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Re: HSC 2015 3U Marathon $Note that $S=\mathrm{e}^{-2t}\left(1 + 2 \left(\mathrm{e}^{-2t} \right) + 3 \left(\mathrm{e}^{-2t} \right)^2 + 4 \left(\mathrm{e}^{-2t} \right)^3 + \cdots \right)$.$ $To evaluate the infinite sum, use this series with $x=\mathrm{e}^{-2t}$: $1+2x+3x^2 + 4x^3 + \cdots...

Re: OFFICIAL PHYSICS HSC 2015 Prophecies!!!!! --> FROM A NOSTRADAMUS DESCENDANT!!!!!! Well according to the Raw Marks Database for HSC Physics ( http://rawmarks.info/wiki/Physics ), in 2010, you needed about 75% raw in the HSC Physics paper to get 90 aligned Examination Mark, and 82% raw in...

And the astronauts, from their point of view, perceive the time and events on Earth as being slower / taking longer.

See http://www.boardofstudies.nsw.edu.au/hsc_exams/calculators.html As it says there, one of the features of calculators that is not allowed in the HSC is "capable of storing, manipulating or graphing functions entered in symbolic form (this includes calculators with a graphic display...

Didn't know graphing calculators were allowed in HSC!

Re: HSC 2015 3U Marathon $No, but remember what the definition of $f_k(x)$ was. It wasn't an arbitrary polynomial, it was a polynomial whose leading coefficient was $\frac{1}{k!}$, since $f_k (x)$ is defined in the question as $1+ \frac{x}{1!} + \frac{x(x+1)}{2!} + \cdots + \frac{x(x+1)\ldots...

Here's a graph of the function (hopefully you'll see how to do it by looking at what it's meant to be)...

Re: HSC 2015 3U Marathon $You assumed $a =\frac{1}{k!}$ in your inductive assumption, as this assumption was that $f_k(x) = \frac{1}{k!} (x+1)(x+2)\ldots (x+k)$.$

Clearly they're meant to be brackets.

Yeah you had to discuss odd/even.

But it wouldn't disadvantage her to help the cohort, and would help her have more of a safety net if she screwed up externally (plus it's good to help out your friends).

Re: HSC Physics Marathon 2015 Just make a separate thread haha.

Re: HSC Physics Marathon 2015 Example of snaking. It's not HSC level stuff of course.

We know that at a speed of v km/h, the time taken (in hours) is T = 500/v (time = distance/speed). Since we know the cost per hour as a function of v, c(v), the total cost is C = c(v)*T = (500/v)*c(v), where c(v) is the given cost function per hour. So now we have C (total cost) as a...

Re: HSC 2016 4U Marathon - Advanced Level $The geometric interpretation is that if we have two fixed circles centred at the origin, and we pick one point each from them (representing complex numbers), the distance between the two points (which is $|z-w|$) is at least the width of the annulus...

$Not necessarily, e.g the function $f(x) = -x^3$ has inverse $f^{-1} (x) = -\sqrt[3]{x}$, and these intersect at two points that are not on the line $y=x$ (the points $(-1,1)$ and $(1,-1)$), but are not the same function.$

$This isn't true, for example consider the function $f(x) = -x$. The inverse is $f^{-1} (x) = -x$, which intersects the graph of $y=f(x)$ in infinitely many places that are not on the line $y=x$.$

re: HSC Physics Marathon Archive 1/Hz is equal to s (seconds), so all you need to do is convert eV to J, and I think the conversion for that is given in the data sheet.

$It can be proved though that any such invertible function $f$ that is continuous on $[a,b]$ (where $a<b$) will cross the line $y=x$, and thus will intersect the graph of $y=f^{-1}(x)$ on the line $y = x$ as well. So if a continuous invertible function intersects its inverse, at least one of the...

$Yes. If the graph of $y=f(x)$ happens to pass through the points $A(a,b)$ and $B(b,a)$ with $|a|\neq |b|$, then the graph of $y=f^{-1}(x)$ will also pass through the points $A$ and $B$, and these aren't on the lines $y=\pm x$.$