help-4

[lost oracle]

well i did try

to use that

but i don’t know how to accomplish it

[pure pulsar]

Okay let me see

[lost oracle]

thanks

[pure pulsar]

I think L = lim inf sn and you gotta prove that L is and upper bound for A

[lost oracle]

right that is one of the things i suggested

but i’m not sure how

[pure pulsar]

I gotta solve it then ig

[somber ore]

fuggy fuggy

[red mulch]

well once u have both, u can conclude that sup A = lim inf s_n. this is the end of the proof for the first equality

do the samething for b

[vale dock]

@lost oracle Has your question been resolved?

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[river shale]

Let the domain of
discourse be the positive integers. The second proposition is
true but the first proposition is false.
I don't quite understand about this part

[steady charm]

the domain of discourse meaning the "for all" statements are meant to be parsed as "for all x in the set of positive integers"

[shut mirage]

continuing off what cloud said, so if we let the two statements be what is said in the answer, the two propositions boil down to this:
1: all positive integers are even, or all positive integers are odd
2: all positive integers are (either even or odd)

[river shale]

Yeah, but aren't they referring to the same stuff?

[shut mirage]

not exactly

the first proposition asserts that either all positive integers are even (as in all of them are even), or all positive integers are odd (as in, all of them are odd)

but we know those premises are not true - some integers are even and some are odd

[river shale]

OH, I get what you mean

So the first statement will be true if all the positive numbers are either all odd or even

[shut mirage]

yup

[river shale]

I see

ty

.close

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[modest hemlock]

yo where would the tension be? the - - - line is kinda confusing me

[light saddle]

It's always along the string, in this case, during the second phase, it goes from the particle to the peg

[vale dock]

@modest hemlock Has your question been resolved?

[modest hemlock]

So where i have ‘a’ theres tension right

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[rose lance]

guys need help
Q4) A 2 kg object is attached to a half meter long string forming a pendulum. At the lowest point of the trajectory, the object has a speed of 5 m/s. What is the tension T in the string when the object makes a 1/4 turn (π/2 with respect to the vertical)? Does the mass make a full turn? If so, what is the tension in the string at the highest point of the trajectory?

[gilded relic]

probably ask in physics discord server?

[pure pulsar]

!status

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pure pulsar]

Dw we can help that person here too

Ofc if they want

Or mods allow us

[gilded relic]

just recommendation

[pure pulsar]

Yeah

[gilded relic]

the issue is also, what do you know / in what context is this question posed?

[pure pulsar]

Tension

[pure pulsar]

@rose lance

[rose lance]

2

[rose lance]

yes but I'm not in the server and I didn't find it

[gilded relic]

#old-network

[pure pulsar]

Can you share your work?

[rose lance]

fa tota la volta means (does a full turn)

[gilded relic]

,rotate

[rose lance]

I think I had to put 3 and 4

[pure pulsar]

I think im gonna learn a couple more languages here

I think you gotta work with energy @rose lance

What do you think the initial energy at the lowest point would be?

If you ask me how to get an idea of how energy equations are required

Then we have an hint about speed at the lowest point which is the initial one

So it energy and speed are related and at the lowest point, only kinetic energy is present

So you can get 2 energy equations and try to equate initial and final energy

You get speed at 1/4 turn

@rose lance If you think im going fast or you dont get any point, feel free to stop me and ask

[pure pulsar]

Law of conservation of energy is what we call it

[rose lance]

so we don't have to think about potential?

[pure pulsar]

At the lowest point, no we dont need to think about it

But at the final point, yes

We need to

For example 2 people playing football

Using law of conservation of energy, we can find velocity, pls find and lmk when you’re done

@rose lance Lmk if you’re not getting it

[rose lance]

OK thanks a lot

I get it now :))

[pure pulsar]

Im sorry but i think its still wrong

v^2 = 15.2

Yeah v is 3.9

Correct mb

Now find tension using formula of force

[vale dock]

@rose lance Has your question been resolved?

[vale dock]

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[finite grotto]

anyone know how to solve this?

(sorry for the bottom image)

[pine prairie]

What do you mean by solve

[finite grotto]

my friend said to solve it

i dont understand

sory

[coral ocean]

https://youtu.be/7Q4h1YTYgnE?si=nP3oJbyiNBQrficP

[pine prairie]

You can’t solve it

[north olive]

It's a formula for solving quadratic equations

[pine prairie]

It’s like say x = 4

What do you mean solve x = 4

[finite grotto]

ohh

formula

tysm!!

.close

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[shell raptor]

help me please 🙏

[shell raptor]

i’m not sure where to go

[vocal tusk]

!status

[vale dock]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[shell raptor]

i tried substituting the x and y

it didn’t work or i did it wrong

[green trail]

!show

[vale dock]

Show your work, and if possible, explain where you are stuck.

[shell raptor]

please bro 😭 i’ll embarrass myself more if i keep sending

my working out is so wrong

[green trail]

That is why you have to show.

[vocal tusk]

we don't judge.

[green trail]

How are you gonna learn eitherway?

[shell raptor]

don’t laugh

i know it’s bad

[green trail]

Do not subtitute both

only 1

for example, use y = x + k

[shell raptor]

ok will i be stuck with a random k tho in my equation

[green trail]

and subtitute

you will handle that later

for now, use only one subtitution not both

[shell raptor]

okay i’ll try

don’t leave me i might mess up

equal 4

[green trail]

instead of equal 4

u can make it equal 0

[shell raptor]

yes i did that

[green trail]

so u have a quadratic equation

and now

[shell raptor]

uhhhhm

[green trail]

for you to have one pair of solutions

what condition do u need?

[shell raptor]

linear

idk i’m 16

[green trail]

do you know the discriminant?

[shell raptor]

oh i learnt that yesterday

so it has to be

equal to zero

[green trail]

right

now find the discriminant

and equal it to 0

[shell raptor]

is my c

k squared - 4

[green trail]

yes

[shell raptor]

i got

[green trail]

good, but u can simplify that

[shell raptor]

okay i did

ohhh

thank you

[green trail]

you are welcome

[shell raptor]

.close

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[rose lance]

thanks :))))

[pure pulsar]

Whats the answer you got

?

For tension

[rose lance]

uhm I think 60.76 or something like that

N

[pure pulsar]

And tension at highest point?

[river shale]

I assume that’s the original question, ping me if I made a mistake

[golden jewel]

please mujhe koi comment me kuch batayega ki ise kese use karte hen

[pure pulsar]

Sorry?

[ashen prawn]

uhh sir write itin english

[ashen prawn]

well @golden jewel :

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@rose lance Has your question been resolved?

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[hazy pivot]

Ah cute

[odd jackal]

$e \in H_i( \forall i \in I) \implies e \in \bigcup_{i \in I) H_i$
\
For some $k \in I$, $x \in H_k \implies x^{-1} \in H_i$ ( As $H_k$ is a subgroup)

[rocky lotus]

wai

$e \in H_i( \forall i \in I) \implies e \in \bigcup_{i \in I) H_i$
\\
For some $k \in I$, $x \in H_k \implies x^{-1} \in H_i$ ( As $H_k$ is a subgroup)
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.1422 ...I) \implies e \in \bigcup_{i \in I) H_i$
                                                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[odd jackal]

I've the proof, just need it checked

give me ~2 minutes to type it

[hazy pivot]

Ah cool cool

Although smth that might make typing easier

[ebon glade]

x^-1 in H_k

[hazy pivot]

Wlog you can rearrange I so that whenever j < k, H_j is contained in H_k

[ebon glade]

no dont do that

[hazy pivot]

For some total order

[shut lion]

Let them type their work lmao

[hazy pivot]

It makes notation easier no?

[ebon glade]

first I am not convinced you can do it. second, you really dont need it

[hazy pivot]

Cool cool

[odd jackal]

$e \in H_i ( \forall i \in I) \implies e \in \bigcup_{i \in I} H_i$
\
For some $k \in I$, $x \in H_k \implies x^{-1} \in H_k$ ( As $H_k$ is a subgroup)
\
Let $x ,y \in \bigcup_{i \in I} H_i$. Then $\exists \alpha \in I : x \in H_{\alpha}$ and $\exists \beta \in I : y \in H_{\beta}$. Then $H_{\alpha}≤ H_{\beta}$ or $H_{\beta} ≤ H_{\alpha}$. Thus, $x,y$ belong to the same subgroup. Thus $xy \in H_{\alpha}$ or $xy \in H_{\beta}$ and thus $xy \in \bigcup_{i \in I} H_i$

[shut lion]

For this implication here, it doesn't really give you what you want.

[rocky lotus]

wai

[odd jackal]

well, yeah

just realised I need to add it;s thus in teh union

[shut lion]

Even the new version doesn't really do it.

[odd jackal]

I'm more concerned about the last part

[shut lion]

You need to prove that e is in the union. Not an implication.

For the last part u pick two elts they lie in two possibly different subgroups

[odd jackal]

as e is atleast in some group, it lies in the union

that's it

[shut lion]

The ordering tells u they lie in the bigger of the two

[odd jackal]

I don't think I need ordering here, strictly speaking?

[shut lion]

I think you do

[odd jackal]

It lies in one of them, which is all I need

[shut lion]

What if they lie in different ones?

You have no closure arg then

[odd jackal]

I don't get how

[shut lion]

The ordering forces them BOTH to lie in a subgroup

In one of the H's

If not u could have x in one H an y in a different H' and then you can't say anything about xy

[odd jackal]

if x in H_a , y \in H_b. Then H_a≤H_b or vice versa

[shut lion]

Yes, that's the ordering.

If you don't know that H_a leq H_b (or vice versa) you're stuck is my point.

You did this right either way.

[odd jackal]

yea, but I've written that, right

[shut lion]

Yah ofc. I'm just pointing out that the ordering is needed.

[odd jackal]

Oh

okay, yea

[shut lion]

For e I don't think your reasoning is quite right.

[odd jackal]

sorry for getting so defensive

[shut lion]

e is in the set but for a different reason than what you wrote.

[odd jackal]

It lies in all of the sets, and thus lies in the union is perfectly valid IMO

I could just say it lies in the trivial subgroup and be done too I guess?

[shut lion]

What if the family is empty? Will e be in the union then?

[odd jackal]

The family is non-empty as {e} is a subgroup of every group

[hazy pivot]

The family is given to be non-empty

[shut lion]

The nonemptyness of the family of subgroups guarantees e lies in the union is my pt.

The implication you wrote is vacuously true in the empty family case. So it's a true implication either way but it doesn't actually tell you e lies in the union.

[ebon glade]

(its not said that the trivial subgroup is contained in the family)

[hazy pivot]

Okay I gotta ask: why can't we do this?

[ebon glade]

I thought about it more. you can but I really see absolutely no reason to do it

I dont see how it simplifies notation in any way

[ruby sleet]

also asks you to use axiom of choice, and the result is true even without axiom of choice

[hazy pivot]

It saves you having to check which subgroup is bigger

[ruby sleet]

better to prove something using the smallest number of axioms at all

[hazy pivot]

What no

Why would this use AoC

I is in bijection with a family of subgroups. That family is totally ordered

The total order on the family will induce a total order on I

Then it's just a matter of rearranging

[ebon glade]

yes AC was also my first fear

[shut lion]

Tbh the way I'd do it is just wlog

[hazy pivot]

That was what I initially said

[ebon glade]

but it offers practically no value

[shut lion]

"Say x in H, y in H' by our ordering wlog let x, y in H"

[proud sequoia]

Um sorry to interrupt but what topic is this/ subtopic

Like the question

[shut lion]

Abstract algebra

[hazy pivot]

Group theory

[proud sequoia]

Ty

[hazy pivot]

Fair enough, it reduces the amount of writing you have to do by one line lol

I thought it would do more

[hazy pivot]

And it adds one line on its own so

[shut lion]

If you do the wlog later you get the same result lol

[hazy pivot]

Yeah I overestimated how much it would help

[ruby sleet]

Ok so there's indeed no need for AoC, but then you add one line talking about the order induced on I, and then "let x in H_j and y in H_k, wlog suppose H_j <= H_k" is replaced by "let x in H_j and y in H_k, wlog suppose j <= k"

so in my opinion it's unhelpful

[hazy pivot]

Yeah like I said, I overestimated how much it would help

[odd jackal]

okay, that's it I guess?

Thanks a lot everyone!

.close

[vale dock]

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[hazy pivot]

Sorry about the lengthy side quest lol

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[mental tartan]

hey im mad stuck on 2. all of it really, i just dont know how to interact with that equation

[mental tartan]

what is differentiating it supposed to give? it just gives 0=0

[pure pulsar]

Euclidean space, mind if i help?

[mental tartan]

just E^3

[pure pulsar]

Did you differentiate the sphere constraint?

Can you share your work actually

[mental tartan]

yeah, but it just reduced to 0=0

gotten nowhere lol just those top few lines rly

[pure pulsar]

x’(s) is the unit tangent right?

[mental tartan]

yeah

[pure pulsar]

x(s) is a curve parameterized by arc?

[mental tartan]

sure yeah any unit speed parameter

course suggests that using s as a parameter implies unit speed

[pure pulsar]

So when we differentiate first

(x-p) * T = 0 for all s

T is unit tangent

[mental tartan]

|x-p|•T=0 right

[pure pulsar]

Differentiate it

uv rule

T^2 =1

In mod

[mental tartan]

so a contradiction

[mental tartan]

because this says T=0

[pure pulsar]

No

Unit tangent is 1 right?

[mental tartan]

oh yeah

but then how does that equation hold?

[pure pulsar]

It only says that tangent is tangent to the sphere

And now when you differentiate, a new constant k will be introduced

For N

Helping with euclidean space shit is kinda tough, you gotta imagine

The situation

[mental tartan]

yeah no worries just trying to get it

[pure pulsar]

Tho i’ll help you for the question

So no need to worry

[mental tartan]

could you explain a bit how that is

cuz like |x-p| is just R

[pure pulsar]

The radius vector from the point p to the point x(s) is orthogonal to the tangent.

[mental tartan]

if it was (x-p)•T=0 i'd see that

but R*T=0 to me means T=0

if R is a nonzero constant

which it is

or should i just accept it and go back to the problem

alr i gotta move on thanks anyway

.close

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[pure pulsar]

Not really

[pure pulsar]

R isnt radius?

[mental tartan]

its not?

[pure pulsar]

What??

[mental tartan]

R is the radius of the sphere

thats given

[pure pulsar]

Thats what im saying

[mental tartan]

ah k

then |x-p|=R right?

[pure pulsar]

Yes

For all s

[mental tartan]

so |x-p|•T=0 implies T=0

[pure pulsar]

No, differentiate that term again bro

[mental tartan]

😭

no wait this

nah im clueless

how tf is the unit normal N a scalar

can you differentiate it because i feel like im messing up

[vale dock]

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[lucid bone]

Hey, I have a question regarding derivatives and l'hopital's rule. I got most of my work done, however, I'm confused on how to find the derivative of ln(-lnx)

[craggy girder]

Doppelganger

[lucid bone]

I got the answer from the answer key, but the work is not shown,

[lucid bone]

im the real one LOL

[craggy girder]

Chain rule

[lucid bone]

yeah if I chain rule tho

i get -lnx/x * (1/x)

do i not?

because derivative of the outside: 1/x * -lnx

• the derivative of the inside 1/x

[craggy girder]

That is incorrect

[lucid bone]

could you explain the steps please? Im lost, i always thought it was the der of the outside * the inside function * the der of the inside function

[craggy girder]

Let's say that u = -lnx

Now calculate dy/du

[lucid bone]

do you mean dx/du?

bc theres no y

or du/dx

[craggy girder]

y = ln(-lnx)

We are basically deriving the chain rule

[lucid bone]

wait so, for the u=-lnx part

would it just be y=1/u?

ohhh i see, so you have to think of those as different, and focus on the first one, find the derivative of that, then move on to the inside?

[craggy girder]

Yess

[lucid bone]

thanks!! I have one more question

[craggy girder]

No wait

Tell me what dy/dx is first

[lucid bone]

find the derivative of y in respect to x

[craggy girder]

I mean what ans do you get

[lucid bone]

ohh oops mb

i got 1/-lnx * -1/x

[craggy girder]

Correct

[lucid bone]

thanks!

also, how do u know if (ln0) is approaching inf or -inf

is it just based on if it is poisitive or not?

[craggy girder]

Do you agree that 1/inf = 0?

[lucid bone]

yes

[craggy girder]

Tell me what e^inf is

[lyric sundial]

By knowing the graph of ln
Or knowing the connection with the function e^x

[lucid bone]

im not sure

[craggy girder]

inf

[lucid bone]

so i would have to just know what the graph looks liike in order to know the behavior>

[lucid bone]

is it because the x value is always apporaching inf

?

[craggy girder]

e^2 means e*e
So e^inf means e*e*e....

So inf

[lucid bone]

ohh okay, so then ln(o) would be inf right?

[craggy girder]

Noo

[lucid bone]

oh shoot

wait so just e^inf is inf

[craggy girder]

Tell me what e^(-inf) is

[lucid bone]

-inf

[craggy girder]

Nope

[lucid bone]

inf?

bc the graph is only going one way?

[craggy girder]

Hint: a^-b = 1/a^b

[lucid bone]

ohh so it would be 0

[craggy girder]

Yes

Take log on both sides

[lucid bone]

for which? sorry

[craggy girder]

Do you remember how log is defined?

[lucid bone]

i dont

[craggy girder]

2^5 = 32

$\log_2{32} = 5$

[rocky lotus]

Wumpus Man

[lucid bone]

oh yeah i know that

[craggy girder]

Nice

[lucid bone]

sorry i didnt know what u meant by defiend

[craggy girder]

Use that in e^-inf = 0

[craggy girder]

Oh lol

[lucid bone]

does it have to be log

or can it be ln?

since we have an e

[craggy girder]

It can be ln

[lucid bone]

ln e^-inf=ln0
-inf=ln0

[craggy girder]

Yes correct

Also do you know the graph of lnx?

[lucid bone]

im supposed to, but i kind of forgot 😭

[craggy girder]

Oh

,w ln(x)

[rocky lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wum.pus.man

[craggy girder]

Here you go

[lucid bone]

okay but isnt the graph going to both -inf and inf?

[craggy girder]

The y value is

Not x

[lucid bone]

x is only going inf tho?

i thought it was -in

ohhhhh wait nvm

ln(0) is the vertical asympotote

and its approaching -inf

I see

[craggy girder]

Nice

[lucid bone]

okay wait, im currently working on a l'hopital question, I simplfied it and got x/lnx

sorry its

lim x--> 0+ x/lnx

would I l'hopital again since its 0/-inf

or would it just be 0?

[craggy girder]

LH is only applicable when limit is of form 0/0 or inf/inf

It would be just 0

[lucid bone]

okay thank you, do u know if u guys have practice questions for l'hopital cases when x^x, x^inf, and stuff like that?

[craggy girder]

I don't think so

Try opening a new channel and asking there

New ppl might visit it and send if they have

[lucid bone]

okay thanks!

what channel can i ask

is discussion fine?

[craggy girder]

Like help-7

[lucid bone]

okay

[craggy girder]

I don't use discussion much, I have no idea what happens there

But first close this one

[lucid bone]

alr alr no worries thanks!

doppelganger u saved me!!

THANKS FOR YOUR HELP

[craggy girder]

[lucid bone]

now how do I close this

[craggy girder]

Lmao

.close

[vale dock]

Channel closed

Closed by @craggy girder

Use .reopen if this was a mistake.

[craggy girder]

WHAT

HOW DID I CLOSE IT?

AN ERROR IN THE BOT?

CAUSE WE HAVE SAME NAMES?

||I'm kidding I can close help channels||

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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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.

[lament oak]

This is leaving cert mathematics 🇮🇪
I have an exam on algebra II soon.
I’m unsure how to continue. (Or if my work is incorrect)
Please help! (Ping me pls!)

[hidden terrace]

Just combine all those equalities

y=-p

t=q

[lament oak]

IM SOS TUPID WHAT

dude thank you so much!!

[hidden terrace]

-ty=r

[lament oak]

[hidden terrace]

You're welcome mate

[craggy girder]

sos?

[lament oak]

So stupid**

This server is my favourite now

Every math noobs dream

[hidden terrace]

Yeah

[craggy girder]

I was jk

[hidden terrace]

I'll advise you to be careful with how you use the word 'stupid' though

[lament oak]

Alr!!

[hidden terrace]

There's some rule that says you can't talk about intelligence level

[lament oak]

Okay so u can just divide both -t and -q by -1?

[lament oak]

Oops !!

It’s alright my exam is on Friday

[craggy girder]

Ohkk

[lament oak]

I’m genuinely not good at maths

[hidden terrace]

Don't you mean helpers?

[craggy girder]

No

[hidden terrace]

Cmon bro

[lament oak]

I feel like if a ping staff for a question for high schoolers i would get kicked

[hidden terrace]

Yep

He's tryna troll you

[wraith heart]

please don't troll

[river shale]

ye, trolling at help channels is a dangerous move

[lament oak]

Hmmmm but it says pq = r
No negatives
t = q
But
Y still is -p

[hidden terrace]

-ty=r

The negatives cancel out

[lament oak]

-q and -p cancel out?

[hidden terrace]

Yes

[lament oak]

I never even knew this

cooked

[hidden terrace]

-p * -q = pq

[lament oak]

I completely looked over the fact that they’re multiplying with each other

[hidden terrace]

[lament oak]

Thank you very much Miyagi. So basically I don’t need to do anything to -t = -q
Because then -q-p turns into qp

Wait can you even do anything to -t = -q ..?

I’m doing illegal maths stuff rn

[hidden terrace]

It's easier if you express it as t=q

[hidden terrace]

Welcome to the math mafia

[lament oak]

Doesn’t that make q positive so it doesn’t cancel out with p anymore?

[hidden terrace]

Everything still works

That's the beauty of math

It's completely consistent

[lament oak]

-qp is the same as qp?

[hidden terrace]

No

Let me list everything

y=-p

-t=-q

-ty=r

[lament oak]

wait. So t = q
-ty = r
Is essentially
-(q)(-p) so they cancel out

[hidden terrace]

Yes!

[lament oak]

1 Neuron activated

[hidden terrace]

You got it

[lament oak]

https://tenor.com/view/floating-levitating-levitation-black-guy-gif-20123481

[hidden terrace]

[lament oak]

.close

[vale dock]

Channel closed

Closed by @lament oak

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[short lantern]

Can anyone tell me what kind of youtube/google search i'd need to learn this? Thanks in advance

[woeful trench]

Most likely "circle theorems" would get videos for a large part

[hidden terrace]

And you need knowledge of cyclic quadrilaterals

[ashen prawn]

yes sir

[hidden terrace]

Wait no

[woeful trench]

(here you may want to add in AO and BO, which might help you...)

[hidden terrace]

Just that one theorem that states:
angle(AOB) = 2x

[fleet burrow]

"circle theorems" / "circle properties"
and more specifically
"tangents from external points"
"angle at center = twice angle at circumference property"
"tangent perpendicular to radius property"

[short lantern]

Thank you guys

.close

[vale dock]

Channel closed

Closed by @short lantern

Use .reopen if this was a mistake.

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[timber saddle]

Guys I need help w this calc question “Given that 𝑓(𝑥)is continuous at 𝑥 = 2 and lim𝑥→2 𝑔(𝑥)= 3, and 𝑔(𝑥)is continuous at 𝑥 = 2, is
the function ℎ(𝑥)= 𝑓(𝑔(𝑥))continuous at 𝑥= 2? Explain why or why not.”

[vale dock]

@timber saddle Has your question been resolved?

[timber saddle]

<@&286206848099549185>

[pure pulsar]

!original

[vale dock]

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.

[clear birch]

The issue is that we dont know f(3) is continuous or not

[timber saddle]

Number 6

[coral urchin]

im assuming this is calc as opposed to rigorous analysis

so, what are your thoughts here

[vale dock]

@timber saddle Has your question been resolved?

[timber saddle]

No idea but I did the quiz and it wasn’t on it anyway

.close

[vale dock]

Channel closed

Closed by @timber saddle

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[ripe wharf]

PLZ HELP ME BEFORE I DIE

[coral urchin]

can you identify the similar angles from this diagram?

[midnight iris]

😭 u have the wrong tag you should use pre university tag ig

[ripe wharf]

rsu and pnq?

[ripe wharf]

yeah that's old my bad

[coral urchin]

pnq isnt an angle here

[ripe wharf]

BRO WHAT

[coral urchin]

this would be pnq

[ripe wharf]

ohhh

ok how about

rsu and npq

[coral urchin]

why

very important to know why

[ripe wharf]

man i forgot the word

hold up

[stark wedge]

the middle letter names the vertex

[ripe wharf]

so

[coral urchin]

not vertical angles

[ripe wharf]

ok how about

[coral urchin]

that means across the point like this

[ripe wharf]

okkkk

makes sense

are they a type of

alternate angle

rsu and npq

alternate exterior angle?

[coral urchin]

yeah

[ripe wharf]

ohhhhhhh

wait so

the answer to the question is angle tsu and tsp

or tsu and rsp

[coral urchin]

adjacent means right next to each other

[ripe wharf]

but

they share the same vertex

[coral urchin]

yes this one

[ripe wharf]

OH

[coral urchin]

i should have specified not tsu and rsp

[ripe wharf]

I UNDERSTAND what ur saying

THANK YOU!

[coral urchin]

you can prove all these from parallel lines and angle sums btw

[ripe wharf]

tysm omg seriously

[coral urchin]

the adjacency theorems all follow from angle sums if you want the proof for it

then alternate exterior comes from parallel lines, and all the others come right out

but ill make you do the proof for them so youll learn it :P

[ripe wharf]

ok can we do one more question

[coral urchin]

yeah

its ur channel ask away

[coral urchin]

what is an interior angle

[ripe wharf]

interior angle is like inside the parallel lines

i think

[coral urchin]

yeah

but not parallel in this case

what about alternate, what does that mean

[ripe wharf]

oh yeah

alternate is like

opposite

[coral urchin]

yeah

[ripe wharf]

so alternate interior angle is

wair idk bc i thought you can only find it with paralllel lines

[coral urchin]

you can only solve them in that way with parallel lines

but they are still alternate interior regardless

[ripe wharf]

how do i solve

[coral urchin]

you need to know the angle between the lines if they arent parallel

cause then youll always get a triangle

but thats not relevant for this problem

[ripe wharf]

hmm ok

[coral urchin]

so, which angles here are alternate interior

[ripe wharf]

is it

lmj and ijm

or

lmj kjm

[coral urchin]

these arent alternate

[ripe wharf]

ok true

[coral urchin]

this is alternate interior

[ripe wharf]

thanks!

[coral urchin]

that all you have for now?

[ripe wharf]

i forgot how corresponding angles work

[coral urchin]

this was like the top one

i have four angles from the intersection

they are said to correspond if the angles are on the same side of those four

if that makes sense

[ripe wharf]

my answer choice wouldn't be on there then

[coral urchin]

one of these is right

[vale dock]

@ripe wharf Has your question been resolved?

[ripe wharf]

ohhhh

do u have more photos like that

[coral urchin]

I just picked the first image off google honestly

Lol

[ripe wharf]

LOL ok no worries

[coral urchin]

so what did you get for an answer

[vale dock]

@ripe wharf Has your question been resolved?

Channel closed

Closed by @ripe wharf

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[fair summit]

Geometry And Congruence, angles, bisects, all that biz. Anyways, I have tried every concievable way I know to solve this question, I KNOW the answer is 56. Typically we would set EBC and EBF equal to each other since we know they're congruent, however I have tried adding like terms in every way imaginable and cannot find the right answer. If someone could describe the formula or what I could be doing wrong would be greatly appreciated.

[unborn sable]

is there some larger context here?

[fair summit]

Anything in particular or just more?

I can send you the full homework page, which might provide larger context

[unborn sable]

please

[fair summit]

this is specifically tied to the question, so it is likely what you mean.

this is the entire sheet (try to read past my scribbles)

I understand everything else in this module aside from this specific type of problem

[rancid laurel]

wow you write exactly like me

[fair summit]

Undiagnosed dysgraphia or just bad handwriting?

Who knows!

[rancid laurel]

handwriting lol

anyway i gtg I have homework

[fair summit]

bro dropped in to say "our handwriting is equally poor" and LEFT

LMAO

[rancid laurel]

LOL

[unborn sable]

alright
so we know that
ABE = EBF = 1/2 ABF
FBH=HBC = 1/2 FBC
and all the angles sum to 180

we want EBF, we have
ABE=2r-20 and EBC=3r+10

what can we do

[fair summit]

There is two things I know we could do. A. Set the equations equal to eachother, something like 2r-20=3r+10 (followed by solving it) or 2r-20+3r+10=180

I have tried the first to exhaustion, the ladder to a similar but not entire degree

[unborn sable]

setting them equal to eachother wouldnt really do anything as the angles arent actually equal
and adding them wouldnt be the entire 180

EBC=FBC+EBF

so 3r+10=FBC+2r-10

since EBF=ABE

any idea where we could go from here?

[fair summit]

No clue, maybe find FBC? Genuinely no idea.

wait

that 10 is supposed to be a 20

lol

[unborn sable]

ive made a little mistake as my diagram was off

[fair summit]

but yeah I have no idea

[unborn sable]

give me a minute ill correct it and get back

[fair summit]

Kk

[unborn sable]

alright ive got it sorted now, scrap all my previous

you were right to be adding them

[fair summit]

This is good news

[unborn sable]

because EBC+ABE=180

so where are you getting lost?

[fair summit]

Okay, so I have been taught to complete the problem by, putting them equal. Completeing like this
3r+10=2r-20 then getting everything to one side via adding like terms, a few ways we can do this, but I fear I may be adding them wrong.

[unborn sable]

you would only set them equal if they were the same angle

thats not the case here

[fair summit]

Ah

SO I SHOULDNT do that

[unborn sable]

not in every case no

[fair summit]

Troubling

[unborn sable]

if it were (EBF and ABE) or (EBH and HBC) then you would do that, as those (pairs) are equal angles

[fair summit]

This make sense with prior equations

So how do I solve this?

[unborn sable]

if you look at the diagram, we can see that ABE+EBC is the entire 180 degrees

so their sum =180

[fair summit]

Yes

I know this

[unborn sable]

thats how we find r

[fair summit]

so setting them equal to 180 and ADDING them is an option

(I always call that kinda variable x (habit))

Something like 3r+10+2r-20=180

[unborn sable]

its an option, just because we know their sum turns over the entire line

[fair summit]

It is the only other way I have been taught

Let met ry

try*

so from adding them I get 35a+43=180

I think

[unborn sable]

you have done some magic concatenation

[fair summit]

WHAT DOES THAT MNEA

!?!?!?

[unborn sable]

actually, im not sure what you did at all

how did you get that

[fair summit]

31a+4a then 45+ -2

35a + 43

how is that wrong...?

[unborn sable]

we're doing 9 now? I thought we were doing 11

[fair summit]

oh Im dumb

sorry I READ THE WRONG THING

[unborn sable]

tut tut

[fair summit]

Big frown

I didn't do magic

Im just a little upid

Its terminal

okay

so I got 5r + -10 = 180

[unborn sable]

looks good

[fair summit]

then subtract -10 from both sides

which makes 190

so 5r=190

[unborn sable]

indeed

[fair summit]

then divide 5 by both sides

giving me 38

do I re-input it now? This doesn't seem right...?

[unborn sable]

its all good thus far

[fair summit]

what do I do from here?

[unborn sable]

What is EBF in terms of r

[fair summit]

I do not know

maybe I do but not in the format you're asking

[unborn sable]

do we know any relations between the angles here?

the bold text above 6 is the kickoff point

[fair summit]

Well I know they all share the same Vertex, that being B.

Uhm

let me think

[unborn sable]

the bisecting parts are the key

[fair summit]

Ah

I see what you're asking

though I do not know whether I should x 2 or divide it by 2

let me see

or that might not be what you're saying

[unborn sable]

as a start
just tell me any relations between
ABE, EBF, EBH, HBC

purely using the bisections

[fair summit]

They're all bisected at the same point?

[unborn sable]

none of those angles are bisected, theyre just part of bisections

read the text and try draw the angles out on the diagram, might help

this text specifically

[fair summit]

BA and BC are opposite rays, basically meaning they go opposite directions on the diagram. That part is easy to understand

[unborn sable]

si

[fair summit]

bisecting is the dividing of two things into congruent or equal parts

[unborn sable]

indeed it is

[fair summit]

So BH bisects EBC into EB and CB

is that correct?

[unborn sable]

EB and CB are rays

EBC is an angle, cant split an angle into rays

[fair summit]

So how does it bisect it?

[unborn sable]

into angles

i cant really tell you without just giving you the answer

[fair summit]

Okay

So an angle is a pair of rays that have a common endpoint

[unborn sable]

but look at EBC, if we split it at BH, what two angles are there

[fair summit]

okay

Okay so bisecting it is splitting it in half, I understand that, but I don't know where to go from here

I basically never had to really comprehend why I was doing the equations or formulas I was, I just kinda did

Barely even knew what bisecting was before this conversation and I did it just a few weeks ago

In this particular diagram I want to know what bisecting here would look like

[unborn sable]

ill tell you this one then

when we bisect EBC, splitting at BH

we are left with EBH and HBC
which must then be equal since its a bisection

so EBH=HBC=1/2 EBC

[fair summit]

Okay

[unborn sable]

(just for me so i dont have to scroll)

[fair summit]

This makes sense in the way of I kinda understand it but will need to work on it further to actually remember

SO from what i get

I cut it in half

the 38 we got earlier

[unborn sable]

afraid not
that 38 is just the value of r

[fair summit]

okay

so we do re-input it then?

[unborn sable]

we will in a minute, but its best to know what to do with it first

[fair summit]

Alrighty

(math is hard and frustrating gosh)

[unborn sable]

we want EBF

does either of the bisections tell us anything about EBF?

[unborn sable]

(it gets easier in time)

[fair summit]

Yeah I just was never taught the proper way of doing things, so I just memorized formulas and threw them away the next day without REALLY comprehending what was put in front of me

Caught the way of doing it more so through patterns than comprehension

[unborn sable]

thats not too uncommon, it just becomes far less sustainable the further you go

maths is like building a tower, need the bricks below to go any higher

[fair summit]

EBF is half of EBC I can see that

like on the graph

[fair summit]

Makes sense

[unborn sable]

im afraid not

[fair summit]

Angle wise like

actual degrees

visually it just looks like that

[unborn sable]

i would never take things as what they look like, it'll put you in the bin

[fair summit]

I see

[unborn sable]

the BE bisection is the one we want here

it it splits ABF into...

[fair summit]

You're gonna have to give me a second NONE of this is clicking🥀

[unborn sable]

its no problem

take your time

[fair summit]

Gonna work this out in my brain

[unborn sable]

(draw it on your diagram)

use different colours for each bisection if you have them

[fair summit]

I don't see how we get EBH or HBC with bisecting

like how does bisecting change it

[unborn sable]

ill draw it, one minute

[fair summit]

It is meant by definition to half something to congruence

All I understand is if something bisects something like BH-EDC that BH has to be involved in the results in some way

pattern recognition :[

[fair summit]

Okay

this KIND of makes sense

like my neurons are rubbing together

[unborn sable]

think you could make a similar drawing for the other bisection?

[fair summit]

Let me give it a shot

also how do you draw so finely on a mouse???

[unborn sable]

ah i have a stylus

my laptop folds into a tablet

[fair summit]

oh cool

Okay

not done yet

but I am looking at it

and understanding parts

why I do the equation I do

Couple questions

So

[unborn sable]

fire away

[fair summit]

One second

working it out

I genuinely

might be dumb

my brain might not function at the correct average level

One sec while I write it but I think I got it

Is it

ABE and FBE

[unborn sable]

you got it

[fair summit]

This is so genuinely peak

but I also might be a tinge dull

an unfortunate ego hit

[unborn sable]

everyone has their own pace, you'll get there if you stick with it

[fair summit]

oh for sure, but goddamn I thought i was decently okay at this whole braining thing

I was LITERALLY feeling my brain fail and finally comprehend something

like it knew the path I had to make it just was

really slow at making it

THERE was a connection

it just took its damn time

[unborn sable]

[fair summit]

So

I know have FBE (or EBF if you do it the opposite of how I did it but yk thats menial thats chill) and ABE (which was in the original question

So

In order to get EBF

from 38

I learned bisecting

I have one of them

I have ABE but I shouldn't get EBF by bisecting...EBF

but

I did get ABE which is one of them

adding ABE (or ABE x 2) would get me my anser

[unborn sable]

(whats the relation between ABE and EBF)

[fair summit]

ABE is halve of EBF?

[unborn sable]

ABE is half of ABF

not EBF

[fair summit]

okay

[unborn sable]

remember ABF was bisected into ABE and EBF

so ABE and EBF are...

[fair summit]

added together into ABF?

[unborn sable]

true, but theres more

remember what bisecting means

[fair summit]

To halve to congruence

so ABE and EBF added together

is ABF

[unborn sable]

yeah, split into two EQUAL pieces

so ABE and EBF are...

[fair summit]

I am confused on what you're asking me, but if got 38 by adding like terms, I believe they're both 19?

split apart?

[unborn sable]

lets set aside the 38, ignore it for now

[fair summit]

okay

[unborn sable]

how do we define a bisection

[fair summit]

The definition is cutting an angle into two congruent (equal) pieces

right?

[unborn sable]

yeah, so if those two pieces we got from the bisection are ABE and EBF
then what must they be?

[fair summit]

They must be equal?

[unborn sable]

bingo

[fair summit]

AH

[unborn sable]

ABE=EBF

[fair summit]

The equation itself is also finding EBF (not ABF) but it hardly matters for the learning process

[unborn sable]

ABF is just important to find that relation

[fair summit]

ABE+EBF=ABF

?

[unborn sable]

indeed so
and ABE=EBF=1/2 ABF

to be more specific

since we have an equation for ABE, theres not much left to do

[fair summit]

ABE=38 right

[unborn sable]

not quite

r=38

ABE=2r-20

[fair summit]

so in order to get ABE we need to re-input 2(38)-20

[unborn sable]

indeed

[fair summit]

Since before we can even DO ANY of this business we need to find r

Im definitely comprehending it better now

[unborn sable]

you could do it in either order honestly

they dont depend on eachother too much

[fair summit]

technically if I just did the equation I probably would have gotten it

but I think its better in general that I start

actually comprehended the math im doing now

omg I did it

Praise the lord

Praise the AZO

thank you Azo

I literally just needed to re-input it

[unborn sable]

you made it my friend

[fair summit]

im a dumbass

but like

But conceptually I understand it better as a whole

[unborn sable]

that wouldnt matter if you didnt know that ABE=EBF, otherwise you would just be finding ABE

[fair summit]

True same answer not the same knowledge gained on how it actually works

Alright

well thanks

thats all I needed

[unborn sable]

no worries

take care

[fair summit]

You too

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[tiny torrent]

to prove this I can just use the hyp to find one limit point? I just need to show that there exists one right

[tiny torrent]

or is this better to be done by contradicition

I can see contradicition possible working here too tho im not sure

maybe ill try both and see

[tiny torrent]

though I want to make sure that to prove this statement this is a sufficient method

actually

its says every

I cant generally describe every bounded infinite set of real numbers

so contradicition is probably the way?

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[turbid portal]

can i put it in terms of x after getting the derivative

[distant galleon]

You could, but that would be a lot more cumbersome.

[turbid portal]

😔

i got shown two different formulas to use one for about x axis and one about y axis and now with being given x = y about x-axis im stuck , i thought maybe i change the terms but i dont wanna make it harder, what do i do

[coral urchin]

you should just be able to rotate this with an integral

[turbid portal]

oh

[coral urchin]

like do revolutions, but on circle perimeter instead of circle area

revo on area = volume
on perimeter = surface area

[turbid portal]

lemme show what i think the setup is

i just need confirmation

uh does this seem right

[coral urchin]

this is good

[turbid portal]

alright let me try and continue with that then

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[tiny torrent]

can S_n be considered a subsequence of x_n?

[tiny torrent]

no it couldn't be

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[tiny sapphire]

How do I do this problem? Do I have to use chain rule or product rule? I’m confused on identifying when to use it

[safe fulcrum]

You use both rules

You use the product rule, but when differentiating both u and v, you also need the chain rule

Try finding u' and v' individually

then sub into u' v + uv'

And factor ofc

[stark wedge]

differentiation rules are recursive

ie when applying one rule, it may be that you will need to apply other rules further down the line

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@tiny sapphire Has your question been resolved?

[tiny sapphire]

So how would I do that in this case

[shut mirage]

respond to the bot first to prevent the channel being closed