help-27

[crystal lava]

like this?

but i can't factor this

[strange adder]

@crystal lava do u have the correct answer available

[crystal lava]

yea

[strange adder]

this would give u about 9.6 and 24.6 which is indeed wrong

@crystal lava actually what u gonna do is equate this quadratic to 180

[crystal lava]

how

what

[last parrot]

AHh i think what is wrong

[strange adder]

[x^2+15x-236=180]

[strange adder]

first apply the alternate segment theorem

[last parrot]

I think the angle 236-15x is this one?

[strange adder]

yes

[crystal lava]

maybe

[strange adder]

u see that angle

[last parrot]

So this is wrong

[last parrot]

You want to find angle beta

Then beta = 180 - alpha

then beta = x^2 by subtending same arc

[strange adder]

just go for beta=x^2 then apply adjacent angles sum = 180

[strange adder]

yea solve this

[crystal lava]

so far so good?

wait

[strange adder]

7 and 8

[last parrot]

Looking good

[last parrot]

There we go!

[crystal lava]

thanks guys

could'nt have done it without you

how do we close this

[hollow vine]

.close

[crystal lava]

.close

[devout snow]

Channel closed

Closed by @crystal lava

Use .reopen if this was a mistake.

[last parrot]

https://tenor.com/view/hu-tao-gif-4781682369701110304

[hasty cargo]

Blud why are you named like ts

[devout snow]

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[plucky siren]

how do i solve this cubic

[azure parrot]

guess a root

[plucky siren]

heh ?

factors of 6 maybe ?

[azure parrot]

try x=-1

[plucky siren]

+-(1,2,3,6) ?

but why do we guess a root and how does that help ?

[azure parrot]

well, here x=-1 is a solution right?

then you can do division

divide that cubic by x+1

to get a quadratic

[plucky siren]

i feel like a clerk atp 🥀💔

[next kettle]

I think you should tell him why guessing the root was useful in this case

[azure parrot]

It's the first thing to do when solving cubic, or any equation tbh

Guessing some roots like try 0,-1,1,2,-2

[coarse flume]

The roots have sums of -6 and product of -6, which combined with the fact 1+2+3=1×2×3=6, a guess would be the roots are -1,-2 and -3
Of course a way that doesn't depend on knowing niche facts would be try plugging in values, once you know one root you can do polynomial division to get a quadratic

[azure parrot]

or this, rational root theorem

[plucky siren]

what if the cubic has complex or imaginary roots ? would you still guess ?

wait

irratonal roots too

[azure parrot]

We don't know if it has complex roots or not, guessing or rational root theorem is just a way to find potential rational root

[thick lotus]

if the root is not nice you cannot guess, its like trial and error for ugly roots you cant find by trivial methods

[plucky siren]

i get it

[thick lotus]

you would need numerical methods

[plucky siren]

mhm

thanks

.close

[devout snow]

Channel closed

Closed by @plucky siren

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[azure parrot]

Cubic formulas 😂

,w cubic formula

[woven radish]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

May 1, 2026, 7:02 AM

[azure parrot]

[thick lotus]

for complex roots you can check the derivate sometimes

if its complex you know you already fucked up

tbh you would not get such cases

just use desmos lowk

[plucky siren]

we can use cuss words in this server ? 😭

[thick lotus]

ye dont be homophobic tho

[topaz axle]

politically neutral ones

[plucky siren]

bet

[proven breach]

brother knows from personal experience

[thick lotus]

ok guys this channel is closed we should move

[devout snow]

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[sharp adder]

<@&268886789983436800>

[devout snow]

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[sand quarry]

what have you tried?

or rather, what do you need help with

[muted folio]

well i dunno about any specific thing i can invoke

like

any property

what property should i use

[sand quarry]

a proof by contradiction definitely comes to mind

[muted folio]

but what is the contradiction am i looking for?

like i said i can't think of any property that would do the job here

[rain summit]

Idk if you learned some derivatives yet

[muted folio]

i do know derivatives

[sand quarry]

Maybe analyse the difference between consecutive terms

[rain summit]

There's a property that squares like 1, 4, 9,... have a difference that differs by 2

1, 4, 9, 16
Differences are 3, 5, 7 that differs by 2

[spiral raptor]

Not to interrupt, but I think a proof would be if you have an arithmetic sequence with common difference d, you can prove that two consecutive numbers a and a+1 have a difference greater than d

[rain summit]

?

[sand quarry]

yeah i think the contradiction gotta be to show that the sequence is bounded when it cannot be

[sand quarry]

this is the way

[versed juniper]

1 25 49 have a common difference of d so you have the right idea but need to flesh out the end of the proof

it's not quite "prove that two consecutive numbers a and a+1 have a difference greater than d" but similar

[spiral raptor]

If the difference in 2 consecutive squares is greater than the common difference, the sequence can no longer have a next term after that point

[rain summit]

But like the squares could be anything

[versed juniper]

the contradiction is constructing an infinite progression from a finite set

[runic prawn]

true

[rain summit]

Not just consecutive

[muted folio]

how do i do that-

[runic prawn]

his point is the difference between consecutive squares goes to infinity

[muted folio]

how

[spiral raptor]

Consecutive would be the best proof I think. Because everything after that has a greater difference

[runic prawn]

so the difference between any 2 squares will eventually be larger than d

[rain summit]

Ah i see now

[versed juniper]

let's start by saying the series is
x1^2, x2^2, x3^2, xn^2
then our difference is xn+1^2 - xn^2 = d

now what?

difference of squares?

[muted folio]

yeaa

[versed juniper]

ok great so we have
(xn+1 - xn)(xn+1 + xn) = d
what does that tell you about
xn+1 - xn and xn+1 + xn

in relation to d

the common difference

[sand quarry]

your difference is going to be [
x_{n+1}^2 - x_n^2 = d \Implies (x_{n+1}-x_n)(x_{n+1}+x_n) = d
]
You may consider the smallest possible difference you could have with something like [
x_{n+1} - x_n \ge 1
]

[muted folio]

both are like factors of d

[versed juniper]

Are there infinite factors of d?

they have to be integer factors

[muted folio]

yea

[versed juniper]

since xn+1 and xn are integers

no it isnt

[versed juniper]

i dont know any number with infinite factors

[muted folio]

uhm how do i get that conclusionn

[versed juniper]

no we wrote xn+1^2 - xn^2 = d

[versed juniper]

both are like factors of d
you know xn+1 + xn and xn+1 - xn are factors of d
and we ASSUME that it is possible to have an infinite arithmetic progression
this implies xn goes on infinitely

so xn+1 + xn and xn+1 - xn being factors of d, where xn infinitely goes on... that would mean d has an infinite number of factors

which is a CONTRADICTION

no integer has an infinite number of integer factors

[muted folio]

so you're saying if we make xn infinitely large?

thats how like we're gonna reach the conclusion?

[versed juniper]

xn as in the terms of the progression

it's an infinite progression so there is no end to the xn's we have

[muted folio]

so lim n->inifinity?

oh wait yeah

its supposed to be infinity

[versed juniper]

not really a limit

its just

[muted folio]

ya ya i missed that part of the question lol-

sorry my brain is kinda frieddd :/

[versed juniper]

lets say n starts at 0 with x0
the progression is x0^2, x1^2, x2^2... xn^2
n -> infinity

[muted folio]

yeah yeah i gotchaa

since its being added

its TOOOOOOOOOO

big

so like

d cannot exist

blah

[versed juniper]

no

[muted folio]

i mean it can

but like

its increasing

[versed juniper]

you dont understand what you concluded yourself

please try to repeat the argument i tried to present

[muted folio]

okay so like

what im understanding is

if we take the common difference for xn^2 and xn-1^2

and then like factor it

so

[versed juniper]

diff of squares yes

[muted folio]

yeah so one term is

(xn+xn-1)

which is well unbounded

cause infinite

right?

[versed juniper]

what is infinite, exactly?

[muted folio]

n

[versed juniper]

n itself is not infinity but there are an infinite number of n's in the progression

likewise xn

(xn + xn-1)
(xn - xn-1)
what is infinite?

[muted folio]

xn + xn-1

right?

[versed juniper]

no like i said n itself is not infinity, nor is xn or xn-1

[hard shoal]

trick question ahh

[versed juniper]

like you need to separate the concept of infinity and infinites

n is a placeholder number okay

infinity is not a number

[muted folio]

yes

sorry

[versed juniper]

there are an infinite number of n's we have in the progression

in other words there are an infinite number of xn's

and xn-1

so, how many xn + xn-1 are there

and how many xn - xn-1 are there

[muted folio]

well an infinite number of those?

[versed juniper]

yes exactly

but hold on...
you claimed earlier that the product of these two is d the common difference

d is a finite integer that is fixed and does not change with the progression

so technically, you just said this integer d has an infinite number of potential factor pairs

is that possible?

[muted folio]

no

[versed juniper]

great so thats the contradiction

in our original assumption that there does exist such a progression, we found an impossible conclusion

hence our original assumption must be false and this progression cannot exist

[muted folio]

but we'll have to assume that d is not zero for that rite?

oh wait yeah

its distinct

i see so like that removes the case that

the factor pairs are equal

which is not possible

meaning all of them are factors

[versed juniper]

i dont quite understand what you mean by that

[muted folio]

was simply stating the obvious that d must not be zero lol

but like thats a given

sorry for the wait was playing a bit around with the factorst trying to see if theres more to be said

[versed juniper]

theres finite factors to extract from d but you need infinite of them to get infinite xn's to sustain the progression

[muted folio]

yeahhh

.close

[devout snow]

Channel closed

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[versed juniper]

now prove that the longest progression can be at most 3 integers!

[devout snow]

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[cosmic lance]

Hi, I am in the midst of a Textile project & I've run into some 3d geometry problems for some of the shapes that I need to cut out of cloth. Currently there are 6 different pieces & for the life of me I cannot understand how to get the area equations for the indivudual pieces.

[cosmic lance]

The larger eclipse section of the object is sort of shaped like a football, with one tip of the triangle ending in a equilateral triangle that is B length on each side

5 & 6 are half circles, I drew them that way just because its something cosmetic I am going to do later in the project

[devout snow]

@cosmic lance Has your question been resolved?

[fair storm]

Is the larger eclipse section you're talking about 1?

[devout snow]

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[mortal temple]

piece 4 is simple, it's just 11 sqin

piece 2 & 3 can be found from the overlap of two square functions, which from desmos approximations, i got as:
\left(x+2.6\right)^{2}+\left(y-1.5\right)^{2}=15
\left(x-2.6\right)^{2}+\left(y-1.5\right)^{2}=15

(image 1)

so, we just need to find the overlapping area of these circles exclusively for where y is positive. we can instead flip the vars x and y and integrate only the +ve area of the \left(x-2.6\right)^{2}+\left(y-1.5\right)^{2}=15 equation which can now be converted into a function, then we integrate from 0 to the intersect points of the plane and then double it.

so now our challege becomes to find the area under the graph of the function \left(y+2.6\right)^{2}+\left(x-1.5\right)^{2}=15

using the circular segment formula, i'm not pasting my math here cause i wrote it down, but it's just plugging in the values gives an area of ~4.76 square units, or, 2x for the piece an area of ~9.52sq units.

for pieces 5 and 6, it looks like what you're trying to say(I THINK) is that the half circumference is 1/2 C or 11/2=5.5, so we'll just operate with r=5.5/pi

area is just \pi\left(\frac{5.5}{\pi}\right)^{2} or approx 9.68 sq. units

and i'm not sure what piece 1 is.. is it two overlapping elipses?

2:9.52
3:9.52
4:9.68
5:9.68

@cosmic lance can you elaborate?

[mortal temple]

bruh

mb

.close

[devout snow]

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[cosmic lance]

yes

[devout snow]

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[cosmic lance]

Its the larger section of 1

[shy osprey]

?

[coarse flume]

Replying to a previous question it seems

[cosmic lance]

Piece 1 is the entire bottom side, its 2 overlapping eclipses

[shy osprey]

i see

[shy osprey]

that help channel is already closed

you can tag them in #math-discussion if you would like to continue it further

[cosmic lance]

I was busy then, would you like me to restart the entire thing in an entire new thread?

[shy osprey]

Well you can continue here

post your question please

ill pin it

[cosmic lance]

^^

I have a solution for everything but piece 1

To find the actual area equation to be able to draw piece 1

[shy osprey]

Hi, I am in the midst of a Textile project & I've run into some 3d geometry problems for some of the shapes that I need to cut out of cloth. Currently there are 6 different pieces & for the life of me I cannot understand how to get the area equations for the indivudual pieces.
(not my doubt, anthropdics)

The larger eclipse section of the object is sort of shaped like a football, with one tip of the triangle ending in a equilateral triangle that is B length on each side

[shy osprey]

.

[shy osprey]

.pin

[cosmic lance]

The eclipses overlap by a length the B variable

[cosmic lance]

Is there any way you could paste the math you had for the overlapping eclipses (Piece 1)?

& for more context, 2 & 3 connect to the lowest point of the large eclipse in piece 1, so then 2 & 3 will end at the intersection of the eclipses to form an equilateral triangle

In 3 dimentional space

[devout snow]

@cosmic lance Has your question been resolved?

[cosmic lance]

Heres a sort of 3d version of what I mean

[devout snow]

@cosmic lance Has your question been resolved?

[devout snow]

@cosmic lance Has your question been resolved?

[devout snow]

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[zealous bronze]

Q 7

[zealous bronze]

I did half diagram

Anyone?

[untold ravine]

did you answer my last q

like what shape do all the points equidistant from AB and CD make

[zealous bronze]

Idk

Straight line ?

[untold ravine]

close

well if it was a straight line how would you identify that straight line

does the angle bisector at O work

[zealous bronze]

Didn't get u

[zealous bronze]

No I can't do that

[untold ravine]

answer this. it will clear things

[untold ravine]

u sure?

[zealous bronze]

If I bisect xy at o then how will it be equidistant ?

[untold ravine]

well take a point on this bisector

and drop the two perpendiculars onto AB and CD

any congruent tringles?

[zealous bronze]

I didn't get ur point

[untold ravine]

draw the angle bisector at O

take a point on it

actually which part did u not follow

[zealous bronze]

Like this?

[untold ravine]

yes

and drop perpendiculars from that point onto the two lines

[zealous bronze]

And label it xy ?

[untold ravine]

sure

[zealous bronze]

Why so ?

[untold ravine]

whats the distance of a point from a line?

its the length of the perpendicular from that point to the line

[zealous bronze]

Yes

[untold ravine]

thats why we drop the two perpendiculars to see if they are equal

[zealous bronze]

Is this right?

[untold ravine]

eah

yeah

[zealous bronze]

But wont it make whole lime equidistant then ,

[untold ravine]

wait no

this is wrong

what was the point you had chosen

[zealous bronze]

Should I have chose o itself ?

[untold ravine]

choose an arbitrary point on the bisector

[zealous bronze]

Ya I did that only

But u said wrong

[untold ravine]

yeah but which point did u choose

[zealous bronze]

Upper portion

[untold ravine]

its not clear to me cus u circled one point and drew lines onto another

alright

<@&268886789983436800>

[crude niche]

<@&268886789983436800>

[untold ravine]

bruh

[zealous bronze]

Check now

[untold ravine]

yeah i understand

but the right angles will be made on ab and cd

not on xy

[zealous bronze]

Wt happened y u called nods ?

[crude niche]

scammer

[untold ravine]

some sort of scam

bruh

[crude niche]

mb

[zealous bronze]

Who scammed ?

[untold ravine]

u missed it

[zealous bronze]

I changed 8t now

[untold ravine]

share

[zealous bronze]

OK?

[untold ravine]

yes

now can you prove that the two perpendiculars are same in length

[zealous bronze]

No

How can I?

[untold ravine]

any congruent triangles?

[zealous bronze]

There is 1 common side

[untold ravine]

yah

[zealous bronze]

2 90 degree

[untold ravine]

ya

[zealous bronze]

I don't see anything else common

[untold ravine]

well remember what XY is

[zealous bronze]

It's a bisected

Or*

So

It splits

Angle equally

[untold ravine]

yessss

[zealous bronze]

Ya so perpendicular of equak length

[untold ravine]

yeah

[zealous bronze]

But wt does this mean with q ?

[untold ravine]

you tell me what it has to do

think about it

[zealous bronze]

That particular point is equidistant

[untold ravine]

well every point on the angle bisector

[zealous bronze]

Yes

[untold ravine]

what about the intersection of the angle bisector and XY (our given line)

[zealous bronze]

But we jst needed 2 points 8ght ?

[zealous bronze]

Didn't understand

[untold ravine]

we have AB and CD

we have somerandom XY

and we have the angle bisector of AB and CD

[zealous bronze]

Yed

[untold ravine]

the angle bisector and XY can intersect at some point P

[zealous bronze]

Wt do u mean by angle bisected?

Isn't xy angle bisector

[untold ravine]

no xy is arbitrary

i mean in ur drawing u labelled the bisector as XY

but im talking of XY in the question

[zealous bronze]

But when I shared pic u didn't say that

[untold ravine]

the name doesnt matter

[zealous bronze]

Huh

Am getting confused

[untold ravine]

just forget everything and use this naming convention

[zealous bronze]

Ok

Next

[untold ravine]

now we label the intersection of XY and the angle bisector as P

does this help

green is the bisector

[zealous bronze]

Yes

[untold ravine]

so P satisfies the condition in the question right

[zealous bronze]

So how did I take perpendicular fro her ?

[zealous bronze]

Yes

[untold ravine]

you took it from P

can you find the second point

[zealous bronze]

Isn't the whole line p equidistant ?

[untold ravine]

p is a point

[zealous bronze]

9h

The whole angle bisected is equid

Istant

[untold ravine]

yes

but im talking about a second point on XY

not on the angle bisector

[zealous bronze]

So is somewhere right side ? Bw c and b

[untold ravine]

yes but how would u find it

tell me one thing

is there only one angle bisector

[zealous bronze]

Oh we ca create one more angle bisector

And similarly prove its equidistant

[zealous bronze]

So by that I proved 2 equidistant points

So that's it done ?

But how do uk that xy comes there onl y

[untold ravine]

yeah done

[untold ravine]

i dont understand

[zealous bronze]

X y should pass from there

How do uk ?

[devout snow]

@zealous bronze Has your question been resolved?

[devout snow]

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[devout snow]

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[tardy carbon]

Kinda confused on ii and iv

[acoustic leaf]

what in particular is confusing you?

[tardy carbon]

kinda the wording

ik thats vague

so for ii

if theres twice as many moniters ig we could start with y < 2z ?

[acoustic leaf]

that's a good start

[tardy carbon]

i meant to put as with eqaul to aswell but my kb doesnt have that

but idk sometimes i find it confusing and it feels like im just guessing icl

[acoustic leaf]

let's just use <= and >= for equal

[tardy carbon]

okay

[acoustic leaf]

the way i think about ii, there are two parts:

• at most tells me the type of inequality
• twice as many monitors as mice tells me what the terms are

so if we said "there are twice as many monitors as mice" that would mean 2y = z

[tardy carbon]

okay okay, so i understand the first point, its the second one whihc i find more confusing

so is it kinda thinking about it oppositely - ik that sound wierd

[acoustic leaf]

yeah the multiplication by 2 does kind of apply to the opposite one

[tardy carbon]

yeah i see when u think about it, it does acc make sense

[acoustic leaf]

so twice as many monitors as mice means that you would have to multiply the number of mice by 2 to match the number of monitors

[tardy carbon]

yeah yeah

hm

i see

i feel like i understand that part now

how about part iv, im more confused on that 😭

[acoustic leaf]

iv is similar to ii but the multiplications happen on both sides

[tardy carbon]

hmm

so like could we just go straight into saying that 3x = 2z

then we see at least so then its 3x >= 2z ?

acc no

it would be more moniters

so 3x =< 2z ?

[acoustic leaf]

yeah

[tardy carbon]

ight

thank you!!

.close

[devout snow]

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[devout snow]

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[devout snow]

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[white elk]

do you know what nCk means combinatorially

(also can I ask if you’re italian?)

[gloomy aurora]

hint #1: tartaglia's triangle = pascal's triangle
hint #2: this is an identity.

[spring oasis]

it's not

[white elk]

it is isn’t it?

[spring oasis]

just because is a triangle is not pascals

[wraith furnace]

Look at Pascals triangle: What does each entry actually mean in Pascals triangle in terms of combinatorics.

Then. look at how you would get that element

[trail eagle]

If not then define what Tartaglia's Triangle is, because Pascal's Triangle is also called Tartaglia's Triangle.

[wraith furnace]

Bro... one google search

[white elk]

regardless, they use the same symbol as n choose k

so it probably is pascal’s

[spring oasis]

I guess i was mistaken

[white elk]

all good

[spring oasis]

I still don't know how to solve the Q

[white elk]

so what do you know about pascal’s triangle

[spring oasis]

is a triangle

was made by tartaglia

[white elk]

true 💀

actually

I believe it’s al karaji

but that’s besides the point

[white elk]

the way it’s usually defined is that the sum of two entries next to one another gives the entry below it

[spring oasis]

how?

[white elk]

So do you see the 10?

it’s the sum of the 6 and 4 above it

and it’s the same for all entries

[spring oasis]

what about it

[white elk]

that’s how the triangle is defined

[white elk]

have you encountered these symbols at all before?

[spring oasis]

wdym

[spring oasis]

what about the 1 at the top

[white elk]

like have you heard of pascal’s triangle, binomial coefficients, or n choose k at all? it just seems about strange question to give without some introduction beforehand

[white elk]

oh the ones on the side are kinda defined to be there

[wraith furnace]

If you have 1 thing, how many ways can you choose 1 thing from that set of 1 thing?

[white elk]

then it’s the sum rule

[white elk]

well it’s 0 choose 0

[wraith furnace]

Wait. Im braindead

LMAO

[white elk]

but I don’t think they’ve seen that

[spring oasis]

ye

[spring oasis]

what about them

[white elk]

oh?

which context have you heard of them?

because these are the same numbers

[spring oasis]

how

[white elk]

I can explain but I need to know where you heard of them from

since there’s about a bajillion ways to interpret them

[spring oasis]

in class

[white elk]

lol

I mean like did you just hear the name or was it defined a certain way

okay I’m just going to go with the combinatorial definition: these numbers are the number of ways to pick k things from n things

for example how many ways are there to pick 2 numbers out of 1, 2, 3, 4?

[spring oasis]

4 c 2

[white elk]

yeah exactly

that’s what these numbers are

[spring oasis]

how

[white elk]

so that’s essentially what this question is asking

here’s a hypothetical:

suppose you knew all of the ways you could pick k things from n things but now you need to find the number of ways to pick k things from n+1 things

[white elk]

so I might list all the ways to pick 2 things out of 4, but what if I wanted to find the number of ways to pick 2 things out of 5 using that information?

[red fractal]

same thing

nCr if there is no restriction right?

[white elk]

uh this is a help channel but no 5C2 ≠ 4C2

[red fractal]

ik

[white elk]

I’m trying to build a recursive formula

[red fractal]

uh huh i see

[spring oasis]

idk

4 c 2 + 5 c 1

[white elk]

what made you say that

good idea but not quite

[white elk]

not this either

[spring oasis]

4 c 2 + 4 c 3

[white elk]

well I’m picking 2 things out of the 5

so let’s break it into two cases

[spring oasis]

5 c 2 = 4 c 2 + ?

[white elk]

how many ways are there to pick 2 total things out of five if I don’t pick 1 vs if I do pick 1?

[spring oasis]

|AnB| = |A| - |AnB^c|

[white elk]

uh let’s think about it for this example first

[white elk]

we don’t really need to think about the most general structure of the problem

[spring oasis]

4C2 vs 5 C 2

[white elk]

so if I do pick the 1, think about how I’ve already completed the action of picking it

so I picked one

now I have one more to pick

so how many ways would that be?

[spring oasis]

w what

[white elk]

can you explain the 5 C 2 part?

[spring oasis]

can you do a drawing

with marbles

[white elk]

um I’m on my phone in a semi-public location

is ascii art okay?

💀

[spring oasis]

ye

but do your best

[white elk]

o o o o o
^5 marbles to pick 2 from.
I can take the first one
o o o o
o
Now there are 4 marbles left and 1 to pick from: 4C1

if I make sure I don’t pick the first one and I throw it away
o o o o
I still have to pick 2 from the 4 remaining so 4C2

[spring oasis]

yeah so

?

5 c 2 = 4 c 1 + 4c2

[white elk]

yes

that’s essentially the crux of the question

[white elk]

what if I had n marbles and had to pick k

how does that generalize here

[spring oasis]

n c k = (n - 1) c (k-1) + (n-1) c k

[white elk]

yes exactly

[white elk]

now how is that related to this

[spring oasis]

we had 5 c 2

where did the marble went

wdym

[white elk]

?

wdym where did the marble went

[white elk]

you mean in the bottom?

it’s gone, reduced to atoms

that’s the case where I didn’t pick the first one

[spring oasis]

wdym

[white elk]

what you answered the question earlier

like if I wanted to pick 2 things

I could either take the first marble or not take it

[white elk]

you got this right? are you unclear how you got there?

[spring oasis]

yes

[spring oasis]

but how

[white elk]

well if I had n things and wanted to pick k, I can divide it into 2 cases:
I pick the first item
I don’t pick the first item

If I pick the first item, it drops into my inventory. how many remaining items are there and how many do I need to pick from them?

[spring oasis]

4c1

[spring oasis]

what if you dont

[white elk]

what if I don’t?

you tell me

[white elk]

in general, this is (n-1)C(k-1)

[spring oasis]

if you don't you still have 5 c2

so 5 c2 = 4c1 + 5c2

[white elk]

look here. the situation does change if I choose not to take it

if I force myself not to take it, it’s exactly the same as if I yeeted it into the sun

[spring oasis]

so the situation is like either I pick the starting marble or one of the others

regardless you always set some marble asside

[white elk]

kind of, maybe I should rephrase.
the starting marble will be part of the group I pick
the starting marble will not be part of the group I pick

[white elk]

yes this is a good way to think about it

you’re making a decision about it either way

[spring oasis]

what group

[white elk]

like the k I pick from

2 in this case

[spring oasis]

how?

[white elk]

how what?

[spring oasis]

I dont understand how you use Binomial of 4

for binom of 5

[white elk]

does this make sense?

[spring oasis]

no

[white elk]

did this make sense?

[spring oasis]

no

[white elk]

did anything make sense 😭

[spring oasis]

no

[white elk]

dam

uhh okay

[white elk]

I guess you have to make a decision on whether to take each marble or leave it

and since we made a decision for the first marble, that leaves 4 more marbles to make decisions on

maybe my explanation style doesn’t work for you (which is totally okay). would you prefer to wait for someone else?

[spring oasis]

you have to pick 2 marbles out of 5
you have to grab 2 marbles at the end of the day
suppose you grab a random marble from the 5 you have
you set it aside, and now there's 4 marbles to pick from
the marble you initially picked either is the one you wanted or it isn't
suppose the marble is the what you wanted then you have 1 marble more to pick from the 4 marbles left
suppose the marble isn't the one you wanted then you have 4 more marbles left and you still need to pick 2

[white elk]

yeah that’s exactly right

it’s like to a tee

[spring oasis]

care to elaborate

I dont see the correlation

[white elk]

what do you mean elaborate you explained it correctly

[white elk]

you mean the formulas?

or the general case?

[spring oasis]

forget it

[white elk]

what 😭

[spring oasis]

is fine I think I got it

[white elk]

okay then lol

[spring oasis]

thanks

wait what is the value of n c k + n c k - 1

[white elk]

ah okay

[white elk]

do you understand this formula now?

[spring oasis]

n c k = n - 1 c k - 1 + n - 1 c k

[white elk]

yup

do you see how that's related to the problem?

[spring oasis]

n c k - 1 = n - 1 c k - 2 + n - 1 c k - 1

[white elk]

that is indeed the same statement

[white elk]

look at the rhs of your (first) equation and this expression

[spring oasis]

2 [(n-1) c (k-1)] + (n-1) c k + (n-1) c (k-2)

[white elk]

uh you don't need to add them

okay here's a hint

[white elk]

replace n with n+1 here

[spring oasis]

n c k = (n-1) c k + (n-1) c (k-1)

[spring oasis]

why

[white elk]

see what happens

[spring oasis]

.close

[devout snow]

Channel closed

Closed by @spring oasis

Use .reopen if this was a mistake.

[white elk]

did we get it?

[spring oasis]

ye

thanks

[white elk]

eyy noice

[devout snow]

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[rough nova]

How to derive this formula for the area of the pedal triangle of a point M wrt ∆ABC with O as the circumcenter using inversion?

[rough nova]

I have seen 2 proofs of this but they both use trig

I think I need to find an inversion that swaps (XYZ) with (O)?

[heady vapor]

ye trig seems fair

I don't exactly get what you're sayin

g

[rough nova]

This is an excercise for inversion, so I need to use inversion

[heady vapor]

sorry might just be language barrier but what's inversion ?

oh ok no I just didn't know this nvm

so if I get it you need to find a function that map from the circumcircle to the pedal triangle

[rough nova]

Well at least that's what i thought

But idk, which is why i'm opening a help channel

Wait... I might have an idea

Circumcevian triangle ~ pedal triangle

[heady vapor]

oh ok I get the idea we use the same tools in optic physics

[rough nova]

So inversion at M with power equals power of M wrt (O) might work

One sec

[heady vapor]

k

[rough nova]

Solved it

.close

[devout snow]

Channel closed

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[lament kraken]

Orz

nice

did you invert about M?

[devout snow]

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[inner star]

Question regarding the constructability of product of numbers with ruler and compass. The definition of D here is the following: $a\in D$ iff the point $(a, 0)$ is constructible from the set of points ${O, I}$, where $O$ is the coordinate origin and $I$ has coordinates $(1, 0)$. The definitions of constructability and allowed opereation are shown in the second screenshot.

[woven radish]

Dedekind

[inner star]

The solution apparently suggests to draw an arbitrary line through $O$ at arbitrary angle, mark $b$ on it and use the similar triangles. My consern is that using only alowed operations, are we allowed to draw an arbitrary angle line? For it seems we can draw only lines between two previously constructed points, and here there is no second one

[woven radish]

Dedekind

[devout snow]

@inner star Has your question been resolved?

[near jolt]

isn't the first line you draw from O already arbitrary?

if you don't want arbitrary angle you can just make a 60 degree angle too. I guess that doesn't adress your question. Undoubtably "random" angle will introduce new length that might not be constructible, so it's best to use a random constructible angle.

[devout snow]

@inner star Has your question been resolved?

[devout snow]

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[inner star]

Sorry for the late responce. No, it's no arbitrarty, it passes through points $O$ and $I$, from the set that is initially given

[woven radish]

Dedekind

[near jolt]

oh so straight edge and compass gives O and I?

but again the point I said was: just pick a random constructible angle

[inner star]

they are the starting points (unit interval lengh) that are given before any constructions

[inner star]

thank you!

[inner star]

btw, could you please recommend some textbooks on ruler and compass constructions with rigorous approach?

[near jolt]

lotta angles can be made

just make buncha circles to get small angles

so you really do have lots of choices

[inner star]

and I guess, the arbitrary circle radii are also not allowed, right?

[near jolt]

😬 I never read on those books or learn much on ruler and compass

[near jolt]

yeah i think so

[near jolt]

except on some application of galois theory

mostly forgotten

[inner star]

still, would like to hear

[near jolt]

that is just lecture I took

no book

[sturdy fiber]

#book-recommendations

[near jolt]

yeah for book recomm you could go there and there are more qualified ppl who can direct you

[inner star]

Thank you!

.close

[devout snow]

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[near jolt]

np!

[devout snow]

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[dreamy spear]

I don't get the solution for this? can someone explain?

[bronze void]

convert 1/3 to decimal?

[dreamy spear]

yeah

[polar chasm]

do you know long division?

[dreamy spear]

yes I do

[polar chasm]

then you can just divide 1 by 3 using long division

you can write 1 as 1.00000000 if it helps

[gloomy aurora]

keep adding more and more zeroes to the write of the decimal point.

[dreamy spear]

so basically it's 10 divided by 3?

[bronze void]

1.0/3

[polar chasm]

with a shifted decimal point

[gloomy aurora]

1.0, to be precise.

you're gonna 'assume' its 10 while dividing, but don't forget the decimal point in the answer

[dreamy spear]

oh

[bronze void]

10/3 is not equivalent to 1.0/3

[gloomy aurora]

similarly 1.00, 1.000 so on so forth. when you're dividing you're 'assuming' the decimal point doesnt exist, but dont forget to put the decimal point in the quotient though

[dreamy spear]

so basically like this right?

[gloomy aurora]

isnt the things flipped here (3 is in place of 1.0, and 1.0 in place of 3, but maybe that's the notation you follow in your country

you're dividing 1.0 by 3, not the other way round

[dreamy spear]

oh wait that makes sense

basically like this?

[polar chasm]

and if you continued like that, the 3's would just go on

so that's why it's 0.3333....

[dreamy spear]

yeah, so the numerator should always be the dividend?

[polar chasm]

fractions are literally division

[dreamy spear]

and denominator will be the divisor?

[polar chasm]

a/b is the same as a divided by b

a/b = a : b

[dreamy spear]

oh

thankyou!!

.close

[devout snow]

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[devout snow]

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[sweet cove]

I’m struggling to understand how we go from raw data to the normal distribution and the probability density function (PDF)
Starting from the beginning:
I have continuous data (like how much customers spend in my shop).
I group it into intervals and make a histogram.
Then I’m told to use frequency density instead of just frequency, I don’t fully understand why.

From there, my book suddenly jumps to something called the probability density function (PDF) and the normal distribution, without clearly explaining the connection

can someone pls help me, I am mostly stuck in the step where my frequency density-class graph is turned suddenly into a smooth curve with something called "probability density" on one axis and the other is empty?! and they call the smooth curve the normal distribution

[sweet cove]

i will be grateful for good

[topaz axle]

continuous data of how much people spend?

i'm like really curious

how that looks

[devout snow]

@sweet cove Has your question been resolved?

[topaz axle]

can you like send the text?

[sweet cove]

what i said is really really what they did

[topaz axle]

well normal distribution is the approximation for this kind of data

[sweet cove]

i mean money is continues data, and then for each chunck from (3-10) for example there is 9 poeple who spent an amount in that range

[sweet cove]

bro this lesson trumatised me for the last 2 weeks

i sleep so i dont study it lol

[sweet cove]

see the paragraph

[topaz axle]

i haven't really studied stats, but there's binomial, poisson, and there's normal, and for some reason real life data is close enough to normal

[sweet cove]

the last one i mean

[topaz axle]

even binomial is close enough to normal if i'm not mistaken

so yes there's actually a leap in logic there

it just works

now this clearly doesn't work

[sweet cove]

thank you man but i guess whatever they are doing in my book needs a person who have taken stats and studied like really well, i am just so confued

[topaz axle]

why is X negative, it's all wrong

but they move the center to the average spending

and it becomes close enough

there's no good reason why they didn't mark vertical axis

[sweet cove]

frowny no man please, one more graph and i would die

lets stick to my question

[topaz axle]

well i literally don't know anything else

[sweet cove]

thanks man

[topaz axle]

oh you said they didn;t mark horizontal

[sweet cove]

the vertical is probability density

what is this?

[topaz axle]

yeah vertical is density

[sweet cove]

what is that

[topaz axle]

it's just probability of getting that X

but because it's continuous it's wrong to say that

you can only take an interval and integrate

[sweet cove]

how they have taken me from the frequency density-class graph to the normal distrubution curve, with new axes, just by simply drawing a curve over my bars?

[topaz axle]

they noticed it looks similar

they need to add a parameter, sigma²

so it's actually correct

well more corect

“Bell curve”

that's this thing if you heard about it

it just magically looks like real life data often

so they kinda aren;t trying to say anything you didn;t get

“look, similar!’

[sweet cove]

bro what

i am thinking of sleeping agai

thats suffering

[topaz axle]

i always tell people to go sleep

[devout snow]

@sweet cove Has your question been resolved?

[devout snow]

@sweet cove Has your question been resolved?

[devout snow]

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[clever sphinx]

I did I and for ii I got ln(coshx) im not sure how to go abt doing iii tho

[primal ferry]

if y=tanh^-1 x, x = tanh y and dx = ... (basically thus forth treat it how you would a usub i.e. changing the limits etc)

[clever sphinx]

Ohh ok

I’ll try that

That works great

.close

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[lost laurel]

this is the formula for cramer rao lower bound I'm usinf

for information lower bound, I use this?

[dim perch]

correct

notation tripped me up for a sec since i usually see $\text{Var}(\hat\theta)$ but yeah

[woven radish]

blanketism

[lost laurel]

Cool

Thanks

.clpse

.close

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[devout snow]

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[sudden pulsar]

Hello can anyone help me

Please help

[heady crown]

show your work

[sudden pulsar]

I’m confused on it all

Ok wait

Is that right

[hollow jolt]

its essentially asking what percentage of the normal distribution curve is above 180 so you would find the z score and find the probability for that z score

[sudden pulsar]

Z score

(X - mew) / sd?

[hollow jolt]

idk how it works in ur curriculum but you should have some way of finding the probability of a datapoint being below that z score in a tables book or something

[sudden pulsar]

Thank u!

.close

[devout snow]

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[devout snow]

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[solemn holly]

i dont know how to find B

so far i found A and C, and AC. but im struggling to visualise this and dk what my next step is

[devout snow]

@solemn holly Has your question been resolved?

[ocean haven]

have you tried drawing it ?

A,B,C lie on the line

and |AC| = |BC|

[pearl thorn]

If u cant visualize it draw it

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[subtle lotus]

how come in here we are allowed to do b times height over 2 even tho is not right angled triangle

[crude niche]

it is because we know the full height of the triangle

[subtle lotus]

I tho we had to use 1/2 ab sinC

[crude niche]

nope that’s for sides

[subtle lotus]

so when do i use this formula

and when do i use abSinc

[crude niche]

when you need side c

[crude niche]

ohh

[fierce heath]

you can use 1/2 bh when you know the length of the base and the altitude (height) from that base

[crude niche]

you use 1/2bh when you need the area of a triangle

for ab sin c is for side c

[fierce heath]

isnt 1/2 ab sin c for area too?

[crude niche]

yeah

ohhh

[fierce heath]

idk what "ab sin c" is for side c is about then

[subtle lotus]

i got it guys

[crude niche]

wait i’m thinking of cosine law

[crude niche]

yes this is for area

[subtle lotus]

thanks for trying to help

[graceful cosmos]

You can use ab sin(c) to get the area of that triangle. I invite you to try!

[crude niche]

if you know height and base this you use 1/2 bh

[mossy cliff]

A=​a×b×sin(ab)/2

[crude niche]

!done

[devout snow]

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

[subtle lotus]

.close

[devout snow]

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[fierce heath]

ill add that 1/2 ab sin c is a form of 1/2 bh

[mossy cliff]

yes

what is the perimetre of elipse?

and why it s diffuclte to calculate it

[marble badger]

idk why this one is sorta chilling, but you may need to grab a mew help channel

for example @mossy cliff use #help-21｜아리스킨충1 is someone isn't already in it by now

[willow helm]

yes

[mossy cliff]

oh im sorry idk exctly how its going here

[devout snow]

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[willow helm]

No problem, #❓how-to-get-help explains the process nicely 🙂

[neon wagon]

can someone help here

[neon wagon]

i have no idea how to proceed

[supple knot]

write out formulas for P(X=1) and P(X=0) and E(X) = 10 using X is binom(n, p) so your answer will be in terms of n and p

[neon wagon]

so 10 = n * P(X=1) or wdym

[supple knot]

no

[neon wagon]

wait this is without caclulator

theres no way wed have to do that stuff

[supple knot]

you should have learned the formula for expectation of a binomial random variable

[neon wagon]

wait i dont understand the difference, there are 2 forumals

one is the (n,q,p)

the other is

E = n * p

[supple knot]

E(X) = n * p yes

[supple knot]

this isn't a formula at all

[neon wagon]

yeah but the p^k stuff

whatever that was

[supple knot]

????????????

[neon wagon]

the n over k

whatever

so i did say 10 = n * P

wait how does this help

we still have 2 undefined

[supple knot]

you've got 1 out of the 3 things done

P(X=0) is simpler

[neon wagon]

oh so

10 = n * P

or

isnt that the same

or 1-p

[supple knot]

this is E(X) = 10 = n*p again

[neon wagon]

yes

[supple knot]

P(X=0) is a probability so it's between 0 and 1. not 10

[neon wagon]

isnt P(X=0) just q

or 1-p

[supple knot]

no

plug 0 into the probability mass function for the binomial distribution

[neon wagon]

nah wait lets skip that this is more importnat

[supple knot]

that's completely different

[neon wagon]

so the possibiliy for P(X=10) is 5,5 and 4,6 and 6,4 right

[supple knot]

5*5 is not 10

[neon wagon]

wait

isnt it +?

the possible events

yeah this doesnt make sense this problem

[supple knot]

product means multiplication *

[neon wagon]

how will u get X=15 if u only throw dice 2 times

[supple knot]

sum means addition +

[neon wagon]

oh my god i didnt read that

oh my god

yeah so

2 * 5

5 * 2

AND 3*5 AND 5 TIMES 3

its the same

[devout snow]

@neon wagon Has your question been resolved?

[devout snow]

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[blissful comet]

Please help me out I dont get this at all

[blissful comet]

61-66

[trail eagle]

For the graphs it's nice to try some values on the axes to see what you should expect the function to look like.
For instance, in 61, the function along the x-axis (y=0) should look like z=sin(x*0) = 0. Along the y-axis (x=0), it should also look like z=sin(0*y) = 0.

Can you spot graphs which seem to be 0 on the x and y axes?

Or at least rule out some of them.

[devout snow]

@blissful comet Has your question been resolved?

[blissful comet]

0 on x and y?

[trail eagle]

Like if you look at the graph along the x and y axes, the function should be 0 there.

If you look at graph A for instance, if you follow the y axis the function is going up and down, so it can't be the graph of 61.

[blissful comet]

for graph A the function is going up and down in xy plane

so shouldnt it be graph of 61

cus sin(xy)

[blissful comet]

hey

[devout snow]

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[simple mesa]

hi

[lyric hornet]

hello! do you have a particular question in mind?

[hasty cargo]

Can you please post your question?

[simple mesa]

doing calc and sequences

<@&286206848099549185>

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[simple mesa]

The question is right their son

[rain summit]

Asking the actual question right away is more likely to get responses.

[simple mesa]

ok

.close

[devout snow]

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[simple mesa]

.reopen

[devout snow]

✅ Original question: #help-27 message

[simple mesa]

halp

[lyric hornet]

what have you tried thus far?

[devout snow]

@simple mesa Has your question been resolved?

[simple mesa]

i dont know how to approach it

[plush quarry]

I'd first target the sum, and then the integral