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do you know the factorising difference of 2 powers rule? xn - yn = (x-y)(xn-1 + xn-2y + xn-3y2 +...+yn) ? well it's like that (I don't know if either of these are actually taught) it's xn + yn = (x+y)(xn-1 - xn-2y + xn-3y2 -...+yn) eg sum of 2 cubes which you would know x3 + y3 = (x+y)(x2 -...

solving questions on here is pretty efficient study imo. I'm not doing the HSC yet so can't exactly study for it anyway

just considering the 5562 + 4162 part, since the 1362 are obviously divisible by 13; (52+3)62 + (39+2)62 (52 and 39 are multiples of 13) = (5262 + 62*5261*3 +(other terms that are multiples of 52) ...+ 362] + (3962 + (other terms that are multiples of 39) + 262) so now we need to only worry...

you know that if x=a, then when @=0, y=0, and when @ = pi/2, sin[y] = 1/rt5 and cos[y] = 2/rt5 (just from looking at the right angle triangle NTO) so subbing into the equation for dy/dt; when @ = 0 dy/dt = ucos(0-0)/(2acos(0) + acos(0-0)) = u/3 when @ = pi/2 dy/dt = ucos(pi/2 - y)/(2acos[y] +...

I wanted to make the expression into the multiple of two differences of 2 squares (so it would become a quadratic that I could complete the square with), so to do this you would make x = v - 3/2, but it's a bit less messy if you just let x = (v-3)/2 so you can get rid of the fractions

how about.. (v2 - 5)2 >= 0 v4 - 10v2 + 25 >= 0 v4 - 10v2 + 9 >= -16 (v+3)(v-3)(v+1)(v-1) >= -16 (v-3)/2(v-1)/2(v+1)/2(v+3)/2 >= -1 let x = (v-3)/2 x(x+1)(x+2)(x+3) >= -1 ???

let x = (v-3)/2 (v-3)/2(v-1)/2(v+1)/2(v+3)/2 >= -1 (v+3)(v-3)(v+1)(v-1) >= -16 v4 - 10v2 + 9 >= -16 v4 - 10v2 + 25 >= 0 (v2 - 5)2 >= 0 :. true

"3! x 2! x2! x2! = 12" um, doesn't it = 48? you do have the exact same arrangement if all couples switch sides - your 'someone [at] the beginning' isn't supposed to move, or else you would have to allow for all people moving 1 to the right etc.etc.

^^, ^ lol yeah already was a great band called carnage, and criticising -core, and then playing some pretty generic thrash with what seems like very strong -core influence.. ehh

from what I can tell.. op has no idea what he's talking about and makes no real good points. it's a bit hard to take the article at all seriously I suppose that the decline of art in certain aspects of music is a good enough topic to write about, but I really think you need to find some better...

there's pretty much nothing hard in the course now, afaik

haha I read it as 8 people aswell ^ wouldn't the answer then be 7!*27 ?

I would say 3! * 2!3 (since it's a circle, there's no top and bottom end)

MX2. As soon as I get the textbook for it.. I probs won't be doing much else. and having a lot more frees

this is extreeeeeeeme

mx2, physics, chemistry, adv english (I think)

by "the same time apart" I just mean that t1 - t2 is the same for both of them, since the angles are taken at the same point in time + the Ucosa and Vcosa come from the vx and the -tana and -tanB are negative since they're downwards angles

where n is the shell number eg. first shell can hold up to 2 second shell can hold up to 8 third shell can hold up to 18 etc

the coefficient of the rth term is given by (I'm letting the first coefficient be r=0) (-1)r25-r 5!/r!(5-r)! we want to find the lowest value of the coefficient such that it is greater in magnitude than the coefficient after it (or however you like to set it up. You can find the greatest...