Search results

Strictly increasing. Monotonic increasing is <=, strictly is <.

I'd say it's strictly increasing on each of the sub intervals of it's domain that have no discontinuities. x > y implies f(x) > f(y) only if x,y <-3 or -3 < x,y < 3 or 3 < x,y

how2staterank

It's far too late for me to do any analysis right now, but it seems a bit basic to me. If you want a good poem for Belonging maybe Ozymandias?

I'm loving them :skip:

Portal reference, anyone?

Don't make lemonade, make life take them back! Get mad! I don't want your damn lemons, what am I supposed to do with these?! Demand to see life's manager, make life rue the day it thought it could give Cave Johnson lemons! Do you know who I am?! I am the man, who's going to burn your house down...

3U half yearly today, praying for that 100 :read: (I swear we keep getting new smileys, I have never seen the gun or the pizza one before)

when you identify as a point form list

Study well throughout the year so you don't have to cram the day before/the night before/the hour before your exam ;)

$Question 2: $\\x^2 - x - 1 = 0\\x^2 = x + 1\\x^4 = (x+1)^2 = x^2 + 2x + 1 = (x+1) + 2x + 1 = 3x + 2\\x^8 = (3x+2)^2 = 9x^2 + 12x + 4 = 9(x+1) + 12x + 4 = 21x + 13\\x^{12} = x^8 \times x^4 = (21x+13)(3x+2) = 63x^2 + 81x + 26 = 63(x+1) + 81x + 26 = 144x + 89\\$Note that the 'trick' was to...

$For question 1, rewrite the equation as $ \left(\frac{1}{2}\right)^{x-1} = 1 + \log_{\frac{1}{2}}x\\$Plotting the two graphs reveals cleary that there is only one solution. $ (x = 1)

I'm doing One Winged Angel by Nobuo Uematso, and I choose it because I feel like it shows off a high level of skill without being too difficult for me to play well. It also features both fast and slow tempos, creating a nice variety and letting me demonstrate my own stylistic interpretation.

You're welcome :)

No, like, that's literally why she responded with 'veracity' lol

Library meat inside joke

I am the life of the party

+1000000000 on Laser Tag yessssss that was my fave thing in year 9 and 10

Haven't done these in a bit but I reckon this is fine: $Let $ w = x + iy$ and $ z = \mathrm{cis}(\theta) $ (note that we can do this since $|z| = 1$.$\\x + iy = \frac{z-2}{z} = 1 - \frac{2}{ \mathrm{cis}(\theta)}\\x + iy = 1 - 2 \mathrm{cis}(-\theta) = 1 - 2\cos{\theta} +...

Notes for maths? People do that? Just practice, practice, practice until you remember everything you need to remember.