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$\noindent Assuming you mean $\left(z-2\right)\overline{(z-2)} = 9$, this is equivalent to $|z-2| ^2 = 9 \iff |z-2| = 3$, so this is a circle of radius $3$ centred at $2$ in the complex plane.$

$\noindent Find the complex number represented by $\overrightarrow{AB}$, multiply this by $i$ (to rotate it $90$ degrees counterclockwise), and this is your answer.$

HSC Physics isn't very useful for university physics. HSC Maths would be a lot more helpful (4U would be the best).

3), Yeah, prime factorisation is unique. So every positive integer has only one prime factorisation. So if n_1 and n_2 have different prime factorisations, they cannot be the same number.

Hahaha was it π(n) that the Q. was actually referring to? Edit: oh, just noticed Drsoccerball's post above. (Wasn't able to see the LaTeX before, I think it's playing up for me at the moment.) Did the Q. provide any fact about the asymptotic nature of π(n)?

$\noindent Note that since $\mathbf{x}(k+1) = A\mathbf{x}(k)$ and $A$ is diagonalisable with eigenvalues $\lambda_{1} = 0, \lambda_{2} = -2$, with corresponding eigenvectors $\mathbf{v}_{1} = \begin{pmatrix}5\\1\end{pmatrix}$ and $\mathbf{v}_{2} = \begin{pmatrix}3\\1\end{pmatrix}$ respectively...

Here's the Wikipedia page for the divergence of the sum of reciprocals of primes (which has a few proofs of it there), if you (or anyone else) want(s) to see it: https://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes . Did they get you to derive some bounds on...

$\noindent We have $\mathbb{P}\left(B\mid A^{c}\right) = 1 - \mathbb{P}\left(B^{c}\mid A^{c}\right) = 1 - 0.85 = 0.15$. Now, $ $$\begin{align*}\mathbb{P}\left(B\right) &= \mathbb{P}\left(B\mid A\right)\mathbb{P}\left(A\right) + \mathbb{P}\left(B\mid A^{c}\right)\mathbb{P}\left(A^{c}\right)...

Re: MATH1231/1241/1251 SOS Thread Yeah that's correct.

Re: First Year Uni Calculus Marathon $\noindent Let $\mathbf{F}\left(x,y\right)$ be a continuous vector field defined on the plane with $\left\|\mathbf{F}\left(x,y\right)\right\| = 1$ at all points $(x,y)$. Let $S$ be the unit square (vertices $(0,0), (0,1),(1,0),(1,1)$) and suppose that...

Intuitively speaking, Variance measures how much fluctuation there is in the values taken on by a random variable. A r.v. with very low variance will generally tend to take on values clustered close to the mean, whereas a r.v. with very high variance will tend to take on more "erratic" values...

Re: MATH1231/1241/1251 SOS Thread Can you recall whether there have been any contradiction proofs in the HSC more recently than 2003? Surely there should've been (though I can't remember any off the top of my head)?

Re: MATH1231/1241/1251 SOS Thread Product rule.

The above shows that memorylessness becomes equivalent to the condition that S(t+s) = S(t)S(s) for all t, s (where S is the survival function of the distribution). So to show that the only continuous distribution that is memoryless is Exponential, it suffices to show that continuous solutions to...

$\noindent To prove exponential implies memorylessness, let $X\sim \mathrm{Exp}(\lambda)$, so the complementary CDF (sometimes called a `survival function' (as usual search Wiki etc. for more details)) is $S(t) = e^{-\lambda t}$. Recall by definition (or an equivalent definition) of...

It's easy to prove that exponential => memoryless. To prove that in fact (for continuous distribution) memoryless => exponential is a little harder – comes down to showing the only continuous function that solves the functional equation S(t+s) = S(t)S(s) will be exponential functions.

Re: MATH1231/1241/1251 SOS Thread Sub. x = 0, which easily lets us find C. Then we can find B by subbing (say) x = 2.

Yeah it will be, reason being we can never achieve an actual cubic (degree 3 poly) in the range of T.

No, remember P_n has dimension n+1. So P_2 (domain) has dimension 3. So the matrix will have three columns. In general, if the linear map T has a domain with dimension d and codomain with dimension c (c,d finite) then a matrix for T will have dimensions c-by-d.

The dimensions of the matrix are based on the dimensions of the domain and codomain, not the image and kernel.