help-36

[small urchin]

EAF are on the same line???theres no angle there

[weary edge]

I have no idea

But their answer is correct

😭😭😭😭

So Idk what now

[small urchin]

😭

id help u out but i have a test in 2 mins

[weary edge]

I got one in 17 mins

[small urchin]

well good luck to both of us

[weary edge]

Welp shi

[small urchin]

i can reach out after my test

[weary edge]

Its fine

[small urchin]

alr then

[weary edge]

Mines gonna start spon anyways

[small urchin]

yh fair

[weary edge]

<@&286206848099549185> help out pls

[vestal kestrel]

The problem with your working is that you assumed that the diagram is symmetrical

[weary edge]

Yea

Js realised my mistake earlier

[soft zealot]

Couldn't find an attached image in the last 10 messages.

Jan 16, 2026, 9:32 AM

[weary edge]

,rcw

[small grail]

,rcw

[weary edge]

<@&286206848099549185>

Please help

[vestal kestrel]

Do you know the property where the angle at the centre of the circle is twice the angle at the circumference?

[weary edge]

Yea

But idk how to use that into effect in thos question

[weary edge]

How do yk its 15 there?

[small grail]

isoceles

[weary edge]

Arent you just guessing that the other side is also 75?

[small grail]

from here you can see that ACF = BCF

[small grail]

no

[weary edge]

Ye

Cus 2 parallel lines

[small grail]

so now you can say angle FAC = FBC

[sand escarp]

i still dont get it

[weary edge]

Btw I dont want to be rude or anything but pls explain faster cus I got a test in 5 minutes 😂

[small grail]

so use trigonometry

[keen chasm]

Bruh

[small grail]

in triangle DCF

[sand escarp]

so what we know ae

[weary edge]

Uh wait why?

[weary edge]

I just need to find out how angle D is 60

Then I can easily do the rest

[small grail]

yes

[sand escarp]

wait

nvm

[sand escarp]

how did you get ae = sqrt2

[keen chasm]

?

[weary edge]

Yo can yall go dms???

[sand escarp]

i know why we need to find ae but how did you get ae

i have no idea

[small grail]

DCB =180

[small grail]

!occupied

[weary edge]

Ya

[keen chasm]

Wait

[final saddle]

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).

[weary edge]

@keen chasm go dms pls

[small grail]

guys the channel is gone

[keen chasm]

Wdym

[weary edge]

Go dms with clifford like stop chatting here

[small grail]

@keen chasm

[weary edge]

Im alr using this channel

[keen chasm]

What

[small grail]

so find DCF

[weary edge]

Ummm how?

Not enough info i thought

[small grail]

the channel you guys were discusssing on was closed and now @weary edge is using it so move to new channel

[keen chasm]

I don't have any doubt, @sand escarp have doubt

[keen chasm]

Ok

[sand escarp]

@keen chasmgo dms please i need explain

[small grail]

DCF +FCB = 180

[weary edge]

Just telling u incase I dissapear I gtg in 2 mins 😭😭

[keen chasm]

Bro what is dms ?

[small grail]

= DCB

[sand escarp]

just tell it here then

[small grail]

FCB is 150

[weary edge]

Bro can yall pls fuck off @sand escarp @keen chasm

[sand escarp]

what does fuck mean

[weary edge]

🙏🙏🙏🙏

[small grail]

@sand escarp claim a new channel

[weary edge]

Uhh wait how?

[sand escarp]

bro what is channel ?

[weary edge]

How did u get that 😭😭

Last question

[small grail]

.

[sand escarp]

@keen chasm

[small grail]

@weary edge

[weary edge]

Ohh

Okok

[keen chasm]

Bro wdym ,why u all mentioning

[weary edge]

Ty

I can figure it out now

I gtg now bye

[small grail]

@sand escarp @keen chasm go to #help-44｜stanley-🌲-v2-dans

[small grail]

alr

[keen chasm]

Bruh why me ,

[keen chasm]

Alr

[small grail]

if you want to help him

[plucky rover]

<@&268886789983436800> this has to be a troll, or someone who was born three days back

Either way, thought someone should inform y'all

[toxic tusk]

Please don’t troll in the help channels, if you’re actually lost look at #❓how-to-get-help

[final saddle]

@weary edge Has your question been resolved?

[final saddle]

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[somber osprey]

Guys can someone verify if the last question is correct?

[somber osprey]

The integral of ln(radical(x)/radical(x))

@ me if someone answers thanks !

[small grail]

substitution looks wrong

[somber osprey]

Can u solve that question using the power rule and not substitution?

[severe canyon]

Where would you apply the power rule? 🤔

[vapid haven]

wht is the diff of lnrootx.

[small grail]

you don't need power rule refer this

btw @severe canyon

should i nosols this

idk i'm new

[somber osprey]

Idk I just wanted to know if that final answer was wrong

[severe canyon]

Yeah usually yes, but in this case it's fine

[severe canyon]

Yours is, yes

If you're referring to ln²√x + C

[somber osprey]

Okay thank you

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[final saddle]

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[proper dagger]

um

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[minor sandal]

How to solve such questions

[minor sandal]

My idea is :-

Let x= limit of sequence.
I put |xn- x| = ε
And then I will get the value where the sequence just touches the epsilon interval of x

[granite thorn]

It is the way kinda

First you let epsilon > 0

And you search for when |xn-x| < epsilon

If this last statement is true for all integer greater than a certain N youre done

[minor sandal]

I was trying to get a minimum real no. Above which |xn -x | is less than ε

So i put it equal to ε

[granite thorn]

It doesnt have to be minimal

[minor sandal]

Yes ....

But it's my way of thinking

[granite thorn]

But that’s an idea

[minor sandal]

Or Solving

[granite thorn]

Yes

[minor sandal]

Butr

[granite thorn]

Try something i can correct

[minor sandal]

I have a doubt

Consider 5 . (c) in the solution manual ....

[granite thorn]

Yes

[minor sandal]

When i solved |xn - x| = ε

I got 13/(4n+ 10)

But they are taking a number greater than this

So my doubt is

[granite thorn]

What they did isnt usefull

[minor sandal]

Since this number 13/4n is greater than the boundary value 13/(4n+10)

This 13/4n will correspond to the xn , which is outside of ε interval of x

Since 13/4n is less than 13/(4n+10)

[granite thorn]

It is not

[minor sandal]

Sorry I wrote opposite

[granite thorn]

They meant 13/(4n)

[minor sandal]

If you consider the difference |xn -x |, then that distance can be maximum 13/(4n+10) , ....above this value , if you take any xn , it lies within ε interval of x

But since 13/4n is greater than 13/(4n+10)....

There might be a term x' which lies outside of ε interval of x , such that
13/(4n+10) < |x' - x| < 13/4n

[spark swan]

all you worry about is showing 13/(4n+10)<epsilon

[minor sandal]

The solution manual confused m

[granite thorn]

13/(4n+10) < 13/(4n) so this last inequality cannot hold

[spark swan]

the idea is after simplifying the first inequality, you get this

[granite thorn]

It happens i got confused too by what you say even tho the solution is clear to me lmao

[minor sandal]

You mean there is no natural number within this range ....?

[granite thorn]

Im saying this last inequality cannot hold at all

You seem to have lost yourself a bit in your reasoning

I suggest you take a pause and stepback, are try again to see what you said

It is ok and common dw

[minor sandal]

I genuinely respect that advice

I'll follow it

But first please

Why doesn't the inequality exist

It gives 0<10

[granite thorn]

If a<b than b<c<a cannot hold

[minor sandal]

Ohkkay

[minor sandal]

I m sorry , I wrote it wrong

[granite thorn]

Thus, as 13/(4n+10) < 13/(4n), what you said cannot hold

[minor sandal]

Corrected

[granite thorn]

Then it is not an issue since in the solution, the answer make n > K(epsilon) so that 13/(4n) < epsilon

So it dosnt matter if |xn-x| > 13/(4n +10) since it will be less than 13/(4n) < epsilon anyway

[minor sandal]

Okay got it

Thank u for helping

.close

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[steep bison]

can someone help me w this one

[thin cloud]

!status

[final saddle]

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

[steep bison]

2

1 actually

smth in between

i tried many ways

[thin cloud]

I feels like similar triangles is the key here

[steep bison]

yep definitely

[thin cloud]

KME and FDE are similar

[steep bison]

i tried angle bisector theorem too

[thin cloud]

What can we do with that

[thin cloud]

huh

[steep bison]

lets see

[steep bison]

like

i mean similar triangles

and it divides the square angle into two parts

where u draw the line from corner to the o point

[thin cloud]

Nah we ain't go that way

[steep bison]

sorry k point

ok lets try ur way

[thin cloud]

Dealing with angle is annoying

[steep bison]

this was what i was talking ab

[steep bison]

let me try this on paper hold on

[thin cloud]

Oh

I would relate DM to ME

[steep bison]

okay

[thin cloud]

And then use Pythagorean to find EK

Note that MK=DM

[steep bison]

js dmed u

[thin cloud]

You can just send it here

No need to DM

[steep bison]

OH i thought u meant direct message here

sorry i have 8 hrs of study time by now..

[thin cloud]

no, not really

EF is known

Use the ratio EK/EF instead

[steep bison]

no its not ef is 442 cms

R of the circle

OH

I READ THAT AS UNKIWN

UNKNOWN

sorry ..

[steep bison]

okay

now im going w a/ x+a i guess?

the biggest issue here is fc i think

this leads no where

can we say me=fc

it feels so right

maybe bcs of rotation?

of the half circle

ef is known

i forgot once again

ek too

im blind by now maybe i shd js sleep

i think im done

yeah okay we found the edge of the square

omg ty sm

[final saddle]

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[arctic sand]

The numbers 1, 2, 3, ..., 99 are written on a blackboard. In each step, we select any three numbers a, b, c from the board, erase them, and replace them with the value abc + ab + bc + ca + a + b + c. This process is repeated until only one number remains. Find the value of this final number.

[arctic sand]

i have found that if ( abc + ab + bc + ca + a + b + c )+1 = (a+1)(b+1)(c+1) and my teacher suggested that the problem had sth to do with invariance or sth

umm wat?

[worldly mesa]

<@&268886789983436800>

[tulip pawn]

lol

[arctic sand]

can u help me 😄

[tulip pawn]

hacked accoun

xd

[worldly mesa]

this means that abc + ab + bc + ca + a + b +c = (a+1)(b+1)(c+1) - 1 and when you use this number next time it will you will be left with a factor of (a+1)(b+1)(c+1)

[royal gust]

The teacher's suggestion is good. You want some kind of procedure you can do to every number on the board, such that you get a single number out of it

[arctic sand]

what do u mean i dont really get it :<

[worldly mesa]

<@&268886789983436800> @junior copper is posting innapropriate photos

[royal gust]

<@&268886789983436800> @junior copper
Posted and deleted nudity

[arctic sand]

so what should i do??

[tulip pawn]

i saw that

[royal gust]

And you want this single number to be "invariant". It shouldn't change after replacing a,b,c with (a+1)(b+1)(c+1) - 1

It's a bit hard to hint at the invariant because it's a bit of a "guess and check" thing

Let's say I suggested multiplying everything on the board together. Why does that not work? Can it be fixed?

[arctic sand]

ok let me try it

[arctic sand]

i dont find anything, why does it has to be this abstract 😭

wait i think i found it

[arctic sand]

thanks for the advice, i will go to sleep now :3

.close

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[tired walrus]

gone before i could even ping mods

who's the new quickest mod-draw in the wild west?

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[tired walrus]

<@&268886789983436800> obvious troll is obvious. nsfw-implied picture and all that.

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[twin pivot]

What if M=2?

How do you do consecutive intersection and union?

[blissful meadow]

Is there any more context?
If M=2 then you just get A_2 in the parentheses.

[twin pivot]

Ok , but what happen i have AU(intersectionB)

How is it done?

[blissful meadow]

$\cap_{i\in I} A_i$ is just shorthand for $A_{i_1} \cap A_{i_2}, \ldots$.

[soft zealot]

Azyrashacorki

[twin pivot]

Yes

[blissful meadow]

You don't actually get an intersection symbol right after the union

[twin pivot]

Ok

But AUB=AB?

[blissful meadow]

AuB = AB doesn't make sense. What do you mean by AB?

[twin pivot]

A*B

[blissful meadow]

AxB is the cartesian product, AuB is the union, AnB is the intersection. They are three different operations.

[twin pivot]

These operations were not the same as Boolean logic or something

Ah AuB=A+B

[blissful meadow]

Yes

Well as it is again A+B doesn't mean anything in set theory, at least not in the sense of boolean algebra...

[twin pivot]

So sets obey Boolean algebra, right?

Like a+not(a)=1--> AUnot(A)=omega

[blissful meadow]

I get what you mean but this is just a fact about ors and ands

Which yield similar properties in sets because unions are constructed with or and intersections with and.

[twin pivot]

Yes

[blissful meadow]

It doesn't mean that you can just and sets together is my point.

Or not a set

They are analogous to intersection and complement for instance, but not the same

And the reason they are analogous is that given some universe of sets omega, you can construct a Boolean algebra on the powerset of omega with operations interpreted as
and := intersection
or := union
not := complement
1 := omega
0 := emptyset

Then in this setting the rules of Boolean algebra you know correspond one-to-one to rules about set operations.

[twin pivot]

Thanks

.close

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[red leaf]

Help?

[red leaf]

Rn I have ax^2-18ax+81a-14

[opal plinth]

It intersects the x-axis, that gives an additional constraint on a

[red leaf]

Must be 1 or greater

[opal plinth]

Why 1?

[red leaf]

Sorry ur right, it has to be greater then 0

[opal plinth]

Exactly

[opal plinth]

What's a+b+c from this?

[red leaf]

a-18a+81a-14

64a-14

[opal plinth]

Right

So can it be -14?

[red leaf]

Sure?

[opal plinth]

What would a be if 64a-14 = 14?

[loud sundial]

hi csp 12

[red leaf]

Wait

Ur him?

[red leaf]

Undefined?

[opal plinth]

Wdym csp 12

[red leaf]

How many alts do u have

[opal plinth]

No

[red leaf]

0?

[opal plinth]

Yes

[red leaf]

,calc 0/64

[soft zealot]

Result:

0

[opal plinth]

So, conclude?

[red leaf]

I dont know, maybe try -23?

[opal plinth]

You don't need to

[red leaf]

Oh

,calc 14+-12

[soft zealot]

Result:

2

[red leaf]

So D?

[opal plinth]

not sure why you did 14+-12

[red leaf]

a needs to be positive

[opal plinth]

I mean ok, D is correct, I just don't really know how you came up with it

If a > 0 then what can you say about 64a - 14

[red leaf]

Well a has to be positive and 14+-14 is 2 and 2/64 is positive

[opal plinth]

14+-14 is 2 ??

[red leaf]

I meant 14+-12

[opal plinth]

But where does that +-12 come from...

[red leaf]

64a-14=-12

[opal plinth]

Oh because D is -12, right ok

[red leaf]

Wait nel

Have u taken the sat?

Or act?

[opal plinth]

I'm assuming those are American things, so no

[opal plinth]

Can you answer that?

[red leaf]

It must be more then 14

-14*

[opal plinth]

Yes

And D is the only answer that is more than -14

[red leaf]

Any tips for problems like these?

[loud sundial]

#bots message

[red leaf]

Wait csp

Have u taken sat/act

[opal plinth]

Uh not really, this problem is a little too easy for me tbh, and I'm not a teacher so...

[rocky tusk]

"what did you get bro"

[loud sundial]

I rawdogged it once while sick on two tamales

[red leaf]

Any tips on studying? like those hard m2 math problems?

[rocky tusk]

what is m2

[red leaf]

Module 2

[loud sundial]

Finish early

[rocky tusk]

noncalculator?

[red leaf]

U get desmos

[rocky tusk]

🤔

i think the SAT is different now

[red leaf]

What?

[red leaf]

Digital now

[rocky tusk]

i took it right before they switched away from paper

[red leaf]

U get desmos for both sections

[loud sundial]

Mate I did comp math in school

[rocky tusk]

the SAT is just repetitive

[red leaf]

Yea thats true

[rocky tusk]

its all the same remedial questions

[sly forum]

"mate"

[red leaf]

No, those m2 math questions tend to be p hard

[rocky tusk]

no

[red leaf]

Yes, since the more questions u get correct the harder it gets

[rocky tusk]

they're restricted to a pretty small amount of things they can ask about

[loud sundial]

@opal plinth csp 12

[red leaf]

But they way the word it is insane

[rocky tusk]

its literally just basic questions about quadratics and systems of equations and their geometric interpretations

[rocky tusk]

eh

ACT math is more involved

much more involved tbh

[red leaf]

The thing is i need to study to inc my scores

I get stuck on m2

[rocky tusk]

the only way to improve is by doing problems

[red leaf]

But i dont know where I can get those problems

[sly forum]

from csp, pretty sure he has alot

[rocky tusk]

im sure there are plenty of books and resources available on the internet

back when i took it we had this huge SAT prep book

fat ass book

[red leaf]

Really?

[rocky tusk]

idk if thats still relevant

[red leaf]

No, those ones are actually easy

Ive seen the paper version and that was very easy

[rocky tusk]

id like to see a digital version

[red leaf]

This problem I posted was sort of difficult

[rocky tusk]

because we see questions posted here

[loud sundial]

50 of these in 6 hours

[rocky tusk]

i swear ive seen the whole which of the following is a possible value of f(9) so many times

[sly forum]

@loud sundial give him your huge pdf

[rocky tusk]

where its like a parabola and they give you the vertex

[loud sundial]

???? what

[rocky tusk]

and you just have to spot the answer that is above the vertex

😭

or ill see like

[sly forum]

I thought you made a huge pdf of questions

[rocky tusk]

find f(1) but they disguise it as a + b + c

[loud sundial]

You mean the old tests like the one I sent above?

[rocky tusk]

routine problems

[sly forum]

yea

[red leaf]

Do u have any similar problems u can send?

[red leaf]

This isnt on the sat 😂

[loud sundial]

the sat math section takes 10 minutes

idt you can even finish that page in 10 minutes, now do 17 more of them

[rocky tusk]

yes this is comp math

[rocky tusk]

don't think so

[sly forum]

welp

[red leaf]

Damn, thats terrible. I guess ill continue to study then. Thanks tho

[sly forum]

ask your teach @red leaf

[red leaf]

Borderline...

I shouldnt say it

[rocky tusk]

did you go here https://satsuite.collegeboard.org/practice/practice-tests

[red leaf]

Actually ass

They gatekeep the hard m2 problems

[rocky tusk]

🤔

[sly forum]

hard '''

do comp math if you want hard

[rocky tusk]

can you show me an example of an m2 problem

[loud sundial]

I’m gonna go listen to 2010s music now

[sly forum]

why not 1950s?

[red leaf]

Sure, give me a 1uick second then

[rocky tusk]

https://www.mometrix.com/academy/wp-content/uploads/2022/11/SAT-Practice-Test-2-with-Answer-Key-and-Scoring-Info.pdf

brother this is all familiar

these are like the same questions they asked 4 years ago

😭

ok i guess this is 2022

lmao

[red leaf]

@rocky tusk

This is one

[rocky tusk]

holy quality

[rocky tusk]

should be clear if you draw a graph

[red leaf]

Another one

10=a^0+b

a^0=1

10=1+9

b=9

325/36 = a^-2+9

1/36 = a^-2

C

54

[rocky tusk]

i thought you said these were hard

[red leaf]

Na, these werent hard. The one i first posted was a bit hard. Also, I will show u the super insane ones

[rocky tusk]

not really

if you draw a graph you'll see it pretty quickly

[red leaf]

Im talking about the one i asked the question for

[rocky tusk]

the linear function has to be above the exponential between the two intersections

[red leaf]

^

This one was sort of hard

[rocky tusk]

omg lmao

no fucking way

i didn't even read the question

and i yapped about f(1) disguised as a + b + c

😭

im telling you

these people aren't original at all

its all the same repeated nonsense

[red leaf]

Hey, thats good for me 😂

[rocky tusk]

like i remember i did a practice SAT before knowing about vietas formulas for the sum/product of roots and i was like omg this is crazy im having to figure this out and im sitting there multiplying the roots from the quadratic formula

meanwhile i shouldve just had it memorized

[red leaf]

Sometimes its crazy

@rocky tusk I found a hard one

The insane one

[rocky tusk]

ive seen this exact question or something similar before. you have a vertex below the x axis with two roots meaning it has to be a > 0 so then f(1) > f(9) and your only answer is D

[rocky tusk]

well if h(m) = 0 then the quadratic inside the sqrt is 0

[red leaf]

Yes thats true

And we know our 0?

If x=0

[rocky tusk]

and you have 4 + 2b + c = 0, c = 266

[red leaf]

,calc -sqrtroot C = -sqrtroot 266

[soft zealot]

The following error occured while calculating:
Error: Invalid left hand side of assignment operator = (char 13)

[red leaf]

So c=266

Right?

[rocky tusk]

oh wait

h

not the quadratic

yea so c = 266

[red leaf]

So solve for b

[rocky tusk]

so b = -135

[red leaf]

Yea

Oh this was easy?

After that we know how to find m

Its whatever

[rocky tusk]

yea i mean it doesn't seem so bad

(x - 133)(x - 2)

m has to be 133

[red leaf]

Dude I swear the problems im talking about were more insane then this

Idk why I cant find them

But its whatever

[rocky tusk]

maybe they look scary until its explained

[red leaf]

😂

Well going to continue to practice man

Take care

[rocky tusk]

you too

[red leaf]

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[lime basin]

For inverse is my answer correct as well?

[lime basin]

wait im about to upload

[worthy wren]

yes both are the same

[opal plinth]

Yes (as long as the domain is (-inf, 1] like you wrote)

[lime basin]

Thank u

[lime basin]

👍

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[jagged flare]

In a Math Olympiad selection test consisting of 4 multiple-choice questions, each question has 3 possible answer options. It is known that for any group of 3 participants, there exists at least one question for which all three gave different answers. What is the maximum possible number of participants?

[jagged flare]

i have no idea where to start

[final saddle]

@jagged flare Has your question been resolved?

[wicked crystal]

What is the maximum number of participants? For what? Is this the whole question?

[jagged flare]

yes

[wicked crystal]

Weird question

[jagged flare]

im guessing the participants to the oly?

[wicked crystal]

What's oly?

[jagged flare]

the math olympiad selection test

[wicked crystal]

Yeah but theres nothing here limiting them

[candid pulsar]

kinda sure whats given is enough

[wicked crystal]

?

[jagged flare]

It is known that for any group of 3 participants, there exists at least one question for which all three gave different answers.

is this not the limiting factor?

[red needle]

hey ,,,, actually i need someone who can help m in maths my exam on 23 jan .. ,<calculus,algebra,geometry>

[wicked crystal]

Yeah I have no idea

[final saddle]

@jagged flare Has your question been resolved?

[sacred night]

!occupied

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

[icy tulip]

this is weird

but I think I know how to do this

you just have to find out how many different combinations of different answers exist for a particular question

oh it’s not groups

mb

[jagged flare]

[final saddle]

@jagged flare Has your question been resolved?

[woven ledge]

I have an idea but it might be wrong. I call the question of a particular triple which they all answered differently as their special question. Theres n choose 3 triples. By php at least one question, say P1 has at least n choose 3/4 triples have it as their special question. Now I define G to be the set of all people who are part of a triple with P1 as their special question. Let a be the number of people in G who answered A(one option of the multiple choices) to P1,b and c are defined similarly. Clearly abc>=number of triples who had P1 as their special question and a+b+c is the number of people in G. So we get n>= a+b+c >= 3(abc)^(1/3) >= 3(n choose 3 /4)^(1/3). This inequality holds true for 3<=n<=26

[raven marsh]

oh this is a famous combi

the numbers are always tweaked around

[raven marsh]

i learnt it before but forgot

[jagged flare]

wtf

what in the world would be the motivation to this

,w expand 8x^3-9(x)(x-1)(x-2)

[soft zealot]

Results provided by WolframAlpha

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

Requested by skill_issue7119

[jagged flare]

hm yeah this is workable

[woven ledge]

leading term of rhs has coeff 3/cbrt(4*6) > 1

so at one point rhs overtakes lhs

[jagged flare]

no like

what would even be the idea of bounding that

normally in bounding problems, there is some sense as to what would get bounded and where that bound comes

but unless im blind, i dont see where the idea of changing it to a+b+c, then am-gm would come

[woven ledge]

i just wanted to see how abc would be maximised in corellation to the number of people

like i was trying a construction with minimum people in G for the triples in G and i realised that am gm works

[jagged flare]

hmm

interesting

okay, thank you!!

.solved

[final saddle]

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[woven ledge]

well we havent proved that 26 is possible

that is whether this is a tight upper bound

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[restive sinew]

how do I check convergence of it?

only two test i have read yet

cuachy nth root and limit comparison

[tired walrus]

is it intentional that it isn't given what the nth term is and you are forced to guess?

[proper dagger]

I assume that the nth term is $\frac{(n+1)}{n} x^{n-1}$?

[soft zealot]

Hanako(x, y); ∂(fox)/∂x

[tired walrus]

otherwise i am thinking about committing malicious compliance with understatement and claim that the coefficients on x^4 onwards are 19 because i said so

the actual series may have been meant as $$\sum_{n=0}^{\infty} \frac{n+2}{n+1} x^n$$ by the looks of it?

[soft zealot]

Ann

[tired walrus]

you don't even have ratio test?

[onyx peak]

if not, i'd just compare it to geometric

[scarlet sequoia]

If you don't have access to ratio test then cauchy nth root isn't that bad

[restive sinew]

i know it

but my book did not introduct yet

[onyx peak]

you can pick any of the 2 tests you have, both will work

[tired walrus]

ah so it is illegal as of yet

anyway yeah

you also need to know about infinite GPs and when they converge

[restive sinew]

not illegal i am just curious over authors

[tired walrus]

one way or another this knowledge will be needed

[tired walrus]

you say it isnt introduced yet

in 99% of math books we are obligated to assume future methods are illegal

[restive sinew]

yes they have introduct it

[restive sinew]

and i will take mod of it?

ohh wait wait i will use nth cauchy now

ohh no i am stuck

surely i need ratio test here

thanks all of you

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[twin pivot]

Why is this a function if at 0 I get multiple values?

[tired walrus]

you do not

you have just stretched it so much it looks vertical but it really is not

it just has very very steep gradient

[twin pivot]

?

Streched?

[tired walrus]

yes

that multiplication by 100 is a vertical stretch by a factor of 100

your trickery is laid bare if we rescale the axes and see your curve never was vertical at all

[gritty chasm]

also as a side note desmos may behave strangely at 0 since $log(0)$ is undefined.

[soft zealot]

Annie Maqionde

[twin pivot]

Oh yes I had to turn off the zoom

But near 1 it always looks the same

????

[gritty chasm]

??

[twin pivot]

Near 1 it's always straight

[gritty chasm]

not vertical. Very nearly, and if only you zoom it in.

[twin pivot]

I remove It look

[tired walrus]

i can repeat myself

it looks almost vertical

but it ISNT

it IS NOT vertical

[gritty chasm]

visual representations can be very misleading at times

[twin pivot]

And how do you say it's not vertical?

[gritty chasm]

Why would it be?

[tired walrus]

you can see it if you change the axis scale

[twin pivot]

How can I tell it's not straight?

[tired walrus]

or you could calculate the gradient at x=1

[tired walrus]

by choosing an axis scale that doesnt stretch it out so dramatically thats how

what you're doing rn is basically like

"oh the number 1000 is so big it LITERALLY IS INFINITY look at me. look at me i cant tell big numbers apart"

[twin pivot]

Look

Its vertical

I put a lot of zoom

[gritty chasm]

Then explain why it would be vertical.

And I don't get your point here. Do you want to know the answer, or argue uselessly?

[twin pivot]

It was the question

[small grail]

gradient is very slighly less than vertical

[twin pivot]

How do I know it's not vertical?

[tired walrus]

YOU PUT A LOT OF ZOOM SPECIFICALLY TO DELUDE YOURSELF INTO THINKING THAT ITS VERTICAL AND THEN YOU ARE AMAZED WHEN IT LOOKS VERTICAL

[gritty chasm]

GRADIENT

[tired walrus]

you can make fucking y=x look vertical with enough zoom dude!!!!

[small grail]

derivative at that point

[tired walrus]

oooooooooooooooooooooooooooo look the function y=x isnt a function anymore!!!! 😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱😱

[twin pivot]

Its not

[twin pivot]

What

Zoom it

[tired walrus]

I AM TELLING YOU TO "ZOOM IT" AND YOU'RE REFUSING

fuck

[twin pivot]

This

I did it

[tired walrus]

nah ok i think this convo is going absolutely nowhere

you're just not understanding me no matter what i say

it is impossible

[twin pivot]

A photo might be clear

[gritty chasm]

SHE LITERALLY PROVIDED A GRAPH

[small grail]

for you

[gritty chasm]

Look you can even algebraically show it's impossible

and mathematical tools are much more reliable than visual ones

[small grail]

i infinitely zoomed

[twin pivot]

I fixed

I clicked on the graph and the numbers change

[leaden moon]

-# why did the graph look so much like a line even tho this is not a linear function

[proper dagger]

look at the scale.

[leaden moon]

i see

[twin pivot]

👍

[onyx peak]

when you hold shift and ur mouse down, you can stretch the axes

this can help you see whether something is/isnt vertical

[twin pivot]

Thanks all

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[plain hemlock]

question

[plain hemlock]

will i get my money back if i buy 27 boxes of cabbage

in every box of cabbage

has 50 seeds

and i bought it for

1.477m

and at the best price it sells for

1,278

will i make my money back and will there be profit

<@&286206848099549185>

[vital crag]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[vital crag]

And

!show

[final saddle]

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

[plain hemlock]

huh?

[silent mango]

this is not a math problem

[vital crag]

You pinged just after 1 minute

[plain hemlock]

can i just get the answer?

[vital crag]

And showing your work means show what you've done so far

[tired walrus]

!noans

[final saddle]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[plain hemlock]

i bought 27 boxes?

why not

[vital crag]

Just rules. Dm Modmail if you don't like it

[silent mango]

cuz then just go to f**king chatgpt and ask it for the answer

[vital crag]

@dusk parcel specifically

[plain hemlock]

then what is the point of u guys

help specialy

[vital crag]

!vol

[final saddle]

Helpers are just people volunteering their time to help you. Be polite and patient.

[shell condor]

wait wait, did you actually go and buy 27 boxes

[plain hemlock]

then answer

[silent mango]

to help people who are stuck on a problem and want to LEARN how to solve it

[plain hemlock]

maybe...

[plain hemlock]

I WANNA LEARN?

[vital crag]

You're free to leave and go elsewhere

[plain hemlock]

its an fr question

[crude island]

doesn't seem like it

[plain hemlock]

well i do

so can u tell me

or the steps

atlease

so i can do it

[crude island]

well to start can you rewrite your question without pronouns so it is less ambiguous

[plain hemlock]

A farmer bought will he make his money back if he buyes 27 boxes of cabbage
in every box of cabbage
has 50 seeds
and i bought it for
1.477m
and at the best price it sells for
1,278
will i make my money back and will there be profit

[tired walrus]

!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).

[silent mango]

I understand that, its just a joke

[tired walrus]

what are 1.477m and 1,278? what currency are these in?

what exactly are they the buy/sell prices of?

[silent mango]

I think they are the same currency but idk.

def a typo

[crude island]

nah frl

[silent mango]

nah m is meters

[tired walrus]

...

what.

oh, you're not op.

[silent mango]

anyways

he went offline

[bleak granite]

presume the OP meant 1.477 million?

[final saddle]

@plain hemlock Has your question been resolved?

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[wintry mountain]

I stumbled upon this integral, and I do not get an intuition or any idea on how to solve this. I looked at the solution, which split it into partial fractions with denominators (e^x +x^2) and (x). Can someone help me get an intuition or a thought process on how to reach this solution or any other natural solution?

[onyx peak]

partial fractions is the go-to technique when the denominator can be factored, and there is no apparent simpler way

in this case, the denominator is already factored

so partial fractions should be one of the first things that comes to your mind

[wintry mountain]

I see, I will think like that from now on

Thanks!

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[hushed imp]

Is anyone able to explain number 54? I don't know how to do a truth table, and I'm confused.

[final saddle]

@hushed imp Has your question been resolved?

[final saddle]

@hushed imp Has your question been resolved?

[hushed imp]

How do you use a truth table? Did I use it right

<@&286206848099549185>

[sly forum]

<@&268886789983436800>

[final saddle]

@hushed imp Has your question been resolved?

[loud sundial]

vacuous truth

[deep condor]

(a) and (b) look good. for (c), i think you inverted a contrapositive on that first statement ("If it is not Landscape Arch...") whereas you should be inverting the original statement (wich is "If it is the largest natural arch in the US, then it is Landscape Arch.")

For (c), I know you don't know truth tables, but F -> T evaluates as true. think about the statement "if i live on jupiter, then 2+2=4." obviously, no one lives on jupiter, but 2+2 still equals 4, so we take this implication to be true as sort of an edge case.
while i don't know what p and q are, the rest of the table checks out logically.

however! if you don't know truth tables, you don't necessarily need them to solve this. you can use the contrapositive to modify statements until you understand them, then explain why they're true, or provide a counterexample to explain why they're false.

i'm not 100% sure what your teacher wants, but that is what i would ask of a student before teaching truth tables

[final saddle]

@hushed imp Has your question been resolved?

[hushed imp]

Thank you :) I just get hung up on the truth tables about what the trues and false are evaluating, but I get it now. I'll stick without the tables for now then do some later. :)

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[misty axle]

guyss can someone pls help 🙏 why did the signs change bcs of the arrow? i get it its like repel and attract but iim confused bcs in the first one its negative and positive but somehow its still the same <- with the like signs?

[old quarry]

sorry i dont understand ur doubt exactly

are u confused on the signs of the electric fields?

[misty axle]

its okayy, the one where E = E1 + E2

and then theres arrows above themm

[old quarry]

yes

[misty axle]

then she switched one of the answers to a negative bc of it

[old quarry]

negative sign is just to denote left direction

[misty axle]

wait do i have to take note of that for electric field or also for coloumbs and chargeee

[misty axle]

ohh

[old quarry]

like find direction of electric field and then u can substitute magnitude of charge in that formula kQ/r^2 instead of having to deal with negatives

[misty axle]

yesss, but in electric charge its absolute v alue right? 🥹

[old quarry]

yes

magnitude = absolute value

[loud sundial]

discord.gg/physics

for future reference, this is not the right server ^

[old quarry]

lmao

[misty axle]

thank uuu

[misty axle]

OH MB I THOUGHT ITS THE SAME CUZ ITS SOLVING TOO 😭

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[barren pebble]

Given n (n≥3) people standing in a circle in some order clockwise. A swap means switching the position of 2 adjacent people. Find the least number of swaps needed for them to stand in that order counterclockwise.

[barren pebble]

Idk how to approach this yet, so i guess i'll try to find a pattern after looking at small cases

Let f be the function with input n that outputs this minimum number of swaps

f(3)=1

f(4)=2

[obtuse hedge]

Is that any two people or specifically two adjacent people? (just wondering)

[barren pebble]

Adjacent, my bad

If they didn't have to be adjacent then it would be kinda trivial

[obtuse hedge]

Haha that's why I was wondering 👍

[obtuse hedge]

A pattern is arising

[barren pebble]

An infinite amount of patterns

[obtuse hedge]

Haha I meant a pattern in the process that I used, keep going 👌

[barren pebble]

Also feel free to check the numbers i put out, i'm just trying by hand

[obtuse hedge]

Those are the numbers I get so far as well

This is in bijection with a nice problem though

[barren pebble]

This is a Vietnamese test from 1990 btw

VN TST

f(5)=4

I guess we should invent a notation to show the swaps

[obtuse hedge]

Yes, I'm interested what order you used to get 4 🙂

[barren pebble]

I'll let the people be 1 2 3 ... n and a sequence of numbers with each nunber k meaning swapping the kth and k+1-th person

With n meaning swap the first person with the nth person

5-3-2-3

[obtuse hedge]

Nice

[barren pebble]

Since it's the first time i introduce this notation, i guess i should showcase it in case someone hops onto this problem too:
1 2 3 4 5 (original config)
5 2 3 4 1 (5)
5 2 4 3 1 (3)
5 4 2 3 1 (2)
5 4 3 2 1 (3)

Note that another sequence that ends with like 1 5 4 3 2 would also satisfies

[whole halo]

9 8 7 6 5 4 3 2 1

1 8 7 6 5 4 3 2 9

1 8 7 6 5 4 3 9 2
2 8 7 6 5 4 3 9 1

2 8 7 6 5 4 9 3 1
2 8 7 6 5 4 9 1 3
3 8 7 6 5 4 9 1 2

3 8 7 6 5 9 4 1 2
3 8 7 6 5 9 1 4 2
3 8 7 6 5 9 1 2 4
4 8 7 6 5 9 1 2 3

4 7 8 6 5 9 1 2 3
4 7 6 8 5 9 1 2 3
4 7 6 5 8 9 1 2 3

4 6 7 5 8 9 1 2 3
4 6 5 7 8 9 1 2 3

4 5 6 7 8 9 1 2 3

f(2n + 1) ≤ 1 + 2 + ... + n + ... + 2 + 1
= (1 + 2 + ... + n) * 2 - n
= n(n + 1) - n (Gauss)
= nn + n - n
= nn

[obtuse hedge]

!nosols ? or is this an upper bound

[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.

[whole halo]

I dont know the solution to this problem

you can see the ≤ symbol

the ≤ symbol means its an upper bound

[barren pebble]

I guess the logical thing to do now is to check the tightness of the bound by testing a case futher, which means n=11

Has anyone found a pattern on how to swap btw?

[obtuse hedge]

That looks like a similar bound to the one I found as well 👌

[whole halo]

1 7 6 5 4 3 2 8

1 7 6 5 4 3 8 2
2 7 6 5 4 3 8 1

2 7 6 5 4 8 3 1
2 7 6 5 4 8 1 3
3 7 6 5 4 8 1 2

3 6 7 5 4 8 1 2
3 6 5 7 4 8 1 2
3 6 5 4 7 8 1 2

3 5 6 4 7 8 1 2
3 5 4 6 7 8 1 2

3 4 5 6 7 8 1 2
```f(2n) ≤ 2(1 + 2 + ... + (n - 1))
= 2((n - 1)(n - 1 + 1))/2
= (n - 1)n
= nn - n

it makes sense that f(n) = O(n^2)

[obtuse hedge]

Yeah that was my original pattern as well, just slide each element into place one by one 👌

[whole halo]

this is our starting cheap solution

already we have to split into even/odd cases

[obtuse hedge]

DId you succeed in finding f(6) = 6; I have f(6) = 8 so far

[whole halo]

theres something optimal here in that in the longer sections, we have to for example move the 4 all the way to its correct spot
if we dont move the 4, we postpone the inevitable
if we do move the 4, we can only ever continuously move it in one direction, as going back will waste moves
and then that direction would just be whichever is shorter to move it into the correct spot

[barren pebble]

I made a mistake, i did find 7

[obtuse hedge]

Oh what was your 7

[barren pebble]

Though i think 6 should be possible

[whole halo]

3 * 3 - 3 = 6, its possible with the cheap way

[obtuse hedge]

This seems promising with what I have

[barren pebble]

6-4-3-4-2-3-4

But yeah i do believe 6 is optimal and possible

[whole halo]

6 5 4 3 2 1:
6 5 3 4 2 1
6 3 5 4 2 1
6 3 4 5 2 1
6 3 4 5 1 2
2 3 4 5 1 6
2 3 4 5 6 1
```tried to do things in a different order, still the same strategy

[obtuse hedge]

Ah I was about to try it the cheap way thanks

Let me write it down on my paper

[whole halo]

as before the cheap way's steps have been counted

since we only need to count the steps, and the pattern of the steps are clear, we should focus on finding better ways around the cheap way

[whole halo]

I dont see much room for improvement in moving a number across, we will have to think of something else

[obtuse hedge]

Interesting

We have 3 nontrivial steps and then 3 steps which is the fastest way to reverse the order of 3 elements in a row

[whole halo]

the 3 nontrivial steps are just the same as those to reverse the order of 3 elements in a row but harder to see

[barren pebble]

These are so clever

[obtuse hedge]

Yeah I'm rewriting it out on the paper for me to see easier

[whole halo]

in a circle, presumably

[obtuse hedge]

I just noticed you reversed the order at the beginning which threw my calculations off 😂 😂

[whole halo]

its easier to notice when we're done when the output is forwards

alr I gtg, this is as far as I can see that

I gotta do my other math homework which is more boring

[obtuse hedge]

One second before you go

Somehow when I reversed the order I only used 5 steps

Ah I see what happened

Somehow missed one

[whole halo]

thats a shame

[obtuse hedge]

Okay 👋 have fun with homework

[whole halo]

well cya

good luck to me

[barren pebble]

I'm failing to see a general strategy

[obtuse hedge]

Ah now that I've fixed it I see what you were saying mtt

So overall you did two switches

(4,2)(5,1)

Yielding 6 5 4 3 2 1 -> 6 3 4 5 2 1 -> 2 3 4 5 6 1

I can see why this becomes more difficult as n grows

Considering the lower cases again

[barren pebble]

So far we've guessed that the formula for f(n) is of order n² and that it has to do with parity of n, right?

[obtuse hedge]

n = 3 required one adjacent swap
n = 4 required one fast-3
n = 5 required one adjacent and one fast-3
n = 6 required two fast-3's

I suppose from that I can conjecture a tighter upper bound

But can't guarantee that there isn't something better

Oh and the adjacent swap can (WLOG) always be the first and last two elements

[barren pebble]

Yep

[obtuse hedge]

So I should be able to find n = 7 with one adjacent swap and two fast-3's, which will give f(7) <= 1 + 3 + 3 = 7

Which would be

Ah no that idea won't work because the distances to the edges are different

I can do it with 3 fast-3's though

<@&268886789983436800>

It's inefficient though

I assume the fastest way to reverse 4 elements isn't the way requiring 2 fast-3's and an adjacent swap

Time to try mtt's algorithm for 7 and see what it looks like

[barren pebble]

So your current best sol for 7 is 9 moves right?

[obtuse hedge]

10 actually

Since I need a adjacent swap as well

Ah no I can do it with 3 adjacent swaps instead

So yeah 9

Which would be

(5,7) + (4,3)(3,2)(2,1) + (2,4)

1 2 3 4 5 6 7:
(3 moves)
1 2 3 4 7 6 5
(3 moves)
4 1 2 3 7 6 5
(3 moves)
4 3 2 1 7 6 5

Where the last 6 moves are all directed at reversing 4 numbers in a row

So if there is a faster 4 then this gets quicker

I think 6 might be the quickest

For now I'll move on to n=8

n = 3 required one adjacent swap
n = 4 required one fast-3
n = 5 required one adjacent and one fast-3
n = 6 required two fast-3's
n = 7 required two fast-3's and three adjacents

[barren pebble]

I feel like we need more key observations instead of more testing because this is getting difficult very quickly

[obtuse hedge]

I agree but I also have to go to sleep in a second so I want to get as much done as possible 😆

[barren pebble]

Also this is an olympiad question intended to be solved in limited time, so there will be a quick sol

[barren pebble]

Relatable

[obtuse hedge]

Yeah I'm fairly certain it's some sort of bijection with permutations of some sort

There's possible a representation that helps with this

I can do 8 with 13 (10 adjacent + 1 fast-3)

[barren pebble]

Mtt's bound does it with 12

[obtuse hedge]

Yeah I had a feeling it can be done with 12

[barren pebble]

Also i don't really understand where mtt's bound come from

[obtuse hedge]

What does his give for n = 9 by chance?

[barren pebble]

16

He presented that above too

[obtuse hedge]

Just rushing to swap each element in place

Basically the greedy algorithm

[barren pebble]

I do see that strategy failing with large enough n though

[obtuse hedge]

Yeah

Number of strict partitions of 2n + 1 having 1 more even part than odd, so that there is at least one ordering of the parts in which the even and odd parts alternate, and the first and last terms are even.
Possible sequence I'm looking at, I would have to think about the correspondence though

[barren pebble]

Can we determine the ending sequence before finding the sol?

Like for n=9, can we tell that the ending sequence will be 4 5 6 7 8 9 1 2 3

[obtuse hedge]

You should be able to

I'd have to look at it a little longer to devise a strategy

Looks like for n = 8 he uses 3 fast-3's and 3 adjacent

n = 3 required one adjacent swap
n = 4 required one fast-3
n = 5 required one adjacent and one fast-3
n = 6 required two fast-3's
n = 7 required three adjacents and two fast-3's
n = 8 required six adjacents and two fast-3's

Theoretically 11 may be possible then but I'd have to look harder

My brain is stopping working so it may be time for bed 😆

[barren pebble]

Gn to you

[obtuse hedge]

Thanks 🙂 I would say look more at what mtt put

The more I look the more I think it might be closing in on optimal

Ah

I see the 12 is trivial

I'm very falling asleep now but

It's two fast-4's

1 2 3 4 5 6 7 8:
(6 moves)
4 3 2 1 5 6 7 8 
(6 moves)
4 3 2 1 8 7 6 5

[barren pebble]

I love how we just came up with new notations and terms

Kinda like with rubik's cubes if you know what i mean

[obtuse hedge]

Unfortunately I don't 😔

[barren pebble]

[obtuse hedge]

I guess this makes 9 nontrivial because we don't know what's optimal for reversing the order of 5 elements (yet) or if it may be more optimal to switch the order of 3 subsets instead of just 2 (which is what we did for n = 4,5,6,7,8)

[barren pebble]

Hmm... a fast 4 is just moving an endpoint to the other side then a fast 3

[obtuse hedge]

Yep

Also noting that the order that it's performed in doesn't matter

Which is why he can always begin with a fast-3 on the last two elements, and then proceeds to move one element over

Well that fast 4 leaves him with needing to reorder the 5 elements

[obtuse hedge]

Now I'm very much feeling this

16 moves, 6 of which for the fast 4, 10 of which for the fast 5

3 for fast 3

It's giving sum of the first n-1 positive integers for a fast n

i.e. (n-1)(n-2)/2

Ah yes because one may just write the series of transpositions greedily

This is what I did at the beginning and wrote off as being inefficient

But that was because I treated the whole thing as one set, when you split it into two you end up with this

So the question is whether there exists n such that you can split into more subsets to yield a solution with less steps

If the optimal number is always 2, then this method is optimal

[barren pebble]

I can't disprove that yet but i don't think mtt's bound is optimal

[trail mango]

slayla core problem