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t=24 i think. ill put up my answer and see what u guys think

don't you have a textbook?

school ranking has no effect on your mark in the test. If you come first and get a certain good mark, u get that mark no matter what. And also, even though it scales to be equivalent to 99+ atar, that is just for 2 units, you need to do well in the other 8 to get 99

i mean this info and solutions to z²+(2+i)z+2-2i=0 are i and -2-2i.

ahh yes good point missed that, also, is there an alternative solution to this that uses the other info, or is that just superfluous?

no worries :)

<a href="http://www.codecogs.com/eqnedit.php?latex=x^2=4ay\\ y=\frac{x^2}{4a}\\ y'=\frac{x}{2a}\\ at~P,~y'=t\\ therefore~the~gradient~of~the~normal~is~m=\frac{-1}{t}\\ point~grad~formula\\ y-at^2=\frac{-1}{t}(x-2at)\\ y-at^2=\frac{-x}{t}@plus;2a\\ yt@plus;x=at^3@plus;2at~-(1)\\...

one normal will have s's in it, one normal will have t's in it. When you solve them simultaneously, your answers will have both s and t's in them. When you solve simultaneously, you are trying to find x and y both in terms of s and t. It is in the second step that you are finding a relationship...

i will give you some steps for the answer, and then come back in a bit, and if you haven't got it by then i will post the answer/. 1. Find equations of both normals 2. Find intersection of normals by subtracting equations from each other 3. once you have subtracted equations, factorise both...

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{9(1-(2/3)^{4})}{1/3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{9(1-(2/3)^{4})}{1/3}" title="\frac{9(1-(2/3)^{4})}{1/3}" /></a>

a) although this is a geometric progression, for this part it is easies to do this I added 9+6+6+4+4+2.67+2.67 like you did. The way to put it into the formula is to take a=9, r=2/3 n=4. Multiply this answer by 2, and then subtract 9. This is because in the motion it moves up and down, doubling...

well done :)

circle with centre (-2,-3) and radius 1. post working in a sec <a href="http://www.codecogs.com/eqnedit.php?latex=|z^2-(12i-5)|=|z-(2@plus;3i)|\\ |(z@plus;(2@plus;3i))(z-(2@plus;3i))|=|z-(2@plus;3i)|\\ |z@plus;(2@plus;3i)||z-(2@plus;3i)|=|z-(2@plus;3i)|\\ |z@plus;(2@plus;3i)|=1\\...

haha i know its true, but definitely wishful thinking

HSC exam experience for accelerated students. I did 2u and 3u accelerated last year and fucked up one exam, but u can just do it again next year. Definitely worth it :)

yep, exactly :) remember to have equations so prove it

amphoteric substances just act as a base in some instances, and as an acid in other instances, there is no restriction upon what type of reactions they have. The example I used just happens to include 2 neutralisation reactions because it is the easiest one to remember :)

nope they are different. Amphiprotic means that a substance can either be a proton acceptor or proton donator. i.e. hydrogen sulfate, HSO4-. This can either accept a proton to become H2SO4, or donate a proton to become SO4 2-. Amphoteric substances are substances that can be act as either an...

So yes, at the poles, g is the greatest. At the equator, g is the smallest, and this is due to the earths rotation. The earth is a fraction larger at the equator due to the tangential inertia of the mass at the equator forcing the earth "outwards" (why people get confused with centrifugal...

yeh they do, i meant split the aluminium up that was on the right side of the equation. and yes put one either side.