help-10

[thick oracle]

ok so can you apply this

[fresh relic]

yes

I now the basic rules

but could you help me with this one instead?

[thick oracle]

with this?

[fresh relic]

no I send it in a while

*will

$-6a \div \frac{-2a}{3}$

[warm shale]

Kaasplank3333

[fresh relic]

this one

[thick oracle]

its the same thing

flip the second fraction

and multiply

[fresh relic]

so it becoms $-6a \mult \fraq{3}{-21?}$

[thick oracle]

bro

read this

[fresh relic]

Yeah I know it becoms *\

[thick oracle]

and stop typing bad latex

[fresh relic]

But I was a bit to fast with the latex

[fresh relic]

Im sorry

[thick oracle]

just write it normally

[fresh relic]

Ok

im a bit new to the latex 😂

OMG sorry

[thick oracle]

pls just apply this😭

[obtuse pebble]

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[obtuse pebble]

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[vapid bobcat]

for part d, im not too sure how to simplify and solve further for t

[latent quiver]

I think your work so far is correct

But like

Setting x = y is not what rhe question is asking

[vapid bobcat]

oh?

[latent quiver]

Its asking for you to equate the rate of increase. You equated the amounts

Can the word amount be plural?

[vapid bobcat]

so I would have to differentiate the two equations and then equate them?

[latent quiver]

Yuh

[vapid bobcat]

mkay brb

[vapid bobcat]

got it thanksss!

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[mossy pier]

Hello I am doing transformations homework and I was wandering what wordz go here?

[mossy pier]

I got line 3 :>

Is 2 & 4 reflection?

[flint hull]

The second and fourth would be "line of symmetry" if I am remembering correctly

I believe the first blank would just be "flips" or something along those lines. But it really would depend on what your teacher uses in class and the vocabulary they teach.

"Line of reflection" would technically be the same thing, but I beileve that "Line of symmetry" is what they are looking for.

[mossy pier]

Ok, thank you!

[flint hull]

You're welcome!

[mossy pier]

Is this correct, and would the letters stay the same?

[hidden compass]

No, the letters also flip

And usually they are indicated with a ' as superscript

[mossy pier]

Ye

[hidden compass]

By the way, you have made the wrong reflection

[mossy pier]

Ok

[hidden compass]

That's along the x-axis, not y-axis as they request

[mossy pier]

Ok, thank you!!! :>

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[fringe breach]

Any hints?

[hidden compass]

What is $e^{\ln{x}}$ equivalent to?

[warm shale]

Alberto Z.

[fringe breach]

X?

[hidden compass]

Yep

[fringe breach]

So 2X?

[hidden compass]

And $2\ln{x} = \ln{?}$

[warm shale]

Alberto Z.

[fringe breach]

X^2

Ohh okay

So the integrand is equal to X^2?

[hidden compass]

u² but yeah

[fringe breach]

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[celest edge]

how does the 2nd red line work?

i dont even know if my expansion in red via linearity is correct because i didnt do the sumtoinf x * f(x) thing

[quick nova]

they basically used linearity of expectation

so $E[X + Y] = E[X] + E[Y]$

[warm shale]

MxRgD

[bronze mica]

(not immediately valid for an infinite sum though...)

[celest edge]

i dont think this is enough to clear my doubts

[old lily]

should the first red line have a capital X?

[celest edge]

is my 4th red line correct?

[celest edge]

er sure ok

[old lily]

the 3rd red line is wrong

[celest edge]

wait then must E[] act on a function instead of a variable?

[old lily]

E(5) = 5

[old lily]

it must act on a random variable

like X

or g(X)

[celest edge]

this is not intuitive at all

after understanding linearity what can i recite as a shortcut?

like, when can we swap and how to swap? is it that if we can factorise x in the summation then we can swap?

[old lily]

ok so this is correct

[old lily]

swapping sums and expectation?

you can always do it when the sum is finite

if the sum is infinite, you can still swap sometimes. this can be explained with measure theory

for ur case it's probably fine to swap

maybe

[celest edge]

is it swapped the same way f(g(x)) = g(f(x)) does?

[old lily]

wdym

[celest edge]

like

how to swap

[old lily]

?

you already did it

swapt what

[old lily]

^

[celest edge]

like this?

[old lily]

YES that's fine

it's correct

you already swapped it

[celest edge]

i plowed thru it i wanted sth concise to follow as a rule

[old lily]

just swap the E and sigma symbol

[old lily]

like in the top line

[old lily]

this is concise

[celest edge]

ok cool

also is this the same for int and sum?

[old lily]

yes sure

you need fubini's theorem for that techincally

which is in measure theory

but if u don't know that, whatever, ignore it

[celest edge]

i heard bounds need careful altering but this isnt like double int where the region is sth on a plane to cleanly identify new bounds

[old lily]

right

[celest edge]

idek if sum and int do the exact same thing considering that one is discrete and one is continuous

[old lily]

sums are a special case of integrals using a particular type of measure space

for example, using a counting measure

on a countable set

with the sigma algebra being the powerset

[celest edge]

i didnt learn that much :sob

but i just observed one thing, if the bounds for int and the bounds for sum are all constants, including +inf and -inf, then can i swap without thinking

[old lily]

yeah sure

[celest edge]

hmm alr

[old lily]

like integrating over a rectangle

[celest edge]

so only if maybe the bounds of one of sum/int contains some variable that is used for defining one another (int/sum) would i need to do some transformation

[old lily]

yur

[celest edge]

very nice

thx for clarifying stuff for me

thats all for now

[old lily]

im dead inside

[celest edge]

er ok

gws?

alr bye cya

[old lily]

bye

[celest edge]

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[latent quiver]

this interaction is so hilarious lmao

[tardy epoch]

typical for probability theory questions

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[steep pike]

proved if, need a headstart on only if (ii)

[keen scroll]

I feel like if i give a hint i'll give the answer

Can you compare $A\cap B$ and $A\cap C$ ?

[warm shale]

Lin Xia

[steep pike]

$A \cap C \subset A \cap B$

[warm shale]

AnitaG

[steep pike]

how I do inclusive subset

[keen scroll]

this $\subseteq$ ?

[warm shale]

Lin Xia

[steep pike]

just assume I mean inclusive

but yh

[steep pike]

this is found in part i anyways

dunno what I'm allowed to conclude from dimensions being equal

or what I can conclude

[keen scroll]

wait are you doing $\Rightarrow$ ?

[warm shale]

Lin Xia

[steep pike]

yeah

[keen scroll]

oh sorry

i though you were doing the other

[steep pike]

thats alright

[keen scroll]

ok so you know that $C\cap A\subseteq B\cap A$ but they have the same dimension

[warm shale]

Lin Xia

[keen scroll]

Do you know a theorem for this case ?

[steep pike]

uhh no not rly

[keen scroll]

do you know incomplete basis theorem ?

[steep pike]

nope

[keen scroll]

it says that if you have an linearly independant family in a finite dimension vector space you can complete it in a basis of the vector space

do you know that ?

[steep pike]

wdym family

I know that a basis is a set of lin ind vectors

and shares the cardinality of the dimension

[keen scroll]

yes a set of linearly independant vecotrs

do you know this theorem ?

[steep pike]

no I dont undrstand

[keen scroll]

ok we will do something else

[steep pike]

r u saying that a set of set of lin ind vectors is a subset of a basis

is thta this theorem

[keen scroll]

yes you can say it this way

ok do you know that if F is a subset of E and dim(F) = dim(E) then F = E ?

F subvectorspace sorry

[steep pike]

nah I don't

[keen scroll]

ok let's prove it then

suppose $F \neq E$

[warm shale]

Lin Xia

[keen scroll]

can you prove that $dim(F) < dim(E)$ ?

[warm shale]

Lin Xia

[steep pike]

yes as $ F \subset E$

is this the contrapositive

[keen scroll]

yes

[steep pike]

ah okay got u

so we have $ A \cap B = A \cap C $

and trivially $ A \cap C \subset C $

So $ A \cap B \subset C $

makes sense makes sense

dunno why latex no work

tyyyy

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[radiant cradle]

how would i do this?

[radiant cradle]

i know s bounded by the 2 funcs y bounded by those 2 funcs and x bounded by 0 and 1

from z upperbound func i get p = 2

unsure how to get the rest now

[obtuse pebble]

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[fathom osprey]

I need help proving this when n is not a multiple of 3

I can write it as 1+w^n+w^(2n), but don't understand how to continue for 'n' not multiple of 3

[deep kraken]

,rcw

[mystic folio]

n not a multiple of 3, so n=3k+1 or n=3k+2 (k is an integer)

[fathom osprey]

Yes

[mystic folio]

you can solve each cases seperately, and use the fact that w^3=1

[ember coral]

Idk how to solve that

I can only do math that’s addition or subtraction

🫩

[urban trench]

Complex number

omega & omega square, use their properties

[fathom osprey]

Got it

[urban trench]

!done

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[fickle latch]

If a, b, c, d are four non-negative real numbers with a + b + c + d = 1, show that ab + bc + cd <= 1/4.
this question is really strange to me because it isn't symmetric in a, b, c, d, even though as far as i can tell adding da to the left side still makes it hold
either way i need help with ab + bc + cd + da <= 1/4 because a bunch of stuff i've tried (AM-QM, replacing the 1 in 1/4 with (a+b+c+d)^2, etc) hasn't worked

[obtuse pebble]

@fickle latch Has your question been resolved?

[uneven schooner]

So my first very big hint would be to factorize $ab+bc+cd+da$

[warm shale]

Annie Maqionde

[uneven schooner]

and yes its a trivial application of am-gm

@fickle latch

[fickle latch]

OH

okay hold on

yeah that worked 😭 idk how i missed that

thanks !!

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[proper beacon]

How many possible endboard of tic tac toes are there if 2 endboards that can be obtained from each other through rotation are identical?

[proper beacon]

I know burnside's lemma but idk how to count the endboards

Obviously the number of X's must be equal or one more than the number of O's, and there can't be more than 1 3-in-a-row

But these aren't easy to deal with

[gilded drift]

does mirroring count as rotation just on a different axis

[proper beacon]

No

Basically we're considering the cyclic group C_4

[gilded drift]

from what I'm thinking you could individually check cases: wins for X, wins for O, and draws, whilst including all possibilities for the positions of O(so 2 in 6 squares, 3 in 6 squares, or 4 in 6 squares) in the case of X winning and vice versa + an additional X and O

although that's probably more complicated than a more proper solution

[proper beacon]

Yeah that's a lot of manual counting

Maybe there's some bijection that turns a type of invalid board into a valid one

[gilded drift]

honestly more like combinatorics

[proper beacon]

Which i wouldn't know how to construct one, because i valid boards can have very uneven number of X and Y

Maybe use a "Catalan number"-ish approach to ensure |X|=|O| or |O|+1

But we only care about the endboards anyways, ugh

[obtuse pebble]

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[proper beacon]

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[vapid bobcat]

im not too sure what else to do with this question, this what I've done so far

[mystic folio]

What are properties of ellipses you've been taught?

[latent quiver]

I think there are two ways to go here

Use the definition of an ellipse, or use the directrices to find PS + PS'

[turbid skiff]

perimeter is 2ae

[vapid bobcat]

if u meant that as ur final answer to this question, you're incorrect

[vapid bobcat]

what definition are u referring to be able to find PS + PS'

[latent quiver]

By definition, an ellipse is the set of all points P such that the sum of the distances to the two foci is always constant, and that constant is equal to the length of the major axis

So...

[vapid bobcat]

whats the major axis?

[turbid skiff]

if u do ps +ps' + 2ae ,u get it .what u did is correct

[latent quiver]

Wdym

Are you asking me to say what that means

[vapid bobcat]

yes pls

[turbid skiff]

its along x axis and a is 6

a>b model

[latent quiver]

Its the longe rside

[vapid bobcat]

oh i didnt know that was the term describing that

[vapid bobcat]

whats the proof to show that SS' is 2ae?

[turbid skiff]

u just add them distance can't be negative

ae+ae

hence distance b/w focii is 2ae

[vapid bobcat]

ohhh

[vapid bobcat]

whats b/w?

[turbid skiff]

between my bad 😭

[vapid bobcat]

ahh kk

[turbid skiff]

and as per PS and PS'

[latent quiver]

Does this really matter

You already knoe a value for SS'

[vapid bobcat]

true

[latent quiver]

You just need to prove that PS and PS'are constants

[latent quiver]

Which goes bsck to this

[turbid skiff]

if ur unable to get it ping me or him

[vapid bobcat]

so just saying that PS + PS' = 4root5? or do i need to show more detail/proof?

[latent quiver]

PS+PS' is NOT 4root5

SS' = 4root5

[vapid bobcat]

oh whoops yes i meant that

[latent quiver]

You still dont knoe what PS and PS'are

So no

You cannot just say SS'= 4root5

[vapid bobcat]

I will come back to this after I have dinner brb

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[latent quiver]

Woops

Ok then lol

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[slim ibex]

i am not sure where to start

[slim ibex]

X~B(n, 0.38)

X~N(np, np(1-p)

X~N(0.38n, 0.38n(1-0.8))

this looks really tricky

idk how to do this

P(X>65) = 0.0438

[slim ibex]

i think i can turn this into the standard normal

so

1-0.0438

0.9562

P(Z<a) = 0.9562

okay now idk what to do

i cant work out a

because to work out a i need:

mean, standard deviation and area

i dont have mean, standard deviation

oh wait

i do know them

because its normal\

a = 1.7

[shy moon]

i dont remember exactly but theres something like Binomial success or something like that

probability of success r times out of n

[slim ibex]

isnt it this formula

[shy moon]

it looks like binomial thorem

[shy moon]

this is the binomial theorem

theres a variation in this for probabilit

wait ill search

yea its caled bernoli trial and binomial distribution

$p + q = 1 \implies q = 1 - p$

[warm shale]

Chetan???

[shy moon]

$P(X = k) = \binom{n}{k} p^k q^{n-k}$

[warm shale]

Chetan???

[shy moon]

@slim ibex does that click smth??

[obtuse pebble]

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[obtuse pebble]

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[vapid bobcat]

back

[vapid bobcat]

backk, ok would this be correct for PS and PS'?

[obtuse pebble]

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[vapid bobcat]

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[trim portal]

cant you just plug it into a general polynomial (with integer coefficients)?

[wide aspen]

I think you can

[trim portal]

what did it simplify to?

[wispy wadi]

1-sqrt(2) is irrational, and sqrt(2) too
is their sum irrational ?

[trim portal]

have you tried actually simplifying it for a specific polynomial instead of using sigma notation?

youd probably notice a couple of nice things

||Namely, that it simplifies to p + q sqrt(2), where p and q are rationals||

[wispy wadi]

you don't know the degree of P

but you do know that 1/sqrt(2) is sqrt(2)/2

yeah but assume it does

and find a problem

(you'll find one, P doesn't exist)

yeah

so you can't write deg P = 2

like here

because then it could exist for deg 3

or 4

[wispy wadi]

1. you have to notice that, you're trying to find P such that P(sqrt(2)/2) = sqrt(3) (I think it's easier to see like this)
   P(x) is the sum of a_n x^n and a_n-1 x^(n-1) blablabla

and the coefficients are integers

so when you look at P(sqrt(2)/2)

you can rewrite it a_n/2^n * sqrt(2)^n + ...

each monomial is either a rational, or a rational times sqrt(2)

depending on when n is even or odd

if you sum rationals and rationals times sqrt(2)

what does the result looks like?

[wispy wadi]

(hint)

it does not, you can say P(1/sqrt(2)) is p + q*sqrt(2) with p and q rationals

even if one of them is 0, well, 0 is a rational

it doesn't contradict what we're saying

for now

not yet

I just want you to notice that P(1/sqrt(2)) is p + q*sqrt(2)

for some p and q rational

for now

like, understand why

once you're convinced, it's just solving p + q*sqrt(2) = sqrt(3)

and it will be much easier to see what happens when you square it

if you square the equation you get p²+2q² + 2pq*sqrt(2) = 3

I'll speak some french, I think we'll understand each other better

l'intérêt de ce que je te propose ici, c'est de réfléchir sur la somme des coefficients avant de commencer à faire des choses comme mettre au carré
un polynôme c'est une somme de monômes, et 1/sqrt(2) = sqrt(2)/2
donc ton polynôme à coeff entiers évalué en 1/sqrt(2), ça va devenir un polynôme à coeff rationnels évalué en sqrt(2)

en regroupant les monômes de degré pair et impair, tu sommes des rationnels d'un côté, et des rationnels * sqrt(2) de l'autre

donc on est bien d'accord que ton polynôme évalué en 1/sqrt(2), il donne un nombre de la forme a+b*sqrt(2)

avec a et b rationnels

donc ton problème ça revient à résoudre a+b*sqrt(2) = sqrt(3)

qui n'a trivialement pas de solution en mettant au carré

je pense que t'es toujours en prépa, mais si tu y réfléchis du pdv des extensions

on est juste en train de dire que Q(1/sqrt(2)) = Q(sqrt(2))

de toute façon les extensions c'est hors programme prépa mais ici c'est juste une reformulation de ce qu'on est en train de dire

a+b/sqrt(2), si tu le multiplies par sqrt(2), ça fait a*sqrt(2)+b

quand t'as a et b rationnels, les nombres de la forme a+b/sqrt(2) et ceux de la forme a+bsqrt(2), c'est les mêmes

ça vient juste du fait que 1/sqr(2) = sqrt(2)/2

ici on te le fait juste remarquer avec une notation polynomiale

càd que a_n (1/sqrt(2)^n) = a_n/2^n sqrt(2)^n

ici on est vraiment sur un exo de théorie des nombres, juste t'as pas encore la vision globale parce que la théorie des nombres c'est pas en prépa

la théorie des extensions, que tu vas voir dans le cadre de la théorie de Galois

donc généralement en L3 ou M1 selon si t'es en ENS ou la localisation de ta fac

en école d'ingé y a pas vraiment de théorie des nombres en général

oui c'est très abordable après une prépa

faut juste avoir fini l'algèbre de prépa avant tbh, c'est le seul prérequis

(à peu près, faut en faire un peu +)

c'est parce qu'il commence pas par l'introduction et définit sûrement très rapidement tout ce qui est extension et corps de rupture etc justement

faut faire de l'algèbre générale

puis la théorie de Galois ensuite

en théorie des nombres y a quand même des postes faut être honnête, j'ai un pote qui va bientôt sortir de sa thèse sur les formes modulaires et il est bien lancé

après oui les math théoriques pour décrocher un job faut être chercheur c'est sûr mais bon

sauf que moi j'ai une thèse en géométrie algébrique, je me suis vraiment faite des amis pendant mes études lol, sur ce serveur bcp de gens qui aident sont ceux qui ont fini leurs études

ou à peu près fini

après ça dépend, pense quand même à regarder le rôle des gens qui répondent, parfois y a des lycéens qui se sentent pousser des ailes et qui vont répondre le truc le plus hors sujet que t'aies vu de ta vie

la prépa c'est l'ensemble des pré requis, tu vois toutes les math importantes pour apprendre le reste ensuite

mais si tu vas vers la recherche c'est que le début

après en école d'ingé t'as parfois des sujets quand même assez théoriques par rapport au fait que ça peut servir en ingénierie

j'ai un pote qui avait eu Centrale et qui m'avait dit qu'ils avaient un cours de physique quantique

mais en général ouais faut bosser avec des bouquins

ou parfois au début, en échauffement

vu qu'il est assez court

en général on te garde le "meilleur" pour la fin

(je parle d'expérience après avoir subi le pire exo de géométrie de ma vie en oral parce que le colleur m'a demandé ce que j'aimais, et j'ai répondu la géométrie)

ah celui-là j'ai la réf il est pas si dur

ça idem c'est fraude, ça ressemble à de la géométrie le dernier que t'as posté

mais si t'y réfléchis, quand tu traces une droite à coeff directeur rationnel et qu'elle passe par un point rationnel, elle donne un autre point rationnel

et au final ça devient surtout de la théorie des nombres

ellipse de Steiner

l'idée de la preuve

c'est que les transformations affines ça conserve les barycentres et les points de contact avec les tangentes

et aussi ça transforme les cercles en ellipses (potentiellement des cercles)

donc t'as juste à considérer une transformation affine qui envoie ton triangle sur un triangle équilatéral

les milieux de côtés deviennent des milieux de côtés

et ton ellipse tangente est toujours tangente (aux milieux des côtés)

et un triangle équilatéral a un cercle inscrit tangent aux milieux des côtés

yes

les milieux de côtés et les points de tangence c'est facile, c'est juste une question de barycentres en général en réalité, les transformations affines sont linéaires sur les vecteurs entre tes points

la géométrie affine c'est juste réécrire des arguments d'algèbre linéaire

et idem pour le cercle qui devient ellipse

c'est assez raisonnable à montrer après réflexion

c'est appliquer ta fonction à ton équation de cercle

et ça donne une équation d'ellipse

bonne nuit

[sand roost]

si vous êtes fini veuillez faire .close pour fermer la chaîne :)

[obtuse pebble]

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[bold marlin]

Hello! I’m a 1st year sciences student. The exercise is to calculate the volume.

[bold marlin]

Specifically in this step my teacher wrote this and said we didn’t have to expand (2-x)^2 because we are about to integrate the area. I understand the logic but I don’t follow. I tried to figure it out by myself and I still don’t see it.

Any help would be greatly appreciated.

[restive gorge]

I dont understand the reasoning but you can simply apply a linear substitution

[vale pelican]

nothing stops you from expanding (2-x)^2, but the integration in V(x) is easier to see with a substitution u=2-x

[bold marlin]

I’m unable to see how this is equal to (2-x)^2/4

[vale pelican]

it's probably just a lot of algebra steps omitted to reduce the clutter

\begin{align}
A(x)&=\int_0^{\frac{2-x}2}(2-x-2y)dy\
&=\left[(2-x)y-y^2\right]_0^{\frac{2-x}2}\
&=(2-x)\left(\frac{2-x}2\right)-\left(\frac{2-x}2\right)^2-0\
&=\frac{(2-x)^2}2-\frac{(2-x)^2}4\
A(x)&=\frac{(2-x)^2}4
\end{align}

[vale pelican]

I know that's different from how you wrote it, but I wanted to minimize the number of y's, as well as collapsing as many constants as possible (writing the antiderivative of 2y as y^2 rather than 2(y^2/2) for instance)

[bold marlin]

That’s this, I’m I right?

[vale pelican]

yeah that looks like a fair grouping

ahh I should've included equation markers huh

[bold marlin]

Wait no o think I didn’t do it right

[warm shale]

Flip

[vale pelican]

alternatively, it's
[\left[(2-x)y-y^2\right]_a^b=\left[(2-x)b-b^2\right]-\left[(2-x)a-a^2\right]]
with $b=\frac{2-x}2$, $a=0$

there's nothing wrong with what you've appended, that's just how I saw it

[warm shale]

Flip

[bold marlin]

This?

But how did you substract those

?

[vale pelican]

[\frac{(2-x)^2}2-\frac{(2-x)^2}4=(2-x)^2\left(\frac12-\frac14\right)]

[warm shale]

Flip

[vale pelican]

and 1/2 - 1/4 is 1/4

[bold marlin]

Okkk!!!! Thank you so much! But I don’t think I can simply it like that if I was given another one?

[vale pelican]

the snippet of what you showed us earlier looks correct as well, it's just the clutter that came from substituting before simplifying

maybe you got concerned and stopped when it didn't look similar?

[bold marlin]

Yes, I stopped because, then I’d have to integrate the expression to calculate the volume, and it would so difficult if it wasn’t so simplified

I guess I don’t have enough algebra skills to simply that or something. Lol crying and screaming rn

[vale pelican]

well calculus provides a decent excuse to practice algebra

if you have a textbook you should try some integral exercises, nothing necessarily about volume, just like integrals of random polynomial or rational expressions

you can even make some up yourself

or do what some helpers do, browse the help channels until you see an algebra/calculus question and try it yourself lol

(don't do their homework for them though)

[bold marlin]

Ok thanks for the help, flip!

Wait, I tried again without looking and got to this?

[obtuse pebble]

@bold marlin Has your question been resolved?

[bold marlin]

I think I made a mistake

Let me redo

[vale pelican]

you missed a ^2 from your rightmost term

it ought to be $(2-x)[y]_0^{\frac{2-x}2}-\left[\frac{\cancel{2}y^{\mathcolor{red}{2}}}{\cancel 2}\right]_0^{\frac{2-x}2}$

pretend there are strikethroughs where you have them lol

it looks like just a typo

[warm shale]

Flip

[vale pelican]

yippee

[obtuse pebble]

@bold marlin Has your question been resolved?

[obtuse pebble]

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[bold marlin]

Yesss thanks!

[bold marlin]

I will keep my eye out for constant, because I've noticed that in some cases it's better to leave them inside of the integral to simplify the expression!

[frail vigil]

Hello

I think you accidently opened a help channel

[obtuse pebble]

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

[bold marlin]

I accidentally closed it by going beyond the time out and then came back. My bad.

.close

[obtuse pebble]

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[merry cairn]

but you want every point, which means we will have to be able to scale the AC as we want

[merry cairn]

I got (2x-1)(x+4)

[trim portal]

wdym?

[merry cairn]

Does anyone know how to factories 2x^2+7x-4?

[trim portal]

that is correct

[merry cairn]

O

Thx

[obtuse pebble]

@merry cairn Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[upbeat pine]

Hey

[upbeat pine]

I need help

[short crescent]

!da2a

[obtuse pebble]

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.

[upbeat pine]

I got 5, -3 and 9

Someone pls double check

[short crescent]

yes

[upbeat pine]

332.7

?

[high lily]

how are you getting that?

that's massive

[upbeat pine]

Are u sure?

Let me re do it

[high lily]

actually seems fine

[upbeat pine]

I got the same thing

Yeaaa

-30.2?

I got number 1

5.611

And

2.369

[fossil pendant]

It's a quiz no?

just submit your answers

wait a minute

[upbeat pine]

I submitted my answer already

It just says quiz

But not really

My teachers is just very lazy

[fossil pendant]

and what is this

[upbeat pine]

Im in vacation

[fossil pendant]

does that redirect you to your teacher

[upbeat pine]

And my cousions pc has that software

No I don’t know what it does never used it

How do I close a ticket I already finished

?

[fossil pendant]

by typing .close

[upbeat pine]

.close

[obtuse pebble]

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[obtuse pebble]

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[tidal parrot]

im too lazy to do it

[wary badger]

4a^2 + 6 has a common factor

and

it should be 4a

[tidal parrot]

thanks

.close

[obtuse pebble]

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[gleaming heath]

hint for this problem pls? idk how to start

[high talon]

i mean primes b/w n and n+1000 gradually decreases

so there must be some n\

\

[gleaming heath]

yea i know this

[high talon]

For primes = 5

[gleaming heath]

what

elaborate more pls

[uneven schooner]

do you have to prove this, or is this a yes-or-no question?

[high talon]

Prove no?

[gleaming heath]

i have to prove

i think answer is no bc answer is usually no for these questions

but like i have to prove it anyways

[high talon]

no its prolly yes actually

5 is quite specific

is there any formula for primes b/w to integers

[trim portal]

its definitely yes and it can be proven relatively easily, the only thing you really need is that there are eventually very few primes between n and n+1000

[trim portal]

think about the window n, n+999

start at n=1 and slide it to the right

[uneven schooner]

I'd say consider a function f such that f(m) is the number of primes b/w [m,m+999]

and as you shift m by one, analyze how f changes

[gleaming heath]

oh ok

[uneven schooner]

well as a fact you may use that the no. of priems b/w 1 and 1000 is 168

you'll also have to find a k such that f(k) = 0

think in terms of factorials here

the behaviour of f will give you the further hints

[gleaming heath]

either stays the same or decreases by 1

[uneven schooner]

the answer is yes; the proof is depending on f

[uneven schooner]

correct.

now follow the steps i've outlined

[gleaming heath]

ok

[zenith raft]

the discrete intermediate value theorem

[gleaming heath]

if i find a valuje of k such that f(k)=0 i would be done

[uneven schooner]

well hint try to think in terms of factorials

quick question

is 1000! + 1 a prime?

what about 1000! + 2

[zenith raft]

[uneven schooner]

1000! + 3,.....1000! + 1000?

[gleaming heath]

1000!+2 isnt

[uneven schooner]

correct

go on

[gleaming heath]

oh wait none of them are prime

[thick oracle]

wow annie you know your stuff

[gleaming heath]

fr

[uneven schooner]

well there you have it

[gleaming heath]

wait maybe 100!+1 is prime

[thick oracle]

didnt think someone else wouldve also watched the cambridge interview video

[uneven schooner]

wwwwait i didnt watch it-

[uneven schooner]

it is.

[thick oracle]

ah shit

nevermind then

[gleaming heath]

wait i ment 1000!+1

[zenith raft]

it is impossible to do ts problem without having seen that video

[thick oracle]

this question is very similar to a question posed at a cambridge interview this is why i was asking

[gleaming heath]

same thing tho bc it isn't divisible by any number from 1 to 1000

[uneven schooner]

also prime

[gleaming heath]

oh

so n=1000! works

[uneven schooner]

i didnt see it 💀

[uneven schooner]

perfect

[gleaming heath]

so f(1000!)=0 and we are done

damn thats smart

did u like just come up with that on the spot

[uneven schooner]

some tricks like the 1000! + 1 i knew

[thick oracle]

annie if you actually did it without watching the video you have my praise

[uneven schooner]

well the discrete intermediate value theorem flashed me
and there you got your sol

now write it down

[gleaming heath]

ok

thx so much for the help!

.close

[obtuse pebble]

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[thick oracle]

ah and of course layla also saw it

[obtuse pebble]

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[past torrent]

hello

[vale pelican]

any ideas?

[past torrent]

no ideas

ive used ai for every problem so i dont know where to start

[elder pebble]

these are chill yeah

[karmic bronze]

do you know how to write a system of equation?

[elder pebble]

how much matrces have you done

[karmic bronze]

!nogpt

[obtuse pebble]

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).

[past torrent]

ive heard of it dont know what it is though

[elder pebble]

a matrix?

[past torrent]

i paid hella attention when we did notes and did the first practice

then completely forgot it

[elder pebble]

fair

[karmic bronze]

do you have the notes?

[past torrent]

no system of equation

[elder pebble]

do you remember how to get an inverse of a matrix

[past torrent]

threw everything away yesterday

[karmic bronze]

you don’t know how to write one?

[past torrent]

and to be honest they were pretty shit

[past torrent]

nah

[karmic bronze]

lol

[past torrent]

heck nah

[elder pebble]

you’ll need that for this i think

[past torrent]

so whats our first step

[karmic bronze]

i would first write out the system of equations

[elder pebble]

acc wait would this problem be done using row reduction

cause i’m not the guy for that

i can do it by calculating the inverses outright

[past torrent]

idk what that is

[karmic bronze]

so if i bought 2 roses and r is the cost of one rose, and i buy 6 carnations, let c be the cost of one carnation, and they total to 17

what would
the equation be?

[past torrent]

i got the answer

i didnt use the most ethical way though

[karmic bronze]

ai?

[past torrent]

maybe

[karmic bronze]

if you did, try and tell it to explain how it got the answer

[past torrent]

why

[karmic bronze]

do you have a test on this?

[past torrent]

took our final last week

and its curved as hell

[karmic bronze]

err what math are you going in next year

[past torrent]

ib A&A HL

[karmic bronze]

yeah u need to know matrices for that

[past torrent]

im joking ts way too hard

im doing

mathmetical reasoning and decision making

easiest math at our school

[karmic bronze]

i mean

u def need system of equations

[past torrent]

my friend told me they llearned how to shop

you dont have to take finals as a senior

[karmic bronze]

i think matrices are like needed but like rarely

[past torrent]

so im just ai everything next year

[karmic bronze]

but you won’t learn anything

• what about tests

[past torrent]

i dont need math

[past torrent]

retake and cheat

[karmic bronze]

💔

[past torrent]

i really wish I had time to learn math

[karmic bronze]

what are your future plans bro 😭

[past torrent]

if i was a freshman or 8th Grader

[past torrent]

law

[karmic bronze]

you might need it for like

taxes

idk

[past torrent]

nah

[karmic bronze]

😭

[past torrent]

law or business though fs

i feel guilty using it but I have to I wish I didnt you know

[karmic bronze]

well you don’t really need advanced math

do you know stats?

[past torrent]

i got a 4/5 on my ACT about it

but i sorta guessed C

[karmic bronze]

wow

anyways do you understand anything that the ai said

for the answer

what was the answer anyway

[past torrent]

i havent my asked my logic is sorta get it right go next

[karmic bronze]

yeah that’s correct

do you understand why it did that though

[past torrent]

havent even read it

[karmic bronze]

!done

[obtuse pebble]

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

[karmic bronze]

?

[past torrent]

.close

[obtuse pebble]

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[obtuse pebble]

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[echo plinth]

can anyone tech me qudr=ati cfrom start in 30 min i dont have much time dont know anything about qudt=rtaics still a nerd

[vital herald]

quadratics are in the form of ax^2 + bx + c

[vital herald]

when a > 0 the graph is a u shape
when a < 0 the graph is a inverted u or n shape

[echo plinth]

I dk

Ionly knwo basic math like class 4 student

just gotta tell you

[vital herald]

i see

alr so help with the qn in the screenshot?

[echo plinth]

yep but i am an nerd

so

pls

tell from start

[atomic hornet]

kahn academy has some great stuff on this

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:quadratics-solve-factoring/v/example-1-solving-a-quadratic-equation-by-factoring

https://www.khanacademy.org/math/algebra-home/alg-system-of-equations

[obtuse pebble]

@echo plinth Has your question been resolved?

[obtuse pebble]

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[quick bone]

Apologies for the Polish, I will translate the question:

On sides AC and BC of triangle ABC there has been 2 points marked D and E where the following is true:

|AD| = 4|DC| and |CE|:|EB| = 1:4

We know that, |AE| = 48, |BD| = 36 and |DE| = 10

Prove that AE and BD are perpendicular to each other

[mystic folio]

Is the subject vector?

[quick bone]

no this is just prep for the polish equivalent of the A level exams

I suck at anything geometry related unfortunately so these types of questions give me a massive headache

[mystic folio]

Ok first step, it would help to have a picture

[quick bone]

one isnt provided but this is most likely what it would look like

AD = 4x CD = x CE = y EB = 4y DE = 10 AE = 48 BD = 36

[mystic folio]

Hint is use Thales theorem

It would also help if you call the intersection point of AE and BD something

[quick bone]

.close

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[daring olive]

hey, can anybody help me visualize that for proving sin, cos, tan for 30* and 60*, no matter what length of triangle we take, the values will always be the same?

[daring olive]

i know that it is true, but how, idk?

[mystic folio]

Like, the value of sin 30 is always the same?

[daring olive]

yes

it is always the same no matter what length of equilateral triangle u take

and i tried to prove it wrong but failed

<@&286206848099549185>

[solar elbow]

!15m please.

[obtuse pebble]

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.

[daring olive]

h sure

[mystic folio]

What's the definition of sin

[daring olive]

sin = O/H

[willow flax]

take an equilateral triangle. drop a perpendicular from any vertex to the other side. you get two right triangles. with angles 90,30,and 60
hypo = a
base = a/2
height = root(3)/2 times a (using pytha)
sin30 = o/h = a/2 / a = 1/2

[daring olive]

yeah, oh so u mean that whatever value of a we take, it will always be this only

got it

thx

[night scarab]

,w soh cah toa

[warm shale]

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[night scarab]

you should try to do the rest yourself

[willow flax]

mhm you can use similarity as well

that's easier and more good looking imo

[daring olive]

pls explain

[willow flax]

dyk about similar triangles

[daring olive]

yes

[azure dagger]

i would like to point out since you know trigonometric functions are ratios, hence any value(s) that give the same ratio will always give the same angle
whether it's 1/2, 2/4, 0.5/1, 5/10, etc

[daring olive]

oh, yeah yeah yeah

got the point

[willow flax]

draw a right angled triangle with 30 degree call it ABC. A=30 and B=90
draw another right angled triangle, A'B'C' . same angles but different sizes.
using AA criteria, both are similar. let's suppose the scale factor is k
side1/side2 = k

in first triangle, sin30 = opposite/hypo
second triangle sin30 = k times opposite/ k times hypo
k cancels out which is same as in the first triangle

[willow flax]

basically this

i yapped all this, because you wanted a "visual proof"

[daring olive]

thx

i got it

[willow flax]

mhm

[daring olive]

.solved

[obtuse pebble]

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[obtuse pebble]

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[ebon shard]

So I'm (finally) learning epsilon delta properly, and I am very stuck here trying to figure out how to proceed from here

[ebon shard]

(Please ping on reply btw)

[abstract vortex]

You actually want to show that x+2 also somewhat limited

Actually $|x+2|<4+\delta$

[wary badger]

you want to introduce |x-2| since you have a bound on this

[ebon shard]

How would one go about that?

[warm shale]

ViNton

[wary badger]

|x + 2| = |x - 2 + 4|

[ebon shard]

Ohhhhh

So I get (|x-2| < \frac{\varepsilon}{|x-2+4|})?

[warm shale]

Ryan + Hexadecimal (She | Hex)

[wary badger]

well you want to use the triangle inequality

but yea

[obtuse pebble]

@ebon shard Has your question been resolved?

[pearl ginkgo]

Did you solce the question? Are you still looking for help?

[obtuse pebble]

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[ebon shard]

No but I'm taking a break, I forgot to get back to everyone lol sorry

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[cyan lodge]

Three bikers move in a circle which is 6 km long. The circle has A,C and B points (they go A>C>B).
Distance between A and C = 1km
Distance between C and B = 2km
Distance between B and A = 3

Bikers move at these speeds:
First - 40kmh
Second - 55kmh
Third - 100/3 kmh

I have to find the shortest time it takes for the First one (40) to get to A, Second (55) on to get to C and Third one (100/3) to get to B if they all start moving from A at the same time

[buoyant star]

The hell happened

In this server

[karmic bronze]

?

[random matrix]

discord is having major api issues, large serves are going to be spotty

oh this is a help channel whoops

[karmic bronze]

rip

[gaunt flume]

i'm not sure if i understand your problem, if the distance and speed both are fixed, the time is also fixed, what's the point of minimum time

unless i'm missing smthg here

[obtuse pebble]

@cyan lodge Has your question been resolved?

[slow quarry]

i think, time will be different as they all have different speeds when they all start from the same point.

[gaunt flume]

oh so the question is meaning to ask in which among these three scenarios, the time required will be minimum?

[obtuse pebble]

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[slow quarry]

kinda...... i think they are asking to find just the time taken for each biker

[gaunt flume]

yeah i earlier was thinking about the sum of these times

well nvm

looks like op figured it out

[slow quarry]

we just have to find the time at each case

[gaunt flume]

the bot accidentally assigned this server to you, you should close it i think

unless you need some help ofc

[slow quarry]

yeah...... i was just helping

.close

[obtuse pebble]

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[slow quarry]

done right? im new

[gaunt flume]

no problem, actually the bot is made to assign any available help channel to the person who messages there first, dw it's closed now

also you are doing a great job helping, keep it up!

[slow quarry]

ohh

[slow quarry]

thank you

[obtuse pebble]

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[obtuse pebble]

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[opaque leaf]

.close

[neon vector]

?

[twilit pelican]

[obtuse pebble]

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[buoyant trout]

Guys I am learning this and idk what to do! I need help learning this please!

[buoyant trout]

Mostly need help on the log stuff

[obtuse pebble]

@buoyant trout Has your question been resolved?

[obtuse pebble]

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[dusty grove]

Hi

[obtuse pebble]

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[gray summit]

What angle do I use here?

[gray summit]

in the law of cosines i know its the side opposite to the side we want but idk what that angle is

and i cant put theta since im trying to find magnitude of Vr

[gilded drift]

Vw is parallel to the y axis

so it's a line intersecting two parallel lines

so the angle between Vw and Vq is equal to the angle between the y axis and Vq

oh wait

[gray summit]

u are saying the angle between Vq which is 25? is the same angle as the angle between Vw ang Vq?

[gilded drift]

no no

my idea was wrong

[gilded drift]

try theta being the angle between Vr and Vq and setting up a right triangle with an extension of Vw to the x-axis(just draw a line segment)

Vr and Vq now have the same base, so cos65 * Vr=cos(65+theta) * Vq

[gilded drift]

so Vr=(cos(65+theta) * Vq) / cos65
now use sin the heights differ by Vw so, sin(65+theta) * Vq= (sin65 * Vr)+Vw

[gray summit]

is this still law of cosines it says do it that way

[gilded drift]

find the bottom one expressed as Vr = something, then let Vr = Vr

[gray summit]

the way u said before works

[gilded drift]

oh no it doesn't work, Vr is 65 degrees not Vq so we can't use that to be the angle

however there is an alternative

[gilded drift]

uhhh, let the angle between Vr and Vq= theta, angle between Vq and the x axis is 65+theta, the angle between Vq and the y-axis and by extension between Vw and Vq=25-theta
now you can use cosine law to figure out the value of Vr, Vr^2=Vw^2+Vq^2+VwVqcos(25-theta)
you can also figure out that cosine of the angle between Vw and Vq, 25-theta, multiplied to Vq, cos(25-theta) * Vq = sin65 * Vr + Vw, then you can do cos(25-theta) = (sin65 * Vr + Vw)/Vq, and substitute that in the original eq

oh lol it makes it quadratic

[gray summit]

but we know what Vq is right from the problem statement

[gilded drift]

the diagram says otherwise

unless I'm blind

[gray summit]

i noticed that but isnt that what the problem is saying?

[gilded drift]

ik it's why I was jumbled up

[gilded drift]

this method works if 65 degrees was the angle of Vq not Vr tho

[gilded drift]

maybe ask ur teacher

[gray summit]

i think your first approach makes more sense

because the Vr angle from the x axis is 57.9

from here

[gilded drift]

the wonders physics teachers do when they discover you can make system of equations in physics 🥀

anyways I'm sorry about confusing you for a bit

[gray summit]

ive been trying to figure out this problem for a while

this is another diagram from the book i think they are all drawn like that?

[gray summit]

nah glad we figured it out tho

[gilded drift]

there is an angle in between

they used sine law

I forgot that existed

[gray summit]

no yeah but on top of that is the cosine law

using the angle as 30

[gilded drift]

ye

[gray summit]

alr anyways thanks

how do i close the occupied status of the channel?

[gilded drift]

type .close

[gray summit]

.close

[obtuse pebble]

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[obtuse pebble]

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[thick oracle]

one of thr grpahs is F, one is f and one is f'. The 4th one is a random extra graph

I need to match the graphs

i was thinking that F=B, f=D, f'=C

[quick nova]

I think D and C don't work

for example, graph D is increacsing at x = 2 but graph C is negative there

[thick oracle]

yeah

so it works

if D is decreasing at x=2

and C is the derivative of D

then C should be negative at x=2

[vale pelican]

graph D looks flat at x=2 to me

[quick nova]

I suppose so

[thick oracle]

i think graph D changes monotony at like 1.8 or something

as is shown in graph C

[obtuse pebble]

@thick oracle Has your question been resolved?

[obtuse pebble]

@thick oracle Has your question been resolved?

[thick oracle]

!helpers

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[thick oracle]

oops

<@&286206848099549185>

[surreal plover]

please dont say hi if you dont think you can help them

[uneven schooner]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[uneven schooner]

I'd love to !15m

What's your question?

[thick oracle]

ugh

[thick oracle]

i just wanna know if this is correct

cuz some friends say it's F=D, f=C, f'=A

[uneven schooner]

well B has a minima at 1 and D is zero at 1 but not sure where this is taking us

[thick oracle]

well yes