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I know 2 syd boy kids who got in last year.

Try looking through 2020 and 2019 statistics for your school. Usually they list out how many people got into a certain degree.

eddie woo

Not CM but: (i): \text{RHS}=(x-1)\left(\frac{x^n-1}{x-1}-n\right) \quad \text{Using the GP sum formula} \text{RHS}=x^n-1-n(x-1)=\text{LHS} (ii): Let: f(x)=(x-1)(1+x+x^2+...+x^{n-1}-n). Since x is a positive integer, the minimum value of x must be 1. The minimum value of f(x) is: f(1)=0 as...

Basically everything in the second dot point of module 6. Off the top of my head: -History of Acids and Bases particuarily arrhenius+Bronsted Lowry Define, Describe and explain,Include conjugate acid/base pairs and general equation for acid and base disassociation. -Strength of Acids and bases...

https://en.wikipedia.org/wiki/Harmonic_mean

Yeah its meant to be the reciprocal

z^4 - 4z^3 + 2z^2 - 4z+1 = 0 z^4+1 - 4(z^3+z) + 2z^2 = 0 z^2+\frac{1}{z^2}-4 \left (z+\frac{1}{z} \right) +2=0 \quad \text{dividing by $z^2$} But z^n+\frac{1}{z^n}=2\cos{nx} So the problem becomes: 2\cos{2x}-8\cos{x}+2=0 which is a 3U problem (remember to take only the first 4 unique...

\frac{d}{dx}\left(\frac{x}{(x^2+1)^{n-1}}+(2n-3)I_{n-1}\right) =\frac{1-(2n-3)x^2}{(x^2+1)^{n}}+\frac{2n-3}{(x^2+1)^{n-1}} =\frac{1-(2n-3)x^2}{(x^2+1)^{n}}+\frac{(2n-3)(x^2+1)}{(x^2+1)^{n}} =\frac{2n-2}{(x^2+1)^{n}} \therefore \int \frac{1}{(x^2+1)^{n}}=\frac{1}{2(n-1)}\int...

Oh yeah idk what I was thinking lol

Do you mean 0.01mL?

Titres that are within 0.05mL of each other are considered concordant so I would only ignore 27.9mL.

Doesn't your school change teachers when you go to year 12?

Wow just wow. Im interested on how thats derived?

Yeah I messaged my opponent as well still waiting...

Damn beat me to it. The proof for why n^2(n+1)^2 is divisble by 4: n is even: let n=2k then the expression becomes: 4k^2(2k+1)^2 n is odd: let n=2k+1 then the expression becomes:4(2k+1)^2(k+1)^2

Yes I think so

The person has to make a net movement of 10 steps to the left and 10 steps to the right. So this is basically just 20 steps in total with 1 or 2 steps so the answer is the F_{21} I think.

This question is very similar to a pervious perms and combs q on this website (I think involving coins?) The question can be rewritten as: x+2y=10 where x is the number of times she takes 1 stair and y is the number of times she takes 2 stairs. Now simply list all integers x and y that...

The Henderson-Hasselbach Equation isn't in the syllabus so a derivation is probably required. It's not too hard to prove: The K_a is defined as: K_a=\frac{[A^-][H_3O^+]}{[HA]} [H_3O^+]=K_a \times \frac{[HA]}{[A^-]} -\log{[H_3O^+]}=-\log\left({K_a \times \frac{[HA]}{[A^-]}}\right)...