Mathematics Question Thread

[RuiAce]

Hello, how would i factorise this completely: x^2 -9y^2 -x - 3y. Apparently completing the square would lose marks here

\begin{align*}x^2-9y^2-x-3y&= (x-3y)(x+3y)-1(x+3y)\\ &= (x+3y)\left[(x-3y)+1\right]\\ &= (x+3y)(x-3y+1)\end{align*}

[sidzeman]

I'm not sure how you got that answer. You assumed that C is the midpoint of AB and D is the midpoint of AE, but I don't see anything that permits this.

I was attempting to use the Proportionality theorem -  "A line drawn parallel to one side of a triangle divides the other two sides into parts of equal proportion". CD should be parallel to BE right?
Edit: I just realised I've been mistaking equal proportional to mean bisect nvm then

[RuiAce]

I was attempting to use the Proportionality theorem -  "A line drawn parallel to one side of a triangle divides the other two sides into parts of equal proportion". CD should be parallel to BE right?
Edit: I just realised I've been mistaking equal proportional to mean bisect nvm then

Yeah, it's really just similar triangles (proportional sides). Bisection would be too good to be true I'm afriad

[pokemonlv10]

\begin{align*}x^2-9y^2-x-3y&= (x-3y)(x+3y)-1(x-3y)\\ &= (x-3y)\left[(x+3y)+1\right]\\ &= (x-3y)(x+3y+1)\end{align*}

Where did the last bit come from? the -1(x-3y) part

[RuiAce]

Where did the last bit come from? the -1(x-3y) part

There's a mistake, sorry fixing it now.

The idea was to factorise -1, but it was factorised incorrectly.

[RuiAce]

Hey all, I just want to say that if you are still up, make sure you get a good night sleep. The exam's in the afternoon so you can sleep in a tiny bit if you want but you definitely don't want to be fatigued for the paper tomorrow later today. Good luck to all of you I'm sure you'll all perform brilliantly!

There should be people online to take last minute questions tomorrow morning so don't fuss too much about that. Just try not making them too last minute.

[winstondarmawan]

Hello!
Seems like the link to the farmhouse question is broken, so I'm just gonna repost the Q here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22713025_1357409827717870_1319962237_o.png?oh=f5adcb772fbeb84dc95709f06ccfc5ae&oe=59EF0120
TIA.

[RuiAce]

Hello!
Seems like the link to the farmhouse question is broken, so I'm just gonna repost the Q here:
https://scontent-syd2-1.xx.fbcdn.net/v/t35.0-12/22713025_1357409827717870_1319962237_o.png?oh=f5adcb772fbeb84dc95709f06ccfc5ae&oe=59EF0120
TIA.

Yep, should be fixed now

[sidzeman]

Hey can someone tell me where i went wrong in part iii of this question - I got an answer of 16 while the solutions says its -6

[RuiAce]

Hey can someone tell me where i went wrong in part iii of this question - I got an answer of 16 while the solutions says its -6

[nattynatman]

Rui, genuine question: how did you get so damn good at maths?

[itssona]

hiii
kinda confused. if a parabola is in the form x^2=4ay then is directrix always y=k-a
where a is focal length
because according to direction I get different answers and idk what to do since my topic test is afternoon

[Opengangs]

hiii
kinda confused. if a parabola is in the form x^2=4ay then is directrix always y=k-a
where a is focal length
because according to direction I get different answers and idk what to do since my topic test is afternoon

If the parabola is in the form: , then the directrix is always y = -a since a parabola is defined as the locus (or the set of moving points) that is equidistant between the focus (0, a) and the line, y = -a

If, however, the parabola is of the form: , we say that the focus is (h, k + a) and the directrix being y = k - a

[caitlinlddouglas]

hey I was wondering how to find when a particle was furthest from the origin of they were to give you the velocity or acceleration graph? Thanks!

[RuiAce]

hey I was wondering how to find when a particle was furthest from the origin of they were to give you the velocity or acceleration graph? Thanks!

You always need to make some kind of deduction based off any information they give you.

You need to go back to the velocity first, or else you're stuck

Rui, genuine question: how did you get so damn good at maths?

Uhhhh, idk probably just too much practice.. Thanks?