Recent content by kawaiipotato

You can also reason it like this: If f takes non-negative values on [-a,0], then the first term is non-negative. The second term will be non-negative because of the absolute value. Then the sum of those two is non-negative, meaning that if we had a function f with the previously mentioned...

Just a note that you should expect 1/2 because the number of desired outcomes are equal in both cases.

Count the ways to fix the boys in a circle first. Then count the ways to place the girls between the boys. This will be your desired outcomes, and your total outcomes is just 7!.

Probably depends on your school's policies as well as your specific class.

Just got mine, zID starts with z511

All MATH provisional marks have been posted on the Math&Stats Portal page.

Do you know how to find the probability if we asked for only one head? How about two heads? Three heads? At least one head means one head, two heads or three heads. So you just need to add up the probability of finding one head, two heads and three heads together. (Alternatively, you...

For some chemicals, I remember that our teacher would have a large bucket where we poured everything into it at the end of the practical lesson.

Yes it is, my bad. You would use this matrix in the same method as how I did it the first time around.

$\noindent The inverse matrix of the transformation is $-\frac{1}{6}\begin{pmatrix}-3 && 0\\0 &&2\end{pmatrix}$. \\ Can you go from here? $

$\noindent I simply solved the equations simultaneously. I.e. I solved $v_1 = 2x$ and $v_2 = -9x$ by ''eliminating'' $x$.

Which part don't you understand?

$\noindent Consider the vector $v = \begin{pmatrix}x\\y \end{pmatrix} = \begin{pmatrix}x\\3x\end{pmatrix}\in\mathbb{R}^2$ (just another way of representing that line $y=3x$. Plug in $x$ values to see this.). Then we apply the transformation to this vector by computing \\...

Oh yes, I meant there is a similar case for sinx = 0. (Try and see the cases using the Unit Circle, it helps a lot).

$\noindent This is a special case. Do you know the Unit Circle? \\ For the ''how?'': it's basically because $\cos$ has roots at $x= 0, \pi/2,-\pi,2$. From what you know, $\cos$ has periodicity $2\pi$. Then since $x=\pi/2,-\pi/2$ are roots, then $x=\pi/2 - 2\pi,-\pi/2 + 2\pi$ are roots as well...