help-19

[mystic saffron]

and

[brittle plinth]

what kind of triangle is AED?

[mystic saffron]

isco

[brittle plinth]

?

[mystic saffron]

nvm'

[brittle plinth]

actually I should mention that BCDE is a rectangle

so BE = CD = 4

[mystic saffron]

ok

[warped glacier]

smart

[odd edge]

@mystic saffron Has your question been resolved?

[brittle plinth]

haiyaa too long also op already solved it that way also it is wrong

[odd edge]

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[quasi shard]

so I know that the text is implying that integral s <= I <= integral T. but all it is saying is that I must be between both integrals. if integral s = 3 and integral t = 7. you wouldn't say that 7 or 3 is between 7 or 3.

[robust lagoon]

It just seems to be an elementary bound, you will take the supremum over all such s and the infimum over all such t, and if the function is nice enough this will end up being the definition of the integral

[quasi shard]

two things what is an elementary bound? And I understand that we show that s has a sup S and t has an inf T and if supS = supT then if the lower and upper integral are equal then I exists. but for all this we need the inequality integral s <= I <= integral T

but this leads me back into how we can say a number falls between two bounds if it is the bounds

[robust lagoon]

I just mean bound. For the moment we don't have any notion of what the integral of f should be, but we know that s(x) <= f(x) <= t(x) for all x. In our natural rectangle sense of area we can see that the integral of s <= the integral of t. For the definition of the integral of f to make sense, we require it to be between these values. I think it's less that we can say that, and more that we want it to be true when we make our definition

[quasi shard]

but in his explanation we use the comparison theorem which explicitly states that integral s < integral t . so s(x) < t(x) and that I is between them. He never states anything about I being equal to.

[robust lagoon]

does your book define the integral? and is this a measure theory course or just basic integration? because there is a more general Lebesgue integral

[quasi shard]

this is just basic integration and what do you mean by define integration. because first he says there must exist an I between integral s and integral t. which we call integral f(x). then he says that a(q) = I where q is the ordinate set for f(x). then he goes on to say that if the sup S = inf T then I exists. And finally he goes on to prove that supS = inf T. But the thing that is getting me hung up is the equals part in the picture I sent

[low locust]

there is no equals part in the picture you sent

[robust lagoon]

When you say 'there must exist an I' I don't understand. Also are you assuming that f is continuous or piecewise continuous because in general supS =/= inf T?

[low locust]

if sup S = inf T, then we define integral f = sup S = inf T

if not, then not

[quasi shard]

well in the photo I sent he starts off by saying that we will apply this to general functions. Later on when he shows how to prove that the integral will exist he says that we are only proving this for bounded monotonic functions.

[quasi shard]

yes I understand that but don't we first need this inequality integral s <= I <= integral T

[robust lagoon]

okay yes these are always integrable

[low locust]

that inequality doesnt appear anywhere

only integral s < integral t

dont make stuff up that doesnt exist

[robust lagoon]

we are trying to define I, it doesn't a priori make sense to talk about it

[amber veldt]

the sup of a set or sequence need not be in the set or sequence

maybe that's why youre confused?

[quasi shard]

but this is before we get to the sup s part

[amber veldt]

then not sure what youre asking

[quasi shard]

he is just saying that using the comparison theorem there must exist an I that satisfies the theorem

[low locust]

he is not saying that

[amber veldt]

whats the comparison theorem?

[robust lagoon]

can you show us this part of the text?

[robust lagoon]

here there is no mention of I right?

[quasi shard]

that was just the definition of the comparision theorem

[robust lagoon]

Ah okay, see the start of the sentence, 'If the integral of f...' this is an explanation of what we would like for our integral to make sense

It is motivation

[quasi shard]

so he is just saying how we would like to define our integral

[robust lagoon]

He is saying 'this is what we would like to be true for it to be a good definition'

[quasi shard]

so if we go on with that chain of thought we either get integral s < I < integral t or integral s <= I <= integral t. we know the latter to be true but I don't see how I can get that from the text when he just says I is between these two integrals

[amber veldt]

maybe you want a book that gets to the rigor more quickly, spends less time on intuition. most students arent like that, but I was like that...

[low locust]

what even is the problem?

the equality signs? <= instead of < ?

[quasi shard]

yes

[low locust]

but why

<= is just a weaker version of <

[amber veldt]

between doesnt mean strictly between

[quasi shard]

if something is between two bounds I thought it cannot be equal to those bounds

[low locust]

its just conversational english

[amber veldt]

that would be the phrase strictly between

[quasi shard]

let say we have bounds 3 and 7

[low locust]

and like, who cares if the equality is ever satisfied

[quasi shard]

how could 3 be between them

[amber veldt]

3 is between 3 and 7

[quasi shard]

but 3 is the bound

[robust lagoon]

between often includes equality

you would say strictly between otherwise

to emphasise

[quasi shard]

ok

[low locust]

if 3<x<7 is true, then 3<=x <=7 is still true

thats just a weaker version of the statement

[amber veldt]

but the fact that you're so hung up on the wording MAY suggest a deeper issue you have with the text

or it may not

but examine it

[quasi shard]

ok

[amber veldt]

whats your background?

[quasi shard]

no background

[low locust]

its also not like you ever will actually have integral s = integral f (if f is e.g. continuous)

[amber veldt]

no analysis knowledge?

[quasi shard]

well I took calculus last year and wanted to learn it from a rigorous standpoint. And I have no "rigorous math background"

[amber veldt]

i didnt mean, like, a degree

just what you know already

[quasi shard]

no I understand the only background I have is just high school math

[amber veldt]

maybe start with a simpler book like

Real Analysis: A First Course by Gordon

[robust lagoon]

I like Abbott Understanding Analysis

[amber veldt]

Abbott has a good reputation, never read it

[robust lagoon]

I used for self study if that's what you're doing

[quasi shard]

my whole objective was too learn calculus in a more rigorous manner and I feel like I have already invested too much time to drop the book now

[amber veldt]

sunk cost fallacy

you can go back to the book later

or not

[quasi shard]

maybe so. but I feel like that from when I have started the book compared to now I feel like my general skills have improved a good amount

[amber veldt]

an oddity of analysis is that the Lebesgue integral is simpler than the Riemann integral for rigor

[amber veldt]

thats great

[quasi shard]

I think another issue lies in the fact that before this I had no experience with any sort of proofs

[robust lagoon]

you should make learning proof writing/logic a priority before anything else you want to do in rigourous mathematics

I learned it in conjunction with an introductory real analysis course

[quasi shard]

well thats what the introduction was about

[amber veldt]

i learned it in lin alg

[quasi shard]

because he made us prove theorems for field axioms and introduced proof by induction

[amber veldt]

mostly

[quasi shard]

so I feel like I have learned some proof skills

[robust lagoon]

the main thing you should be comfortable with is given a statement 'what do I actually need to do here'

to prove it is true

or disprove etc

[amber veldt]

beginning of Knuth's TAOCP is good for that actually

before he gets to computer stuff

thats discrete math,but still

[quasi shard]

for proofs?

[amber veldt]

yeah

[quasi shard]

ok will check it thanks

[odd edge]

@quasi shard Has your question been resolved?

[odd edge]

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[low aurora]

Hello, I need help with statistics.

Addt'l Instructions:
Identify the following;
a.) Variable
b.) Type of data (if quantitative, classify as discrete or continuous)
c.) Level of measurement
d.) Range
e.) Class intervals
f.) Interval width

[low aurora]

I'm confused on how I should state the variable

[amber veldt]

its difficult because it kind of straddles the border of quantitative and qualitative

how would you describe the possible values the RV can take? lets start with that @low aurora

[odd edge]

@low aurora Has your question been resolved?

[odd edge]

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[swift steeple]

May I have some help with the questions attached?

The first image is a subproblem that came about in my and my friends' attempts to try to solve the related problem in the second image. My belief is that there are infinitely many such triples, but finding a proof for whatever the case may be has proven difficult. I have been able to explicitly find triples that work i.e (2,8,5), (3,3,7) (and of course any shuffling of these triples works) but that has only come about by being more specific, which doesn't answer the most general case.

[odd edge]

@swift steeple Has your question been resolved?

[hidden token]

The terms in the r.h.s remind me of the relation between the coefficients and roots of cubic polynomials

[swift steeple]

I agree, but the problem is that we can set up our -b/a, c/a and -d/a as usual, and make them all positive integers without the individual roots being positive integers... eg x^3+x-1

Or at least, that felt like a problem to me

[odd edge]

@swift steeple Has your question been resolved?

[odd edge]

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[odd edge]

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[ebon oar]

why?

[narrow crypt]

cuz null vector doesnt contribute to the span at all

wait

lemme

[ebon oar]

But why it is LD?

[nocturne brook]

because 1 * 0 = 0

[ebon oar]

how can we sure at it?

[narrow crypt]

nah wait

[nocturne brook]

linear independence demands that whenever a sum of scaled vectors is the null vector, the scalar terms themselves are 0

but in your field, you have the scalar 1, and 1 * 0 = 0, but 1 is not 0

[ebon oar]

i did not get it properly

[narrow crypt]

a_1v_1 + a_2v_2 + a_3v_3 + ... a_nv_n = 0

there should only be the trivial soln

[nocturne brook]

but in the set of real numbers, you have the real number 1, and 1 * [the null vector] = [the null vector], but 1 is not equal to 0

[narrow crypt]

if they are linearly independent

[ebon oar]

i guess you are saying that

a1v1+a2v2+a3v3+....=0 so here suppose v1 is null vector

and i am bound that a1 is not 0

but the multiplication will make it zero

so yeahhhhhhhhhhhhhhhhhhhhhhh

hm

thanks

the proof is like HELL

https://tenor.com/view/fire-eyes-elmo-hell-gif-13483060728962175532

[nocturne brook]

it's really not so bad

[ebon oar]

i meant they forced it

LOL

hmm but yeah i loved the logic

maths proofs are funny

https://tenor.com/view/so-good-wink-smile-oh-yeah-its-true-gif-24163876

[narrow crypt]

i dont think they are..

[nocturne brook]

existence-based ones can be unintuitive in general, sure

[ebon oar]

Both of you looks master degree?

[narrow crypt]

no

[nocturne brook]

nah

[ebon oar]

ahh shhh

my bad

high school>

[narrow crypt]

look at my mistakes with linear alg

when tryna explain it to u 😭

[nocturne brook]

I graduated and have a bachelors in math, that's it thus far

[ebon oar]

so you studied only maths?

along with physics, chem?

[nocturne brook]

physics, or maybe just some choice of science, was mandatory for the degree

[ebon oar]

scalers are real numbers?

or they can be complex"

[nocturne brook]

but my last four semesters consisted solely of math courses, plus some random elective for elective's / variety's sake

[narrow crypt]

they are the elements of the designated field

[nocturne brook]

your vector space can choose R, C, or any other "field" as its set of scalars

[narrow crypt]

can also be 0,1 in mod 2 field

[ebon oar]

nice to gaining knowledge

i like to ask people

discuss things

[nocturne brook]

in early linear algebra, you don't define a field, and instead just assume all vector spaces choose the reals as its field

[narrow crypt]

@nocturne brook is this right? 😭

[ebon oar]

it sticks me to topic

[nocturne brook]

maybe the complex numbers if you need them

[ebon oar]

addtion and multiplication both

thanks

.close

[odd edge]

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[karmic stirrup]

how would i do this assume no calc allowed

[karmic stirrup]

i dont know where to start

[tribal ingot]

have you learned l'hopitals rule

[narrow crypt]

sinx/x -> 1

[karmic stirrup]

i know that

[narrow crypt]

it wokrs but plz dont

[tribal ingot]

whoops

you can rewrite

sinx/sin2x

[narrow crypt]

not beautiful

[tribal ingot]

think about what k said

[lean yew]

can't you rewrite sin(2x) as something else

[narrow crypt]

no need for trig mani

[tribal ingot]

l'hopital was my saving grace

[tacit wasp]

Then use it

[karmic stirrup]

how

[narrow crypt]

definitely a tool in exam rooms for speed running 🙏

oh do u know that ap allows desmos now?

[tribal ingot]

if you rewrite sinx as sinx/x

[tacit wasp]

In this case it's pretty quick and useful, I'd say

[tribal ingot]

then what can you rewrite 1/sin2x as so that you end up with the original expression

[wooden python]

sin(2x) = 2sin(x)cos(x) kills this problem in particular

[tribal ingot]

wtf

[narrow crypt]

sure

[tribal ingot]

deadass

[tacit wasp]

By dividing numerator and denominator by 2x

[wooden python]

though it isn't generalizable

[tribal ingot]

like starting next year?

[narrow crypt]

deadass, i sat the ap calc ab this year.

[karmic stirrup]

that doesnt make sense

[tribal ingot]

oh

say less

[wooden python]

do you want easy but ad-hoc, or do you want harder but generalizable

[tribal ingot]

amazing news

[tacit wasp]

Lol, then what way would you do it?? Let's listen...

[karmic stirrup]

harder

[wooden python]

ok

then let me try to explain the general idea that hopefully you can get mileage out of

namely that you can go on to apply it somewhere other than this specific limit question

[karmic stirrup]

great

[wooden python]

so to start off with

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

[clever fjord]

Ann

[wooden python]

you know this, right?

[karmic stirrup]

yea

[wooden python]

this is the cornerstone

so one useful corollary of this

[bold elbow]

Can someone help me with this question?-How many ways can you pick three different numbers from 1 to 1999 so that any two numbers in the chosen set are at least 9 apart?

[tribal ingot]

!occupied

[odd edge]

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).

[wooden python]

!occupied

[odd edge]

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).

[wooden python]

is that given a constant $k$, you have $\lim_{x \to 0} \frac{\sin(kx)}{x} = k$

[clever fjord]

Ann

[wooden python]

you can also view this as $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$ [same fundamental limit as before] and then multiplying both sides by $k$

[karmic stirrup]

ig that makes sense

[clever fjord]

Ann

[wooden python]

yeah ok

[tribal ingot]

oh shit i see where this is going

wait this is beautiful

i didnt think of it like this

[mystic nova]

Indeed

[wooden python]

you can also say $\lim_{x \to 0} \frac{x}{\sin(x)} = 1$

[clever fjord]

Ann

[wooden python]

do you see why?

this is basic/simple but i wanna take it slow here

[tribal ingot]

[wooden python]

@karmic stirrup

[karmic stirrup]

is it becuase

both sides have to be the same for it to be 1

so u can flop it?

flip

[wooden python]

mmm no you're kinda overthinking it

it's just that lim 1/f(x) = 1/(lim f(x)) assuming the limit of f(x) exists and isnt 0

and the reciprocal of 1 is still 1

that make sense?

[karmic stirrup]

yea

[wooden python]

lim[x->0] x/sin(x) is the same as the cornerstone limit of sin(x)/x just reciprocal'd

alright.

now when it comes to $\lim_{x \to 0} \frac{\sin(x)}{\sin(2x)}$

[clever fjord]

Ann

[wooden python]

as of yet, we don't have any sin(x)/x that we could squeeze out of this

so you'll see that it looks like we can't apply the sin(x)/x limit yet

yes?

[karmic stirrup]

yeah

[wooden python]

right

but heres the big reveal

we can make it happen.

[karmic stirrup]

O:

[wooden python]

and the way to make it happen is:

$\frac{\sin(x)}{\sin(2x)} = \frac{\sin(x) \cdot x}{\sin(2x)\cdot x} = \frac{\sin(x)}{x} \cdot \frac{x}{\sin(2x)}$

[clever fjord]

Ann

[narrow crypt]

"squeeze out of this" (reference?)🔥

[karmic stirrup]

wait so is the answer 2?

becuase 1 * 2

[wooden python]

no, nothing about squeeze thm @narrow crypt

[wooden python]

x/sin(2x) will go to 1/2, not 2.

[narrow crypt]

):

[karmic stirrup]

oh

[wooden python]

the answer will be 1/2

but hopefully ive shown how you can like

[tribal ingot]

wow this is pretty nice

[wooden python]

basically kinda introduce the missing x factor for sin(x)/x in this way

[tribal ingot]

more beautiful than how i wouldve used l'hopitals here

:kek:

[wooden python]

lhop is completely soulless

[lean yew]

TIL, taking notes

[wooden python]

and also unnecessary heavy machinery

[karmic stirrup]

i like your solution cuz it doesnt use trig identiys i wouldent know

[wooden python]

watch and learn, y'all

[leaden yarrow]

Is the answer 1/2?

[tribal ingot]

anytime i see a trig limit in an indeterminate form i pull out my good pal l'hopital

[karmic stirrup]

how do u use l'hoptial

[tacit wasp]

But also knowing the identity sin(2x) = 2sinxcosx is very useful, not only for limits
@karmic stirrup (just a side note)

[wooden python]

gonna second alberto here

knowing trig identities is pretty useful

but also, the following intuition is useful:

for small values of x, sin(x) ≈ x

[tribal ingot]

if you have a limit in an indeterminate form you can differentiate the numerator and denominator and take that limit

wait lemme write it out

[wooden python]

@karmic stirrup have you seen the "engineers think sin(x)=x" meme?

[lean yew]

should probably add for radians?

[karmic stirrup]

not realy

[wooden python]

yes obviously radians!

in calculus you need to put your big girl pants on and use radians all the time

[tacit wasp]

By checking if you have 0/0 or ∞/∞ form and applying the theorem

[narrow crypt]

i love sin(x) ≈ x ≈ tan(x)

[narrow crypt]

the ol' reliable

[tribal ingot]

@karmic stirrup

this is my lord and savior

[wooden python]

important: lim f'(x)/g'(x) has to exist

[lean yew]

i really need to learn to know when to introduce an x/x factor in all situations that call for it

[karmic stirrup]

does it work all the time?

[wooden python]

no, you need to be very very careful checking for applicability

[narrow crypt]

1. indeterminate
2. the derivative exists

[tacit wasp]

There are few cases where it doesn't. Let me send you a video I've watched some time ago

[tribal ingot]

in all honesty though what Ann said is right you should learn the manual way instead of relying on l'hopital

[tacit wasp]

https://youtu.be/KZ90b3n3NVI?feature=shared

[karmic stirrup]

it looks quite simple to do considering most trig functions derivatives loop in a circle

[tribal ingot]

oh dr peyam i love that guy

[leaden yarrow]

sin(2x)=2sin(x)cos(x)
sin(x)/2sin(x)cos(x) = 1/ 2 cos(x)
now take the limit as x-->0
lim x tends to 0 (1/2cos(x)) = 1/2cos(0) = 1/2(1) = 1/2

[tribal ingot]

yea thats what a lot of people use it for

well if you're in a time crunch l'hopital could be easier and better

but in general

knowing how to normally do the limit is the way to go

[wooden python]

this is correct but late to the punch

[karmic stirrup]

yea im in a time crunch got my ap calc final in 3 days and i dont know a lick of it

[wooden python]

also @karmic stirrup l'hôpital is slow-acting brain poison

[narrow crypt]

i mean they work just wrong application, no?

[karmic stirrup]

my favorite kinda poison

[tribal ingot]

I can agree

[wooden python]

there's like... a select few situations when l'hop is appropriate

but generally

99% of the time there is something better

like the x/x trick

[mystic nova]

Too overpower ikr

[tribal ingot]

like highschool course?

[lean yew]

overusing l'hopital sends you to the hospital i suppose

[karmic stirrup]

yea im testing out of ap calc ab monday

[tribal ingot]

ohh

goodluck

[karmic stirrup]

ill need it

[lean yew]

glgl

[narrow crypt]

be careful tho

if u misuse it, shit happens

as always

[karmic stirrup]

also

for a deravtive

[tribal ingot]

just know your trig limits thats all i can say

[karmic stirrup]

yea

so i could use le hospital on something like this right

x->3

[tribal ingot]

yes you could

[wooden python]

not recommended

[tribal ingot]

or you can rationalize and solve it normally

[karmic stirrup]

are there situations i wouldent be able to use l'hoptial and would be forced to use reg trig rules

[tacit wasp]

There's no such rule for this, I'd say

[tribal ingot]

you can technically use l'hopitals as long as the limit is indeterminate and the limit with the derivative of the numerator and denominator exists

[narrow crypt]

[ \lim_{x \to \infty} \frac{e^x}{x^3} ]

[clever fjord]

k

[narrow crypt]

l'hopital gives u the algebraic thingy

but commonsense gives u it diverges

[karmic stirrup]

if the derivative doesnt work could i just keep taking the deravtives untill it does work

[tribal ingot]

yes you could

butttt i would strongly recommend that unless you need to solve something on a time crunch don't use l'hopitals

[karmic stirrup]

ok yes master ):

[tribal ingot]

and don't think of l'hopitals as your first option when you see an indeterminate limit

because you will see a lot of those in ap exams whenever limits show up

[narrow crypt]

(nvm expansion also works)

[karmic stirrup]

quick question how would i take deravitive of this

[tribal ingot]

differentiate each term separately

[narrow crypt]

power rule

• chain rule (kinda)

[karmic stirrup]

well 2- would turn into nothing right?

then

[tribal ingot]

no

2 would turn into nothing

[karmic stirrup]

oh

oh ok

[tribal ingot]

the - sign applies to the root(x+1)

[karmic stirrup]

how do u do the power rule for (x+1)^1/2

[tribal ingot]

well you use the chain rule

d/dx f(g(x)) = f'(g(x)) g'(x)

[tacit wasp]

Well, if you don't either know derivatives, de l'Hôpital for you is FORBIDDEN...

[karmic stirrup]

i know deravtives 1/10 of the time

[wooden python]

$\dv{x} (x^p) = px^{p-1}$ works even when $p=\frac12$ or any other non-integer

[clever fjord]

Ann

[tacit wasp]

Lol

[tribal ingot]

i would suggest u brush up

[wooden python]

then lhop is still very much off the table for you

[tribal ingot]

considering you have your exam in a few days

there are a lot of derivatives

[tacit wasp]

And you think you can apply l'Hôpital? How? 😅

[tribal ingot]

for derivatives just learn your rules

thats lowkey all you need to know

[karmic stirrup]

yeah

[tribal ingot]

pretty much anything can be differentiated by some rule that exists

unlike integrals 😔

[tacit wasp]

$$\lim_{x \to +\infty} \ \frac{x\ e^x}{e^{2x}}$ $

[karmic stirrup]

i wouldent know where to start i havent deal with limits to infity yet

[clever fjord]

Alberto Z.
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[narrow crypt]

domination?

[tacit wasp]

This is ... very weird

[tribal ingot]

try expanding e^(2x)

[tribal ingot]

wait what

[tacit wasp]

@karmic stirrup Do you think l'Hôpital works and/or it is useful?

[karmic stirrup]

ive worked with em ofc but whenever i compute em i just used a calculator

[tribal ingot]

[lean yew]

oh dear

[karmic stirrup]

i bet i could im built dif

[tribal ingot]

if you're weak with limits and derivatives testing out of ap calc in 3 days might not be the move

[narrow crypt]

quiz

or test?

[karmic stirrup]

like test to test out of the class

[lean yew]

idgi. what do you mean "testing out" of ap calc?

[karmic stirrup]

like skipping a year in math

[lean yew]

oh so you are taking a test to jump years?

[karmic stirrup]

yeah

[lean yew]

um

[narrow crypt]

ap can let u jump years??

[karmic stirrup]

no

[late sinew]

Why r u jumping without knowing the content for the jump test

[karmic stirrup]

well i was planning on studying

for the past 6 months

then i lost my minecraft account

and got in depresion

[narrow crypt]

😭

[tribal ingot]

[late sinew]

[karmic stirrup]

and now its sorta late

[lean yew]

then uh...

[tribal ingot]

you can test out of the course in some schools

[lean yew]

all the more reason to NOT jump years

don't you think?

[tribal ingot]

yea wait i would highly recommend not jumping

considering you are pretty weak with derivatives and limits

[quasi sparrow]

jump backwards

[lean yew]

you're already struggling here, and you wanna jump ahead?

[tribal ingot]

which are really foundational

[karmic stirrup]

yeah

[lean yew]

you'd be skipping so much stuff

[karmic stirrup]

ik

[lean yew]

why?

[karmic stirrup]

i like harder classes

even if im dumb in them

[lean yew]

but you can't deal with the basics?

[tacit wasp]

Lol

[late sinew]

Thats illogical

[tribal ingot]

well you needa know what you're doing to understand the "harder classes"

i'd suggest you just take your ap class

[lean yew]

wait.

[tribal ingot]

you have a headstart anyway

[karmic stirrup]

nah for the most part i just fail the harder classes but i still end up learning more so i take em anyway

[tribal ingot]

😭

[lean yew]

are you going in harder classes just for the prestige of being in them?

[tribal ingot]

thats not a good strat

just take the ap class

you have a headstart anyway because you know some calculus

[lean yew]

no, see

[tribal ingot]

bits and pieces

[lean yew]

yes, you learn from classes harder than your current standard

BUT

if the classes are so hard you understand those as well as you understand arabic

which is to say, you don't

[narrow crypt]

u kinda need to pass classes for colleges..

[karmic stirrup]

i dont plan to go to college

[lean yew]

then what are the benefits of going to those harder classes?

[narrow crypt]

ok

[karmic stirrup]

less stupid people and i can learn more and they are often easier cuz its more test based instead of assigements

[lean yew]

i... don't get the logic here

maybe because i don't take AP

[karmic stirrup]

well in normal classes tests are worth 50% of my grade so its impossible to pass cuz i dont do assignenemnts but in ap its like 70% so i can pass em

[lean yew]

"i don't do assignments"

personal opinion, but i think that's gonna hurt you more in the working world, good sir

[karmic stirrup]

i relize

[lean yew]

if you realize all this and still choose this route, i suppose i can only wish you luck

you have been warned

[karmic stirrup]

well yeah i appreciate it dont worry i have a plan in life alr

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[quasi shard]

can someone help me understand why absolute value is introduced here

[quasi shard]

what I understand from this image is that if we have proven that upper I = lower I. Then I exists and it equals upper I and lower I. We also said that integral of f(x) = I. So we can replace Upper I with I and lower I with integral f(x). To get 0 <= I - integral f(x) <= c/n

[odd edge]

@quasi shard Has your question been resolved?

[odd edge]

@quasi shard Has your question been resolved?

[orchid torrent]

The idea is that because the two numbers are both in the interval, the distance between them (hence absolute value) cannot be more than the length of the interval (which is C/n)

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[pulsar tiger]

How to solve this?

|1/f(x) - 1/g(x)| = |g(x) - f(x)|/|f(x)g(x)|

[pulsar tiger]

Relevant theorem and definitions used:

[cerulean flare]

$\left | \frac{1}{f(x)}-\frac{1}{f(y)}\right |=\frac{\left | f(x)-f(y)\right |}{|f(x)||f(y)|}<\frac{\epsilon }{k^2} =\epsilon '$

[clever fjord]

TargetVN

[cerulean flare]

numerator is smaller than epsilon, according to the definition, and denominator >= k^2 because the problem gives |f(x)| >= k

then set epsilon/k^2 to another arbitrarily small variable and done

[pulsar tiger]

I still don't see why K^2

[quasi sparrow]

k * k = k^2

[pulsar tiger]

Why is there two k's

[quasi sparrow]

How many f's are there in the denominator

[pulsar tiger]

two

But they are different f's

f(x) and f(y)

[quasi sparrow]

Replace x with y

It's just a dummy variable

[pulsar tiger]

So, why have 1/f(x) - anything?

[quasi sparrow]

Wut

[pulsar tiger]

?

[narrow crypt]

This is literally from the definition of uniform continuity😭

I havent done real anal yet

But it seems to follow a similar trend to ur generic epsilon-delta proof

[narrow crypt]

|f(x)| and |f(y)| are both greater than k are both implied from the given inequality

[narrow crypt]

Epsilon here is taken from the epsilon used to prove that |f(x) - f(y)|

Whatever it is, we don’t know

But we know that it exists

Which is sufficient

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@pulsar tiger Has your question been resolved?

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[honest rivet]

Would i be able to receive some help interpreting interaction plots?
I am trying to identify some but I am stumped on the significant variables for th ebottom 3. I think I am just genuinely confused what variable A affects and what variable B affects

[honest rivet]

This is how my prof explains it but I am having a hard time understanding it

To interpret an interaction plot:

the x-axis variable influences the slope of a line connecting the average y-axis value across both levels of the x-axis variable (disregard the line-type variable)
the line-type variable influences the average vertical distance between the lines (disregard the x-axis variable)
the interaction effect is assessed by comparing the slope among the lines (i.e. are they parallel or not).

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[frozen barn]

Why is it when you reflect a point (a,b) to the line y=x
The image is (b,a)

[frozen barn]

Pls help

[lean yew]

i'm thinking of how best to explain to you in visual form

[wooden python]

if a ≠ b,

• the line joining (a,b) with (b,a) always has slope -1 and thus is perpendicular to y=x
• and these two points are at the same distance from the line y=x

[frozen barn]

Source?

[wooden python]

you can prove this yourself

for the second point you will need to know how to find the distance from a point to a line

[frozen barn]

the 2nd point i obviously know

[wooden python]

i cannot give you an academic source on this, sorry.

do you wish to dismiss me as making an invalid and unsubstantiated claim?

i can disappear if so.

[frozen barn]

Nah don't do that

The first point you made i also understand,

Respectfully Seia you've been typing for 15 minutes

[lean yew]

so y = x is the line where all points on it are of the form (a, b), a = b
but if a != b, then to reflect it along the line y = x, we first have to find another point (c, c) on the line y = x

you can choose this point in one of two ways - moving along horizontally until you hit the line, or moving vertically until you hit the line
once done, you move perpendicularly to the first direction chosen the same number of spaces you moved earlier.

so suppose you chose to move horizontally from (a, b) to (c, c) on the line. but on the line y = x, we know that the x and y coordinates are equal.
since horizontal movement only changes the x-coordinate, we can then reason that the point we moved to must be (b, b) (changing the x-coordinate).
the number of steps moved, n must therefore be b - a.
now, we are moving vertically from the line the same number of steps |b - a|. since we are moving vertically, only the y-coordinate changes, and the new y-coordinate is b - (b - a) = b - b + a = a
so the new point is at (b, a), and thus the coordinates have swapped

the image shows what happens if you move vertically first

[lean yew]

my bad, i was trying to make sure i could be understood the best i can

i had to rewrite this two times

sorry

i hope i made sense

[frozen barn]

Yeahhh im sorry but how is moving horizontally until you hit the line and then moving vertically the same amount of distance from when you moved horizontally from the object to the line(mirror) will end you up in the exact point where the image of the reflection is
May you explain

[lean yew]

good question. i know that you're asking about the definition of reflection, but let me think of some way to explain to you

may need some time, apologies in advance

[frozen barn]

Thank you

[frozen barn]

Take as much time as you need

[lean yew]

ok, so let me probe for understanding a little

what do you know about reflection?

how would you define reflection?

[frozen barn]

A transformation where the image and the object has the same distance to the mirror
With the line connecting them being perpendicular to the mirror

[lean yew]

good definition

so that means the object and image must have the same perpendicular distance to the mirror

[meager juniper]

Easy mode, the reflection across the line y=x induces a linear transformation that sends the point (0,1) to (1,0) and vice versa. This linear transformation can be represented by the matrix A = {{0,1},{1,0}}. If we take a point v = (a,b) and treat it as a column vector we can run the linear transformation forward by multiplying the matrix on the left Av.

This gives us the point (b,a) back.

[frozen barn]

Yes

[lean yew]

now, say you have an object

like this

you know the image must lie on a line intersecting the object as well

but there are infinite ways to draw a line through A with only this constraint

but we have another constraint - the line must be perpendicular to y = x

and we know that the gradients of perpendicular lines are connected by
m1m2 = -1

so whatever line we need to draw here, it must take the form y = -x + c

[frozen barn]

Ok

[lean yew]

if we take the point in the image as an example

A is (3, 5)

so solving for c:
5 = -3 + c
c = 8

our new object-image line is thus y = -x + 8

agreed?

[frozen barn]

Yes

[lean yew]

now we have our line

the next thing is to find where the image is on this line

[frozen barn]

How?

[lean yew]

to do that, we need to find out how far A is from the black line

then, we need to find the point on the other side of the black line that is the same distance away

[frozen barn]

Yes correct

[lean yew]

so zooming in a little

if you don't wanna use the distance formula, let's count steps

so A is one vertical and one horizontal away from the black line. agreed?

[frozen barn]

Its 1 × square root 2

[frozen barn]

Yes

[lean yew]

yeah if you take the straight line distance, but visually, it's (1, 1) away, yeah?

[frozen barn]

True

[lean yew]

so knowing that, and knowing that the image is the same distance away

we can go another 1 right and 1 down

[frozen barn]

We can find where the imageus

Is*

[frozen barn]

From the grey dot right

[lean yew]

however, notice what we have done here

[frozen barn]

Yes i do

[lean yew]

we have went two spaces down and two spaces right

in total*

[frozen barn]

Correct

Yes from the objecf to image

[lean yew]

so that means, we take the y-coordinate and subtract 2 (moving down)

and we take the x-coordinate, and add 2 (moving right)

[frozen barn]

Yes

[lean yew]

so we have essentially taken some of the y-coordinate and given it to the x-coordinate

[frozen barn]

Yes you're correct

[lean yew]

but the final important thing that makes this possible is the coordinate of the mirror point

that mirror point has a coordinate of (x + (n/2), y - (n/2)), where n is the number of steps made
/2 because the mirror point is half the journey

[frozen barn]

What is a mirror point?

[lean yew]

that grey dot

in the image

[frozen barn]

Ok

[lean yew]

the one circled in green

[lean yew]

but here's the thing about that point

it lies on y = x

that means that the x and y-coordinates at this point are equal

so x + (n/2) = y - (n/2)

[frozen barn]

Yes that's true

x + n = y?

[lean yew]

so if we now also take the second half of our journey

[lean yew]

yes!

this tells us that no matter how we chose to move

we will have reduced one coordinate by some number of steps

and increased the other coordinate by the same number of steps

[frozen barn]

y-n = x
So from the first journey from (x,y) going horizontal for n x+n=y you'd end up in (y,y) and then going down from (y,y) by n you'd end up in (y,x) which is the image?

[lean yew]

yes!

excellent reasoning

let's use a number example

suppose our object is (4, 7)

we have two ways of moving for our first half

we can either move right three spaces to reach (7, 7) (x-coordinate +3)

or we can move down three spaces to reach (4, 4) (y-coordinate -3)

[frozen barn]

Yes correct

[lean yew]

but then we have to move perpendicular to our direction to the other side of the line

so if we chose to move to (7, 7), we cannot move up, since that would keep us on the same side of the lline

so we have to move down 3 spaces (remember, same number of spaces)

moving down 3 spaces reduces the y-coordinate by 3

[frozen barn]

We'll get to (7,4)

[lean yew]

correct!

now notice one thing here

if we chose to move one way in the first half

the second half must be done using the other way we didn't choose

otherwise we would remain on the same side of the line and it's no longer a reflection

in this case, we chose to move right 3 spaces first

so the second half we must move down 3 spaces

conversely, if we chose to move down 3 spaces first, the second half we must move right 3 spaces

[frozen barn]

I understand
Right-down vise versa or left-up vise versa

[lean yew]

yeah

that's the nutshell of why the statement in your question is true

[frozen barn]

Ok thank you for explaining
Im gonna close this and review your explanation
I'll reopen if i have some trouble
I appreciate your help
Bye✋

[lean yew]

nps, glad to help!

[frozen barn]

.close

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[pseudo rover]

I would like to know where did i go wrong while taking this integral maybe it might be related to the Si(x) and Ci(x) functions idk since i dont know much about em

[pseudo rover]

The actual answer turns out to be this

oh wait it spoilered it when i saved it too-

i took it from my friend lol

[pseudo rover]

I can explain what i tried to do if i wrote it in a messy way-

The top right part is basically for the partial fraction decomposition stuffy

[odd edge]

@pseudo rover Has your question been resolved?

[odd edge]

@pseudo rover Has your question been resolved?

[leaden karma]

Are you sure this integral is supposed to be indefinite

The definite integral is much cleaner

[shell rivet]

Do you know the values of sin(i) and cos(i) ?

I think your answer is correct

[pseudo rover]

My friend sent the question as an indefinite integral so uh idk

[pseudo rover]

No D:

[shell rivet]

Ah ok

[shell rivet]

Do you know euler's identity?

[pseudo rover]

Ye

[shell rivet]

e^i*(theta) = cos(theta) + isin(theta)

Right?

[pseudo rover]

Ye

do i just try to write i for theta?

And simplify from there?

[shell rivet]

I mean

I don't that will work just yet

[pseudo rover]

D:

[shell rivet]

You need to find the values of cos and sin in terms of powers of e

[shell rivet]

Substitute -theta instead of theta here

You'll get two equations

[split summit]

Because the line y=x runs diagonally through the plane so when you reflect an object diagram it swaps the horizontal and vertical distances from the line

[pseudo rover]

the e^ix - e^-ix/2i = sinx stuffy?

[shell rivet]

Uh

[shell rivet]

Yupp

[pseudo rover]

For thjs one

[shell rivet]

Now just put i in for x

[split summit]

wait a second

oh

[shell rivet]

For coa and sin

[shell rivet]

Yeah

[split summit]

i looked too far back in the help channel

mb

[shell rivet]

😭

[pseudo rover]

lol

[pseudo rover]

Thank uu

[shell rivet]

Np

[pseudo rover]

do u think it would become like the answer when i do that?

(Imma close the channel if so)

(Because i dont wanna makw the channel open for too long idk D:)

(panik)

This is actually very cool

Imma try to plug em in now

Oh wait i forgot to write i for x

Yee

It looks like it equals to the same answer

[pseudo rover]

(Definitely didnt get lazy to do the whole thing wdym)

Thank u so much

https://tenor.com/view/yippee-happy-celebration-joy-confetti-gif-25557730

close.

oh

.close

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[errant token]

hey looking at practice questions and got lost here, isnt this step wrong?

[wooden python]

wrong how

[errant token]

since the y should be squared before the integral, it should be y^3 /3

after

[vivid wyvern]

Yea

[wooden python]

oh. yeah you might be right..

[errant token]

alright thanks

[wooden python]

can you show the entire thing just in case

[errant token]

sure

[narrow crypt]

Ye seems wrong

[wooden python]

oh yeah ok it's wrong

[narrow crypt]

Should be y^3/3

[errant token]

alright thank you, i thought so

i have to close this channel now right? i forget command

[vivid wyvern]

.close

[errant token]

.close

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[tardy wedge]

How do i make it zero without a calculator?

are there any numbers that are predetermined that i should know?

[vivid wyvern]

Wdym

[latent scaffold]

Yes, there are common angles you should know from the unit circle

[tardy wedge]

the function should equal zero

[vivid wyvern]

Yes but u dont need to use a calculator for that

Wait do u mean sinx=0?

[tardy wedge]

so i just need to have the number in mind?

[tardy wedge]

no sinx = -1/2

[vivid wyvern]

U can hv common ones in ur mind

[latent scaffold]

Well you really only need to know the first quadrant to rebuild the whole unit circle if you need

But when starting you can also just refer to the unit circle directly

[vivid wyvern]

Sinx is negative in 3rd amd 4th quadrant so to get rid of the negative u could represent in (pie) form as in 180 form

Sinx=-1/2
We know sin30=1/2
Amd sin is negative in 3rd or 4th quadrant
So let's say it is in 3rd quadrant
180+ refers to 3rd quadrant
So sinx=sin(180+30)
U can do the rest from here

[tardy wedge]

ohh i think i get it

sin30 equals 1/2

and everything over sin180 equals something negative?

[vivid wyvern]

Sinx is also negative in 4th quadrant

[vivid wyvern]

Uh that's not how it works

[tardy wedge]

from 320 to 360

[tardy wedge]

ah i understood it like that mb

[vivid wyvern]

So do u know where the values of sine,cosine,tangent r positive and negative

[tardy wedge]

not really i didnt need that before

i think there is an easier way to solve the exercise

[vivid wyvern]

There is
basically calc input

But this isnt hard

U see in 1st quadrant every single trig function is positive

In 2nd quadrant
Sin,cosec is +,rest -

3rd
Tan,cot +,rest -

4th cos,sec + ,rest-

[tardy wedge]

ahh okay i see

[vivid wyvern]

There r reasons behind these ofcourse

But let's move with what we have

So
1st quadrant means 90°

Within 90°

So sin(90-angle) refers to 1st quadrant ryt?

[tardy wedge]

okay hold on you reminded me of something i have

[vivid wyvern]

Yes but u dont need to memorize all of them

[tardy wedge]

no i get this sheet in the exam too

[vivid wyvern]

Ohh

[tardy wedge]

but my original question was this

how i get the extreme points of this function

[tacit wasp]

By computing the derivative and studying its sign

[vivid wyvern]

U alrwady did half ,didnt u

[tacit wasp]

So that you can see where points of max and min occur

[vivid wyvern]

U can find the value of x from here ryt

[tardy wedge]

how do i get x to stand on its own when i have sin though?

cause i need one value for x

[vivid wyvern]

R u allowed to use calc

[tardy wedge]

nope

sadly

[vivid wyvern]

U will need two if u want to fiond both extreme points

[sand horizon]

on which interval ?

[vivid wyvern]

I see
Then u will be given this sheet ryt

U could identify from that

[sand horizon]

On R so ?

[tardy wedge]

R?

[tardy wedge]

yes but i need to keep the x somehow

[sand horizon]

you solve it on an interval neccessarly

which one

[vivid wyvern]

It isnt a must,is it

If there is notthinh we consider R ig

[sand horizon]

you can't invoke differentiation then

[sand horizon]

yeah ok

[vivid wyvern]

See
Sinx=sin210
U eliminate the sins u get x=210

Du u get it?

[tardy wedge]

yeah i figured it out now i think

[vivid wyvern]

So how many values of x can u get

[sand horizon]

since you're supposedly on R you gotta report that angle with a "360" multiple

[tardy wedge]

i find x through arcsin

[vivid wyvern]

That's same

[sand horizon]

what are thoses two

[vivid wyvern]

U should get two values tho

[tardy wedge]

yes there are two

from the table

[sand horizon]

arcsin can't send back two values

[vivid wyvern]

So i think u do know how to get 7(pie)/4

[sand horizon]

it goes to -pi/2, pi/2 so eliminate the evil one

[vivid wyvern]

U gotta mention that one of them is for 3rd quadrant and other is for 4th

[tardy wedge]

i looked at my solution from a couple of months ago and it seems i forgot how i did it back then

[vivid wyvern]

I see

So uk how to get 2 values now?

[tardy wedge]

you mean it should be the two that you had circled in the other picture?

[vivid wyvern]

Yes

[tardy wedge]

so 11pi 6

[vivid wyvern]

Yes

But will u be able to find it

In exam without actual maths but js a table

[tardy wedge]

man why did i write -pi 6

but i get it now

thank you

[tardy wedge]

yes

[vivid wyvern]

Cuz that's -30

[vivid wyvern]

Extreme points?

[tardy wedge]

no value of x

[sand horizon]

that is wrong, you can't apply arcsin and get two values

[vivid wyvern]

Okay so find f''(x)

[tardy wedge]

idk why the teacher marked it as correct

[vivid wyvern]

Yeah she didnt get 2 values from arc sin

[tardy wedge]

ill do it

[vivid wyvern]

She got one and from there on she found two values vasing quadrants

[sand horizon]

maybe they don't count it as wrong but it will be in a certain future of math classes

[tardy wedge]

well the teacher did note that i forgot periodicity of sin

has to do with that?

[vivid wyvern]

Btw @tardy wedge
If they dont ask u to specify which extreme point is maximum or minimum ,u dont need to find f''(x)

[sand horizon]

somehow

[vivid wyvern]

Ig

[sand horizon]

but they should have precise you that the arcsin use was also not good

if they were taking it in the count

[tardy wedge]

yeah i know I we just need to use that to make sure its an extreme point

if it doesnt equal zeri

zero

[tardy wedge]

okay thank you for telling me

[vivid wyvern]

Arc sin gets u to -30
So u can find two values from there sin is neg in 2 quadrants

[sand horizon]

sin(x) = sin(a) has two solution, (x = a mod 2pi) and (x = pi - a mod 2pi)

[tardy wedge]

thats what I saw in the table too yeah

so i just read

[vivid wyvern]

One of the reasom yea
The other is to know whether it's max or min

[tardy wedge]

yeah i know

[vivid wyvern]

Alr find it then

[tardy wedge]

give me 5 mins ill get back to it

[sand horizon]

the problem is the use of arcsin and the writing of it which leads to presumably the periodicity being forgotten and also arcsin is bijective by definition ! so having two output values (and one of them not being in the range of arcsin) is an error, i don't know if their teacher take it in the count but someday it will and the bad habits will be "anchored"

[vivid wyvern]

Yes the presentation is wrong
But otherwise u'll get ryt answers,i suppose they will be taught the proper way later onwards

[sand horizon]

why not start today ? :)

[vivid wyvern]

Write it down or type it out
Might help her out

[sand horizon]

i did it here

the answers follows

[vivid wyvern]

She missed that ig

[tardy wedge]

i really did miss it my bad

thank you for explaining it i will write that down

[vivid wyvern]

I js noticed but u already found it here

f''(x) i mean

[tardy wedge]

you mean -2cosx ??

i thought ur talking about the extreme points

by using f''

[vivid wyvern]

That's what i meant

[vivid wyvern]

But did u find this

[tardy wedge]

im doing it righgt now

okay this was an easier task than i thought

I had just forgotten the methods and that table

but is the arcsin use fine like this?

and translation the first one is a minimum the second maximum @vivid wyvern

[vivid wyvern]

Um did u plug in the values of x in f(x)

[vivid wyvern]

Not rlly ,u gotta write it in the form he showed

[tardy wedge]

i did

the function is short like that

i only look up its value for cos

in the table

[tardy wedge]

well i guess its fine for now

cause honestly i have never seen that and the teacher never mentioned it

so they wont have a problem with it

and this is my last math module so i hopefully wont come across it again

[vivid wyvern]

U deducted the x too from 2cosx ryt?

[tardy wedge]

why should i deduct it?

[vivid wyvern]

Hopefully

Here

U need to place the values of x which u got in f(x)

[tardy wedge]

i placed the x into f'(x)

so the -x turns into -1

or hold on it disappears completely cuz i used f''(x)

so it is correct

[vivid wyvern]

Lemme tell u u find value of xusing f'(x)
Cuz that's 0 for extreme point

And then u plug in the x in f"(x) to determine whether the extreme points exist ang which one is max and which on is min

Ok ryt i misunderstood sorry

I thought they asked u ro determine the extreme vallues of the function

[tardy wedge]

yes haha i got confused for a moment

[vivid wyvern]

U r good,u did it ryt

[tardy wedge]

yeah

and thanks again

was easier than i thought

one min

was this what you two meant?

[vivid wyvern]

Similar ig
What does the n mean here

Any positive integer ig

If u multiply with any positive integer ,it refers that the the angle is in 1st quadrant where every function including sin is positive

[tardy wedge]

yes for any positive

yeah and so you get every value for sin because it has 'multiple' every 2pi

[vivid wyvern]

Yeah

The multiple could be 1,2,3.....

[tardy wedge]

yeah i finally got it now

the last missing piece

[sand horizon]

since you on R (presumably) the n can be negative aswell

i.e in Z

[tardy wedge]

yeah thats right

[wanton bison]

Ja stimmt

[vivid wyvern]

For this sin will be positive only if n is positive
Or am i tripping

[sand horizon]

it depends the values of x

[tardy wedge]

erklärung auf deutsch wäre schneller gewesen lol

[tardy wedge]

i guess so

[vivid wyvern]

Anyway our real prob was solved ryt

[tardy wedge]

yep thanks to you all

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[manic grail]

Anyone got any source with problems in math for knowledge up to calc 3 ? Or ODEs/PDEs too

[manic grail]

Like any problem that utilizes calc2-3 and differential equations at any degree might be straightforward asking to calculate an integral or be a general problem thats solved through this

If anyone has a good source online thats free

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[tepid beacon]

I am stuck with this task i do not know how to go about it, im sending what i did in a sec

[tepid beacon]

Don't really know what to do from here, or even if i started correct

[mystic nova]

U can't solve $2t^2-7t-4<0$ ?

[tepid beacon]

yes

[clever fjord]

Alexis_Fx

[narrow crypt]

ts factorable

[mystic nova]

Hmm have u learnt how quadratic function behaves?

[tepid beacon]

i don't think so?

[narrow crypt]

have u learned the + - + -

thingy

[tepid beacon]

it sounds rly familiar but i don't know so im better off just saying no

[mystic nova]

What about derivatives?

[narrow crypt]

https://www.youtube.com/watch?v=Fd5ys4PQ-aM

[narrow crypt]

i dont think op knows derivative 😭

[tepid beacon]

i went through it but i forgot it all

[narrow crypt]

https://tenor.com/view/everywhere-at-the-end-of-time-its-just-a-burning-memory-backrooms-the-caretaker-trollge-gif-24774320

[tepid beacon]

🥹

[narrow crypt]

maybe briefly review this video

and ur sleeper agent might be activated

who knows

[late sinew]

k with the 2s orbital pfp and a status saying chemistry is for nerds 💀

[narrow crypt]

chemistry is for nerdds

electron is in physics

[late sinew]

[tepid beacon]

okay will do

[daring frost]

Dont think you ead this to solve the question?