help-39

[sacred nimbus]

upon opening the bracket

u get

(m^4)^-2

which is?

u wrote m^-6

[brazen night]

😭

[sacred nimbus]

🤣

[brazen night]

Would it be m^6

[sacred nimbus]

no dude

see

[brazen night]

Homie I can’t

🥲

Yes

[sacred nimbus]

it will be m^-8

[brazen night]

What the flurp

[sacred nimbus]

when exponents are on bracket

like in this case

they multiply

think it for urself

(m^-4)^2

means

[brazen night]

4 x 2

😞

[sacred nimbus]

u multiplying (m^-4) 2 times

so m^-4 * m^-4

which is -8

u got it right?

[brazen night]

Yes 🙂‍↕️

[sacred nimbus]

no issue

try again

[brazen night]

Do I move onto the next steps or do I continue rechecking my steps

[sacred nimbus]

i mean

there are many things wrong

[brazen night]

Yes 🙂‍↕️

[sacred nimbus]

i would seriously reccomend u to

do each step

[brazen night]

Okay bet

[sacred nimbus]

instead of jumping

write more steps

more simple steps

[brazen night]

Math is tew hard

[sacred nimbus]

just practice

enough practice

then everything easy

trust me

[brazen night]

Yes yes yes

[sacred nimbus]

speak what u write

and do slowly each step

speaking helps sometimes notice that u did something wrong

i also do many silly errors 🤣

its nothing uncommon

[brazen night]

Thank you bro 🥲🥲🥲

[sacred nimbus]

do tell me if u get answer

[brazen night]

I will sir 🫡🤧

Okay

Yes u were right there was a lot of mistakes

😂

I will I’ve on to the next steps

Okay I did it

[sacred nimbus]

damn

nice

[brazen night]

HAHA YES

[sacred nimbus]

keep going all the best

[brazen night]

I got the answer too

Thank you sir

[sacred nimbus]

yea i can see that

[brazen night]

😛

[sacred nimbus]

np

[brazen night]

Also a question

[sacred nimbus]

yea

[brazen night]

Do yall do this voluntarily for fun???

The math helping thing 👍

[sacred nimbus]

yeah

[brazen night]

Wow 😮

[sacred nimbus]

i am anyways in grade 12

[brazen night]

What the Flurp

[sacred nimbus]

i already know this stuff pretty well

so i thought an easy problem

[brazen night]

😀😆😃🫰

Thank you for saving me

[sacred nimbus]

haha

no issue

i mean u did it urself

[brazen night]

Yeah but you helped out with pointers

So thanks bro

[sacred nimbus]

hmmm yeah yeaa np

[brazen night]

I wish you well in ur future college 🙂‍↕️

Peace out

[sacred nimbus]

haha thanks 🙏

byee

[brazen night]

✌️🕊️

.close

[pearl pond]

Channel closed

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[pearl pond]

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[unborn charm]

I need help in proving that for any linear operator diagonalizabilty implies semi-simplicty however the converse is only true for an algebraically closed field.

[unborn charm]

I know what all of this means individually but not why this statement is true, so i guess i do not know where to begin

okay nvm, I got diagonalizabilty -> semi simplicity

I need the converse

[tropic saddle]

the only thing holding you back from diagonalizability is the potential lack of eigenvalues, your algebraically closed field fixes that

basically, start with any eigenspace (which exists due to your field) and then consider the invariant subspace which is in direct sum to that subspace and then induction

[pearl pond]

@unborn charm Has your question been resolved?

[pearl pond]

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[olive vine]

i need a help. i’m confuse on how will i solve this worded problem, i believe it’s a combination of projectile motion and newton’s law (kinetic force??) (this is physic)

[pearl pond]

@olive vine Has your question been resolved?

[olive vine]

<@&286206848099549185>

[rough meteor]

i cant see the word kinetic force first of all

but yea mechanical energy consevation must b in play here no doubt

[olive vine]

ohh

im just gonna close this it’s kinda hard...

.close

[pearl pond]

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[spring oracle]

I am stumped on what should I begin from

the right side is me trying integration by parts with a t substitution

left side is me trying different substitution methods

[steep saddle]

what did you get stuck on for t-sub then int by parts

[spring oracle]

I'll rewrite it

stuck on ∫[t^5 *[∫e^(t-4)]]

[steep saddle]

i have no idea how you got that

[spring oracle]

yeah

I make a lot of sign mistakes

rewriting it

[steep saddle]

the 2nd integral... is that 1/t^4?

[spring oracle]

I'll rewrite and send it

consider that one

[steep saddle]

ok

[spring oracle]

?

[humble root]

Your working out here is a little messy sorry, so I just would like to clarify:
Let $I$ be the integral you are trying to find. At the top, you essentially have:
[ I=\int \frac{e^{\left(t-4\right)}}{t^{4}}dt-4\int\frac{e^{t-4}}{t^{5}}dt ] Which is good. You then apply IBP, but it is a little unclear as to how you apply it.

Oh I see, you apply it to the first integral.

[spring oracle]

I'm doing it integral by integral

first one only then I move to the second one

[humble root]

Okay, I see. So yeah, you apply it to the first integral, and so you get: [ \int_{ }^{ }\frac{e^{\left(t-4\right)}}{t^{4}}dt=\frac{e^{t-4}}{t^{4}}+\int_{ }^{ }4t^{-5}e^{t-4}dt ] This looks good so far.

[spring oracle]

the right term means that you'll have to apply IBP again? []

[humble root]

Actually, you won't need to :)

[spring oracle]

wait you made a mistake

[humble root]

If you substitute your result into here, what do you get?

[spring oracle]

x is t-4

not t+4

[humble root]

Oh yeah

[spring oracle]

my t should be looking like a plus

[humble root]

Good catch.

[jolly parrot]

Redfern Station

Redfern Station

[humble root]

Anyways yeah, so what should we do next?

[spring oracle]

when I put it back and begin with the first step for IBP in the 2nd integral

[i'm sending a photo, I can't use latex]

[humble root]

wait wait

Don't use IBP again

Trust :)

I would recommend you write out the entire equation.

[spring oracle]

I am doing that

which means -8 [the second term ]

[humble root]

Wait, you have a sign error

The second term should not be negative

[spring oracle]

you thought it would cancel out?

[humble root]

yes because it does :)

[spring oracle]

it won't
the original equation in the numerator is xe^x

when I put t = x+4
I get x = t-4

splitting and solving each brings us here

[humble root]

hold on hold on

[spring oracle]

while the negative remains

[humble root]

hold on a moment.

[humble root]

Can you explain to me

[spring oracle]

here

[humble root]

Why there is a plus sign in front of this integral?

wait

I'm only realising now

You have the wrong integral by parts formula

[spring oracle]

because this is actually the first term in the

[spring oracle]

what

[humble root]

[ \int_{ }^{ }f\left(x\right)g\left(x\right)dx=f\left(x\right)G\left(x\right)-\int_{ }^{ }f'\left(x\right)G\left(x\right)dx ]

[jolly parrot]

Redfern Station

[humble root]

Look at your formula

You have a plus sign

There should not be a plus sign here

[spring oracle]

huh

[humble root]

okay

Can you tell me

[spring oracle]

yep so it DOES cancel out

[humble root]

👍

[spring oracle]

was taught the wrong way the first time, now its' stuck in my head

thanks mate

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[spring oracle]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[spring oracle]

if I have more questions, do I do this or there's something else?

.close

[pearl pond]

Channel closed

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[pearl pond]

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[torpid wolf]

for argand diagrams how do i show the points of intersection do i need to show it as (x,y) or just write the value (x+yi)

[humble root]

I don't think it matters too much, but I would write it as x + iy just to he safe.

[torpid wolf]

ok

[pearl pond]

@torpid wolf Has your question been resolved?

[pearl pond]

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[pearl pond]

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[ivory basin]

. @brisk pike You can put a dot before your message and it won't claim the help channel that way

. This way you can leave a final comment to an old conversation without claiming the channel

[ruby wharf]

Can i please get help with this question? My solutions don't match the multiple choice i don't get it.

[prisma kernel]

on it

[ruby wharf]

TY

[prisma kernel]

isn't sqrt(28) = 2sqrt(7)

and not 2sqrt(3)

[ruby wharf]

Wait how did i get 28/2 = 24 🤦‍♂️

i accidentally wrote 24 instead of 14

[prisma kernel]

It's fine

[ruby wharf]

Tysm!! ❤️

[prisma kernel]

No problem mate, is your answer matching now?

[ruby wharf]

Yesss! tysm ❤️

.close

[pearl pond]

Channel closed

Closed by @ruby wharf

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[prisma kernel]

Yes?

[ruby wharf]

My answer was right now thanks! ❤️

[prisma kernel]

That's great!

[ruby wharf]

Tyvm

[ruby wharf]

what is this image XD

[prisma kernel]

Well, their way to make things fun ig😆

[bitter herald]

I dont even get it

LOL

[pearl mauve]

🙏

[ruby wharf]

replace discriminant with like something that u would find XD

[pearl pond]

Available help channel!

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bitter herald]

.eh?

[pearl mauve]

Like saying this is something ahead

[pearl mauve]

Use it

Kinda

[ruby wharf]

Yea

[bitter herald]

oh

ok

oh yeah im helpful

i can do this now

.close

[pearl mauve]

Oops

[pearl pond]

Channel closed

Closed by @bitter herald

Use .reopen if this was a mistake.

[pearl mauve]

.close

.close

[prisma kernel]

You can reopen it if you have a question @ruby wharf

[pearl mauve]

Its not closing

.close

[prisma kernel]

.It's closed already

[ruby wharf]

Ty i dont! but if i do ill ask

i really appreciate it

[prisma kernel]

.Sure mate

[nocturne grail]

you're welcome 🤩

I got credit even tho i did nothing 😭

[pearl pond]

Available help channel!

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[midnight haven]

@slow oak

[pearl pond]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[midnight haven]

@fluid oar

[slow oak]

Why are you pinging random mods

[midnight haven]

bcuz i need help with math bruh

[fair scarab]

just send your sh you ll get help

[slow oak]

Then you should wait for a helper, it’s not our job to help people with math

[midnight haven]

https://tenor.com/view/fadding-dog-traumatized-gif-10801559770686219962

[slow oak]

Also this channel will close shortly

Don't randomly ping individual users next time you need help, wait for someone to come into your help channel and help you @midnight haven

[prisma kernel]

You can reopen it if you have a question

oh god, with a mod here too

[pearl mauve]

Bro infront of the mods?

[midnight haven]

i do bro

[pearl mauve]

This guy so stupid 😭

[prisma kernel]

@slow oak

[midnight haven]

who

[prisma kernel]

The scammer

[pearl mauve]

The scammer

[slow oak]

was it the image spammer?

[midnight haven]

his brain is like a joker rn

[prisma kernel]

Yeah, the usual scams

[midnight haven]

@prisma kernel shall we continue now?

with the question

[prisma kernel]

.reopen

[midnight haven]

.reopen

[prisma kernel]

type this

oh wait

wait the channel is closed, why isn't it working

[midnight haven]

lets try different one

[slow oak]

original message was deleted

[prisma kernel]

ahhhh, so we abandon this one?

[sharp vigil]

https://cdn.discordapp.com/attachments/1064643750493691967/1212186065809440768/attachment.png?ex=691323ff&is=6911d27f&hm=7f4f7813413c63b3ec4a034089fbeccfe780356c164303379649fbcd6cf4ec99&

[midnight haven]

ig

[midnight haven]

https://tenor.com/view/test-gif-2341822946696656132

[prisma kernel]

Alright, open a new one then @midnight haven

[midnight haven]

.close

[pearl pond]

Available help channel!

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[proper nova]

if i had $(-m)^{-n} = (-n)^{-m}$ then what should i do ($m, n$ are positive integers)

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

i received some help earlier related to this, and they said that it is just a matter of doing $m^n = n^m$, i don't understand how is that related

[jolly parrot]

1 divided by 0 equals Infinity

[pearl mauve]

Are u trying to find solutions for m and n?

[prisma kernel]

log?

[snow sail]

surely you can get to $(-m)^n = (-n)^m$?

[jolly parrot]

jan Niku

[proper nova]

reciprocal right?

[pearl mauve]

Pretty sure u can just use exhaustion with pos ints

[snow sail]

yea

now look at $(-1)^n m^n = (-1)^m n^m$

[jolly parrot]

jan Niku

[proper nova]

ooh

[snow sail]

exhaustion works, yea

but think it through, what cases are allowed here

[proper nova]

what's that?

haven't heard that before

[snow sail]

just try every case

'exhaust' the possibilities

[proper nova]

$m \equiv n \mod{2}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

and $m \nequiv n \mod{2}$

[snow sail]

its more signs here

[jolly parrot]

1 divided by 0 equals Infinity
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[snow sail]

you notice the (-1)

[proper nova]

yeah

[snow sail]

so if (-1)^m = (-1)^n then

[prisma kernel]

oh yeah hell no, I'm out

[proper nova]

if m is congruent to n mod 2

[snow sail]

oh i see what you mean

[proper nova]

then (-1)^m = (-1)^n

[snow sail]

whys that haha

[proper nova]

otherwise, one of them is -1 and other is 1

[snow sail]

yea

if its true then we can just remove them

and you get n^m = m^n

[proper nova]

but if it's not true

[snow sail]

so what if its not true

[proper nova]

then -m^n = n^m or vice versa

[prisma kernel]

I'm not familiar with this depth, like I should be but I'm not, scares me a bit, prebably won't if I actually give it some time

[pearl mauve]

Why are u using congurence eq

[snow sail]

its fine

[proper nova]

i don't even know how you actually pulled out the logs

[snow sail]

i get what they mean

[snow sail]

yea, exactly

one side will have a -1 left over

okay so lets say its the case

$-m^n = n^m$

[proper nova]

this should clear your question @pearl mauve

[jolly parrot]

jan Niku

[prisma kernel]

Well it's all magic✨

[snow sail]

is this allowed

is this a real case

do we have to worry about this being true

[proper nova]

it's the matter of proving it's false

[pearl mauve]

Ohh ic now

[proper nova]

earlier i said m and n are positive integers

[snow sail]

yea

[proper nova]

so the LHS < 0 but the RHS > 0

contradiction

[snow sail]

yea\

so we conclude

what has to be true

[proper nova]

alr

thanks for all of yall's help

.close

[pearl pond]

Channel closed

Closed by @proper nova

Use .reopen if this was a mistake.

[snow sail]

[pearl mauve]

Oh lol the answer is just m is congurent to n mod 2 😂

[proper nova]

there's also m different from n

which i proved earlier it's not possible

[pearl mauve]

M cant be different from n?

Why not

[proper nova]

oh, it's uhm

original question issues

[pearl mauve]

I just checked for example n=2 and m = 4 works

[proper nova]

oh hell no

💀

how did i miss that case 💀

[pearl mauve]

😭

[proper nova]

original question was x^y = y^x (x differents from y and x, y integers)

[pearl mauve]

I thought it was (-m)^n=(-n)^n?

[pearl pond]

Available help channel!

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[pearl mauve]

Ohhh ok

[proper nova]

CLOSE IT

[prisma kernel]

.close

[pearl pond]

Channel closed

Closed by @prisma kernel

Use .reopen if this was a mistake.

[pearl mauve]

.close

[prisma kernel]

.finally got to use my acquired powers

[pearl pond]

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[midnight haven]

Let (G,*) Be a group and A a non empty finite part of G Stable by the law *

i don't understand question 2?

2. deduce that x^-1 is in A because A is a subgroup of G *

it sounds contradictory

how did we know it's a subgroup

to begin with

[cursive wraith]

It's written "deduce that x^(-1) is in A, and THEN A is a subgroup of G"

it's not "puisque"

it's "puis que"

[midnight haven]

i read puis que puisque

thank!

but still

how do i get x^-1 in A

[cursive wraith]

use question 1

[midnight haven]

i tried

[midnight haven]

How??

[cursive wraith]

what does non-injectivity mean

[midnight haven]

mean there are images with multiples preimages

[cursive wraith]

can you write it more formally

in the case where phi is our non-injective function

[midnight haven]

$\exists (x_1,x_2) \in \mathbb{N}, f(x_1)=f(x_2) And x_1 \neq x_2$

[jolly parrot]

Drk

[cursive wraith]

ok, few things

first, if you want to write there exists a couple of natural integers

the couple itself belongs to N^2, not N

otherwise, get rid of the parentheses

[midnight haven]

oh yeah

[cursive wraith]

second thing

when we'll apply it to phi(n) = x^n, probably not the best to call them x1,x2 when the base is called x

[warm current]

I always just write $\exists(n_1,n_2\in\bN)$

[jolly parrot]

SWR

[midnight haven]

okay i'll try to use it and what i gett thanks

[cursive wraith]

This triggers some death impulses in me

(/j)

[midnight haven]

i can't seem to find the way to use that info

hello

may i get some guidance

[cursive wraith]

ok

so, there is $n < m$ such that $x^n = x^m$

[jolly parrot]

Raphaelisius Maximus MMIII

[midnight haven]

yeah

[cursive wraith]

recall that we're looking for x^(-1)

which means we're looking for some element 'y' such that $xy = e$

[jolly parrot]

Raphaelisius Maximus MMIII

[cursive wraith]

so might as well transform this equality into '.... = e'

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

so

[cursive wraith]

yes, unfortunately you did it the wrong way around xddd

[midnight haven]

i think we should switch the diffination

[midnight haven]

yeah

lol

[cursive wraith]

but it's alright

[midnight haven]

this is for n>m

[cursive wraith]

do (LHS)^-1 = (RHS)^-1

[cursive wraith]

ok we can do that too

let's say we had n > m from the beginning

then x^(n-m) = e

[midnight haven]

what does LHS and RHS mean btw? i'm not familiar with the notation

[cursive wraith]

left hand side

right hand side

[midnight haven]

oh

[cursive wraith]

coming back to this

either we write x * ... = e

or we can immediately multiply by x^(-1)

x^(-1) = x^(n-m-1)

[midnight haven]

why.? are they equal ughhhh

OH

OHHH

I GET EVERYTHING NOW

tTHANK YOU

[cursive wraith]

de rien 👍

[midnight haven]

.close

[pearl pond]

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[wary vault]

Hello. Suppose we have an open set A (or open from above), and its supremum (let it be s and a real number). Is it possible to prove completely formally that no matter how many elements of A we take away, the supremum shall stay the same? I hope the question makes sense lol

[tropic saddle]

well thats obviously false

A=(0,1) has sup 1 but if you take away [1/2,1) then the new sup 1/2

[wary vault]

okay i see my mistake

let me rephrase

[warm current]

Countably many?

[wary vault]

if we take away up to n elements. (meaning the set of members we take away can have a cardinality up to aleph null)

[tropic saddle]

this sounds like another xy question

[wary vault]

yeah

[tropic saddle]

!original

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wary vault]

okay sorry

.close

[pearl pond]

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[rough forge]

[cursive wraith]

"what's the original?"
closes

[wary vault]

im not taking chances

ill repost my question in the form i meant it

[pearl pond]

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[torn willow]

can someone help me with this q pls

[torn willow]

i have no idea how to approach this

[somber adder]

try sketching it out to help you visualise it

[pearl pond]

@torn willow Has your question been resolved?

[torn willow]

ik what the graph looks like...but the thing is im meant to do it algebraically

[somber adder]

okay

do you know what's integration?

[torn willow]

well integration is antiderivative, its used to find the area under a graph

[somber adder]

right.

can you see how that might help us here?

[torn willow]

i was thinking of setting the two eq together and finding the points of intersection...

but idk how to solve the eq

[torn willow]

also idek if that'll work

[somber adder]

hmm

[steep saddle]

you're finding the area between curves, which is what integrals do

[torn willow]

ye i get that, i was just talking abt finding the points of intersection

[steep saddle]

why do you need to do that?

curves dont have to intersect

find the area between the curves y = 10 and y = 5

they dont intersect

[torn willow]

hmm

[somber adder]

this is our graph rn

[torn willow]

is it not x=1 and x=10?

[torn willow]

yes

[steep saddle]

its an example of why you dont need to find the intersection

[somber adder]

finding the intersection points is a little tricker than i thought...

[torn willow]

but how will i approach this without the knowledge of the graph i.e. algebraically

[steep saddle]

damn they do intersect

[torn willow]

ye

[steep saddle]

ok my bad i didnt think they would from estimation lmao

[torn willow]

np

[torn willow]

^

[steep saddle]

good point its transcendental which is why i didnt think theyd make it intersect for you as a problem

[torn willow]

i see

[somber adder]

have you worked with these sorts of equations before?

[steep saddle]

must be a typo man can you ask your teach or something

[torn willow]

uhmm i dont think so,,,i've worked with similar ones tho

[steep saddle]

even wolfram alpha cant work it out algebraically

[torn willow]

lol i dont think so, its from a textbook

[torn willow]

💀

[steep saddle]

is calculator allowed for this section?

[torn willow]

no wonder i was stuck

[steep saddle]

i rly dont see how someones supposed to do this

[somber adder]

I think it could be solvable with lambert-W

[torn willow]

ye obv

[somber adder]

dont quote me on that though

[torn willow]

huh 😭

[steep saddle]

yes i tried changing it to e^(-x^2 + 100) = 6x

dont think it can tho

and no way they ask you to use lambert w

[torn willow]

idek what it means, im in S5 (11th grade lol)

this is the textbooks answer

[steep saddle]

yeah ur definitely suppoesd to use a calculator

[torn willow]

ye

[torn willow]

idk that

[somber adder]

it's like

😭 there's a lot to explain

[torn willow]

so im not meant to use it

[somber adder]

probably

[torn willow]

the textbook doesn't cover it

[somber adder]

well you can always approximate the intersection point

to like 6 or 7 s.f. would be fine

[torn willow]

i dont think im meant to appromate like that asw 😭

well u can app ur final anwer obv but i dont think u can use app as a method

there has to be a way

hmm

[somber adder]

does your calculator have lambert-W on it?

like

a W button

or smth

😭

[torn willow]

no 😭

[pearl pond]

@torn willow Has your question been resolved?

[torn willow]

@somber adder @steep saddle

[pearl pond]

@torn willow Has your question been resolved?

[pearl pond]

@torn willow Has your question been resolved?

[tulip ore]

@torn willow you here?

there is a major problem in the textbook answer

you only get 536.8 by doing this

this is not how youd correctly calculate the area

if you graph the functions, this is what it looks like from x=1 to x=10

536.8 ends up finding the upper-left area then subtracting the small lower-right area

the area also never needs to consider the x-axis, because, at the lower-right, the red curve 100 - x^2 intersects with x=10 at y=0

[pearl pond]

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[woven river]

Hi, I have a question. This is a multivariable calculus course.

For number 8) on both sides, I do not get why the constant C needs to be a function of that specific variable? Why not just write C?

[sharp vigil]

it's not a constant, it's a function of that variable

maybe a better notation would be f(x) or something

[woven river]

but its gonna be a constant in the end

what...

[sharp vigil]

C(x) could be like 3x^2 + cos(x) or something

similarly C(y) could be 6y^3 - tan y

[woven river]

OHHH

OH

understood

thanks

.solved

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[pearl pond]

@void grail Has your question been resolved?

[pearl pond]

@void grail Has your question been resolved?

[void grail]

I'm trying to figure out what to prove and what the components are

[void grail]

Here is a sketch:

1️⃣ Axioms: f: I→I, μ>2+√5, Itinerary(x,w) defined.
2️⃣ Lemma1: Itinerary injective → each word → unique point.
3️⃣ Lemma2: Cylinders = intervals → points with same prefix are contiguous.
4️⃣ Lemma3: Lex order preserved → w₁<w₂ ⇒ x<y.
✅ Conclusion: Order of Γₙ points in I is μ-independent.

1️⃣ Assume order depends on μ: ∃ μ₁, μ₂>2+√5,
words w₁<w₂, but points x₁>y₁ in I for μ₁, or x₂<y₂ for μ₂.
2️⃣ By injectivity, each word → unique point.
3️⃣ By cylinders, points with same prefix are intervals → order cannot flip inside interval.
4️⃣ By lex order, w₁<w₂ → LeftOf(x,y).

⚡ Contradiction: assumption μ changes order violates lemmas → order must be μ-independent.

If μ changed the order, two points from nested cylinders would swap left↔right inside an interval.

But Lemma2 says cylinders are connected intervals, and Lemma3 says lex order = left→right.

So a “flip” is impossible → contradiction → order independent of μ.

[pearl pond]

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[autumn meadow]

can someone help me on this problem. i stuck on this. am i missing there any logarithm rules ?

[blissful cloak]

(\log_c(a)+\log_c(b)=\log_c(ab))

[jolly parrot]

ΠαϳαμαΜαμαΛλαμα

[blissful cloak]

note that all the logs are of base 2

[autumn meadow]

im pretty sure doing it wrong

[blissful cloak]

oh wait the logs are squared

[autumn meadow]

wdym

[blissful cloak]

log^2_2(10) I read as just log_2(10)

[autumn meadow]

oh

[blissful cloak]

well log_2(10)=1+log_2(5)

see if you can rewrite the expression in terms of log_2(5)

[autumn meadow]

huh

how

[blissful cloak]

(\frac{\log^2_2(10)+\log_2(10)\log_2(5)-2\log_2^2(5)}{\log_2(10)+2\log_2(5)}=\frac{(1+\log_2(5))^2+(1+\log_2(5))\log_2(5)-2\log_2^2(5)}{1+\log_2(5)+2\log_2(5)})

[jolly parrot]

ΠαϳαμαΜαμαΛλαμα

[autumn meadow]

oh okay

is it correct?

[pearl pond]

@autumn meadow Has your question been resolved?

[autumn meadow]

no

[lean oak]

,rccw

[pearl pond]

@autumn meadow Has your question been resolved?

[autumn meadow]

No

[bright imp]

Is the last term in the numerator (log2(5 squared)) all squared?

[pearl pond]

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[proud epoch]

Im doing task b
I did x1=4 into y=0.25x^2-2x+3 and got the coordinates of A1
Now how do i get the coordinates of the rest? (B1C1D1)

[pearl pond]

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[lapis wagon]

How is this not -51???

[warm current]

,rccw

[pearl pond]

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[shut flicker]

what / where is the triangle bounded by the lines x = a/(a+1), y = 1/(x/a) and y = 0?

[pearl pond]

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[unborn beacon]

Hey guys, can someone help me understand how the solution to this exercise using the well-ordering principle for integers can be considered a valid proof?
It seems to me that the set S not being constructible using a even value for m shouldn't be enough to imply that it must exist when m is odd. Why would that be the case?

[pulsar flax]

do you understand why n = 2^k * m for some k, m where m is minimal to this property?

[pearl pond]

@unborn beacon Has your question been resolved?

[unborn beacon]

Wait, I see now. The proof abuses the fact that 2^0 = 1 to transitively show that r = m, but because it could be the case that there's only a finite amount of integers that can be constructed through an odd value for n, and every other integer must use an even value, it uses the well ordering principle to show that this edge case is not possible, or something like that

I think I had in my mind that the first line was "Let S be the set of all integers n", so I thought that the **second and third line was just a formality to show that S is not empty so that the WOP could work. Because I didn't pay attention to it I couldn't follow the logic from then on

[pulsar flax]

you are over complicating it

the set of r's as described is a set of natural numbers so it has a minimal element m

and this m must be odd unless it wouldn't be minimal

[pearl pond]

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[eager jewel]

idk how to start

i tried assuming some parameters

but NOTHING was simplifying

[modern rivet]

rearrange for x=y^2/6

4a=6

So a=6/4=3/2

Also to simplify the english, L is parallel to the x axis and passes through C

Try to work through

[eager jewel]

yes that was obvious

[eager jewel]

work through what

i got through the diagram

what do i do after that

looks like u have been typing for a long time..

[modern rivet]

Let A be (x1, y1) where y1^2 =6x subscript 1
B (x2,y2) where y subscript2^2=6x subscript 2
C same as above but 3

Since L is parallel to the x axis and passes through c, L is y=ysubscript 3

To find D where AB meets L:
D is where AB intersexts y=y subscript 3
AM = y subcript 1 minus y subscript 3 which is a vertical distance from A to L
And BB is ysub2 - ysub3 which is a vertical distance from B to L
AM times BN is just those two multiplied cus theyre perpendiculars on a horizontal line

[eager jewel]

oh ok i tried this but didnt assign paramater for C

let me try that

[modern rivet]

From D on line L where AB meets it, CD is the horizontal distance from C to D

Incase im not around when ur done, solution is 1 if im not wrong

[eager jewel]

uh no its 36

[modern rivet]

Cus AM BN should equal CD

[eager jewel]

why

[warped violet]

I am sorry

I have to ask

[eager jewel]

yea?

[warped violet]

What is with these arbitrarily sized letters 😂

They are massive

[eager jewel]

oh its from a website lol instead of taking a picture from my book i just search it up and take a screenshot of the question text

better?

[warped violet]

😂 no worries. Sorry for distracting you. I just found it funny

[eager jewel]

@modern rivet im not able to find CD length in terms of parameters

<@&286206848099549185>

[eager jewel]

also if someone could pin this

[modern rivet]

Nvm im wrong yeah its 36

[eager jewel]

can u tell how

what will i do with the answer 💀

[pearl pond]

@eager jewel Has your question been resolved?

[modern rivet]

This guy has a video, sorry for getting back late to you
Im unsure if im allowed to send links, so ill send it as a pic, just google it

[pearl pond]

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[inland laurel]

[
\text{The value of } x \text{ satisfying the equation }
\sqrt{2}^{\log_2 (2)^{\log_2 (2)^{\log_2 (2)^{\log_2 (2)^{\log_2 (x)}}}}} = 5
]

[jolly parrot]

BlackidoZΣ

[inland laurel]

A) 5 B) 16 C) 25 D) 32

i tried letting log_2 (2) = t as it is repeating

then it became a power tower

[worthy lance]

That is false

log_2(2) is 1

1^whatever = 1

sqrt(2)^1 = sqrt(2) not 5

you might have written the question incorrectly

[inland laurel]

i wrote it correctly

there are 4 times log_2 (2) repeating

[worthy lance]

That doesn't matter

log_2(2) is 1

1 raised to any number is 1

[ashen ivy]

even though you're evaluating top down, at some point you get 1^(log x) as the top is evaluated, which is 1 for any real input

(and complex, but ehatever)

[pearl pond]

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[pearl pond]

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[dreamy wind]

How do i do this>

[worthy lance]

move aside a countable subset of (0,∞)

[dreamy wind]

Wdym?

[worthy lance]

1, x = -1; 2, x=-2;

and the rest of the piecewise function is similar to the previous exercise

[dreamy wind]

So something liket his?

[worthy lance]

yes, and non-natural same position

[dreamy wind]

Wait did u reverse this? or

Oh wait nvm mb

[worthy lance]

You have to add the interval in the last line

[dreamy wind]

discrete

[worthy lance]

because if you have non-natural negative you are adding them also

so you have to add (0, inf) in the last one

hope you understand what I mean

[dreamy wind]

Is that the same as N+?

[worthy lance]

No

real numbers can be also negative

so you have to put the interval

also, N is already positive, no need the +

you can use instead {1, 2, 3...}

[dreamy wind]

instead of just N ?

[worthy lance]

Like this. x+2, x ∈ N = {1, 2, 3...}
x, x ∈ (0, inf) \ N

That's good, but too long

x, x ∈ (0, inf) \ N this is so much shorter

yes

A ∖ B= A−B ={x : x ∈ A and x ∉ B}.

[pearl pond]

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[pearl pond]

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[wary vault]

,tex
Hello, i wanna prove that $$\forall n\geq 2, n\in\mathbb{N}: 10| ( 3^n + 3^{n-2} ) $$ via induction. For the n+1 part, we can let $3^n +3^{n-2} = 10k , k\in\mathbb{N}$ and then: $$3^{n+1} +3^{n-1} = \
3\cdot 3^{n} +3\cdot 3^{n-2} = 3\cdot 10k $$ and boom proof done. Is this correct?

[jolly parrot]

fijokazż

[wary vault]

the 10|| part is "10 divides"

for sum reason it printed two thingies

[grim fractal]

yes

[wary vault]

yay thanks

.solved

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[shadow reef]

I know I did something wrong but idk what

[tulip ore]

try not to write uppercase V and lowercase r so similarly

that way its easy to see that r' = 2/4 = 1/2

[cyan sail]

r' is 2/r?

[tulip ore]

its right there in the question, yes r' is 2/r

the radius (r) increases at a rate of 2/r cm/s

[cyan sail]

oh yes

my fault

i think you just forgot to replace r for 4 then right?

[pearl pond]

@shadow reef Has your question been resolved?

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[tired forum]

Can anyone explain to me how this is partial credit?

[tired forum]

I recently took a 6th-grade test, and I got partial credit on this question. However, I'm not able to understand why.

[lofty fossil]

6th grade?

[tired forum]

Yea

[lofty fossil]

wtf

who learns this in grade 6

[tired forum]

I'm in a privite school so like,

ummm

yea

[lofty fossil]

I believe that partial credit is given because you selected the three points that dont satisfy the equation?

who tf learns this in 6th grade even in a private school is this bait 😭

[tired forum]

Well I know that equation simplifies to y is smaller or equal, to 2x+2

[lofty fossil]

yep

[tired forum]

Which means the only points in the solution that statifys this is the points below that equation.

So like I don't know how this is wrong.

[ripe lantern]

I think there may be one more point you omitted

[lofty fossil]

wait I think you're right

sorry

[tired forum]

The only points are the ones in blue.

[lofty fossil]

I'm not sure then

[tired forum]

So it is either me or there is something wrong with the question.

[ripe lantern]

The correct answer is the points in blue?

[tired forum]

No I picked the blue ones

And got partial credit.

[ripe lantern]

Yeah

So, I think one point you forgot

[tired forum]

?

Which one tough

There is onyl 6 points

[lofty fossil]

isn't it correct

[tired forum]

*only

[lofty fossil]

ah

[tired forum]

I got partial credit,

[lofty fossil]

the top one

the one on the top right

is part of the solution set

[tired forum]

(5,12) ?

[lofty fossil]

doesn't look like it but it is

yeh

[ripe lantern]

Well OP could have figured that

Yep

[lofty fossil]

we got y <= 2x + 2

if we plug in 5 for x and 12 for y

we get 12 <= 2(5) + 2

so 12 <= 12

which holds true

[tired forum]

Bruhhhhhhhhhhh I'm dumb. Yea ok that makes sense

[ripe lantern]

Also, 6th grade and learning this is impressive

[tired forum]

At least I'm in honors

[lofty fossil]

are you in 8th grade

[tired forum]

This is what honors learn.

[tired forum]

No

[lofty fossil]

this guy is a ragebaiter bro

😭

[ripe lantern]

Lmao

Hard to believe 6th graders do this

[tired forum]

(typo)

[lofty fossil]

doing this in 6th grade is wild work

[ripe lantern]

Welp, good for him if he’s being honest

[tired forum]

I want to go to public shcool,

[lofty fossil]

I woulda got that shit wrong as well 😭

[tired forum]

they have like easier stuff

and not a bunch of rules

plus they are less strict- we have homework everyday and even during summer.

.sovled

.solved

[pearl pond]

Channel closed

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[lofty fossil]

dang bro do I feel bad for this kid

[magic iron]

hey wassup can one of you guys help me on a pre-cal problem I am having trouble on?

this is the problem

[pearl pond]

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[sharp smelt]

I know this is quite simple , but I'm learning a new mrthod , so just wanna be sure

Firstly I believe I use this method

the joint pdf would be 1/16 I believe

with U=X; V=Y here

the jacobian is 1

and f( R,S)= 1/16

so the density functions is given by 1/16 over [-4,4]

I see where I messed up

I''l do this after lunch

.close

[pearl pond]

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[pearl pond]

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[torn willow]

Sry I was a bit busy

[torn willow]

@tulip ore

[torn willow]

How would you do this question then

[pearl pond]

@torn willow Has your question been resolved?

[pearl pond]

@torn willow Has your question been resolved?

[pearl pond]

@torn willow Has your question been resolved?

[pearl pond]

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[tulip ore]

.reopen

[pearl pond]

✅ Original question: #help-39 message

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[tulip ore]

.@torn willow
theres a few possible areas that they could be talking about, but they gave too many lines

you saw earlier that there are three areas, but each area doesnt use every line

• B: the bottom area between the curves y = 100 - x^2, y = ln 6x, and the x-axis
• L: the large left area between the curves y = 100 - x^2, y = ln 6x, the x-axis, and x = 0
• R: the small right area between the curves y = 100 - x^2, y = ln 6x, and x = 10
  we'll calculate B, L, and R

first, you use a graphing calculator to find when y = 100 - x^2 and y = ln 6x intersect
this happens around 9.7942, we can call the exact value w = 9.7942049...
it's not possible to solve for the value exactly, you havent learned the lambert W function yet
you can just use 9.7942 for w

area of R is ∫ w to 10 (100 - ln 6 - x^2 - ln x) dx
you end up getting 0.4199

to find the area of B, keep in mind ln 6x crosses the x-axis at (1/6, 0)
so area of B is ∫ 1/6 to w (ln 6x) dx + ∫ w to 10 (100 - x^2) dx
you end up getting 30.6903

to find the area of L, you can use that area L + area B would just be the area under 100 - x^2
area L + area B = ∫ 0 to 10 (100 - x^2) dx = 2000/3
so area L = 2000/3 - 30.6903 = 635.9764
(its possible to figure out that area L is exactly 2/3 w^3 + w - 1/6, you arent expected to know this)

[tepid root]

hiw do i calculate the area

i was trying to construct parameters for a spheroid but i didnt get anywhere

[unkempt yacht]

[tepid root]

Soo

[warped violet]

What is happening here

[eager jewel]

wtf

i feel like i should ping mods

but i dont want to

[tepid root]

Id give my Reasoning but it isnt relevant * and im stuck so Please treat it like any other question and thank you

[eager jewel]

do u have any info to work with

[tepid root]

yeah

312px vertical profile, ~200px half width

[warped violet]

[eager jewel]

try approximating it as a circle

and calculate radius using pythagorus theorem

[tepid root]

there's a nonneglibile error in curvature and area if i do that

im working in 3d

[eager jewel]

how accurate do u need ur result

[tepid root]

i don't need pinpoint accuracy but a circle wont do

[eager jewel]

ask an ai tbh

[warped violet]

!nogpt

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[warped violet]

What is even the question

[tepid root]

I wanna try solving it on my own, and i don't think an ai would get it right

[warped violet]

Or am I just stupid

[tepid root]

the area of the

circled

[warped violet]

What circle

[tepid root]

section

her chest

[eager jewel]

area of the red part circled bro

come on

[warped violet]

Oh!

I swear I didn't see that

[eager jewel]

that was the first thing i saw

[warped violet]

I was confused as to why that image is there

[eager jewel]

ohh

[eager jewel]

ok u help then

[warped violet]

Okay then another question, what is this

[tepid root]

that's a previous question

unrelated

[warped violet]

And how did you get the ownership of the channel

[eager jewel]

he paid the mods

[tepid root]

it was open

[eager jewel]

500$

[tropic saddle]

the other message started with a .

[tepid root]

there's a 1 hour 30 minute gap between my message and his

[cobalt oasis]

starting a message with a . suppresses opening the channel.

[tropic saddle]

that way you dont claim ownership

[warped violet]

Oh I didn't know that

Interesting

Thanks

[eager jewel]

can we all pls try to solve the question

[tropic saddle]

what even is the actual question. you can just measure the area with an image editing program (I imagine) so thats not it

is it the best way to 3d model boobs?

or what is going on

[eager jewel]

tbh i just thought he was trolling and went along with it

[craggy pecan]

I can feel math pros here

[tepid root]

that gives the 2d area

[tropic saddle]

area is 2d

[eager jewel]

lmao

[tepid root]

as in it

only gives the area of one face

[craggy pecan]

I m a 10th grader lol

[tropic saddle]

ok so you want to have a 3d model of a boob and then compute the surface area?

[tepid root]

yes the surface area sorry

[tropic saddle]

but why

[tepid root]

i need it

[tropic saddle]

for what

[hollow sonnet]

what questionable motives

[tepid root]

no questionable motives

it's just artistic curiosity

[tropic saddle]

sure

[tepid root]

there is a demarcation in size

a really obvious one

so i'm trying to find how big it really is

[heavy onyx]

a front view may help

[tepid root]

yes, hold on

[eager jewel]

..

[runic shoal]

I'm defeated

[tepid root]

i don't know i don't even know how to approach it mathematically

[tropic saddle]

buy a bra in RL and measure it

[atomic oracle]

If you have to do it this way you have to count pixels of something you know the size of in the image and compare it

[tepid root]

I used 15px/cm as a guesstimate

[atomic oracle]

Or just do a little research on the internet and get the size or something

[tepid root]

I was planning to get the pixel result and then figure out pixel conversions

[tropic saddle]

its not like two images from different perspectives will actually be to scale

[warped violet]

May I ask something

What will you do with the value, if you get it

[cedar sentinel]

you know, OP. for all of your visual novel skills, you sure didn't think to look up the concept art of this particular character, if it exists?

[tepid root]

wasnt listed

[radiant terrace]

This is not an appropriate use of the help channels.

[tepid root]

nothing

[radiant terrace]

You should instead find a way to rephrase your problem without it being inappropriate.

[tepid root]

okay

[heavy onyx]

maybe like surface area of revolution

[eager jewel]

is it not

[radiant terrace]

I am telling you it is.

Lemme see if I even remember the purge cmd.

[pearl pond]

Channel closed

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[tepid root]

the images r gone

do i start a new help channe?

[heavy onyx]

yes

[radiant terrace]

Yep, just rephrase the q in a way that's not about anime tits and open a new channel.

[tepid root]

lol Sorry

[pearl pond]

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[fading ledge]

I am curious to understand what (x+y+z)^n total terms formula mean
i want to understand it with stars and bars

[autumn fossil]

u wanna understand how to get the formula for expansion of (x+y+z) and u wanna use stars and bars?

[fading ledge]

i just want to understand the real life meaning

[autumn fossil]

oh how many terms are there?