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School ranks are not irrelevant. Aligning occurs which, although reflects your cohort strength, does affect your marks. I was at a school which was ranked about 100 and now I'm in one ranked in the top 10. You can't say that, if I put in the same effort in either school, I would get the same...

Not really, in Australia most courses are solely dependent upon your ATAR.

Re: Math help (Because Carrot made me do this) Anything to an even power is positive. So (-x)^2 = x^2 in the same way that (-2)^4 = 2^4.

I think you may be misinterpreting the question. For i): There are 5 people, and 5 hotels. So in each hotel, ANY of the 5 people can stay at them, since there are no restrictions. So it's 5 x 5 x 5 x 5 x 5 = 5! = 3125. I'm not to sure why you're using cases (addition). ii) We both agree on...

1: i) Each traveller can stay in each hotel, so 5^5 = 3125 ii) 5 can stay in the first one (any one) then there are 4 remaining that can stay in the second, then 3, then 2, then 1. So its 5! = 120 iii) You group the husband and wife as one set. So say there are 4 'individual' items now, since...

Obviously focus on Eng. Adv and Maths.

It's because the volume and concentration are considered in the calculation of the equilibrium constant. Because temperature isn't considered, it will affect the value of the constant.

\dpi{150}%20x^{3}\%20term\%20in\%20(2-5x)(x^{2}-3)^{4}%20\\%20\\%20Expanding:\%20(2-5x)\left%20\{%20\binom{4}{0}(x^{2})^{4}\%20-\%20\binom{4}{1}(x^{2})^{3}(3)^{1}\%20+\%20\binom{4}{2}(x^{2})^{2}(3)^{2}\%20-\%20\binom{4}{3}(x^{2})^{1}(3)^{3}\%20+\%20\binom{4}{4}(3)^{4}\%20\right%20\}%20\\%20\\%20S...

Yes, you can change your integral to be in terms of dx by doing what you said. Just be careful that you change the limits.

AB = 4 [given] ABC = -4 [products of roots] therefore, c = -1 ------ (A) A + B + C = 17/3 [sum of roots] Using (A): A + B = 20/3 ---- (B) Sub (B) into AB = 4: B(20B/3 - B) = 4 Using quadratic formula, B = 2/3 or 6 Therefore the roots are 2/3, 6 and -1.

I'm not sure but I wouldn't think that it would have much of an effect on dissolution rate. pH is only the H+ concentration so doesn't really affect the inter-molecular bonding which is the basis of dissolution.

Pretty sure you don't, but you have to know the one with electric fields.

Wouldn't a reason be because CO2 is released (you connect the apparatus to limewater to prove this), and once all of the CO2 is released the mass stops decreasing. C6H1206 ---> 2CO2 + 2C2H5OH

Always have your integral table in front of you when doing integration, (especially for inverse trig and those special logarithmic integrals). Always draw a diagram for mechanics.

The question tells you that the cation is B2+. Because there's a reaction between B and the E3+ cation, the E3+ become E and the B becomes B2+. It's a simple redox reaction.

I + Al --> AlI3 It's ionic, and Al is 3+.

1. C, A, E, B, D. Just count how many reactions each metal has. 2. i) 3B(s) + E3+(aq) ----> 3B2+(aq) + 2E(s) We know this because it is a redox reaction. States can be found because it is given that B is a solid, and ions are aqueous. ii) B is more reactive so it is oxidised. So the...

I do about 1300 words in 40 minutes. Of course, it's barely legible though.

Agreed. Done Complex Numbers, Conics, Resisted Motion, Polynomials and Integration so far, and nothing really is too challenging.

It's not about the hours, trust me. Different people take different amounts of time to understand concepts. Your ATAR is also highly influenced by your school. Although if you're aiming for 95, you should try to get an hour a night in.