help-36

[tiny gorge]

if you know the values of f(x) for all rational x, then you know f(x) for all real x, by continuity

(moreover, you can't specify f(x) freely for all rational x, either)

[manic leaf]

hmm

[tiny gorge]

also, consider that the constant functions are continuous

that tells you something about the cardinality

[manic leaf]

wait

is it similar

Let $F$ be such a set. Observe there is a bijection $g: \mathbb{R} \to F$, where $g(r) = f_r$, and $f_r : \mathbb{R} \to \mathbb{R}$, where $f_r(a) = r$ for all $a \in \mathbb{R}$. Therefore, $|F| = |\mathbb{R}| = 2^{\aleph_0}$.

[soft zealot]

toast

[tiny gorge]

that's not a bijection

[manic leaf]

wdym by thqt

o

[tiny gorge]

there are continuous functions that are not constant

[manic leaf]

oh

no this is just proving set of constant functions

[tiny gorge]

it is an injection, though

[manic leaf]

\textbf{(j) ${f: \mathbb{R} \to \mathbb{R} : f$ is constant}.}

[soft zealot]

toast

[tiny gorge]

ok, you see that (j) is a subset of (k) right?

[manic leaf]

ye

it is

[tiny gorge]

so the cardinality of (k) is at least the cardinality of (j)

[manic leaf]

ah

so thats the lower bound

[tiny gorge]

yep

[manic leaf]

so now i need to figure out the upper bound

[tiny gorge]

yep

[manic leaf]

do i just need an injection

[manic leaf]

soemthing like this

if i can show an injection from continuous functions -> real numbers am i good

[tiny gorge]

yes that would do it

then you have injections both ways, so by schroeder-bernstein there exists a bijection

[manic leaf]

ye

wait

so you said

a continuous function can be uniquely identified by the rational nmbers it takes up?

[tiny gorge]

it's uniquely identified by the rational subset of its domain

if you know f(x) for all rational x, then you know f

simply because for arbitrary y, you can take a sequence x_n of rationals such that x_n -> y, and then by continuity, f(y) = lim f(x_n)

[manic leaf]

ohh

so we have a set of rational numbers

we want to maybe inject a subset of Q to R

[tiny gorge]

well, it gives you an injection from your set to the set of all real-valued functions on Q

see if you can determine the cardinality of the latter set

[manic leaf]

that is aleph_0^aleph_0 = 2^{aleph_0}?

[tiny gorge]

what's your reasoning?

[manic leaf]

well set of all real valued functions on Q seems just like {f : Q -> R}

any function f is in

|R|^|Q|

.solved

[final saddle]

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[sweet prawn]

What does it mean by 108 are either made of clay or have coloring? Based on my understanding it's saying 108 artifacts are either clay or colored and no between???? Then I look at the 81 and 34 and that doesn't add up to 108

[supple pawn]

do step by step

that means there's a total of 108 artifacts with either clay or colouring

for example, 60 artifacts have colouring and 48 made of clay

[sweet prawn]

So 115 artifacts they take out 150 and of that 150 we have the 108

That are like either this or that

😭

[supple pawn]

try eliminating

if 34 of them has colouring

[sweet prawn]

Huh 😭

[supple pawn]

how many do NOT HAVE colouring?

which means
No colouring or made with clay

[sweet prawn]

Oh

Wait

[ornate token]

You already knew that Clay is 81 and Colouring is 81.

[sweet prawn]

No color and made of clay plus no clay but colored are equal 108

[ornate token]

Do you know Union rule?

[sweet prawn]

No

[supple pawn]

set theory

[ornate token]

A U B = A + B - A cap B

A unions B here is 108, as given.

Also knew A and B, then how many artifacts both coloured, and made out of clay?

The table here is perhaps unnecessary in my opinion.

[sweet prawn]

What is a cap b

[ornate token]

A intersects B

[sweet prawn]

Oh

[ornate token]

Upside down symbol for U

[sweet prawn]

My answer is 7

[topaz kite]

sorry for interrupting but perhaps OP may have heard of this rule referred to by its proper name, inclusion-exclusion.

[ornate token]

7 is the number of the artifacts having both property, you would need the probability of them.

[sweet prawn]

Oh

7/108 then

[ornate token]

Thank you for your explanation, I used Union rule term in my country

[ornate token]

Correct.

[sweet prawn]

YAY

https://tenor.com/view/roblox-panic-laugh-cry-wave-gif-14569362056240409740

[ornate token]

Congrats!

[radiant igloo]

fahhh

[ornate token]

!done

[final saddle]

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

[sweet prawn]

.close

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[calm ether]

Is my proof correct? \noindent \textbf{Problem.} Prove that $\frac{1}{n+1} < \log(n+1) - \log n < \frac{1}{n}$.

\begin{proof}
Note that
[
\log(n+1) - \log(n) = \log\left(1 + \frac{1}{n}\right) = \int_1^{1+\frac{1}{n}} \frac{dx}{x}.
]
Now, for all $x \in \left(1, 1 + \frac{1}{n}\right)$, it holds that
[
\frac{n}{n+1} = \frac{1}{1 + \frac{1}{n}} < \frac{1}{x} < 1.
]
Therefore, due to the continuity of $\frac{1}{x}$ on the interval $\left[1, 1 + \frac{1}{n}\right]$, and by Theorem 13-7, we have:
[
\int_1^{1+\frac{1}{n}} \frac{n}{n+1} , dx < \int_1^{1+\frac{1}{n}} \frac{dx}{x} < \int_1^{1+\frac{1}{n}} 1 , dx.
]
Evaluating the integrals of the constants:
[
\left[ \frac{n}{n+1} \cdot x \right]_1^{1+\frac{1}{n}} < \log\left(1 + \frac{1}{n}\right) < [x]_1^{1+\frac{1}{n}}.
]
This simplifies to:
[
\frac{n}{n+1} \left( 1 + \frac{1}{n} - 1 \right) < \log\left(1 + \frac{1}{n}\right) < \left( 1 + \frac{1}{n} - 1 \right),
]
[
\frac{n}{n+1} \cdot \frac{1}{n} < \log\left(1 + \frac{1}{n}\right) < \frac{1}{n},
]
which yields
[
\frac{1}{n+1} < \log\left(1 + \frac{1}{n}\right) < \frac{1}{n}.
]
This completes the proof.
\end{proof}

[soft zealot]

YuyuHenry

[calm ether]

This is the theorem I'm citing

[rugged merlin]

Seems legit

You can also prove it without srtting up an integral by means of the mean value theorem

[calm ether]

But like, it's without strict inequalities

[calm ether]

so the thing is that I don't know if my cite here is correct, I need it with the strict inequalities

[calm ether]

really?

[rugged merlin]

Yeah i guess you need to modify your logic slightly

U can still use it

But like

f(x) = 1/x is strictly decreasing and constant on [1, 1 + 1/n] which means the bounds are not reached and ur inequality is strict

[tiny gorge]

you could try to prove that if f is continuous and if there's some point in [a,b] where it's strictly between m and M, then the inequalities are strict

[calm ether]

wait, a friend of mine found smth

this problem which I can cite

[tiny gorge]

yea, that's what you want basically

[rugged merlin]

Its like a four liner woth mvt yea

[calm ether]

can I see it?

[rugged merlin]

Its in my head and I dont want to latex rn sorry

[calm ether]

okay, so let me try

by MVT of $\log x$ on $[n,n+1]$ there exists $c \in (n,n+1)$ such that $\frac{\log{n}-\log{n+1}}{1}=1/c \Rightarrow \log{n+1}-\log{n}=-1/c$, and we have that $\frac{1}{n+1}<- \frac{1}{c} < \frac{1}{n}$ we're done

[soft zealot]

YuyuHenry

[calm ether]

@rugged merlin like this?

[rugged merlin]

U got rhe right idea

But you messed up ur signs

It should be log(n+1) - logn

[calm ether]

That's why I multiplied by - after the Rightarrow

oh wait

I see

[topaz kite]

sorry to interrupt but wouldn't you get in a little bit of nitpick trouble for applying the MVT the wrong way in the first place?

[calm ether]

I forgot the signs lol

[rugged merlin]

But ur denominator is not right then

[rugged merlin]

Elaborate

[topaz kite]

never mind, I was just thinking that even though OP multed by -1 to 'fix' the signs, the application might be wrong to begin with so some pedantic examiner could dock him for misuse of the MVT.
maybe it's not a concern though, sorry for interrupting.

[rugged merlin]

This is what i was pointing out yes

[topaz kite]

ah, sorry for interrupting once again then.

[rugged merlin]

I mean bt that logic -1/c would be a negative number which would be apparently sandwiched by two positive fractions

Which doesn't make a whole lot of sense

[calm ether]

ohhh I see

[rugged merlin]

You're good

[calm ether]

No way I forgot the mvt

lol

[topaz kite]

(though if I may comment, please bracket the n+1 in the log, so as not to confuse it with log(n) + 1 when reading back later!)

[calm ether]

thanks

[rugged merlin]

Although

I dont recommend using mvt here for pedagogical reasons

[calm ether]

by MVT of $\log x$ on $[n,n+1]$ there exists $c \in (n,n+1)$ such that $\frac{\log{(n+1)}-\log{(n)}}{1}=1/c \Rightarrow$, and we have that $\frac{1}{n+1}< \frac{1}{c} < \frac{1}{n}$ we're done

[soft zealot]

YuyuHenry

[rugged merlin]

From what im understanding you are building up yhe theory on integrals from the ground up, so you are likely formally defining the logarithm by its integral form

[calm ether]

yes

Probably if I write the MVT my teacher will say "didn't I teach y'all how to integrate"?

[rugged merlin]

The fact thst the derivative of logx is 1/x is proven later so you probably cant (shouldn't) utilise that uet

[topaz kite]

not to mention, if I may comment again, isn't that Spivak's Calculus?

[calm ether]

Yes

[rugged merlin]

U got it now

[calm ether]

we're in sequences

so we're already past that

[rugged merlin]

But did you prove FTC

[calm ether]

FTC?

Ohhh fundamental theorem of calculus

yes we did

[rugged merlin]

If I apply the MVT for derivatives on log(x) to solve an integral problem before the FTC is established, I am using circular logic and relying on maths that I dont have access to yet

Oh ok

[topaz kite]

are you in chapter 18?

[calm ether]

22

almost done with the book

"almost"

This is supposed to be introductory calculus

[topaz kite]

in that case, just for context for Lex, OP probably has access to many theorems.

[rugged merlin]

Oh lol

Yeah never used that book

Guess its free game then

[calm ether]

I feel kind of sad. All I see is that MVT is broken

it needs a nerf

[rugged merlin]

Upload your suggestions for when maths 2.0 drops

[calm ether]

Fr

Just for mantaining the topic

I ended up citing this problem

part (c)

[rugged merlin]

Yea

Sorks

Works

[calm ether]

Okay

Thanks tho

[rugged merlin]

Ofc

[calm ether]

.close

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[copper roost]

hello can someone explain how you approach these kind of problems

[copper roost]

i mean

do i just do like 1-3/62/61/6

[drowsy epoch]

did bro sleep 💀

[drowsy epoch]

Which one

[copper roost]

I WILL FAIL MATH TOMORROW

[copper roost]

the 2.1

[hollow wave]

anyone here know k map?

[drowsy epoch]

For the first one you have 6 total and your goal is P(RRG)+P(RGR)+P(GRR)+P(RRR)

Like when you pull 3 times you want at least 2

[copper roost]

But i mean how do you differ between a green and red gummy

or is that not necessary

[drowsy epoch]

You nees to compute the current probability

First try red is what

Also let's do P(RRG)

[copper roost]

so

thats

4/6 * 3/6 * 2/6

[drowsy epoch]

Oh well you did Gegenwahrscheinlichkeit

[drowsy epoch]

You take one away everytime

You dont put it back

[copper roost]

yeah

yeah i did

[drowsy epoch]

Ok lets do your way

At least 2 red means at most 1 red

[copper roost]

ohhh

so

[drowsy epoch]

1-P(RGG)-P(GRG)-P(GGR)

[copper roost]

P(RRG) P(

yeah

[drowsy epoch]

Hmm doesnt seem better

[copper roost]

no there are only 2 green gummies tho

[drowsy epoch]

Oh right

Ok one less probability

So first try red is 4/6

Then twice green is 2/5 * 1/4

So P(RGG)=4/6 × 2/5 × 1/4

[copper roost]

so 1-P(GGR)-P(GRG)-P(RGG)

[drowsy epoch]

Yea

[copper roost]

but why do i have to consier

the order

is that important too

like green first then red

[drowsy epoch]

Oh here yes I think cause we dont put it back

Like P(GGR)=2/6 × 1/5 × 4/4

[copper roost]

yeah

man fuck

for the 2.2

let me thinn

[drowsy epoch]

What do u first want in 2

[copper roost]

n?

wdym

hmmm

id say some stuff like

2 * something with n but

we dont even know

n-1 also

but in the denominator too?

[final saddle]

@copper roost Has your question been resolved?

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[clever roost]

i need help

[clever roost]

so im learning the vector cross product

a x b = magnitude of a x b x sin(x) x n

i understand when i do a cross b i get a vector,

what is that vector?

is it n?

what does the rest of the formulae mean

[timber leaf]

that is an orthogonal unit vector

[covert tree]

yeah it is n,

[timber leaf]

Or unit normal vector

[clever roost]

a what now

[covert tree]

with magnitue the same as absinx

[timber leaf]

It is a vector that orthogonal to the plane containing both a and b

[clever roost]

so n = absinx

[clever roost]

magnitude

[clever roost]

i have no idea what orthogonal means

[covert tree]

perpendicular

[timber leaf]

Perpendicular

[clever roost]

so n is perpendicular to ab

what about the absinx part thats what i dont get

[covert tree]

https://tenor.com/view/feel-me-think-about-it-meme-great-minds-gif-15937839

[covert tree]

perpendicular to a and b. with a caveat

[clever roost]

idk what caveat means either

[covert tree]

u have to follow the right hand rule

[clever roost]

yes ik thsg

i just dont get the

absin bit

[covert tree]

u have to follow the right hand rule for direction

[covert tree]

i just mean it needs some nuance

[covert tree]

because there are two vectors perpendicular to a,b

like it can point up or down (will be perpendicular either way)

[covert tree]

that just gives u how long the vector is

it just so happens to be the same as the area of parallelogram a,b forms

[final saddle]

@clever roost Has your question been resolved?

[final saddle]

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[copper roost]

so here right

[copper roost]

im on the 3.2

F! is loaded

So you basically do

[supple jolt]

du brauchst P_gezinkt(fair)

oder?

[copper roost]

Pf!(Z

[copper roost]

ja aber

also P_F!(Z!)

oder???

[supple jolt]

nein

du weistt schon dass er gezinkt ist

und du willst wissen

[copper roost]

aber Z! ist doch das es fair ist

[supple jolt]

was die wahrscheinlichkeit fuer fair ist

[supple jolt]

nein

Z hat nichts mit fair zutun

oh

ha warte

doch

[bold turtle]

Z ist, dass es gezingt eingestuft sei

[supple jolt]

also das ist sicher nicht richtig

[bold turtle]

Nicht dass es IN WAHRHEIT schon so ist

[supple jolt]

genau

[copper roost]

hä also

[supple jolt]

.

oder besser gesagt

er wurde bereits als gezinkt eingestuft

nun willst du wissen, mit welcher wahrscheinlichkeit dieser gezinkte wuerfel fair ist

denn gezinkt eingetuft heisst nicht immer unfair

[bold turtle]

Du willst die Wahrsch. dass es eigentlich fair ist, in Anbetracht dessen, dass es so klassifiziert ist

[copper roost]

ein zufällig herausgegriffener würfel, der als gezinkt eingestuft wird, fair ist

das heist doch einfach, Die wahrscheinlichkeit von einem fairen würfel unter der bedingung das es gezinkt ist

[supple jolt]

ja

[bold turtle]

genau

[supple jolt]

.

[copper roost]

aber

[bold turtle]

ngl ich kenne diese Notation nicht

[supple jolt]

ur not german right

[bold turtle]

(vielleicht ist die etwas Deutsches was ich nicht verste-

yh

[copper roost]

wtf

[supple jolt]

yes in germany we use different notation for conditional probability

[copper roost]

bro speaks better german than me

[supple jolt]

we dont use P( A | B)

[bold turtle]

ye

[supple jolt]

We use P_B (A)

[bold turtle]

Ich verstehe aber was gemeint wird, 'so es ist mir problemlos

[supple jolt]

krass du sprichst aber gut deutsch

so formell

haha

[bold turtle]

hab in der Schule gelernt

[copper roost]

ich hab die aufgabe falsch verstanden

[bold turtle]

Ist ein bisschen besonders ausgeschriebn

keine Sorge

Wenns "eingestuft" vermisst wird, ändert sich das Problem

[bold turtle]

aber doch jetzt einverstanden, wie mans löst?

[copper roost]

Ich überlege gerade

[copper roost]

ja aber die aufgabenstellung ist sehr komisch irgendwie

[bold turtle]

Der Würfel "sei" gezinkt, dh wir sind mit einem "Z" beim... Baumdiagramm?

-# nennt man es so?

Davon ausgegangen, dass wir also ein Z haben, was wird die Wahrscheinlichkeit sein, dass wir auch ein F (fair) haben?

-# hier kommt Star mich zu korrigieren

[supple jolt]

ja

[drowsy epoch]

Ops sorry bruh, P(orange)=1-P(nicht orange)=1-n/(n+1) × (n-1)/(n-2)

[bold turtle]

cool

[supple jolt]

also ist es dir klar warum wir das berechnen wollen

[drowsy epoch]

Hallo Schildkröte und Wlan

[supple jolt]

hallihallo

[copper roost]

i basically thought, that Z! means not gezinkt which means fair

[supple jolt]

no

Z and F are different things

[bold turtle]

WLAN was zum

[copper roost]

yeah

but

[supple jolt]

if we know that Z, it could be fair or not

we dont know

[bold turtle]

but yeah, Z ist "gezinkt klassifiziert", nicht "gezinkt in der Tat"

[supple jolt]

Z is kinda like a Vermutung

if it helps to understand

[copper roost]

oh so the dice should be Fair, but classified as gezinkt

[supple jolt]

yes

[bold turtle]

genau

[copper roost]

MAN

we got 60 minutes

for 6 Problems

this is inhumane

[supple jolt]

oh i have 60 minutes for 24

[copper roost]

and no calculator allowed

[supple jolt]

haha

yes and no calc

[copper roost]

is bro at harvard or some shit 😭

[supple jolt]

nah but im doing the TMS

[copper roost]

man theres just so much content that could appear

[bold turtle]

ich meine, braucht man honestly nen Taschenrechner?

[copper roost]

honestly i must neglect some stuff snd gamble

[bold turtle]

Use fractions and make your life a little simpler

First, what's the probability that it's classified as loaded?

(fok I was gonna acc write "klassified as zinked" )

[copper roost]

Wait so P(F and G)/P(F)

[supple jolt]

ja

[bold turtle]

Ist G Z, dann yeah

warte mal

[copper roost]

yeah

[supple jolt]

nein

eh

[bold turtle]

korrecter Ding zu sagen

Falsch geschrieben

[supple jolt]

ja der nenner ist falsch

[copper roost]

oh then P(Z)

[bold turtle]

on the bottom, yeah

[supple jolt]

the bedingung goes in the nenner

[copper roost]

B is the condition right

bedingung

[supple jolt]

yeah

[bold turtle]

sorry that's just funny to read lmao

[supple jolt]

[bold turtle]

Weil's fast alles schon auf Englisch war, hab ichs mit nem englischen Akzent lesen müssen

[copper roost]

okay so 1.1 i could just say that for value bigger than 3 there is a negative root which isnt possible right

[bold turtle]

yeah

nun

[supple jolt]

smaller than -3 and bigger than 3

[copper roost]

but

[bold turtle]

Outside of the range, x^2 > 9

[copper roost]

if i were to insert -4

[bold turtle]

THIS makes the square root contain a negative inside it, which is not solvable

[supple jolt]

then you get sqrt (-7)

[copper roost]

wait i never properly understood

[sinful spindle]

it is solveable

[bold turtle]

Not in the reals

-# so shhhh

[sinful spindle]

yeah

can anyone translate the question

[copper roost]

-3^2 is -3*-3 which is 9, but isnt (-3)*(-3) the same basically

[supple jolt]

https://tenor.com/view/pff-loser-silly-bad-mood-débile-gif-23814348

[bold turtle]

Currently translating dw

[copper roost]

so when is () making a difference when using potences

[supple jolt]

-3^2 is just -1 * 3^2 = -9

[bold turtle]

Translation into English:

1. Given is the function f with f(x) = sqrt( 9 - x^2 ). The graph of the function f is a semicircle.
   1.1. Explain why f is only defined for x in [-3,3].
   1.2. The graph of f is rotated in the interval [-3,3] around the x-axis. Evaluate the volume of the body of revolution, and describe its shape.

[supple jolt]

and (-3)^2 = (-3) * (-3) = 9

[copper roost]

but -3*-3 is also 9

[supple jolt]

yeah thats what i said here

[bold turtle]

yes, but when you say "x ^2", you mean "take whatever x is and multiply it by itself". So if x = -4, say, then the minus is counted INSIDE the bracketed expression

That is, if x = -4, then x^2 = (-4)^2 (because I take -4 and multiply it by itself) - so this evaluates as 16

[copper roost]

yeah shit

yep

[drowsy epoch]

Is tmrw exam

[copper roost]

man you see the 1.2 this is what im talking about

[bold turtle]

,w x = -4, calculate x^2

[soft zealot]

Results provided by WolframAlpha

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

Requested by waes_wires

[copper roost]

its so much content

but its so rare

everything

so muchhhh

[copper roost]

yes

[drowsy epoch]

Ts u do in one semester lwk

[bold turtle]

I mean, it's a spun semicircle around the x-axis along that interval

What shape does it take, first?

[copper roost]

man im boutta restart this year 😭

[bold turtle]

(that might help you find the volume more easily)

[drowsy epoch]

But yeah u have other subjects too

[copper roost]

i mean

a semi circleV

we learnd how to do this stuff before i just gotta relarn it

[bold turtle]

Try and spin it in your head around the x-axis

What 3D shape does it cover then?

[copper roost]

man i dont know how you guys just immediately know how to solve stuff directly even when ur probaly not doing this in uni or whatever

[copper roost]

like

wait

like a bowl

?

[bold turtle]

Almost, keep spinning it

-# is a sentence I never thought I'd say today

[copper roost]

or like a cone maybe

or

i cant visualise this in my head ngl

yeah idk

[bold turtle]

Try spinning a circle instead

[copper roost]

i mean

thats like

a cylinder

[brazen breach]

I'm assuming they mean spinning a circle on its center

kind of like making a coin 180 on itself

[bold turtle]

https://www.geogebra.org/m/RyGPhBtU#material/NPpFNnWA has a good animation of this btw

[copper roost]

oh a globe basically

how do u call that

[brazen breach]

I thought ur finals was today damn
I alr took mine and was cooked

[copper roost]

tomorrow

[bold turtle]

ne Kugel?

[brazen breach]

-# damn, study well dawg I was all confident and shi saying my country's education ain't shi I was wrong

[copper roost]

i still gotta go through hypothesis tests, some plane vector stuff and some probability

[copper roost]

yeah like a sphere

[bold turtle]

A sphere

ye

[copper roost]

but i mean

it said semi circle

does that still make it a sphere

yes right

because it spins like that

yeah

[bold turtle]

ye

because it's gonna double around itself

Right, so we've got a sphere with... what as its radius?

[copper roost]

3?

[bold turtle]

ye

Lastly, volume of a sphere given its radius?

[copper roost]

weneed to intrgrate something ive done this before but i forgot

[copper roost]

shit like

1/2 pi *

some stuff like tha

yo theres no way i gotta memorize this hell nah

[bold turtle]

(4/3) pi r^3

[copper roost]

damn

[bold turtle]

oh you can do this by integration yh

hold up lemme see if I can get a textbook screenshot of that

(wird aber auf Englisch)

[copper roost]

let me check my notes

[bold turtle]

soll irgendwo da sein, yh

Gefunden

[Source: Edexcel AS and A Level Further Mathematics Core Pure 1/AS textbook, page 72]

[drowsy epoch]

Wire or Fire

[supple jolt]

1 sec

i have the german math book

[copper roost]

so its just

yep

i have it

bro

dw

[bold turtle]

If nothing else, the blue box

[supple jolt]

yeah

[bold turtle]

See also: Doctor Who

[supple jolt]

they always do some useless ass calculations before the blue box

for no reason

[copper roost]

i swear bro 😭

[bold turtle]

oh there's a timelord below this channel anyways lol

[copper roost]

they show the herleitung

[supple jolt]

yeah

but for 99% of the students its too difficult to understand

and everyone only reads the blue box

[bold turtle]

The idea is that, when you're not rushing to learn the content, this is supposed to motivate the definition

[copper roost]

we had to do the background of integrating once

[bold turtle]

When you're pressed for time, that blue box gets you quite a good way there though

[copper roost]

i gave up on that like i dont even care

but tysm guys

for the help

[supple jolt]

yeah us too i think

[copper roost]

ill probably open a help chanel some time later when im stuck on something again ty

.close

[supple jolt]

yo ure welcome

[final saddle]

Channel closed

Closed by @copper roost

Use .reopen if this was a mistake.

[final saddle]

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.

[atomic bramble]

My theory, i want ur thoughts

📜 THE SIMPLICITY IMAGINARY STANDARD (V1.0)

A Simplified 2D Simulation Logic

🧭 1. THE COMPASS

• Standard: I = 90^\circ
• Anchor: 360^\circ (Top/North) is the Origin.
• Rotation: Always Clockwise.
• I = Right | 2I = Bottom | 3I = Left | 4I = Reset Top.

🚫 2. THE GOLDEN LAW

• Linear Only: Use Addition/Subtraction for movement.
• FORBIDDEN: SQUARING (I^2) IS ILLEGAL. * Logic: Prevents simulation crashes and keeps calculation speeds maxed out.

🔍 3. THE ZOOM & SNAP

• Micro-steps: 1I contains 90 sub-units (degrees).
• The Snap: Round complex decimals (e.g., 45.123) to the nearest 0.1 for efficiency.

🌊 4. THE \pi LINK (Curve Distance)

• Formula: \text{Dist} = (\text{Turns} / 360) \times (2\pi r)
• (Lazy Mode: \pi = 3)
  Purpose: Designed for high-speed game logic and "Lazy" physics engines. EPIK. Logic > Complex Trig.

This used ai, but the idea came from me. Ai is built to help and assist human. So i used it to assist me

[ornate token]

So... What do you want to ask?

[tropic crest]

!noai

[final saddle]

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

[topaz kite]

you seem to have merely described a bearing system. what's your question with it?

[timber leaf]

Is that factoid necessary?

[atomic bramble]

I want ur thoughts

Opinions

[proper dagger]

if you're looking for opinions it might be better to ask in #math-discussion.

help channels are for problems and/or questions that you need help with

[tropic crest]

It is prolly

[ornate token]

#study-discussion perhaps?

[tropic crest]

Ehh no

[ornate token]

Since it is academic suggestion.

[atomic bramble]

Oh

[atomic bramble]

Btw no

!close

[proper dagger]

.close

[final saddle]

Channel closed

Closed by @proper dagger

Use .reopen if this was a mistake.

[atomic bramble]

Alr

.close

[final saddle]

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.

[sonic cairn]

\textit{Task 22.}

\textit{A triangle $ABC$ is given, in which $|AB| = 6$, $|AC| = 4$, and $\angle CAB = 60^\circ$ (see the diagram).}

\textit{Task 22.2 (0--1)}

\textit{Complete the sentence. Choose the correct answer from the options below.}

\textit{The length of side $BC$ of triangle $ABC$ is equal to:}

\textit{A. $\sqrt{28}$ \quad B. $\sqrt{40}$ \quad C. $8$ \quad D. $\sqrt{76}$}

[soft zealot]

Addison Tarna

[sonic cairn]

https://cdn.discordapp.com/attachments/903481102591750184/1501197374188556428/image0.jpg?ex=69fb3280&is=69f9e100&hm=ab5637dd33e1ceeed27f4ac1b82f8c5b6c73a709065a7923ccf77639890e48bd&

[cloud linden]

What's this lang

[ornate token]

Don't just drop your question here, expect people to solve and leave, OP.

[sonic cairn]

Ya

[tropic crest]

you need to find bc

[sonic cairn]

Can you stop?

[tropic crest]

then use it

[sonic cairn]

With the attitude

[tropic crest]

this is just a reminder that we are here to help you and not give you the answer

[sonic cairn]

b/sin(B) = a/sin(A) = c/sin(C)?

[sonic cairn]

Stop whining

[ornate token]

It is your attitude, you dropped it here, not asking for help, or showing your work.

[sonic cairn]

@viscid stirrup

[tropic crest]

please dont be rude

[sonic cairn]

Mhm and it works here?

[sonic cairn]

Area = 1/2 * a * b * sin(a)?

[tropic crest]

nope

look at the 2 pictures i sent

[sonic cairn]

So it’s this?

@viscid stirrup

But you said it also works

[final saddle]

@sonic cairn Has your question been resolved?

Channel closed

Closed by @sonic cairn

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• 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.

[sour copper]

Consider the unit circle $C$ defined in the real plane $R^2$ as:

$\C={(x,y)\in R^2 : x^2+y^2=1}

\Let $T_\theta$ be a rotation of this circle in the counter-clockwise (positive) direction by an angle $\theta > 0$. It is given that $\theta$ is not a rational multiple of $\pi$ i.e. $\theta \notin \pi Q$
Let $M (1,0)$ be a point on this circle.

• Prove that $T_\theta^k(M) \neq M$ for all $k \in Z^*$
• Use the Principle of Mathematical Induction on $n\in N$ to prove the following statement:
  For every $n\in N$ there exists an integer $m\in N$ such that the length of the arc of the circle extending between $M$ and $T_\theta^m(M)$ does not exceed $\frac{\theta}{2^n}$
• Prove that any segment (arc) of the circle $l$ o matter how small, will contain an infinite number of points from the orbit $l$, where:
  $l={T_\theta^n(M),n\in N}$

[glossy zephyr]

$\pi&

4th line of text

Oh yea, and {} is usually problematic.

If you want to make sure it renders properly, \{ \}

[soft zealot]

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

[glossy zephyr]

Consider the unit circle $C$ defined in the real plane $R^2$ as:

$$C=\{(x,y)\in R^2 : x^2+y^2=1\}$$

Let $T_\theta$ be a rotation of this circle in the counter-clockwise (positive) direction by an angle $\theta > 0$. It is given that $\theta$ is not a rational multiple of $\pi$ i.e. $\theta \notin \pi Q$\par
Let $M (1,0)$ be a point on this circle.

- Prove that $T_\theta^k(M) \neq M$ for all $k \in Z^*$\par
- Use the Principle of Mathematical Induction on $n\in N$ to prove the following statement:
For every $n\in N$ there exists an integer $m\in N$ such that the length of the arc of the circle extending between $M$ and $T_\theta^m(M)$ does not exceed $\theta\over2^n$\par
- Prove that any segment (arc) of the circle $l$ no matter how small, will contain an infinite number of points from the orbit $l$, where:
$$l=\{T_\theta^n(M),n\in N\}$$

youre missing the $ on the end of the C ={...} part.

[sour copper]

lol

[glossy zephyr]

Anyways, to the problem

[sour copper]

yeah

i am stuck at part 3

[glossy zephyr]

I assume Z* is the integers?

[sour copper]

yeah

Z \ {0}

[glossy zephyr]

Well, the "infinite" part has to do with the fact that the intersection is not just the end points meeting.
And this is somewhat equivalent to proving that, for the rotation T, with theta being a non-rational multiple of pi (that is, in some sense, theta and pi are linearly independent for n in N rotations)

Which in simpler words means that, for an arbitrary amount of rotations for an angle that is not a rational multiple of Pi, you will cover the whole circle once again.

[sour copper]

my works is:

let $\varepsilon > 0$ be the length of any given $C'$ of the circle
\for any given $n$ we find $m\in N$ whereas the length between $M , T_\theta^m(M) > \frac{\theta}{2^N}$

[soft zealot]

Amer

[sour copper]

let $\alpha = dist(M , T_\theta^m(M))<=\frac{\theta}{2^n}$

[soft zealot]

Amer

[glossy zephyr]

\leq for less or equal btw

[sour copper]

$\alpha < {\theta\over2^n} < \varepsilon$

[glossy zephyr]

\over overarches the whole expression, use {} to enclose it

[soft zealot]

Amer

[sour copper]

m here is an amount of rotations right?

for $T_\theta ^m (m)$

[soft zealot]

Amer

[glossy zephyr]

yes?

[sour copper]

then i can say something like

is that correct?

[glossy zephyr]

Ig. You know practically nothing about 2m, 3m, etc...

but continue ig

[sour copper]

this proves the points are infinite?

[glossy zephyr]

Im pretty much sure it doesnt

Im not giving it enough brain power rn tho.

[glossy zephyr]

I was more of going on this line of thought

lemme try to put it better

With T a positively oriented rotation by θ ≠ πq, q ∈ ℚ, and L (arc) a continuous subset of our unit circumference C; We want to prove that the set ∪_N T^i(L) = C

Aka, from our arc L, by applying T an arbitrary amount of times, and joining all the resulting arcs L', we get back our original circumference C.
If this is true, then obviously, our original L has to have an intersection with this new "reconstruction of the circle".

∪_N should be read as

$\bigcup_{i=1}^\infty$

[final saddle]

@sour copper Has your question been resolved?

[glossy zephyr]

We know that theta can be written as a θ = xπ with x being part of the irrationals.

As such, we can divide 2π and θ both by π and get 2 and x ∈ R\Q

Equivalently, our rotations now become addition under mod 2.

Soo, we want to know that an interval of the form { ([0,a] + nx) mod 2, n ∈ N, x ∈ R\Q} = [0,2]

I hope youre following with the idea

[sour copper]

a here is alpha right?

[glossy zephyr]

not really, just some arbitrary real number less than 2

ig i should have clarified

Well, ig we can start to play with what x is in relation to a too

if x < a, then we know that [x,a+x] (aka n=1) will have an obvious intersection with [0,a]
(same idea repeats for every single rotation, and since they all are in an intersection with eachother, then they will eventually fill in the circumference)

if x = a, then we know that [x, a+x] will share an endpoint with the previous interval, and as so, once a+nx > 2 for some n ∈ N, then in mod 2 the interval will have an intersection with [0,a]

So they basically tile one after the other until the last one.

and with x > a we have to deal with the fact that x and 2 are linearly independent w/ respect to natural scalars.

[sour copper]

i am type of guy who loves solving by equitation and numbers

but i think this part is more on the talky side

i am drawing blank on how to write it down

[glossy zephyr]

ig, we are exhausting options

oh, and just now i noticed, the x < a and x = a can be condensed into one single step.

aka x ≤ a

Okay, i assume i never mentioned why this makes sense, but basically we are redefining our problem in the (r,θ) space instead of (x,y)

Where a "continuous subset" of the circumference is much more obvious.

[sour copper]

each segment from M to $T_\theta ^m (M)$ is smaller than ${\theta\over 2^n}$

[soft zealot]

Amer

[glossy zephyr]

Not quite as of a statement.

It isnt "each segment"

[sour copper]

i'm having an english moment

[glossy zephyr]

Just ensures you that theres some m such that dist(M,T^m(M)) < theta/2^n

For any arbitrary n

[sour copper]

but basically if we make C smaller than that, it's guaranteed that some point will fall in C

[glossy zephyr]

The problem doesnt revolve around the fact if the arc falls in C

Since by definition its part of C

The problem is asking if the "orbit" of said arc will ever intersect the original one.

[sour copper]

if we prove L is dense then that solves it right

[glossy zephyr]

Because we havent proven that the orbit is itself the whole C

Ohhhh, yea, i got what you were trying to do previously, the problem is showing that this rotation is positively oriented w/ respect of the original arc position.

[sour copper]

my principle is each step is smaller than our segment

[glossy zephyr]

Yea, thats the problem, thats not needed at all

Like, the rotation could be near half a circle

while the segment be tiny.

The proof of part b) tells you that after some amount of rotations you will eventually get to a rotation thats in magnitude smaller than the length of the arc.

But not its direction

[sour copper]

it will be bigger than the rotation if $n$ in $\theta/2^n$ is big enough

[soft zealot]

Amer

[sour copper]

i suppose

[sour copper]

oh.. that makes sense

...

i am having too much info rn i am scrolling up so often

[glossy zephyr]

Thats kind of why trying to show that the orbit of L is in reality the whole circumference.

Since at that points its obvious that there must be some intersection

Oh, yea, I found a neat way to tie both ideas. @sour copper

[sour copper]

O.o

please do tell

[glossy zephyr]

Okay ill try to have it all as short as possible.

First of all, and for the sake of my sanity, i will have this be expressed again in terms of angles and not precisely as subsets of R^2

We have from your work on b) that, for some T^m(M), the difference in angle between M and T^m(M) (which we think of as a length of the arc)

Must be lower than θ/2^n

for any arbitrary n we may choose.

Notice, since θ/2^n is quite obviously lower than θ, we can extend this inequality as α < θ

where α is again this difference in angle.

Now, if θ itself represent a shorter arc than the arc we choose as "L"
Then we follow the x ≤ a i previously mentioned

Aka the rotations T by themselves will start to reconform the whole circumference.

Now, if x > a, we redefine with the idea of T^m, T^2m, T^km... you previously mentioned, as some form of "composite rotations", that, either positively or negatively will reconform the whole circumference again.

You can say that you create a new T' which rotates the arc by α instead of θ

Which ties back to the x ≤ a logic.

Btw, for the sake of writing this algebraically, try to deal with the length of L as an angle itself.

[sour copper]

okay

oh btw i omitted a part during translation, r = 1

don't think it's relevant

i think it's obvious

[glossy zephyr]

It said unit circle yea

[sour copper]

so in the end i should go like

the entirety of C is inside L

[glossy zephyr]

the other way around

[sour copper]

my work not looking too good

so after we play with x, i can conclude that

i create $T'_\alpha ^m$

[soft zealot]

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

[glossy zephyr]

You can avoid the ' notation if you clarify that alpha is just lesser than theta

And the m should be reserved for T_theta

Youd have T_theta^m = T_alpha

[sour copper]

we can say that right

[glossy zephyr]

Its not precisely the distance, but yea

[sour copper]

when you said we divide 2pi and theta by pi

you meant for theta = xpi + 2pi

[glossy zephyr]

Not really

If we say that θ ≠ πq, q in Q

Then we can say that θ = πx, x in R\Q

Aka its an irrational multiple of pi.

And by dividing by pi, we just get x, an irrational number.

[sour copper]

i am not that good in math i am not sure how theta = xpi transaltes to 2 and then [0,a]

[glossy zephyr]

The full circle is 2pi

thats where the 2 comes from.

Is just measurements in radians

[0,a] is the "length" of our arc measured by radians.

consider that a isnt alpha, its just some arbitrary measurement

[sour copper]

and nx is a number that gets smaller and hence smaller arcs

or wha

[glossy zephyr]

if you assume that x is the irrational multiple that gives place to theta

Then + nx is just the arithmetic representation of applying the rotation T "n" times

if youre more comfortable with it, you can keep in the π for all the operations.

[sour copper]

so theta = xpi + 2pi is not incorrect

[glossy zephyr]

nope

Well, its geometrically equivalent inside the circle

but id just keep it as xpi

[sour copper]

after we say

we need to elaborate on x whether it be >=a or <a

i got a little confused

or do we say T_\alpha^m = T_\theta

[glossy zephyr]

Im a bit occupied rn, but my overarching idea was to show that the "orbit of L" = C, as sets of points.
For that, we start from the definition of C itself.

C = { (x,y) ∈ R^2 : x^2 + y^2 = 1 }, which can be rewriten as:
C = { (r,θ) ∈ R^2 : r = 1, θ ∈ [0,2π) } (polar coordinates)
In this frame of reference, we can ignore the radius r and focus on the angles.

We have that an arc L is a subset L = { (r,θ) ∈ C : θ ∈ [0,πa] }, a ∈ R

So the angle that the arc L has goes from = 0, to = πa, measured in radians

With some rotation T / T_w, were w = xπ, x ∈ R\Q we have that
The "orbit of L" can be defined as the union of all the "rotated arcs"

"orbit" = U_N T^i(L) = { (r,θ) ∈ C : θ ∈ [0,πa] + wn, n ∈ N}

Consider that what we used to call "theta" i just wrote as w, since it would conflict with the notation

You can keep it as theta anyways

And we want to show that this union of rotated arcs is in reality equal to the whole circumference.

For that, we divide into two cases.

1. w ≤ πa: Where its easy to show that one single rotation by T_w generates an arc with an intersection with the previous one, so they reconstruct the whole circumference without any extra consideration.

2. w > πa: Where you use the conclussion from b) to create a new "composite rotation" T' / T_α, which is what you originally proposed. We have that with some [T_w]^m you get a lower angle than that of the original rotation by "w", so you create [T_w]^m = T_α

This is why i previously divided by π, you can see it appears for all angles.

And from the use of these smaller rotations T_α, you can apply logic as the case 1.

The idea is that from doing multiple rotations with angle theta/w, you can get a smaller "composite" rotation with angle alpha.

[sour copper]

brain not in service

[glossy zephyr]

lmao.

[sour copper]

i need a LOT of time to process that

is there any simpler method by any means?

can i suppose L is a subgroub of C without showing a proof?

or is there something i am missing

[glossy zephyr]

L is a subset of C by definition

the orbit of L also is, but we want to show that in reality it eventually becomes the entirety of C

Which ensures that the orbit of L intersects L itself.

just to be sure you follow, i want to make clear that
L, C, and the orbit of L are 3 "entirely different" objects.

[sour copper]

the orbit is all the arcs where L can be right?

which is = C

[glossy zephyr]

More precisely, if you move L by a certain angle theta/w, you get a new arc, which we will call L'

The collection of all L' (depending on that angle theta/w) is the orbit of L

[sour copper]

you're giving me all the info i need i just need a LOT of time to get everything as a single solution

[glossy zephyr]

We are interested in showing that for irrational multiples of pi, the orbit of L is equal to the entire circumference

[sour copper]

which we do with not equations, right?

i am like, really confused how

do i put it down on a piece of paper

like everything here makes sense but if i wanna type it out i go fully blank

we let there be C' an arc >0 -> 'a' is mod 2 theta -> theta = xpi

[glossy zephyr]

You can write it down algebraically, yea

[glossy zephyr]

Yea, C' or L

We use mod 2 or mod 2pi (depends if you wanna keep in the pi for radians), since angles that go over 2pi can be reduced to a smaller angle.

like, 380º = 20º for us.

3pi = pi, etc...

[sour copper]

have we assigned a name to the decreasing arc between T(M) and M

[glossy zephyr]

Not really, its just some arc. Doesnt really matter, since we only care for the new angle it creates

btw, if you wanna see how this idea operates, https://www.desmos.com/calculator/wvsgwjgmkr

Its just a way to visualize the concept.

[sour copper]

that makes it way less vague, i needed that

so arcs intersect

[glossy zephyr]

If you set it up for n = 1, w = 0.7, you can see how the first arc rotation deviates to about a third of a circle.

Now, if you dont change w, you make it n = 3, you can see the third arc rotation (L3), is actually just a smaller rotation from the original arc.

On that logic, once you can construct any angle that is shorter than that of the original arc L, you can then cover the whole circumference.

[sour copper]

that makes perfect since

but not for me writing it down 😭

i am made for equations

[glossy zephyr]

rn i have to study for a chem exam, DM me if you want, once i have more time ill try to write it out if you didnt manage to do it by then.

But i assume that the idea should be pretty clear as of now, its just being able to write it down algebraically.

[sour copper]

i will try as hard as i can to word it, but i am likely to fail and dm you 😭

[glossy zephyr]

np. good luck

!done

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[sour copper]

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[stone cradle]

how do i derivate this?

[stone cradle]

i keep failing

[willow cliff]

you mean as in take derivative right?

[stone cradle]

f'(x)

i mean that

[willow cliff]

oh ok yeah so u just use the product rule

(fg)' = f'g + fg'

so take each derivative separately and then just combine em

[stone cradle]

oh wait i might got it

[willow cliff]

i think it should be f'(x) = (2x-2)e^{-x} - (x^2 - 2x + 1)e^{-x} if im not mistaken

[stone cradle]

ye i found what i did wrong

thanks

[willow cliff]

np

[stone cradle]

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[smoky night]

Hi everyone

[worldly spruce]

Hello Amy

[worldly spruce]

Can we be of assistance?

[smoky night]

Can you help me with that number?

I have 5 more similar numbers as well

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.close

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[keen hare]

no clie how to even start on this

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[fallen valve]

Ah sorry for the late reply, I see that makes sense yes. Since if $f$ is continuous on a compact set $E$ then $f(E)$ is compact, (so closed and bounded) and hence it attains its minimum and maximum on it, say the maximum is $c$. And then like you say later, since $B(0, r) \subseteq \overline{B(0, r)}$, so then we know the supremum must be $\leq c$ and since $B(0, r)$ is non empty the supremum exists.

[soft zealot]

LXDL

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[halcyon night]

[radiant igloo]

please ask your question.

[halcyon night]

Hi

So if we have a formal system

Can we prove that it can't have self referential statements

[rugged merlin]

isn't it the opposite

[halcyon night]

?

[rugged merlin]

you cannot prevent it from having self-referential statements

that's what Gödel's whole thing was about

[halcyon night]

Only if it's strong enough right

I'm talking about proving weak systems as weak

[rugged merlin]

nvm

[halcyon night]

Yea it only specifically talks of strong systems

[rugged merlin]

yeah welp

[halcyon night]

[slate tiger]

dibya's a funny name lol

[halcyon night]

[slate tiger]

https://cdn.discordapp.com/attachments/1447461885355229285/1497948518961385532/caption.c688d2a2.gif

[upper leaf]

thank YOU buddha

[halcyon night]

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@halcyon night Has your question been resolved?

[halcyon night]

What I understand so far is that if a system is incomplete it has to be undecidable but inverse isn't true

So in this case the undecidable language is the set of all theorems expressible in the system

So like I know how to prove a system is decidable

I know how to prove it's incomplete(this would prove undecidability)

But how to prove it's complete but undecidable

[trail mango]

[halcyon night]

[halcyon night]

[halcyon night]

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[alpine dove]

can anyone help me with these two questions and explain

[alpine dove]

thxxxxx

can you help me with the other one

[worldly spruce]

!nosols

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[orchid coral]

Just calculate the value of:

50% of 120 litres

[worldly spruce]

What does 50% mean?

[sharp herald]

im pretty sure thats what the question is, yea

well done for reading it out

[orchid coral]

Calculating the value of "50% of 120 litres" is easier than answering the question

[alpine dove]

thx

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[orchid coral]

np

[worldly spruce]

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[blissful meadow]

Is equal to what?

And what have you tried?

[autumn agate]

unfortunately nothing...

i didnt understand where i can begin

[autumn agate]

um thats what we have to find

[rugged merlin]

Seems like the question is worded poorly

[blissful meadow]

Oh I read it as f(1) + f(2) + f(3)

I see

[rugged merlin]

It is just trying to find the functions such that f(1) + f(2) = f(3)

[autumn agate]

yeah

[blissful meadow]

Yeah I just saw the equal sign.

[autumn agate]

can you tell me a beginning step?

like what i can try

or smth

[rugged merlin]

Just go through like each possible value of f(3) and see the possible sums you can have

There is a pattern to be found

[autumn agate]

i didnt understand

[blissful meadow]

Well f(4) is free to be anything.
And f(3) is determined by f(1) and f(2).

[autumn agate]

really sorry

[rugged merlin]

If f(3) = 2, how many (f(1), f(2)) pairs do you have that satisfy the equality

[autumn agate]

ohh

i think i got it

1s

[autumn agate]

so something like this?

[rugged merlin]

Yuh so

Count all of them up

How many in total

[autumn agate]

15

[rugged merlin]

Now back to the first point here

for f(4) = 1, you have 15 pairs
For f(4) = 2, you have 15 pairs
For ....

How many pairs does it imply in total

[autumn agate]

15 * 6

[rugged merlin]

Congrats; you're done

[autumn agate]

90

ohh

btw

i thought of this

if i do something like this

will i get the solution

or did i mess smth up

[rugged merlin]

I dont really understand whats going on in that image I'll be honest

[autumn agate]

nvm

[autumn agate]

well ignore the first 1

but what i tried is

in this i took f(3) as 3

and the values in circles are the possible numbers

it can be

but i blundered

f(1) and f(2)

[autumn agate]

yeah it worked too

tysm for the help

ill close this now

.close

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[autumn agate]

.reopen

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✅ Original question: #help-36 message

[autumn agate]

another doubt

oh wait i had to use division into groups

[patent compass]

Combi?

[autumn agate]

yeah

pnc

[patent compass]

Leme trynthis1

[autumn agate]

i mean i can do smth like this but theres a condition

[patent compass]

Size means no. Of s_n right?

[autumn agate]

S1 cant be in x

n stuff

[proper dagger]

!nosols as a gentle reminder, in case you're going to provide the whole answer

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[autumn agate]

wym

[patent compass]

No nothin i got it

[autumn agate]

ight

[patent compass]

I understood the ques

[autumn agate]

what exactly do i do

like can i use set theory in this

or smth

idk

[patent compass]

I don't think so

[autumn agate]

wht did u do exactly

like the initial step

[patent compass]

I didn't got the soln

I'm trying

[autumn agate]

ah ok

ill do a diff q till then

[patent compass]

We can say s_3, s_4, ...., s_9 can in any of the teams.

[autumn agate]

yeah

[patent compass]

Les leave s_2 and s_1 for now

[autumn agate]

and what to do with the other 7

[patent compass]

Then we will have 7•6 = 42 ways to make X team.

[autumn agate]

yes

[patent compass]

60 for Y team

And Z we will have

4 seats

[slate tiger]

wsup dawg

[patent compass]

Fine

[slate tiger]

ah

[autumn agate]

no that was the old 1

[slate tiger]

.