help-28

[glossy valve]

Raphaelisius Maximus MMIII

[rapid rain]

Even without the bonus problem, I hope that was clear enough

[signal solar]

Yeah that's super helpful

[signal solar]

Ok so would the original students approach work for this version

[rapid rain]

I'll let you take a guess

what are the "problematic" points for this integral?

(the bound(s) which make(s) the integral improper)

[signal solar]

Infinity and 0?

[rapid rain]

yes

in those integrals, the only improper bound was 0

in here, it was only infinity

to sum up, you've only seen what happens with one improper bound

so, the p tests that we used were around 0, or around infinity

there definitely isn't a p test that works with both

the values of p that work around 0 are p < 1

and the values of p that work around infinity are p > 1

so a single p test won't get us out of this

[signal solar]

So do you just do multiple?

[rapid rain]

the question is, how are we going to do multiple p tests

we would need to have multiple integrals

for example, one where only 0 is the improper bound, and another one where only infinity is the improper bound

Do you see what we're gonna do?

[signal solar]

Ah

Yes I see

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[royal dew]

I am asked to find the area of the shaded region

[royal dew]

so is this integration correct

,w integrate root(x) - x/4 from 0 to 4

[glossy valve]

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[royal dew]

i think there should be a plus there, is that true?

[next cedar]

you are subtracting a negative, so yes there should be a plus

[royal dew]

yeah

,w plot x=-y and x=2-y^2

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[royal dew]

to find the area between these two, i have to do integration wrt x

but idk how to write the second function in form of y=f(x)

[nimble crane]

you can just rearrange the equation x = 2 - y^2 for y

and choose the positive root since the graph you’re given has a y-intercept of 2

[royal dew]

why do i have to choose the positive part?

the graph is going under the x axis too

[nimble crane]

because the graph here has a y-intercept of 2, not -2

[nimble crane]

hold on, they flipped the x and y axes

[royal dew]

yeah its different of desmos

tf

[nimble crane]

well, either way, you need to pick the function that matches the graph

use the x-intercept to do that

[royal dew]

the intersection points are -2 and 1

so i just integrate now?

,w integrate root(2-x)+x from -2 to 1

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[royal dew]

is this gonna be correct?

[full forum]

@royal dew Has your question been resolved?

[royal dew]

is this the correct integration for area between y = -x and x = 2-y^2 guys

[robust plinth]

ye\

[woven cave]

it’s not different

[royal dew]

aight thx guys

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[royal dew]

find volume of solid of revolution using shell method

[royal dew]

okay guys for this one i used integral of 2pi y f(y) dy

,w integrate 2 pi y (2y-y^2) from 0 to 0.5

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[royal dew]

i got this

am i doing it right

[foggy vapor]

Why are you integrating up to 0.5 lol

Also its 2y- 2y^2 not 2y-y^2

Otherwise its fine

[royal dew]

lmao i mixed up the x and y dimensions

its gonna be from 0 to 1

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[stone meteor]

Since there is no finite-dimensionality assumption, we must explicitly construct the isomorphism. I am looking for hints on how to do that.

[full forum]

@stone meteor Has your question been resolved?

[stone meteor]

We can at least begin by stating some basis of V/U, but either way, I'm not sure where to go from there.

[topaz valley]

to build a map U x (V/U) -> V, you can start by mapping (u, 0) -> u

how do you use the basis to map elements of the form (0, v)?

[stone meteor]

i guess we map them to

$a_1v_1+\dots+a_nv_n$

[glossy valve]

holo_morph

[topaz valley]

what are v1, ..., vn?

[stone meteor]

Let $v_1+U,...,v_n+U$ be a basis of $V/U$. We claim that
[
V = U \oplus \text{span}(v_1,...,v_n).
]
We know that $V = U + \text{span}(v_1,...,v_n)$ since for all $v \in V$,
\begin{align*}
&v +U = a_1v_1+\dots +a_nv_n+U\
&\Rightarrow v-(a_1v_1+\dots +a_nv_n) = u, \text{for some } u \in U\
&\Rightarrow v = u + (a_1v_1+\dots +a_nv_n).
\end{align*}
Moreover, suppose $v \in U \cap \text{span}(v_1,...,v_n)$.
\begin{align*}
&v = u = a_1v_1+\dots+a_nv_n\
&\Rightarrow v+U = 0+U = a_1v_1+\dots+a_nv_n + U\
&\Rightarrow a_j = 0\ \text{for each } j =1,...,n\
&\Rightarrow v =0\
&\Rightarrow U \cap \text{span}(v_1,...,v_n) = {0}.
\end{align*}

[glossy valve]

holo_morph

[stone meteor]

this is my current progress

[topaz valley]

this is pretty much the argument, but you should rephase it in terms of proving an isomorphism

[stone meteor]

but this isn't done

now I have to construct the argument

now I have to construct an isomorphsim from V to U x V/U

although I guess that is quite easy to see now

because I just decompose elements v into u + a_1v_1+...+a_nv_n

then send

[topaz valley]

you've described the map

[stone meteor]

u to u

and

a_1v_1+...+a_nv_n

to a_1v_1+...+a_nv_n + U

[topaz valley]

yes

[stone meteor]

no I haven't lol

I just did here

but not in the proof

but I think I can do the rest anyways

.close

thanks

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[topaz valley]

[stone meteor]

.reopen

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✅ Original question: #help-28 message

[stone meteor]

I'll keep this open actually

because I may find trouble with showing injectivity and surjectivity

or maybe i should instead opt for the possibly better method of just constructing an inverse

okay I am kind of confusing myself

Let $v_1+U,...,v_n+U$ be a basis of $V/U$. We claim that
[
V = U \oplus \text{span}(v_1,...,v_n).
]
We know that $V = U + \text{span}(v_1,...,v_n)$ since for all $v \in V$,
\begin{align*}
&v +U = a_1v_1+\dots +a_nv_n+U\
&\Rightarrow v-(a_1v_1+\dots +a_nv_n) = u, \text{for some } u \in U\
&\Rightarrow v = u + (a_1v_1+\dots +a_nv_n).
\end{align*}
Moreover, suppose $v \in U \cap \text{span}(v_1,...,v_n)$.
\begin{align*}
&v = u = a_1v_1+\dots+a_nv_n\
&\Rightarrow v+U = 0+U = a_1v_1+\dots+a_nv_n + U\
&\Rightarrow a_j = 0\ \text{for each } j =1,...,n\
&\Rightarrow v =0\
&\Rightarrow U \cap \text{span}(v_1,...,v_n) = {0}.
\end{align*}
Now, we define a linear map $T \in \mathcal{L}(V, U \times (V/U))$ by
\begin{align*}
T(v) = T(u + (a_1v_1+\dots+a_nv_n)) = (u, (a_1v_1+\dots+a_nv_n) +U).
\end{align*}
It is self-evident that this map is linear and well-defined; we now direct our focus to showing that it is indeed an isomorphism. Suppose $T(v) = (0,0)$,
\begin{align*}
&u = 0, (a_1v_1+\dots+a_nv_n) \in U\
&\Rightarrow (a_1v_1+\dots+a_nv_n) + U = 0 +U\
&\Rightarrow (a_1v_1+\dots+a_nv_n) = 0\
&\Rightarrow v = 0 + 0 = 0.
\end{align*}
This shows that $T$ is injective. Furthermore, for any $(u, v'+U)\in U \times (V/U)$,
\begin{align*}
&(u, v'+U) = (u, (a_1v_1+\dots+a_nv_n) +U).
\end{align*}
Thus, $\exists v \in V$ such that $Tv = (u, v'+U)$ since we can just define $v$ by
\begin{align*}
&v = u + (a_1v_1+\dots+a_nv_n)\
&\Rightarrow Tv = (u, (a_1v_1+\dots+a_nv_n) + U) = (u, v'+U).
\end{align*}
This shows that $T$ is surjective, establishing an isomorphism of $V$ onto $U \times (V/U)$.

[glossy valve]

holo_morph

[stone meteor]

How is this looking?

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[peak ermine]

im doing part e) and i honestly don't understand how to do it after sayin that f(-1) - integral (from -1 to -4) of f'(x) dx = f(-4), how do i go about it? i was thiking maybe find the area of the interval? How would you solve it?

[peak ermine]

this is where i am at

[forest trail]

Good! Yeah this looks about right.

You don't know what f', so its reasonably difficult to find a precise value for f(-4). I'd say use any estimation that you like.

For example, you can create rectangles.

[peak ermine]

ok brb lemme try that

ok so now i have that f(-1) - 4.7ish =f(-4) and since f(-1) =3 that is 3-4.7ish = -1.7=f(-4)

[peak ermine]

is that good?

[forest trail]

Yeahh 4.7 looks good since it does look over the x axis

Yep! Fine to me.

[peak ermine]

bet thx

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[maiden mist]

.

[maiden mist]

@umbral dome can u explain here?

[umbral dome]

yes

[maiden mist]

tysm

[umbral dome]

so you said that the centroid of suape 2 is at y = 75

but that's not correct, it would only be true if the second shape went all the way to the left edge

[maiden mist]

I’m having a similar problem here 💔

[maiden mist]

So how would you go around that

[umbral dome]

the second shape as you defined it is a rectangle whose left edge is at y = 6 and whose right edge is at y = 150

[maiden mist]

And how can u tell if it is or isn’t

[umbral dome]

it's up to how you decide to split up the shape into rectangles

[umbral dome]

based on this information, what should the average y coordinate of this shape be?

[maiden mist]

So for Qyy I DONT include the first shape in my calculations?

[maiden mist]

Hmm

One sec

72 for the second shape?

[umbral dome]

not quite

remember we removed the left section, so the centroid should not be further left

[maiden mist]

R we talking about the vertical or horizontal for the second shape/ right side?

[maiden mist]

I meant this along the Y axis so horizontal

Ohhh wait

Yeah I was gonna go for 72💔

Sorry🥹

I’m terrible at section properties😭

So when we split the shape we never include the other parts of the shape in our calculations?

[umbral dome]

your shapes should not overlap

[maiden mist]

OH WAIT

78?

@umbral dome

[umbral dome]

yes

[maiden mist]

I think I get it now

I’ll be back soon tho

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[chilly kernel]

Do u guys know someone's explaining How to solve a linear programming problem using the graphical method
I looked in YouTube but like no one's explaining it in a good ?way

[rancid basalt]

hello, infinity

i can help youe\

send DM to me

[chilly kernel]

I'd be thankful mate

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@chilly kernel Has your question been resolved?

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@chilly kernel Has your question been resolved?

[restive geyser]

I would much recommend you keep the assistance to this channel.

While not disallowed, it's more practically beneficial if you can get people to cross-check your work

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[stiff valve]

hi,
I am wondering if any tutors would be able to take a couple hours tonight and help me study in a call for my differential equations final exam. I am not sure if this is against the rules of your server but if it is please feel free to close the ticket. Id be willing to pay for the time as long as the price named isnt outrageous.

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[celest fulcrum]

I am having a hard time understanding why this is not closed using the S\A definition (the complement must be open). I understand that it doesnt include 0 so its closed via limit points but my book has not introduced this definition yet

[celest fulcrum]

More plainly, A^c is just the union of all sets (1/n,1/(n+1)) which is clearly open.

and also (-inf,0) for the first set which is also open, then (1,inf) for the last set with is also open

hense, the complement of A is open so A must be closed? am i missing something?

[rare dock]

are you sure about that?

[celest fulcrum]

this part yes

but also

just realized

doesnt include 0

so thats the closed part

[nimble crane]

the complement includes the point 0. think about the open sets containing 0

[celest fulcrum]

Yeah

[lime ether]

🤔

[celest fulcrum]

0 must be in A^c

[lime ether]

so the complement is the union of empty sets?

[celest fulcrum]

so its (-inf,0]

[lime ether]

🤔

[celest fulcrum]

no

[nimble crane]

right, are there any open sets containing 0 that lie entirely in A^c?

[lime ether]

(1/n, 1/(n + 1)) is empty since 1/(n + 1) < 1/n

[celest fulcrum]

I ment the other way around

but yeah its becase the first set would be something like

(-inf,0]

which is not open**** (whoops im just learning this lol)

and by alkamedian there is always some 1/N between 0 and 0+r which will be in the r neighborhood

if we just append 0 that would be a closed set right

[lime ether]

this is the entire argument

i’m not sure i understand what your question is

did you answer your own question?

[rare dock]

yea

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@celest fulcrum Has your question been resolved?

[celest fulcrum]

I didnt know how the complement was open but i figured out i made a mistake not including the 0 in the complement

which was the question

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[chrome canyon]

this is beckmann rearrangement right?

[robust slate]

discord.gg/chemistry

[chrome canyon]

go to other doubts mr pigeon

whats funi beckmann then?

[blissful sail]

Idk why this happens so frequently, people ask stuff other than math here

[sudden chasm]

because it isn't against the rules to do so actually

[blissful sail]

Fr?

[chrome canyon]

one guy asked 10/2 when 5*2=10 was written in the qn

[sudden chasm]

yesh

[sacred yarrow]

This one?

[chrome canyon]

yes

[sacred yarrow]

Or this

I found them by googling

[chrome canyon]

mb this one actually

[sacred yarrow]

Not sure if it’s what you’re looking for

[chrome canyon]

ts one mate

im js asking whats its name

beckmann rearrangement or funi beckmann rxn?

[urban jacinth]

Yeah it is

Was that all you wanted to ask?

[chrome canyon]

yea

[urban jacinth]

Couldve just googled it...

[chrome canyon]

any ideas whats funi beckmann

[urban jacinth]

Unfortunately I dont

[chrome canyon]

ik that very well but wanted to do it specifically in this sv

i wish to make it a kinda norm

[urban jacinth]

If you wanna learn more, theres certain exceptions too

Those reactions are called abnormal Beckmann

[chrome canyon]

u hv sm material relating to it? if yes do dm me

[urban jacinth]

Ah i used to do this a year ago

My notes are probably just crumpled up now

[chrome canyon]

ic

im doing all that now

beckmann fravorskii fetizon hvz etc

[urban jacinth]

Ah yes i remember favorski

[chrome canyon]

i learnt fetizon very recently

[urban jacinth]

Havent heard of fetizon

Are you a chem major?

[chrome canyon]

a highly chemoselective reagent

[chrome canyon]

im in 12th grade

[urban jacinth]

JEE?

[chrome canyon]

ill also hv to understand chemoselectivity regioselectivity n enantioselectivity thoroughly

[chrome canyon]

right

wbu

[urban jacinth]

Already done

: )

Lets take this to dms

!done

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[chrome canyon]

.close

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[summer creek]

If pipe A fills 9 hour faster than pipe B

What would be the equation?

A=B-9
A=B+9

[sacred yarrow]

Neither, afaik

[onyx glen]

you have to be clear first in what the letters A and B mean in your equations

cause if they mean the flow rates of the pipes, then yeah, neither

also just to make sure: you are saying pipe A and pipe B are connected to the same pool and we are comparing the fill times from each pipe open individually, yes or no? @summer creek

[summer creek]

Pipe a and b are connected to the same pool

Yes individual

[onyx glen]

and, once again: in your equation, the letters A and B represent the flow rate of water through each pipe, yes?

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@summer creek Has your question been resolved?

[onyx glen]

op gone

let him say it

[stone matrix]

mb

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[desert sonnet]

hi yall, I need help with this question. we solve it using taylor series method, I did it for sinx and got my answer as 1/6, for tanx and the denominator x^n idk how to deal with them, I tried puting tanx as sinx by cosx [which is 1], but that is not working

[desert sonnet]

the method given in solution is insanely long compared to 60 seconds time limit we are given, so like anybody knows any short cut to solve it quickly?

[delicate torrent]

Try plugging and checking

Since it's multiple choice

[desert sonnet]

but if I do direct substitution without cchanging the equation then I plug 0 in x, so in denominator any power of 0 is just 0, soo like how wld that work?

[delicate torrent]

Plug n in and find the limit

It's multiple choice?

[desert sonnet]

yes

[desert sonnet]

wdym?, [I got introduced to limits tdy, I'm finding it difficult with terms]

[solid rapids]

2?

[solid rapids]

he's telling sub all the answers and check manually what works

[delicate torrent]

For each answer they gave you, substitute n in and find the limit

[solid rapids]

@desert sonnet

expand

and cancel

[desert sonnet]

its 4

[solid rapids]

yeah

i got it

expand and cancel

@desert sonnet sub the expansions of sinx , tanx , multiply the leading coefficient and that is = to ur n

[delicate torrent]

Im not really used to this trig thing in algebra, but i'd be better with no trig

[desert sonnet]

same

[solid rapids]

for the question to be satisfied , the condition is the denominators power shud cancel the power in the numerator

x^n = x^4

n = 4

[desert sonnet]

so tanx is an odd function that we have to take odd values. got it thankss

[summer creek]

A is someone B is someone

[desert sonnet]

.close

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[summer creek]

@onyx glen

[summer creek]

.

[sacred yarrow]

@summer creek Can you add those descriptions to your original question?

[onyx glen]

those "A is someone B is someone" make no sense to me

[summer creek]

A,B represent a name

[onyx glen]

if they represent a name then they are of type String at best & cannot fit into any equation!

!xy

[full forum]

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.

@summer creek Has your question been resolved?

[full forum]

Channel closed

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[lost canopy]

Explain me fractions

[next cedar]

what about fractions are you confused, and what grade are you atm, if you don't mind me asking?

[fiery cedar]

Status

[full forum]

@lost canopy Has your question been resolved?

[full forum]

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Closed due to timeout

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[proper pewter]

I need help with a lesson coming in my Geometry midterm

[blissful sail]

Just post your question

[proper pewter]

ok one sec

i have mulitiple questions if thats fine

[next cedar]

that's always fine

[proper pewter]

this is the first one

???

[next cedar]

ok, so what have you tried for this one?

afaict, you are asked to merely describe the relationship of these lines. you're not told to calculate anything.
so a qualitative answer would do.

specifically, you should aim to describe the relationship of the escalator to the floors it passes.

[proper pewter]

i dont understand what they mean

[next cedar]

which part?

[proper pewter]

the releationship of the lines

[next cedar]

do you know what transversals are?

[proper pewter]

a line that cuts between two or more linrs

[next cedar]

ok, almost correct

but two or more what kind of lines?

[proper pewter]

parralel?

[next cedar]

there we go.

now look at the floors. they're all horizontal. they're parallel to each other, aren't they?

[proper pewter]

yea i do see that

[next cedar]

and you see how the escalator cuts through those floors like a transversal?

[proper pewter]

yes

[next cedar]

that is the relationship needed

and I believe that's all they are asking for - merely to describe the floors as parallel lines and the escalator as a transversal to all of them.

[proper pewter]

so the escalator acts as a transversal cutting through parralel lines?

[next cedar]

cutting through the floors*, which are parallel to each other

[proper pewter]

ok

alr can i show my next question?

[next cedar]

no permission needed mate. go ahead.

wish I had pin perms to pin each question as they come though, but welp

... this is an English question.

[proper pewter]

oops

[next cedar]

got a math one for the math server?

[proper pewter]

by accidemt

[next cedar]

nps

[next cedar]

right, so what have you learnt about angles formed by transversals and parallel lines?

though I will give you a spoiler: you don't need any knowledge of transversals whatsoever to find x.

you do need some knowledge to find y, however.

[proper pewter]

i think that there are diffrent types of angles

[next cedar]

not wrong. but what kinds?

specifically, there are three main kinds of angles you should look out for when dealing with transversals

can you name all three (or more/less depending on how your syllabus phrases this) and the corresponding properties?

if you can draw a diagram for each, that'd be best!

[proper pewter]

like corresponding alternate interior/exterior interior

[next cedar]

ok that is cool, but do you know the properties of each?

[proper pewter]

linear pair vertical angles consecutive interior/exterior

[next cedar]

let's first separate those to make them easier to recognize.
go in order of: corresponding angles, alternate angles, interior angles.

[proper pewter]

i think all of them are equal except linear pairs and consecutive interior which equal to 180

[next cedar]

linear pairs and vertical angles have nothing to do with transversals directly

[proper pewter]

idk but my geometry teacher said some questions will have them.

[next cedar]

oh ok I was a bit confused as to what consecutive angles are, but I see what you call consecutive, I call cointerior.

[next cedar]

yes, no doubt questions will have them.

in fact, it is used in that very question you shared

but it is not exclusive to transversals. linear pairs exist any time a straight line intersects another straight line

[proper pewter]

oh ok

[next cedar]

ok so

from the diagram you shared, can you find x directly?

[proper pewter]

i dont think so

[next cedar]

why not?

the 40 and the 5x have one very straightforward relationship

in fact, I was just talking about that specific relationship

[proper pewter]

because idk how to know what the angle relationship is from figures like that, and my mind goes blank...

[next cedar]

well, going blank should be the last thing on your priority list, but let's make it this way

is line B a straight line?

[proper pewter]

i think so

[next cedar]

well you don't have to think

they are straight

and so is C

there's not a single curve to be found here

[proper pewter]

Alright

[next cedar]

more importantly, B and C are straight lines

and they intersect each other.

what kind of angle(s) did we just say is/are created by two intersecting straight lines?

[proper pewter]

linear pair?

[next cedar]

mhm. and what are the properties of two angles in a linear pair?

[proper pewter]

supplement

[next cedar]

apply it here.

[proper pewter]

5x=40=180?

• 40 i mean

[next cedar]

there are two equal signs here because?

oh ok

yes

solve for x

[proper pewter]

28

[next cedar]

,calc 140/5

[glossy valve]

Result:

28

[next cedar]

good. one down, one more to go

now there are two ways I can immediately see to find y

but I want to hear your ideas first.

[proper pewter]

ok

[next cedar]

note that lines A and B are parallel. use that to your advantage

so what attack plan do you have in mind?

[proper pewter]

first to find the angle

[next cedar]

ok for this one, you should probably mark the angles you want to find

[proper pewter]

wdym?

[next cedar]

since there are no letter names to help describe the angles you mean

well you said find the angle

there are literally eight angles in this image, and even excluding the three marked angles there are still five of them

so find which angle?

[proper pewter]

Alternate exterior

[next cedar]

so which angle is the alternate angle here?

(that's why I told you to mark the angle you're referring to!)

[proper pewter]

40 and 3y-1 are alternate exterior

[next cedar]

no, unfortunately not

40 is not an exterior angle when considering the perspective that allows 3y-1 to be an exterior angle

[proper pewter]

then idk...

[next cedar]

but now that you mentioned that alternate exterior angles are available to you, I see three ways of doing this

the first way is to indeed use alternate exterior angles, but not using 40

you'd use the 5x angle instead

[proper pewter]

140?

[next cedar]

yes

[proper pewter]

3y-1=140

let me calculate one sec

47

3y=141

so 47

[next cedar]

,calc 141/3

[glossy valve]

Result:

47

[next cedar]

mhm

now, a brief overview of the other two methods

[next cedar]

so that you don't get hopelessly stuck as you do here

[proper pewter]

ok

[next cedar]

the second method is to use the fact that 40 and the blue angle are interior angles. find the blue angle, then 3y-1 is vertically opposite that. and boom.
the third method is to use the fact that 40 and the red angle are corresponding angles. find the red angle, then 3y-1 is a linear pair with that. same shit.

but from what I can tell now, it seems like you need more exercise on identifying corresponding, alternate (interior or exterior), and cointerior angles, as well as maybe linear pairs, in any given diagram

you should probably get a few diagrams to practice this

[proper pewter]

ok

last question

its similar to this one

[next cedar]

I immediately see two attack plans

I'll give you a hint on what types of angles are involved in both

[proper pewter]

ok

how about u give one and i find one?

[next cedar]

I'll give you the harder one to spot first then

[proper pewter]

ok

[next cedar]

the method I'm talking about involves forming a system of two equations: one with a linear pair, and the other with cointerior angles

that being said, it is the harder method. there is a way easier method. try finding that.

(or surprise me with another method)

[proper pewter]

hmm

ok

8x-10=7x

ima add 10 to each side

[next cedar]

by what property?

[proper pewter]

wdym

[next cedar]

I mean exactly what I said.

what allows you to equate them?

[proper pewter]

they are alternate enterior angles

[next cedar]

since you're not that hot in identifying angles, I want to make sure you get this right, so you can explain your steps if you are asked to.
and trust me, you do not want to not explain your steps in an exam.

[next cedar]

you mean interior?

then yes

proceed

[proper pewter]

yes my bad

so 8x=7x+10

i subtract 7x from both sides to get x=10

[next cedar]

cool. keep going

[delicate torrent]

Genius

Now find y

[next cedar]

I can handle this, btw

just informing

[proper pewter]

6y=20 and 7x are linear pairs

[next cedar]

hope you meant a + there instead of an =

[proper pewter]

yes i forget sometimes to click shift

[next cedar]

but yes, 6y+20 and 7x form a linear pair

keep going

[proper pewter]

so 6y+20 +7x=180

[delicate torrent]

Good

[proper pewter]

i think i substitute x with 10

[delicate torrent]

That's the key

Do it

[next cedar]

I'll leave this channel to 1/0 then

[proper pewter]

bye and thanks for ur help @next cedar

ok so 6y+20+70=180

6y+90=180

6y=90

division

so 15

anyone here

[delicate torrent]

😭

[delicate torrent]

Yes

[delicate torrent]

at least ping me for that 😭

[proper pewter]

alr thats all

[delicate torrent]

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[next cedar]

why should I if you were watching

[proper pewter]

thank you @next cedar and @delicate torrent

[next cedar]

I mean I said I got this, yet you insisted on jumping in

[delicate torrent]

Im hopping into a load of other channels too

[proper pewter]

.close

[full forum]

Channel closed

Closed by @proper pewter

Use .reopen if this was a mistake.

[full forum]

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[lunar creek]

Really struggling with this one and not sure where to start

[lunar creek]

Well I'm pretty sure the last one is equal to it

but idk what else

[sick karma]

remember that switching bounds negates an integral

[queen crater]

Since $f$ is odd, what is $\int_{-a}^a f(x) \dd x$?

[glossy valve]

Nel

[lunar creek]

0

[lunar creek]

0

but what does that have to do with it?

[real vector]

cassie can i try with my tools?

[lunar creek]

sure

[queen crater]

Can you split $\int_{-4}^2 f(x) \dd x$ for example?

[glossy valve]

Nel

[lunar creek]

why is ur name invisable

[real vector]

the problem is the tool i use has a watermark you have no issue with that?

[robust slate]

hi

[lunar creek]

this feels weird

you give off weird vibes

I like Nel

[real vector]

cassie

you need to believe i am a human!

[lunar creek]

sorry, what is splitting it?

[queen crater]

Like transform it into a sum of two separate terms

[lunar creek]

Ohhh

I mean

would that be f(2)dx - f(-4)dx

[queen crater]

You might want to learn a bit of LaTeX

[lunar creek]

what the fuck is LaTeX

no offense lol

[real vector]

Cassie on a scale of 1 to 10 how much would you rate my preformance!

[queen crater]

The notation to render these

[lunar creek]

commit alive not /10

[lunar creek]

Oh

Uhhh maybe eventually?

I'm going into chemical engineering so I might have to learn it eventually

idk

I have to learn coding and whatnot

[queen crater]

I'm saying that because I'm not sure what this means

[lunar creek]

Oh

it's the second fundamental theorem i think

[real vector]

Nel its okay you tried!

[lunar creek]

I'm not really sure how to do this problem

[real vector]

and thats what truley matters..

[lunar creek]

nel should we get a mod in here

Ig I can just block it

[queen crater]

Feel free

[lunar creek]

<@&268886789983436800> There's some weird bot here, @real vector

[queen crater]

Do you understand that $\int_a^c f(x) \dd x = \int_a^b f(x) \dd x + \int_b^c f(x) \dd x$ ?

[glossy valve]

Nel

[real vector]

bro im real

and u dont need to learn katex

to use me

[lunar creek]

oh

yeah

I kind of forgot

but I do now lol

[real vector]

this is a detailed explaination bruzki

..

[lunar creek]

sybau

[queen crater]

Intuitively that should be obvious if you look at areas under the curve

[lunar creek]

anyways

[real vector]

......

[rare dock]

yea sybau

[real vector]

like atleast look at it///?

[lunar creek]

Yeah I've just been doing math homework for 5 hours now so I'm a bit lost in the sauce if you get me

[real vector]

....ong

[umbral dome]

@real vector do not use AI to answer other people's help channels.

[real vector]

its not AI!

its a script

like

[lunar creek]

you should ban it fr fr

[real vector]

hmm how do i explain it

[rare dock]

it literally says ai on it

and you have an ai server tag

[real vector]

thats my ai website

where i have a different product

[lunar creek]

Bro I just want to do this problem so I can eat lunch it's 2pm

okay so

[devout valley]

And @lunar creek has made it clear enough that she doesn't want you to help her, so please just leave the channel

[real vector]

do yk what html scripts are?

[lunar creek]

cool

[umbral dome]

in either case i do not think cassie wants a canned response

[real vector]

oh okay!

[rare dock]

you’re not fooling anyone when you say that it wasn’t ai generated

[real vector]

whats a canned

[lunar creek]

How do I apply this to this problem?

Just not 100% sure

[real vector]

1. its not?
2. if its incorrect its my fault

thats insane nel back me up here bud

[queen crater]

For $\int_{-4}^2 f(x) \dd x$, given you know $\int_{-a}^a f(x) \dd x = 0$, you should be able to figure out appropriate bounds to split and simplify

[glossy valve]

Nel

[queen crater]

I'm not your bud, and you are annoying

[real vector]

1. my personality does not matter this is a professional chat to explain mathematical concepts

2. is my explaination wrong? i want to know for my understanding

[umbral dome]

@real vector this channel is for helping cassie. since she does not want your help, please leave rather than continuing to argue about what your response is or is not

[real vector]

alright

but if my explaiantion is wrong do tell me cuz i attempted it and my bad if thats the case and stuff ok byes

[lunar creek]

I'm gonna get some paper and pencil

and try and work this out with those guidelines

brb

[lunar creek]

fourth from the top is correct

and sixth from the top

and second from the bottom

Not sure what else tho and there has to be something else

[queen crater]

You said last one was too

There's one more

[lunar creek]

oh yeah

well now I'm not sure why it's the last one

[queen crater]

.

[lunar creek]

oh I know

that's why it's second to last one

okay so

I got the correct answer

but why is the third from the top correct

[queen crater]

Right let's just go through them one by one

[lunar creek]

I mean I understand why it's the last one

if it's the third one

or vice versa

just not sure why it's one of those

you get me?

Oh

can you just negative both a and b

[queen crater]

So like you're no longer sure about this?

[lunar creek]

I mean I think i get it now

You can negative the upper and lower limits

and then if you can do that

the bottom one makes sense

I got the question correct

[queen crater]

So, the first one equals the opposite, because the bounds are swapped

The second one equals the opposite again, but this time because the function is odd

Third one is the opposite of the second one, so valid

Fourth one is just the original + a part centered on 0 that equals 0

Fifth is that but bounds are swapped, so the opposite

Sixth is from -4 to 2 but with swapped bounds, so the opposite of going from -4 to -2, plus a zero part

which is the opposite of the second one, so it's valid

Seventh is the same as sixth but the bounds are swapped (or like sixth has swapped bounds and this one doesn't)

so it's the opposite of sixth, so not valid

Eigth is the opposite of seventh, so valid

And finally ninth is the opposite of second, so valid

[lunar creek]

yep

thank you

I get it now

that should be all for now, I've been doin homework for a bit over five hours so I'm gonna finish one more quick assignment then take a break

have a good rest of your day!

I fuckin hate that tilde is so close to exclamation mark istg

.close

[full forum]

Channel closed

Closed by @lunar creek

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[royal dew]

Do 9 and 10 imply that all isomorphic graphs have euler & hamiltonian circuits?

[queen crater]

Isomorphic is not a property of one object

Two objects can be isomorphic, it's a relation

An object can be isomorphic to some other reference object, that would be a property

[royal dew]

wat do 9 and 10 mean then

[queen crater]

They are invariants for graph isomorphism

If some graphs are isomorphic, they either all have the given property, or none of them have it

[royal dew]

so if one has euler circuit then the other shuld hav it too?

[queen crater]

Read the definition again: for example, if G has an Euler circuit, and G' is isomorphic to G, then G' has an Euler circuit

Just replacing "has property P" with "has an Euler circuit"

[royal dew]

well then all of the listed invariants are satisfied by these 2 graphs

does that mean that the two graphs might be isomorphic?

[hallow walrus]

Yes, they may be

[queen crater]

Yes, might

[hallow walrus]

If one satisfied some invariant and the other didn’t then you know they are not isomorphic

[royal dew]

welp

using invariants was useless

they aint isomorphs

[queen crater]

That's kind of the point

[hallow walrus]

Well the point is that if you can show two graphs are isomorphic, then you can prove one of them has some property, and that property automatically translates to the other

(If the property is invariant)

[queen crater]

An invariant is like one part of what makes an object. You can have 1000 different invariants, but if the object requires 1001 to be completely defined, then you can't say that another object with the same 1000 invariants is actually the same object

[grave elm]

kinda handwavy, but in general, 2 graphs are isomorphic if they can be drawn in a way, they look exactly the same (just with different labels at vertices). So for example this graph would be isomorhic to H (it's the same thing, just rotated)

and so would this graph be isomorphic to H

[royal dew]

yeah but using the invariants doesnt even help

G, H and these two graphs satsify all listed invariants btw

[grave elm]

They can sometimes help

[royal dew]

and yet G is not isomorphic to H but it is to your graphs

[queen crater]

Invariants can make disproving something a lot easier

[grave elm]

If 2 graphs satisfy all the invariants, then they might be isomorphic
If they dont satisfy just one, then they are 100% not isomorphic

[queen crater]

No, H is isomorphic to @grave elm 's graphs, G isn't

[royal dew]

okay well if i am given a question to determine isomorphism, should i start satisfying these invariants?

[queen crater]

You can start by looking at some easy invariants

Like a different number of vertices is easy to spot

[royal dew]

well yeah the starter ones are easier

[queen crater]

But if you can't show two graphs are not isomorphic that way, then you need to look deeper

[royal dew]

but after i check all 10 properties and then end up satisfying all of them

[grave elm]

If they are given by a picture and u dont need to prove it, then I'd say just use your eyes.

If you are supposed to prove it, then looking at invariants can help (saying "it doesnt satisfy this invariant" would be certainly lot easier than the proof they did in ur pic)

If the graph isnt given by picture, but perhaps by a more abstract definition, then invarians are probably a good idea

[queen crater]

I wouldn't bother checking these 10 properties

[royal dew]

wat would u check?

[royal dew]

wym use ur eyes 💔

wat if i get a complicated shape

[grave elm]

then use the invariants

[queen crater]

Easy ones would be: number of vertices, degree of each vertices (like, number of vertices with degree k for all k), number of connected parts (just 1 if the whole graph is connected)

[grave elm]

but e.g. here it's pretty clear that they are not the same graph, that they are not isomorphic

[royal dew]

aight wat next

[royal dew]

yea ig

[queen crater]

Next you try to find a way to move the vertices of one graph around to match the other graph

[royal dew]

💀

i forgot to use my eyes

how do i use my eyes on this one

i think they are

[queen crater]

Both graphs have two vertices of degree 3

[royal dew]

if i shift the v2

[queen crater]

Keep them in the same place

Try to move the other ones

[royal dew]

and take the v3 down

alright they are isomorphic

but how do i write this on paper

[queen crater]

Show a correspondance between vertices of both graphs

Like u4 - v5, u2 - v3, and then for example u3 - v4, u1 - v6, u5 - v1, u6 - v2

[royal dew]

oh

so f(u1) = v1

like that

[queen crater]

Yeah if you write it that way then f is your isomorphism (from G to H)

[queen crater]

.

Same thing

[royal dew]

so i start reshaping them to look the same in my head?

[queen crater]

Yes

[royal dew]

ts kinda hard 💔

[queen crater]

It might help to see that they are both made up of a triangle and a quadrilateral

[royal dew]

but the triangle is inside the quadrilateral

in the second one

[grave elm]

just move b outside

[queen crater]

Doesn't matter

[grave elm]

it doesnt really matter what is inside and outside in graphs (at least not here)

it only matters how they are connected

[queen crater]

A graph is defined by vertices and edges, specifically which vertices are connected by edges

It's not defined by how it's drawn

[royal dew]

ok then what is the correspondence

wat do u think

[queen crater]

From left to right, c -> b

Follow the edges, the rest is easy

[royal dew]

does a->b not work

[queen crater]

No, on the left a is connected to b and e, but b and e are not connected

On the right, b is connected to a and c, and these are connected

[royal dew]

i was going thru the slides and discovered some adjacency matrix method

is that useful in this case

[queen crater]

Graph isomorphism is a complex problem: https://en.wikipedia.org/wiki/Graph_isomorphism_problem

[royal dew]

but you did these questions in like 2 min

[queen crater]

Because they are small graphs

[royal dew]

wat abt these

are these ez for u too

[queen crater]

Just look at vertex degrees

v3 and v5 are degree 4

So are u2 and u4

v2 is the only other vertex that connects only to v3 and v5

u3 for the other graph

v1 and v4 are connected, so are u1 and u5

It's really not that hard once you classify the vertices into degree classes

It becomes hard when there are many vertices in the same degree class, or when there aren't enough degree classes (both things usually happen together)

[royal dew]

at how many vertices does it become hard

[queen crater]

That depends entirely on your ability and on how you define hard

Give me two graphs with 10 vertices and 30 edges and I'm giving up

[royal dew]

okay well can you teach me this method

[royal dew]

start with this

wats the first step now

[queen crater]

.

[royal dew]

look and then what

[queen crater]

Classify them

[royal dew]

aight i see u1 with degree 3

all have degree 3 aexcept u3

in graph 1

[queen crater]

Wrong

[royal dew]

ok

u2 got 4

u4 too

ok now i do the same with graph 2

ok

v1 v4 got 3

v3 v5 got 4

v2 got 2

now wat

[queen crater]

Well if the graphs are isomorphic then vertices of degree k in one must correspond to vertices of degree k in the other

[royal dew]

yea they do

the degree set is same

3,3,4,4,2

[queen crater]

Right so v1 and v4 must correspond to u1 and u5

[royal dew]

damn

[queen crater]

v3 and v5, to u2 and u4

[royal dew]

aight lemme do da mapping

[queen crater]

v2, to u3

There's only one option for v2 and u3

[queen crater]

Two options for v3,v5,u2,u4: either v3 is u2 and v5 is u4, or v3 is u4 and v5 is u2

Same for v1,v4,u1,u5

So 4 options total

At this point you can just check them all

[royal dew]

wym check all

make the graph?

[queen crater]

Yes

[royal dew]

how do i make a graph with this info

[queen crater]

Take the second graph, redraw it to look like the first one

using the vertex correspondance you found

[royal dew]

waaat

how

[queen crater]

Do I have to spoon-feed you

[royal dew]

i dond get it

how do i make it look like the first

using the corresondence

[queen crater]

I just chose v3 to be u2 and v1 to be u1

Then you actually draw the edges from the second graph

If it gives you the same graph as the first, then it's an isomorphism

This is one option out of the four, but FYI all four options work in this case

[royal dew]

yea

i did dis

is it okay to just write the correspondences

without showing any working on how u found them

it would look like u did hit and trial to get there

[queen crater]

Depends on what the question is and what your teacher expects

[royal dew]

wat do i even write in the working

[queen crater]

I would just write, for each vertex of the first graph, which vertex of the second graph it transforms into, and maybe list all the edges to show they are the same after transformation

If you fancy drawing a bit more, you can draw what the first graph looks like after moving one vertex, and repeat until you're actually drawing the second graph

So for example if you take the bottom graph here and move just v1, you can get this, which is already pretty much what you drew on paper

[royal dew]

wow

but i would go with the correspondence trick

that's quite nice

@queen crater lmao

let me do this one

@queen crater what is this supposed to mean

[queen crater]

That this one is tricky

[royal dew]

should i go for invariants

[queen crater]

You already know all vertices have the same degree, so probably not

I would try to gradually morph the graph on the right because it looks uglier

[royal dew]

there were also some euler path invariants

[royal dew]

morph mean change the shape?

[queen crater]

I mean you can do that if you want

Yes

[royal dew]

If it was on a computer then we could do that

its hard on paper

[queen crater]

Find a 5-cycle in that second graph and draw it like the outer regular pentagon of the first graph

[royal dew]

ima just satisfy all the variants and conclude that they might be isomorphs

[queen crater]

Then draw the other vertices inside

[royal dew]

i mean you arent 100% sure that they are isomorphs rigth

[queen crater]

I am

[royal dew]

how

the question says determine

not show

[queen crater]

It still expects a yes or no answer

[royal dew]

i'd say might

[queen crater]

That's neither yes nor no

[royal dew]

dang

do u think that adjacency matrix thing is gonna help

but i feel it is the same as finding the perfect correspondence

trying all combos

[queen crater]

Not sure what you're refering to

[royal dew]

you have not heard of that?

like when you draw the adjacency matrices of two graphs

and if they are the same then graphs are isomorphs

you have to manipulate the matrix in some way to make it look like the other

idk bout that in detail

[queen crater]

Sure, you can try

It's not really easier than just gradually morphing the graph by drawing

[royal dew]

alright, thanks Nel

ty

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[iron sand]

With respect to $n\in\mathbb{N}$ find using $\mathcal{Z}$-transformation sum of this finite series [1\cdot 2+2\cdot 2^2+\cdots + n\cdot 2^n]

[glossy valve]

Slowaq

[iron sand]

so far i have this but i got stuck because if i try to plug z=1 its outside of region of convergence so this doesnt hold

kappa is series whose sum im trying to find

<@&286206848099549185>

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@iron sand Has your question been resolved?

[iron sand]

<@&286206848099549185>

[gritty rose]

What's the definition of z transformation

[zenith kernel]

You are not allowed to simply directly calculate it? Have to use this transformation thing?

[full forum]

@iron sand Has your question been resolved?

[zenith kernel]

Anyway, write one 2 first row
two 2^2 second row, three 2^3 third row … then sum each columns and add up. No idea what this transformation is you are talking about

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[edgy glade]

For a ring for multiplication, does identity hold

[rapid rain]

Can you reformulate that?

[edgy glade]

Uh

In a ring,

Does the set with the multiplication operation only

Have identity property

[rapid rain]

wait so

(R,+,x)

and you're looking at (R,x)?

[edgy glade]

Yeah

Does that have identity

[rapid rain]

I mean... by definition of the ring (R,+,x)

there is a neutral element for x

so there is a neutral element in (R,x)

but it doesn't make it a group

[edgy glade]

Why not

[rapid rain]

well, many things

inverse most notably

(I gotta stop saying things hastily, it's 3AM I have to give myself more time to think)

[edgy glade]

[rapid rain]

Anyways

[edgy glade]

So it does have identity

[rapid rain]

0 doesn't have an inverse

[rapid rain]

in a ring, you always have a "0" and a "1"

so in (R,x), "1" is your identity I would presume

[edgy glade]

I see

And if this 1 isn't there

[rapid rain]

1 is in R, because it's a ring

[edgy glade]

What about like {...,-2,0,2,4...}

Under standard addition and multiplication

[rapid rain]

wait, are we talking about the same rings

[edgy glade]

?

Theres different rings ?

[rapid rain]

does ring for you not mean unitary ring

[edgy glade]

Never heard the term unitary ring before

[rapid rain]

ok, what's your definition for a ring

[edgy glade]

That's what I'm trying to figure

Some places it's mentioned that identity exists

Some places not

[rapid rain]

well Ig it depends on the places then

[edgy glade]

[rapid rain]

some sources are gonna say a ring is (R,+,x) where (R,+) is a commutative group and x is associative and distributive with +

[edgy glade]

Okay

[rapid rain]

some other sources are gonna add the condition that there's a "1"

so a neutral for x

[rapid rain]

and those first sources are gonna say "but when there is a 1, we call it a unitary ring!!!!!"

so very tough luck. Use the definitions you have in your own book/lectures

[edgy glade]

I may be wrong, so check this: So (R,+) is a commutative group
And (R, x) is a monoid or semigroup depending on source

[rapid rain]

yep,

[edgy glade]

And what about (F,x)

[rapid rain]

"(R,x) is a monoid" means "there's a 1"

"(R,x) is a semigroup" means "there's not necessarily a 1"

[rapid rain]

F as in field?

[edgy glade]

(F \{0} , x) is a commutative group

[edgy glade]

Yeah

[rapid rain]

that is true

[edgy glade]

Yeah so what about (F, x)?

[rapid rain]

(F,x) is only a monoid (there's a 1 because F is a field, but no inverse)

[edgy glade]

Okay

[rapid rain]

but (F \ {0}, x) is a group

[edgy glade]

Hmmm

Monoids can be commutative right

[rapid rain]

(F,x) in this case is a commutative monoid

[edgy glade]

(F, x) is commutative methinks?

Okii

Thanks a lot

❤️

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[half lava]

hi

[half lava]

so i was learning abput AP

and there is a property where the sum of terms equidistant from starting and end are equal

like 1,2,3,4,5,6,7 be an ap so it says 7+1=^=2=5+3.....

now my doubt is like is it applicable to 3 terms also like let a1,a2,a3,a4.....an be ap is a1+a7+a8=a9+a5+a2