Search results

I believe you are generally allowed to interpret things metaphorically.

Note that the midpoint of a chord of fixed length on a given circle will always be a fixed distance from the centre (in this case, that distance is 6, which follows from Pythagoras and recalling that the line joining the circle centre and the chord's midpoint is perpendicular to the chord). The...

It's easier to just see it intuitively since the numbers are easy to calculate mentally here. The new volume is 100 times the old one, so the concentration must become 100 times smaller. The original concentration was 0.1 mol/L, so the new one is one-hundredth of this, i.e. 0.001 mol/L (shift...

$\noindent The answer is indeed a), but the calculation should be $0.10\times \frac{10}{1000}$. The $v_1$ is the initial volume (10 mL) and the $v_2$ is the new volume (1000 mL).$

Basically you should do three cases as I outlined above. I.e. Find the points where those functions inside the absolute values change sign (which happens when they cross the x-axis, i.e. when they're equal to 0), and consider the three intervals that these two points split the real line into.

$\noindent First one here: just find the value of the derivative at $x=3$, i.e. answer is $6\times 3 = 18$.$ $\noindent Second one: note that $x^2 -4 = (x+2)(x-2)$. So for $x \neq \pm 2$, the given expression equals $\mathrm{sgn}\left(x+2\right)\cdot \frac{1}{x-2}$, where sgn is the...

$\noindent First one: we have $2\sin ^2 x -\sin x = 0$, so $\sin x = 0$ or $\sin x = \frac{1}{2}$. Just solve these two now.$ $\noindent Second one: Let the roots be $a,2a$. From sum of roots: $k+2 = 3a$. From product of roots: $2k + 4 = 2a^2$. Solve these two equations in two unknowns to...

$\noindent We need to split the cases up into intervals which dictate the signs of the functions inside the absolute values. So the intervals to use are $x < \frac{3}{4}$, $\frac{3}{4}\leq x \leq 1$, $x > 1$. When $x < \frac{3}{4}$, we have $4x-3<0$ and $x-1<0$, so the LHS is $3-4x + 1-x$. When...

This isn't an inductive way but the result follows immediately from the fact that the given expression is n choose 3 and this is the no. of ways to choose 3 points from the n-gon.

Re: MATH1231/1241/1251 SOS Thread Well c is strictly between a and x, so it's in (a, x) or (x, a) (depending which of a or x is bigger).

Is it MIF? I think that textbook often has incorrect answers.

We're not meant to sub. the x-value into the cubic, we are meant to do it in the line equations (which you did originally I think).

Re: Discrete Maths Sem 2 2016 Those would mean two different things. One of those (the first one you wrote) is "A does not imply B", the other is "A implies not B" (assuming 'means' refers to 'implies'. It may have instead referred to something like 'is equivalent to'.) .These are different.

Can't see a Q.7. And wouldn't you increase the sample size to improve the margin of error?

You would probably need to prove these things, otherwise it's assuming too much I think (like the pq = -1 thing is equivalent to saying the chords are perpendicular, which is what we're trying to prove, so assuming pq = -1 for a focal chord is too big an assumption I think).

$\noindent Note that the sample size is $\binom{40}{5}$.$ $\noindent i) We can choose the three $4$'s in $\binom{4}{3}$ ways. We can choose the two $7$'s in $\binom{4}{2}$ ways. So the number of ways to obtain three $4$'s and two $7$'s is $\binom{4}{3}\times \binom{4}{2}$. So the required...

Yeah, rank 1 gets the cohort's highest external mark for their internal. And for your last question, no, your internal mark that you currently have (~68) will get moderated first before contributing to your overall HSC mark for that subject.

Jacaranda

Re: MATH1231/1241/1251 SOS Thread $\noindent For this one though it'd be better to just prove from first principles that the given set is spanning and linearly independent (and hence a basis). This would in fact be an elementary way to \emph{prove} that $M_{22} (\mathbb{R})$ is...

Re: MATH1231/1241/1251 SOS Thread Yeah in general if you have a vector space of dimension n (n finite), then S being a set of n linearly independent vectors implies it is also spanning, and hence a basis. Another fact is that S having n vectors and being spanning implies it is linearly...