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Re: HSC 2017 MX2 Integration Marathon By inspection

Re: Statistics $\noindent The final answer will be $\frac{q}{p^{2}}$, where $p$ is the ``success'' probability parameter, and $q = 1-p$. In your case, $p = \frac{2}{3}$.$

Re: Statistics $\noindent One way is to essentially first derive the formula for $\sum_{k\geq 0} k^{2}z^{k}$ ($|z| < 1$) by applying the operator $z\frac{\mathrm{d}}{\mathrm{d}z}$ a couple of times to the geometric series $\sum _{k\geq 0} z^{k} = \frac{1}{1-z}$ ($|z| < 1$).$

It's a classic trick to integrate sec(x) (the multiply top and bottom by sec + tan). The question in general is mainly just slightly tedious calculations I think. The method for doing it is standard and expected to be known by 4U students. (Write numerator as a*q'(x) + b for some wisely chosen...

Use log laws and so log|A/2| = log|A| – log(2). The – log(2) is just a constant so can be absorbed into the +C. The net effect is you can ignore the '/2' inside the log.

Re: ATAR Notes vs Bored of Studies Maybe they just want notes or something.

They are the same, but when you found the infinite product with your method, you changed around the order of multiplication, and for infinite products, this can change the answer (like how it can for infinite sums (see: https://en.wikipedia.org/wiki/Riemann_series_theorem); note that an infinite...

$\noindent Basically. He's finding an expression for the partial product up to a finite number $k$ (obviously an integer) and then taking the limit of this as $k\to \infty$, as the original question asks (and other methods can fail as we have seen above).$

$\noindent The differential equation is $\ddot{x} = -4^{2}x$, so the solution is $x = A\cos \left(4t - \phi\right)$ for some constants $A,\phi$. Since $x(0) = 3$, we have $A \cos \phi = 3$. Also, $\dot{x} = -4A\sin \left(4t - \phi\right)$. So $\dot{x}(0) = 16 \Rightarrow 4A\sin \phi = 16...

Re: HSC 2017 MX2 Marathon $\noindent With the question written the way it is, it looks more like $\cos\left(x^{2}\right)$ inside the radical than $\left(\cos x\right)^{2}$ (if the the latter were intended, it would normally be written like $\cos^{2} x$).$

If you only did General Maths though you would probably need to go through a lot of topics in order to catch up (probably even relatively basic algebra).

It doesn't generally work like third rank gets third highest mark (if you meant third highest external mark becomes your internal mark if you're third internally), and it's not true that the only thing that matters internally is rank. (It's also not the case that your assessments' marks mean...

Yes, you can get in.

In your example, the two rectangles have the same aspect ratio, so yes, they're similar.

Two rectangles which have different aspect ratios (the ratio between longer and shorter sides) to each other will not be similar (they are different in shape). Two rectangles will be similar if and only if they have the same aspect ratio.

Don't need angles. You can see the definition of similarity here: https://en.wikipedia.org/wiki/Similarity_(geometry) .

The answer is 'true'.

Re: International Baccalaureate Marathon The area values you got are all very close. Haven't checked your calculations but if they're all right, the small discrepancies would just be due to the fact that you used rounded values for the angle. $\noindent To calculate $\sin \theta$ exactly as a...

Re: International Baccalaureate Marathon $\noindent Oh yeah, they accidentally did $-2-2k$ instead.$

Re: International Baccalaureate Marathon What do you mean? The vectors b and c have no k, so vector BC would have no k.