onto the LHS you're left with (9m+8) / 4m < 0 so then, you now have 2...

for the parabola x^2 = -4y, a = 1 therefore P = (2p, -p^2) dy/dx = (dy/dt) / (dx/dt) = -2p / 2 = -p tangent at P y + p^2 = -p (x-2p) y = -px + 2p^2 - p^2 y = -px + p^2 through x^2 = 4y x^2 / 4 = -px + p^2 x^2 + 4px - 4p^2 = 0 x = [-4p +- root(16p^2 + 16p^2) ] / 2 x = -2p +- 2p root2 x = 2p...

1. gradient of PQ = (ap^2 - aq^2) / (2ap - 2aq) = [ a (p-q)(p+q) ] / [ 2a (p-q) ] = p+q / 2 = m that is, (p+q)/2 = m now the midpoint of PQ, x = (2aq + 2aq) / 2 = a (p+q) y = (ap^2 + aq^2) / 2 = a (p^2+q^2) / 2 Now, x = a (p+q), but p+q = 2m therefore, x = 2am, which is a constant and...

lol soz iuno how to use latex but,, u = x root x = x^3/2 therefore, du = (3/2) x^(1/2) dx then integral x^(1/2) / [ x^(3/2) + 2 ] dx = 2/3 integral 1 / (u+2) = 2/3 ln (u+2) = 2/3 ln [ x^(3/2) + 2 ] + c :)

+1 lol XD

Mayank is a very dedicated person, who helped me with Maths all throughout high school. He will be able to explain concepts clearly and in a way to suit your ability. Highly recommended. He's very hard working. I mean, just look at his marks =P

for the notes, what my physics teacher made us do was we got a book and wrote ONE dot point on ONE page and we just filled in that dot point whenever we could having one page means you can fill in extra information later because there'll probably be extra space and its convenient having all your...

chemistry definitely needs more memorising, and phys needs more understanding but choosing which one,, it really depends on you some people find memorising really easy and thats just how they get through the hsc both have great scaling lol BUT,, i worked my ass off for chem and i only ended up...

doing 13 units is a pain in the ass because you have like 3 units of extra study which arent gna count i did 13 units, but i was considering dropping physics like all year round because i really didnt like it, except that in the first assessment and half yearlies i did REALLY well, so i was...

Re: TUTOR 2u/3u/4u Maths | 99.90 ATAR | 3u 99 4u 97| State Rank for Jap | Western Syd haha thanks guys :) bump~

Because you cant find the square root of a negative number. If its in any other quadrant than the 1st quadrant, then either x or y is negative, and you cannot evaluate them in the real number system.