Login | Register

Welcome, Guest. Please login or register.

February 23, 2020, 03:06:47 am

Recent Posts

Pages: [1] 2 3 ... 10
1
Hello everyone ,

for this question i was wondering how i would do it using x' and y' notation so i could states the sequence of transformations from one graph to the other.

Thanks  :)
2
Monash University / Re: Monash General Chat
« Last post by Springyboy on 3 hours ago »
The unit is ENG1005. Some of the lectures have their descriptions left blank whereas others say 'learning capture/live streaming enabled' so I'm not sure whether they'll all be recorded or just those ones which have mentioned learning capture.

No problem nah according to this all lectures are recorded don't stress. You only need to attend the applied classes. I can see what you mean after looking here though. Some lectures definitely don't mention LC. However, as all lectures are held in C1 which is the second largest lecture theatre on campus, then they should usually record. However I have had technical faults before in C1, so do try and attend if you can.
3
VCE Chinese SL & SLA / Re: Chinese SL + Chinese CLS
« Last post by vox nihili on February 22, 2020, 11:40:59 pm »
Apparently Iím only allowed to do one Chinese, i called VTAC and VCAA and they said I can do both and it will count towards my final scores, then my teacher says she got confirmation from the chief assessor that Iím only allowed to do only one either Chinese SL or Chinese CLS... She is correct or is VTAC correct?

Why do you want to do two?
4
HSC Mathematics Extension 1 / Re: Need help with these induction questions
« Last post by fun_jirachi on February 22, 2020, 10:39:18 pm »
Thanks for that. I have been trying to work on question 6 (x^n - y^n  / y - n) but the long line of xs and ys keep catching me off guard. Also with the one about being divisible by 6, Im a bit stuck because after factoring the equation of n = k+ 1, 6 was not a common factor.

Sorry for the late reply!

With the divisibility question, I'm going to put the answer up, but I'm going to give a few hints for Q6.

Skipping the first step, since I'm sure you've done that


I think it's safe to assume you got up to here, then realised hang on, that's a 3, not a 6. But consider k(k+1) for a second. In any two consecutive integers (we gave the condition that k was an integer!) one must be even, the other must be odd. That means we can actually express two consecutive integers as 2c(2c+1) for some integer c, or if k was odd, (2c+1)(2c+2) for some integer c. Essentially, what this all means is that k(k+1) must be even, and thus 3k(k+1) is an integer divisible by 6. Does this make sense? I think it's relatively straightforward from here since we now have some integer multiplied by 6 :)

For Q6, here's the hint (which basically gives away the answer, so look at your own discretion):
Spoiler


5
Just to answer your follow-up:

To clarify what I said, here's a diagram:
Spoiler
Basically, we are given the side lengths parallel to the coordinate axes, 1 and tan θ. We can fill in the length of the hypotenuse using Pythagoras' theorem/trigonometric identity. Recall that for any number x+iy on the complex plane that the tangent of its argument will just be y/x. Hence, the tangent of the angle the hypotenuse makes with the real axis (opposite/adjacent !!) will be tan θ/1 as given. This way of thinking eliminates the requirement of using arctan - which really just simplifies things.

To answer the additional question:
You can't always assume that arctan and tan will cancel each other out - recall that arctan(tan x) = x if and only if \(x \in \left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right)\). Here, this assumption is okay because the domain we're given is a subset of the domain for which this identity holds.

Something to also note is that this problem is simplified a lot because of the assumption that θ lies in the first quadrant. This means that every trigonometric ratio is positive and we don't have to worry about minus signs or anything. Basically, it means we don't really have to consider the domain of any of the functions to a rigorous degree.

Thank you so much. Really appreciate the detailed explanation.  :)
6
The VCE Journey Journal / Re: jeyda's journey through VCE ʕēᴥēʔ
« Last post by Geoo on February 22, 2020, 10:25:46 pm »
Super happy for you that you are enjoying all of your subjects! It really does make such a difference when your doing stuff you enjoy. That's awesome that you got you learners. Your not alone it getting them late. I turned 16 in April 2018, and I only got mine in February 2019, so welcome to the club. This may seem like a stupid tip, but I really liked using L plates that had suction cups on then, that way you can just jump in and out of the car and take them on and off quickly without leaving any residue or what not. This makes it so easy when other people have to use the car as well.

When I did attend mainstream school, I ended up getting a middle locker and being tall really sucked with it, so as someone who is 174 cm tall, I can 100% relate. (but then again I didn't know how some of the 180+ cm guys managed a bottom locker...)

Have a great week :)
7
The VCE Journey Journal / Re: ✩Chocolatemilkshake cartwheels through VCE and beyond✩
« Last post by Geoo on February 22, 2020, 10:18:32 pm »
Quote
Iíve volunteered to go first in my class
That takes bravery my friend! Good luck on your SACs coming up, i'm sure you do great with all the prep you have done. Since I haven't done a language, your french SAC kinda dose sound scary to an outsider....

Did you enjoy like a house of fire? I haven't gotten around to reading it yet, and I have heard mixed things and I was wondering what you thoughts are.
8
VCE Specialist Mathematics / Re: Converting Cartesian Form to Modulas-Argument Form
« Last post by fun_jirachi on February 22, 2020, 10:13:47 pm »
Just to answer your follow-up:

To clarify what I said, here's a diagram:
Spoiler
Basically, we are given the side lengths parallel to the coordinate axes, 1 and tan θ. We can fill in the length of the hypotenuse using Pythagoras' theorem/trigonometric identity. Recall that for any number x+iy on the complex plane that the tangent of its argument will just be y/x. Hence, the tangent of the angle the hypotenuse makes with the real axis (opposite/adjacent !!) will be tan θ/1 as given. This way of thinking eliminates the requirement of using arctan - which really just simplifies things.

To answer the additional question:
You can't always assume that arctan and tan will cancel each other out - recall that arctan(tan x) = x if and only if \(x \in \left(-\frac{\pi}{2}, \ \frac{\pi}{2}\right)\). Here, this assumption is okay because the domain we're given is a subset of the domain for which this identity holds.

Something to also note is that this problem is simplified a lot because of the assumption that θ lies in the first quadrant. This means that every trigonometric ratio is positive and we don't have to worry about minus signs or anything. Basically, it means we don't really have to consider the domain of any of the functions to a rigorous degree.
9
The VCE Journey Journal / Re: Evolio's VCE Journey
« Last post by Geoo on February 22, 2020, 10:11:02 pm »
Yay, an update! It's nice to hear from you again!
Quote
I donít want to ask my teacher everyday because I feel like thatíll be pretty annoying,
This, so much! I feel like I am really annoying, or that I am becoming a bit of a teacher's pet when i'm really just trying to find out where I went wrong with my questions. I kinda hope my teachers don't hate me by the time year 12 finishes... But yeah, hopefully giving your teacher a bit of space before the next bombardment will hopefully help with that.
Quote
I also thought that the pain of getting that 38 and 41 would fuel me to just study with so much fire and motivation whenever I sit down to do it but it hasnít, I thought that would allow me to reach the top, but whenever I study...I have no feeling.

I 100% relate to you on that statement. Too be honest I don't really think it is a great motivation source, but everyone is different so you do what works for you. For me, I feel just looking back on my score makes me feel kinda like a failure, and that I'm not smart enough to achieve anything. So for me, I stopped using it like I must strive harder so that I avoid that, and I now get my motivation from just knowing that I am doing my best and am learning from my mistakes from last year. Whatever you motivation is, i'm sure you will reach you goals :)

BTW, have you checked out NileRed on youtube? I know you said you watch educational chem videos but have you checked out his? They are kinda educational and really fun at the same time. Good luck studying!
Pages: [1] 2 3 ... 10