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$\noindent That would be a plane through the origin (assuming the two vectors are non-parallel and non-zero of course).$

Re: HSC 2016 3U Marathon $\noindent He means the ``$\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}$ rule''.$

$\noindent Wait, why is it $\cos 100^\circ$? Shouldn't it be $10^\circ$?$

Haha, that's like the non-lazy way, isn't it? The lazy way is what I did (go straight to base 2 :P).

Yeah, I intentionally used the base 2 form because it's more convenient when given the half-life. If they had instead said something like a "decay constant", we would have done it Silly Sausage's way with base e (the k there is known as the decay constant).

$\noindent Let time $t$ be measured in years. The half-life is given to be $\tau = 1600$. Let $A(t)$ be the amount of radium at time $t\geq 0$. Since we are only asked about percentages, we can just assume $A(0) = 1$. The standard exponential decay formula is then $A(t) = 2^{-\frac{t}{\tau}}$...

Re: MATH1131 help thread $\noindent You need to find an $M\equiv M(\varepsilon)$ (so it depends on $\varepsilon$) such that if $x>M$, then $|f(x)-L|<\varepsilon$.$

Paradoxica didn't say 'cusps', he said 'cups' (in reference to the shape of the graph).

$\noindent As they say, the determinant can be found by expanding along any row or column. Here, it is best to do it down the second column, since it is mostly 0's, so we only need to evaluate one determinant. Remember that the sign associated with the entry in the first row and second column is...

54. Yes. By definition of the limit (using delta = epsilon/2), the limit is 3. 55. The limit is + infinity, by definition of right limits equalling +infinity (use delta = K/(K – 2)). 58. (b) This is just a numerical version of 58 (a). If you found the equilibrium concentration in (a) in terms...

To show C = 0 in Q 9(c), sub. t=0. Then the area becomes 0 since the point in the hyperbola is on the x-axis. Also, t/2 = 0 when t = 0. Hence C = 0.

Re: HSC 2016 2U Marathon $\noindent Well done!$ $\noindent \textbf{NEW QUESTION}$ $\noindent Let $a$ be a positive constant. Find $\frac{\mathrm{d}}{\mathrm{d} x}\left(a^{a^x}\right)$.$

Here are some hints. $\noindent 6) (a) We can use calculus to find the stationary point, and sketch the individual graphs of $\ln x$ and $x$ to determine the sign of $\frac{\ln x}{x}$ at differing $x$-values. We can use the behaviour of $\ln x$ and $x$ near $x = 0$ to check whether there's an...

What I said before still applies to Senior Science (not sure how the TAFE exams work though so not sure about those). Incidentally, just because no-one got 99.95 with a certain subject in a UAC table, it doesn't mean it's impossible to get 99.95 with that subject, or that doing that subject...

It is hypothetically possible. Imagine if suddenly one year everyone doing Aboriginal Studies aced or did well in all their other subjects and most people doing typical "high scaling" subjects flunked all their subjects (or maybe it was just how well they do in English that determines the...

$\noindent We have $0\leq x-y<3$. Firstly this means $x-y \geq 0$, so $y\leq x$. So the entire line $y=x$ is included. Also, any remaining parts must be below this line. For the second part of the inequality, $x-y<3 \Rightarrow y>x-3$. So draw the line $y=x-3$ (as a dotted line, since we won't...

$\noindent Call the first vector there $\bold{v}_1$ and the other one $\bold{v}_2$. Consider the line segment $\ell$ joining the origin to the point with position vector $6 \left(\bold{v}_1+\bold{v}_2\right)$ $\big{(}$this line segment can be written as $ \left\{\lambda...

$\noindent iii) (First question) (a) Recall that $\cos \xi \leq 1$ for all $\xi \in \mathbb{R}$. Let $x\geq 0$. Integrating both sides of the inequality from $0$ to $x$ gives us $\Big{[}\sin \xi \Big{]}_0 ^x \leq x \Rightarrow \sin x \leq x$.$ $\noindent (b) We showed above that $\sin \xi \leq...

$\noindent Oh right, sorry, I misread it (Q5). In that case, since $\frac{1}{x}$ tends to $\pm \infty$ as $x \to 0^{\pm}$, $\cos \left(\frac{1}{x}\right)$ will just oscillate infinitely often near 0, so the limit does not exist.$

$\noindent When we only have two non-zero vectors, showing that they are linearly independent amounts to showing that one of them isn't a constant multiple of the other (if one of them is the zero vector, they're automatically linearly dependent). In this case, it is clear that $\vec{u}$ is not...