# Recent content by kawaiipotato

1. ### Integration MC Question - North Sydney Boys 2017 Trial

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.
3. ### Probability

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!.
4. ### Is it ok to use other peoples code with reference in a school assessment ?

Probably depends on your school's policies as well as your specific class.
5. ### UNSW 17/S2 Official Results Thread

Just got mine, zID starts with z511
6. ### UNSW 17/S2 Official Results Thread

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

8. ### Just realized my school disposes of lead containing solutions down the sink?

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.
9. ### VCE Maths questions help

Yes it is, my bad. You would use this matrix in the same method as how I did it the first time around.
10. ### VCE Maths questions help

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

14. ### Inverse Cosine Function

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).
15. ### Inverse Cosine Function

$\noindent This is a special case. Do you know the Unit Circle? \\ For the ''how?'': it's basically because$\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...