help-10

[glossy yacht]

I know how to find MLE

and to be honest I'm not even sure what a sufficient statistic is, I'm assuming it's just what i get from Factorisation Theorem

[meager swift]

Yeah I've misunderstood

First step is to separate x from alpha

[glossy yacht]

i think I've done that right?

[meager swift]

What you can't separate is what you can choose to be T(x)

Yeah this seems right to me too

[glossy yacht]

sum of cubes

i needed clarification before exams but now I'm just interested

[meager swift]

Well youve indeed shown that there exists functions g, h, T s.t. $l(\theta) = g(T, \alpha) h(x)$

[warm shale]

edwardborn

[meager swift]

In this case T was the sum of cubes and g(T, alpha) = exp(-T alpha)

[glossy yacht]

why can i ignore the product of squares?

[meager swift]

That's your h function

To make it clear that you've satisfied the theorem it might be good idea to define these functions in order to make it more apparent that a factorization like this does exist

Sorry, kinda sleepy atm

[meager swift]

Also add the (3alpha^3)^n factor

[glossy yacht]

so push as much as possible into h?

[meager swift]

It would probably make it easier to see what T needs to be so yeah

Not a bad idea

But there is no single method for finding T, you just have to experiment a little sometimes

Notice also that T isn't unique

[glossy yacht]

what an example of this ?

[meager swift]

In this case we chose g(T,alpha) = (3 alpha^3)^n exp(-T alpha)

But we could also have said that T is the sum of cubes times -1

And then define g as g(T,alpha) = (3 alpha^3)^n exp(T alpha)

[hidden compass]

<@&268886789983436800> how old could they be?? I don't think they can stay here in Discord

[glossy yacht]

ah amazing, thank you

.close

[obtuse pebble]

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[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[glossy yacht]

?

[distant shadow]

Nah bro it wasnt meant for you

[obtuse pebble]

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[fossil spoke]

How to solve |x-4|-|x+2|=6 algebraically for x?

[primal rover]

just solve for three domains x<-2 , -2 < =x < 4, x<=4

[fossil spoke]

i have this

[primal rover]

yeah, add those functions in those domains

for example, x < -2
4-x +x+2 = 6
6 = 6
so all x smaller than -2 is solution

[fossil spoke]

but for x<-2, i have 4-x+x+2=6

simplifies to 6=6

[primal rover]

yeah i edited

[fossil spoke]

oh k

[primal rover]

next, when -2 <= x < 4
4 - x - x - 2 = 6
x = -2 which indeed lies between that domain so x = -2 is also solution

when x >= 4
x - 4 - x - 2 = 6
-6 = 6 which is not true therefore no solution exist in that domain

[fossil spoke]

so it's x <= -2

[primal rover]

yeah

[fossil spoke]

ah thanks

so i need to see if the solved x lies inside the domain which im solving for

[primal rover]

yeah

because you defined the function in that domain, value should also satisfy in that domain you defined one by one

[fossil spoke]

i see

also if x has a coefficient that's not 1 then is it easier to factor it out?

[primal rover]

for example?

[fossil spoke]

e.g. |2x-5|-|4-x|=10
2|x-5/2| - |4-x| = 10

[primal rover]

you can only take positive value out of modulus

[fossil spoke]

how about if |-x+1| = |(-(x-1)|

= |-1|*|x-1|

[primal rover]

yeah

that is what i was saying

so if |-2x -5| - |4-x| = 10
then it would be |-2|*|x-5/2| - |4-x| = 10
which is 2*|x-5/2| - |4-x| = 10

[fossil spoke]

would factoring out this -2 make it easier to solve the equationfor x

since then i can just look at |x-5/2|

[primal rover]

I would say it wouldn't matter at all, because at last you have to solve it.
To make it easier, you can draw a number line and plot the domain in which you define the modulus

[fossil spoke]

also |4-x| = |x-4| right

[primal rover]

yeah

[fossil spoke]

when expanding the piecewise

thanks for the help

.close

[obtuse pebble]

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[granite prairie]

just wondering if A is the correct anwser or is it C

[obtuse pebble]

@granite prairie Has your question been resolved?

[granite prairie]

<@&286206848099549185> sorry to bother but whenever u can

[cunning topaz]

Idk what dilated means
But I think it should be a

[granite prairie]

can i get more than i think pls

[reef pier]

The correct answer is (A) f((x+2)/3) - 4

Translated left 2 units: This is achieved by replacing 'x' with 'x + 2' inside the function.
Horizontally dilated by a factor of 3: This is achieved by dividing the 'x' term (now x + 2) by 3.
Translated down 4 units: This is achieved by subtracting 4 outside the function.
The other options:
(B) f(3(x+2)) - 4: This dilates the graph before the translation, changing the order of operations.
(C) f(x/3 + 2) - 4: This translates the graph up 2 units before dilating, again incorrect order.
(D) f(3x+2) - 4: This combines the dilation and horizontal translation incorrectly

[granite prairie]

thanks so much

i love you

[reef pier]

^^thanks

[granite prairie]

.close

[obtuse pebble]

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[steel urchin]

I solved the system and got 0=0

Which i believe shows independence

So I have showed that for 8, this is independent

However, how do I show that this only works with 8

[rocky goblet]

what's F?

[steel urchin]

R or c

I just solved for c

And got 8

Tyty

.clos3

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[valid crater]

In this exercise i’m asked to rotate this triangle 90° about centre (-2,0)

[valid crater]

How do i know when to stop while rotating 90°

like how can i accurately rotate 90°

i use tracing paper btw

<@&286206848099549185>

[buoyant axle]

huh

draw a point on where the point of rotation is

[valid crater]

It is (-2,0)

my question is how do i know how to exactly rotate 90°, since the triangle is overlapping quadrant one and two

[undone wing]

uh

btw do you want one of the points at the coordinate (-2,0)

[valid crater]

wdym

[surreal forge]

your center is (-2, 0) so there's no overlap

[valid crater]

the triangle itself is overlapping, im confused because i don’t know when to stop while rotating 90°

the triangle is between quadrant one and two, it’s overlapping

[surreal forge]

rotate each point individually, then draw in the line segments

[valid crater]

let me try wait

[undone wing]

which grade math is it?

college or school

[valid crater]

it didn’t work

it’s school @undone wing

anyone???

[undone wing]

bro

can you elaborate your question

or send me the screenshot of the question

i will try to solve

[valid crater]

dude i literally said what my question is 😭😭 when i’m rotating the shape right, how do i know when to stop rotating? it asks me to rotate it 90° clockwise, since the original triangle is between quadrant one and two of the graph, it’s difficult for me to know when to stop rotating the shape because it keeps overlapping quadrants, my question is when do i stop rotating so that it can be rotated 90° clockwise

[undone wing]

send me the screenshot of question

[valid crater]

how

[undone wing]

camera

[valid crater]

oh

[undone wing]

:/

[valid crater]

like the original question

ok

[undone wing]

yes bruh

[undone wing]

what are the options

[valid crater]

wdym options

[undone wing]

is it mcq?

[valid crater]

idk what mcq is

i’ll show u the answer wait

[undone wing]

multiple choice question

[valid crater]

oh

it’s not

[undone wing]

bro

this question is taking time to understand

i mean i understood

but how

[valid crater]

what do you mean how bro

[undone wing]

OH

OH

[valid crater]

😭😭

[undone wing]

I UNDERSTOOD

bro

[valid crater]

LMAO

[undone wing]

simple trick

take a compass

[valid crater]

please tell me 🙏

YES OMG PLEASE TRACH ME THE COMPASS THING ITS SO CONFUSING

Teach*

i literally rely on tracing paper it’s so annoying

[undone wing]

place the uh

what do u call

the needle thing

[valid crater]

uh huh

at a point

[undone wing]

the sharp one

[valid crater]

yes ik

[undone wing]

what do u call

?

[valid crater]

idk

needle

[undone wing]

ok

place the needle at the point (-2,0)

[valid crater]

mhm

[undone wing]

make a circle

[valid crater]

from what point

like yes (-2,0)

[undone wing]

yeh

[valid crater]

but where do i put the pencil

[undone wing]

on b point

[valid crater]

ok yeah

can it be any point btw

[undone wing]

i didn't get you

[valid crater]

like can i put it at c point or a point for example

[undone wing]

wait

yes you can

[valid crater]

ok then what

[undone wing]

wait

let me assemble the words in my brain

[valid crater]

okk

[undone wing]

now make 90 degree from any point

clockwise

from the point (-2,0) to A, B, or C point the angle is 0 degree

[valid crater]

what do you mean

[undone wing]

did you understand this?

[valid crater]

nope

[undone wing]

ok suppose the point (-2,0) is O

[valid crater]

Ok

[undone wing]

now imagine

what is the measure of the angle from O to any point on triangle (take only 1 point)

[valid crater]

i’m not sure

do i have to use a protractor?

[undone wing]

try

yes u can

no one is stopping you

[valid crater]

i don’t know how

[undone wing]

0 degree

[valid crater]

why

[undone wing]

uh how do i explain lol

ok leave it

uhm

[valid crater]

ok

[undone wing]

do u have protractor in ur hand?

place the centre of protractor on point (-2,0)

[valid crater]

ok i did

[undone wing]

which point on the triangle do you want to take?

a b or c

[valid crater]

we can do b

[undone wing]

ok

now

place the sharp end of protractor

on b

then rotate it 90 degree

got it?

[valid crater]

ok wait

[undone wing]

done?

[valid crater]

@undone wing hey man i just asked my friends if the teacher said if we’re gonna have this exercise in the exam and she said no, but i appreciate your time and effort thank you 🙏🙏

[undone wing]

bro did you understood it?

thats the question i asked

[valid crater]

kind of but i don’t have to know it anyway since it’s not in the exam

[undone wing]

lol

dont be lazy

[valid crater]

anyway i have to move on now to other exercises because my exam is tmw but tysm again 🙏

.close

[obtuse pebble]

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[undone wing]

ok

bye

.close

[obtuse pebble]

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[slow sonnet]

I should probably use laplace transform

I'm stuck

can someone verify if Y(s) = (s+2)/((s+1)(s^2-2s+2))

the transformed function

[opaque dome]

https://www.youtube.com/watch?v=QvpE3BrOGlQ

[opaque dome]

this is the way to do it

[opaque dome]

and this is verified in the video

[slow sonnet]

thanks

[obtuse pebble]

@slow sonnet Has your question been resolved?

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[heady turtle]

a1 here is chosen randomly?
It's also ok if we pick a random m ∉ ｛an｝?

[obtuse pebble]

@heady turtle Has your question been resolved?

[heady turtle]

<@&286206848099549185>

Anyone😭

[fossil crag]

I think it's a bit not well worded

but yeah you pick any upper bound you want

like

you want to show that after some time you can't get more than you already have

a1 is just the first term in the sequence

but since a1> 0 and lim(a_n) = 0

[fossil crag]

after some time all terms will be below a_1

here we picked N such that all terms after a_N are below a_1

so all terms afterwards can't be the max

since they're not bigger than a_1

[obtuse pebble]

@heady turtle Has your question been resolved?

[obtuse pebble]

@heady turtle Has your question been resolved?

[obtuse pebble]

@heady turtle Has your question been resolved?

[obtuse pebble]

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[lethal charm]

.close

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[hoary onyx]

Hi, I need to make a parabola with a maximum height of 9.5m (so vertex is 0,9.5) and a width at "the ground" of 16m, I have come up with the following equation for this: (y-9.5)^2=11.281x. Is this correct?

[grizzled shore]

That’s not even a parabola in the normal sense

This guy is sideways

[hoary onyx]

Yes I am aware, but we need to use the formula Y^2=4ax or (y-k)^2=4a(x-h)
I know it's werid and doesn't make sense to me either, but it's the criteria.

[hoary onyx]

So can we solve it using the formula I gave above?

[grizzled shore]

Well I’m not sure what “maximum height” means for a parabola that goes sideways

[hoary onyx]

the vertex would be at (0,9.5)

[grizzled shore]

Unless you swap the axis…but then why have it be a quadratic in y

[hoary onyx]

Yes, x should represent the height and y the width

[grizzled shore]

Huh

Sure

[hoary onyx]

I know, i dont like it either, its just how the criteria says it has to be done, which is annoying

[grizzled shore]

Ok

If x represents height

What does the point (0, 9.5) mean

[hoary onyx]

I am confused too. Let me get back to you

[grizzled shore]

Well what does x = 0 mean

[hoary onyx]

Okay, lets do it with a downwards opening parabola, and hopefully that will suffice, as I really cant get my head around this being sideways

[grizzled shore]

You literally just swap the x and y at the end

[hoary onyx]

So what would the equation be now, for this parabola?

[grizzled shore]

What conditions does it need to satisfy?

[hoary onyx]

maximum height of 9.5m and a width "at ground level" of 16m

[grizzled shore]

What does that mean in terms of the quadratic equation?

[hoary onyx]

Well, part of the critiera is I can't use y^2=Ax^2+B

I have to use Y^2=4ax

[grizzled shore]

That’s not even a parabola

[hoary onyx]

Okay, so help me out here

[grizzled shore]

Let’s do the max first

Use the vertex form to write out the general equation of a parabola with a max at (0, 9.5)

[hoary onyx]

Okay, so y=ax^2+9.5?

[grizzled shore]

Don’t worry about the y = ax² form

Use the vertex form

[hoary onyx]

Yes, so doesn't (0,9.5) represent h,k?

Can you advise the equation? Then i can try and make sense of it

I have been trying this for hours.

[grizzled shore]

Ok right

I see

No you’re right, I was just expecting y = a(x - 0)² + 9.5

Which is the same as what you wrote

Now one more thing, we want this to be a maximum, what does that mean for a?

[hoary onyx]

This though is the form that the criteria said not to use though, y^2=Ax^2+B

I learnt with y^2=4ax.

[grizzled shore]

But we just swap the x and y at the end it’s the same thing

[hoary onyx]

Okay lets, keep going then

Heres the whole question for your reference though: A design has a parabolic cross-section. The design will
have a maximum height of 9.5 metres and a width at ground level of 16 metres.
Find a mathematical model for the cross-sectional shape of this alternative design.
Use the You must use the equation for the parabola in the form y2 = 4ax
The model y = Ax^2 + B is not sufficent.

[grizzled shore]

The other way would be to first consider all 3 points you require to be on the parabola then solve for the unknowns

[grizzled shore]

Ok fine we’ll do it this way then

[hoary onyx]

[grizzled shore]

Pick 3 points that satisfy the conditions

x is height and y is width

Pick 3 points your parabola has to go through

I’ll give you the first one

It needs to pass through (9.5, 0)

[hoary onyx]

Yes, and (8,0) (-8,0)

[grizzled shore]

Nope

Wrong way

[hoary onyx]

Do you have a link to the graph on desmos so I can visualise this?

[grizzled shore]

It’s just the other way

(0, 8) and (0, -8)

At height 0, the sides are 8 and -8

Then they’ll be 16 apart

Does that make sense

[hoary onyx]

Yes, kind of. Can we graph it please?

Desmos perhaps?

[obtuse pebble]

@hoary onyx Has your question been resolved?

[obtuse pebble]

@hoary onyx Has your question been resolved?

[obtuse pebble]

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[celest musk]

If V is a vector space of dimension 6, W is a subspace of dimension 4, then the complement of W has dimension 2

[celest musk]

True or false? This was a question in my exam right now

I wrote dim(V) = dim(W) + dim(W's complement) - dim(intersection between W and W's complement)

So we get 6 = 4 + dim(W's complement) - 0

And therefore the complement of W has dimension 2

Is this correct or wrong?

I think it's wrong but idk

W + W's complement is a direct sum

So their sum must have dimension = 6

I think

[shut lagoon]

I mean it depends how you define complements.

In the usual set theoretic sense, the complement wouldn't even be a subspace to begin with, so that's that.

In the more likely case where we define this as a complementary subspace, then it is defined exactly so that the dimensions add up to 6, so you would be right.

[grizzled shore]

Shouldn’t you be talking about the orthogonal complement?

Rather than a set complement?

In which case you need a notion of orthogonality

Which means you need an inner product space

[celest musk]

My prof defined W's complement as V\W

[grizzled shore]

That’s not a subspace

It doesn’t contain the 0 vector of V

You took it away when you did V \ W

[celest musk]

Oh my god

I wanna cry

[grizzled shore]

$W^\perp \coloneq {x\in V| \langle x, y\rangle = 0,, \forall y \in W}$

[warm shale]

Frosst

[celest musk]

So I got it wrong

Because its not even a subspace to begin with

🙂

[grizzled shore]

No I think your professor got it wrong

[celest musk]

Are you sure? Lol

That would be funny

[grizzled shore]

The what complement of W

[wild swallow]

a vector space complement of W is a subspace U such that V is a direct sum of U and W

in this context, it's most likely to be interpreted to be vector space complement

[celest musk]

Yes and that's what I wrote, so dim(W's complement) should be 2

[grizzled shore]

Ah there’s also that one

[celest musk]

But if its not a subspace...

[wild swallow]

no a vector space complement is a subspace

[celest musk]

He wrote W's complement = V\W

[wild swallow]

are you sure it's not V/W

[celest musk]

Pretty sure it was V\W

[grizzled shore]

Sussy

[celest musk]

Idk results are in 4/5 days

I'll update you guys

[wild swallow]

i'm very sus of that

the complement of W is isomorphic to V/W

[grizzled shore]

Is that just taking everything in W to be an equivalence to 0

[wild swallow]

it's a vector space quotient

also, you don't need orthogonality to define complements

complements are in bijection with projections

[grizzled shore]

I forgor about the vector space complement

[celest musk]

If it's V/W, then dim(W's complement) = 2?

[wild swallow]

yes

[celest musk]

Ok thank you!

Thanks @grizzled shore

.close

[obtuse pebble]

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[tawdry hinge]

Hi
I have a question in calculus I've been stuck on:
"a_n and b_n are 2 series, b_n != 0. lim(a_n) = infinity, and (a_n + b_n) is bounded. Prove that the limit of a_n/b_n exists (can be ±infinity), or show it doesn't exist"

I tried doing the following:
M > 0
|a_n + b_n| =< M
|a_n + b_n| = |a_n - (-b_n)| >= ||a_n| - |-b_n|| = ||a_n| - |b_n|| >= |a_n| - |b_n|
|a_n| - |b_n| =< M
|a_n| =< M + |b_n|
|a_n| - M =< |b_n|
infinity - M = infinity =< |b_n|
which is why lim|b_n| = infinity
and so lim(a_n/b_n) is non-existent

However, a friend of mine proved that the limit exists by doing the following:
|a_n + b_n| =< M
-M =< a_n + b_n =< M
-M - a_n =< b_n =< M - a_n
a_n/(-M - a_n) =< a_n/b_n =< a_n/(M - a_n)
a_n/(-M - a_n) = ((-1)*(-a_n - M) - M)/(-M - a_n) = -1 - M/(-M - a_n)
Same for a_n/(M - a_n)
And so lim(a_n/b_n) = -1

Which of us solved it correctly?

[obtuse pebble]

@tawdry hinge Has your question been resolved?

[tawdry hinge]

<@&286206848099549185>

I'd really appreciate help, I've been stuck on it for 1-2 days

[meager glade]

my guess is that youd have to do something with (a_n+b_n) is bounded but lim a_n =infinity

which gives some restrictions on b

[tawdry hinge]

yeah which is why I feel like b_n->(-1) doesn't make a lot of sense

[meager glade]

yeah, b_n has to go to -infinity aswell to keep a_n+b_n bounded

but showing that in general? no clue

[tawdry hinge]

Yeah this is what I was thinking
But I can't seem to see where my friend made a mistake

[meager glade]

-M - a_n =< b_n =< M - a_n
a_n/(-M - a_n) =< a_n/b_n =< a_n/(M - a_n)
i see that here that he did 1/x, if you do that, theny ou have to inverse the inequalities

[dreamy cove]

your friends proof seems mostly right

apart from, as flappie just mentioned, the inequalities need to flip

the other issue is that they then say lim bn = -1

[tawdry hinge]

oh he may have done it and i forgot to write that in my message

[dreamy cove]

but what they actually show is lim an/bn = -1

one minor note though, you should first restrict to n >= N some N where a_n > 0

[timid silo]

Hi

[dreamy cove]

just to be sure that -M - a_n is <0

[timid silo]

Hello

[tawdry hinge]

yeah we did that

[dreamy cove]

hi

all good then

[timid silo]

Que tal

[tawdry hinge]

wait so he's right?

[dreamy cove]

yeah

[tawdry hinge]

where did i get it wrong in my proof?

[dreamy cove]

with this bit they show that an/bn is squeezed by -1, but then forget that in the next line haha and talk about lim bn

[dreamy cove]

you are right that lim |b_n| is infinity

[tawdry hinge]

it might have been my bad writing it down from the picture he sent me

[dreamy cove]

but consider a_n = n, b_n = -n

an -> infinity, an + bn = 0 is bounded and an/bn has limit -1

[tawdry hinge]

i understand

[tawdry hinge]

but this part feels a bit wrong to me, doing a_n/(-M - a_n) -> -1

[dreamy cove]

basically the key is that you can have (and in fact this is very common) lim |an| = inf and lim |bn| = inf but lim an/bn exists

[tawdry hinge]

i understand how he got there, but it still feels wrong

[dreamy cove]

this is perfectly valid

[tawdry hinge]

huh

Ok well thank you
I think I understand

I guess I should just practice more cause it didn't occur to me naturally

[dreamy cove]

nws ! best of luck

[tawdry hinge]

thx

.close

[obtuse pebble]

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[neat yarrow]

I'm trying to figure out the amount of unique algorithms you can make on a Rubik's Cube with any n moves The algorithms must be unique, meaning that if two algorithms result in the same computation, they are the same algorithm.

[neat yarrow]

I need help

[zenith raft]

what does computation and algorithm mean here?

[neat yarrow]

Computation is any order of peaces

Algorithm is any set of moves

Computation is more of an arrangement

[old lily]

can you move the middle pieces

like M or E

[neat yarrow]

Not on 3*3

[old lily]

i mean middle moves

[neat yarrow]

No

[old lily]

we're talking about 3x3 right

[neat yarrow]

Yes

[old lily]

ok what's the answer if n=1

[neat yarrow]

12

But if n is 2 then 132

But when n is 3?

It gets annoying

Cuz is ur 1st and 2nd move was r, your next move can't be r or r'

Cuz 3 right clockwise is 1 anticlockwise

[old lily]

so the 12 are r,r',r2,l,l',l2,u,u',u2,d,d',d2 ?

i'm thinking some recursion thing

[neat yarrow]

We are on a quarter turn metric

So r2 isnt

[old lily]

so b,b'

[neat yarrow]

Or u2 or b2 exetra

[neat yarrow]

It doesn't count not together at least as they undo each other

[old lily]

yeah

but seperate moves

for n=1

[neat yarrow]

Yes

12

6 faces 2 ways of rotating each

[old lily]

ohhhh

valid

f,f' as well

anyways

[neat yarrow]

Yes

Separate

R,u,f,b,l,d and their inverse

And n=2 is 12×11

[old lily]

mhm

[neat yarrow]

But n=3 could be 12×11×11 or 12×11×10

[old lily]

so if you have a combination from n-1 moves, there are 12 possible moves you can make

[neat yarrow]

I think I'm missing something

[old lily]

i think 12x11x11 not sure

[neat yarrow]

Yes

[old lily]

i guess call the sequence so far c

then like cr is equivalent to c'u for some other c'

c' done in n-1 moves

[neat yarrow]

I'm confused

C represents which move of the algorithm?

[old lily]

i mean like c is an algorithm itself

[neat yarrow]

Ok

[old lily]

like r,u,l,... n-1 times

idk there might be a better way to write it

[neat yarrow]

What is the purpose of c?

What do you think the amount of unique permutations is when n=3?

[old lily]

not sure if c' will have n-1 moves tbh

[old lily]

idk

[neat yarrow]

I would do to the power of n-1

[old lily]

yh i was thinking you could do some group theory thing as well

[neat yarrow]

But that presumes that it's always the same choice amount

Just for reference my maths isn't too good

[neat yarrow]

Which isn't the case if ur last 2 moves where the same

[old lily]

i see what u mean

like rrr is the same as uu'r'

[neat yarrow]

Rrr=r'

[old lily]

yh

[neat yarrow]

No

It's not commutitive

[old lily]

isn't uu' the identity

like you get the same thing

[neat yarrow]

No no change to the state is the identity

So uuuu

[old lily]

But u' reverses u

[neat yarrow]

Any algorithm repeated enough cycles back to the identity

[neat yarrow]

Yes

Anything combine with its inverse gives e

The binary operator is function composition

[old lily]

so every element of the rubix cube group has finite order

[neat yarrow]

Yes

It is a finite permutation group

4.3252×10^19

[old lily]

oh a finite group

[neat yarrow]

Yes

[old lily]

so the elements are moves

[neat yarrow]

No, the permutation

[old lily]

and binary operation is composition

[old lily]

okay

[neat yarrow]

Denoted by moves that get u there

[old lily]

oh yeah

[neat yarrow]

So permutation rrr=r'

[old lily]

i was thinking sequences of moves

upto equivalence

same thing

[neat yarrow]

I just wanna know how many permutations you can get with n moves and its annoying me

Cuz it seems simple till u try

[old lily]

sorry i'm not too sure tbh

i'd probably learn more about the rubix cube group

if i were to approach this

good luck

[neat yarrow]

Thanks

[dark stirrup]

.close

[obtuse pebble]

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[worn ermine]

I’m confused on the equation down below

Because

It makes sense

But if I plug that same equation into

This

It doesn’t work

[meager glade]

does D_0,2 mean you only scale y and not x?

[worn ermine]

O means origin

Ignore that

[meager glade]

ah okay

[worn ermine]

D Is dilation

[meager glade]

here its asking the inverse

[worn ermine]

So we are dilating it by a scale factor of 2

[meager glade]

P' is what you get after you dilate P

P' is given

you need to find P

[worn ermine]

Oh that’s why

I didn’t see

[meager glade]

happens

[worn ermine]

So I would really need to do

1/3

[meager glade]

exactly

[worn ermine]

Or that equation down below

[worn ermine]

And

[meager glade]

so it would be -1,2

[worn ermine]

For this one

I would just multiple x and y by 2

Because it is normal

[meager glade]

yes

you have to see which one you are given and which one you need to find

[worn ermine]

Ok

[meager glade]

if you are given the dilated one

[worn ermine]

Thanks

[meager glade]

you need to inverse the dilation

[worn ermine]

I just need to read next time

[meager glade]

indeed

[worn ermine]

One more thing

Could you help me with?

[meager glade]

!da2a

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[worn ermine]

Ok

Lemme get picture

[meager glade]

this is similar triangles

how did you try to solve this?

[worn ermine]

How do I know the order

Like CD/CA = DE/AB

[meager glade]

it uses the fact that the proportions of the big and small triangle are the same

so, take the proportions of the small triang

triangle

which is 8/15

or 15/8

you can choose either one, but you have to be consistent and take the same side when using the big triangle

[worn ermine]

How do I know it is not AB/CD = AD/DE

In that case it would be

[meager glade]

this doesnt really make sense

[worn ermine]

12/15 = x/8

[meager glade]

the 8 in the small triangle is the same side as the 12 in the big triangle

[worn ermine]

So the cross multiplying would not give the same answer

[meager glade]

do you follow this?

[worn ermine]

Hm

O

Ye

[meager glade]

so, that proportion

is the same proportion as the other side

[worn ermine]

8/12

.667

And

Small triangle is CD

[meager glade]

so we can take smallside 1/largeside 1 = smallside 2/largeside 2

[meager glade]

does this make sense?

[worn ermine]

Big triangle side is x + 15

[worn ermine]

Yea

So

15/x+15

=

[meager glade]

exactly

[worn ermine]

Yeah

Cool

[meager glade]

continue

you wrote the =

[worn ermine]

Ok

[meager glade]

what is it equal to

[worn ermine]

8/12

[meager glade]

yes!

[worn ermine]

Yeay

[meager glade]

so, how would you solve it?

[worn ermine]

Cross multiplying

And then algebra

[meager glade]

so, do that

[worn ermine]

Ok

x = 60/8

x = 7.5

[meager glade]

yep

[worn ermine]

Thx

[meager glade]

👍

[worn ermine]

Ima ask another question

In this channel

[meager glade]

sure

[worn ermine]

You don’t have to stick around to answer if you want to move on

[meager glade]

its oaky

[meager glade]

i cant stop myself anyway

[worn ermine]

lol

So

[meager glade]

the first one?

[worn ermine]

I got these right

But like I kinda guessed on both

[meager glade]

okay, so whats the question

ah okay

[worn ermine]

Like how do these things work

[meager glade]

okay

lets go through it one by one

what are the sides AD, AE, and EB?

okay, mb

one first

what side is AD?

[worn ermine]

AD/DC = AE/EB because of angle bisect theorem??

[worn ermine]

Uh

[meager glade]

big triangle, small triangle, left, right, up, down, etc

[worn ermine]

Ok

It is the left small triangles

Side

[meager glade]

its part of triangle ADE right?

[worn ermine]

Yes

Mb

[meager glade]

and its the left one in the picture

(i would consider the other ones down and right)

[worn ermine]

Yes

[meager glade]

okay, now what is AE

[worn ermine]

Looks like it is an angle bisect or

[meager glade]

that doesnt matter here

thats not the question

[worn ermine]

Ok

[meager glade]

we're talking about the triangles

not the angles

[worn ermine]

Splits triangle adf in half

[meager glade]

.

same question

just a different side

[worn ermine]

It is part of the two small upper triangles

It is a side of both

Shared side

Of

[meager glade]

which one do we care about?

[worn ermine]

ADE

And

[meager glade]

(e.g. which triangle have we already looked at and want to know more of)

[worn ermine]

AEF

ADE

we care

We love

I think

Because we have more info

On

ADE

[meager glade]

exactly

we already have identified the left side of ADE

whcih is AD

and now we have identified the right side of ADE

which is AE

now, what is EB?

[worn ermine]

He is a side of

ACB

and

AEG

[meager glade]

which one do you think we care about

[worn ermine]

ACB

[meager glade]

triangle ACB or AEG

[worn ermine]

we have more info

[meager glade]

well, we dont have info about ACB

but it shares an angle with ADE, which we do know

[worn ermine]

AE + EB = Ab

Which is one of ACB’s sides

[worn ermine]

It shares

Angle

d

And c

And

Then

E and b

[meager glade]

angle DAE=angle CAB

[meager glade]

watchout for the order here

angle DAE is not the same as angle EAD

[worn ermine]

Yes

[meager glade]

anyway

so

we have identified AD to be the left side of triangle ADE

we have identified AE to be the right side of triangle ADE

and

we have identified AB to be the right side of triangle ACB

which other side are we missing?

[worn ermine]

Ac

[meager glade]

yes!

[worn ermine]

Cb

De

[meager glade]

okay so,

i made a mistake

i misread

i thought EB was AB

[worn ermine]

😱

[meager glade]

so then it would be AC

[worn ermine]

Your good

[meager glade]

but the question is EB

then what is the side we're missing?

[worn ermine]

Dc

[meager glade]

exactly

maybe if i colour it like this, it becomes clearer

[worn ermine]

Is this part of the angle bisect or theorem

[meager glade]

nonono

dont think of that

just think of similar triangles

[worn ermine]

cause that’s part of that lesson

[meager glade]

the angles in these two triangles

are the same

but one triangle is scaled up/down from the other

[worn ermine]

How do I do pt 2

We have

AD

As

[meager glade]

i coloured them for you

you can do the same

[worn ermine]

ADE side

Ok

[meager glade]

is it clear from this image what side is missing?

[worn ermine]

Yea

So

Is this like

A logic thing

Cause this feels like a logic

Question

[meager glade]

its a similar triangle thing

and in this case

[worn ermine]

Idk

[meager glade]

here you use angle bisector

[worn ermine]

What similar triangles is it proving here

[meager glade]

since, this would not be true unless that angle A is split in 2

[meager glade]

triangle AED and AEF

[worn ermine]

Ok

[worn ermine]

Wait aef?

[meager glade]

look at the angle DAE, and see that it is the same as FAE

[worn ermine]

Why do we need DC then

[meager glade]

we are talking about part 2 right?

[worn ermine]

Pt 1

[worn ermine]

What similar triangles are those

[meager glade]

ive become confused

im not sure what we are talking about anymore

[worn ermine]

Pt 1

[meager glade]

we already solved part 1

[worn ermine]

I switched back to pt 1

[worn ermine]

But like

[meager glade]

whats your question about part 1?

[worn ermine]

What similar triangles

In

Pt 1

[meager glade]

in part 1 the similar triangles are DAE and CAB

[worn ermine]

Ok

[meager glade]

do you see why thats the case?

[worn ermine]

That is why we need DC?

What about

DE

And

Cb

What about those sides

Do we not need to figure those out since the image shows they are parallel

[meager glade]

you can also write that

but in this question its not asked

[worn ermine]

Because it shows they are parallel ?

[meager glade]

but if you want it $\frac{AE}{EB}=\frac{DE}{CB}=\frac{AD}{DC}$

[meager glade]

thats so that you can "prove" the triangles are similar

since by F-angles, the triangles ahve the same angles

[worn ermine]

Now we would be missing AD and Dc

[warm shale]

Flappie

[meager glade]

happy now?

[worn ermine]

Ok

Thank you

[meager glade]

but its unnecessary to write it out

[worn ermine]

Ye

[meager glade]

if you have any more questions i recommend you close this channel and open a new one

[obtuse pebble]

@worn ermine Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[fluid snow]

Does it always follow that if we have a figure with 540° interior angle sum, then it's a pentagon?

[haughty coyote]

there is a proof that the sum of the angles always sum to 180(n-2)

Silly idea: prove it for regular polygons, then argue that deformations of the plane that regularize polygons can always be taken to be holomorphic and therefore angle preserving, and conclude. That might also work

[obtuse pebble]

@fluid snow Has your question been resolved?

[obtuse pebble]

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[elder coral]

yo

[elder coral]

i need help with a limit concept

[obtuse pebble]

@elder coral Has your question been resolved?

[rigid plaza]

This seems like a complicated question to answer in general

@elder coral

What video did you watch

The usual method of proving a limit exists is an epsilon delta proof

If we don't have continuity there is no guarantee that the limit even exists, even though it can

[elder coral]

i didn’t understand anything you said

it was khan academy

the video

[rigid plaza]

Short answer: this is a university level topic typically discussed in an analysis course

[elder coral]

oh

[rigid plaza]

If you link the video I could take a look for you

[elder coral]

it’s on the khan academy app

it’s called theorem for limits of composite functions when conditions aren’t met

maybe it’s on yt

[rigid plaza]

I found it

So they actually have explicit graphs for what the functions are

And those functions are sufficiently nice

They are piecewise linear

What they are doing is looking at those graphs

And analyzing the behavior the left and right limit separately

If the left and right limit exist and are equal, then the limit exists and it is that value

So what they are implicitly doing is directly looking at f(g(x))

Instead of f and g separately

[obtuse pebble]

@elder coral Has your question been resolved?

[elder coral]

mmh

[obtuse pebble]

@elder coral Has your question been resolved?

[obtuse pebble]

Channel closed

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[blazing vale]

-4 > |x| > 4
is that right??

ignore what i had in the box in the picture

[gilded needle]

|x| is always nonnegative, it can't be less than -4

[blazing vale]

wait wha

so what goes in the

absolute value

[gilded needle]

also, -4 is less than 4, so nothing can satisfy -4 > something > 4

[blazing vale]

box then?

4

im so lost bungo

wym -4 is less than 4

[marsh geyser]

The question is too simple to give hints

[blazing vale]

fr

-4 > x > 4

am i onto something with this?

[marsh geyser]

That doesnt use absolute value first of all

[blazing vale]

okay

[marsh geyser]

Do you know what is distance?

[blazing vale]

7?

[marsh geyser]

Imagine yourself opening hands

[blazing vale]

word

[marsh geyser]

One side to the other

[blazing vale]

im doing it rn

mhm

yes

[marsh geyser]

That distance from your center

[blazing vale]

the space between them

[marsh geyser]

(Head)

[blazing vale]

yeah so 4

[marsh geyser]

Imagine each side is 4

But you want each side is >4

Because as u can see in the picture

The blue line is not your arms

Is after it right?

[blazing vale]

so 4 should be in my

absolute value

because

5

4

and negative 4

and it is

nvm nvm im gonna go on khan academy or somethign

im really lost

thanks tho

[marsh geyser]

Here, 0 is your head, the white space are your arms

[blazing vale]

yes

[marsh geyser]

The blue is outside

[blazing vale]

YES

yes

yes

[marsh geyser]

You want bigger than 4

[blazing vale]

okay

yea

[marsh geyser]

Absolute value >4

|x|>4

[blazing vale]

ohhh

[marsh geyser]

👍

[blazing vale]

soo

is that it?

or do i need ot

add the

-4>|x|>4

[marsh geyser]

No, that expression makes no sense

The absolute value will handle both sides

[blazing vale]

okay

OHHH \

[marsh geyser]

I recommend you to study the absolute values

[blazing vale]

okay

thanks

[obtuse pebble]

@blazing vale Has your question been resolved?

[blazing vale]

@marsh geyser

i studied a little

and i think im onto something

|x-0|>4

is this correct

[hearty terrace]

is this not just |x|>4

[blazing vale]

yea

if i simplified it

it would be

that

right

so that would be my final

answer

[hearty terrace]

in this ur saying "the absolute value of x is greater than a negative number"

[blazing vale]

i guess

[hearty terrace]

which is like

obvious because it has to be

[blazing vale]

yea no ignore the

what i put in the box

[hearty terrace]

but yes you're right |x|>4

[blazing vale]

i had no idea what i was doing

cool thankyou

[hearty terrace]

you dont need this

[blazing vale]

i locked in

yea i dont know what i was doing

there

thank you @hearty terrace

your the best

[hearty terrace]

yea np

lol ty

[blazing vale]

@hearty terrace

im lost again

so basically

should i

[hearty terrace]

yea i gotchu one sec

[blazing vale]

word thanks

so

am i solving for x