hold up
#help-33
still temple
that was my first step :/
I just completed the square
and did -8/2 with the whole thing squared
which would be 16
OH WAIT
nvm I see what I did wrong
ok
new final answer
is 6(x-4)^2 - 7
is it right?
@late geode
marsh citrusBOT
marsh citrusBOT
Available help channel!
weak lintel
I am factoring polynomials, and I wanted to make sure I’m doing this one correctly. I have more lesson notes I’m fallowing but I want to just have someone check this one.
marsh citrusBOT
I am factoring polynomials, and I wanted to make sure I’m doing this one correct...
knotty trellis
no, thats incorrect
first try to factor out 8
then do the difference of squares
cobalt sentinel
Try multiplying (x-4)(x+4), and see if u get 8g^2 - 8
weak lintel
This is new and they passed me along in grades, so not clearly
knotty trellis
or 3a^2+6a=3a(a+2)
weak lintel
Could I have a example closer to the original one? Because I’m confused with the only one g and unsure
gray shale
i have to answer true or false :
E a vectorial space
u and v both endomorphism of E, Im(u+v) = Im(u) + Im(v)
Idk what i can use about the fact that u and v are endomorphism to answer
marsh citrusBOT
i have to answer true or false :
E a vectorial space
u and v both endomorphism o...
gray shale
i mean idk how to use the information i was given
marsh citrusBOT
@gray shale Has your question been resolved?
gray shale
i see
night mica
gg
gray shale
i have another one : uov = 0 <=> Im v is include inside Ker u
same problem : don't know how to use endomorphism
night mica
i have another one : uov = 0 <=> Im v is include inside Ker u
yes
u(v(x)) =0 implies v(x) stay in ker u right ?
gray shale
True
night mica
so if y stay in image of v then y=v(x) and
for the previous reasoning you have that y=v(x) stays in ker u
night mica
its used because you can compare im v with ker u because theyr are in the same vector space
night mica
because an endomorphism ist a linear map from a vector space V to V
gray shale
Ah true
frosty linden
Let n integers a1,a2 ..., an be given. Prove that there exists a subset of {a1, a2, - - -, an} whose sum of elements is divisible by n. They say that we can use pigeonhole principle but don't know how
marsh citrusBOT
Let n integers a1,a2 ..., an be given. Prove that there exists a subset of {a1, ...
night mica
set this notation: let A a sub set of {a1, a2, - - -, an}, then $$s(A)=\sum_{a\in A}a$$ is it clear ?
elfin berryBOT
everg
night mica
because there are only n possible rest from division by n ... and there is 2^n subset of {...} ...there are a least two set A and B such that the rest of division by n of S(A) and S(B) are equal
frosty linden
I don't get it
glacial hedge
I don't get it
do you know that there are 2^n different subsets (as in, different indices) you can take ?
frosty linden
no
glacial hedge
because for each number, there's 2 choices: you take it or you don't
There's n numbers so you make that choice n times. Since they're independent, you multiply them to get all the outcomes (think of it like a decision tree). Therefore there's 2x2x2x2.. = 2^n different subsets you can create
frosty linden
so If I take two numbers : 1,2 I can make 4 subsets ?
glacial hedge
yes
Empty
1
2
1 2
frosty linden
ah ok we count the empty one yes
glacial hedge
Let n integers a1,a2 ..., an be given. Prove that there exists a subset of {a1, ...
in this case I'd assume you don't otherwise, as the empty sum is 0 you always have a solution
frosty linden
yes of course
glacial hedge
but 2^n - 1 > n still holds for any n >= 2 (and n = 1 isn't very hard to solve)
frosty linden
so there is more like 2^n -1
glacial hedge
yes
frosty linden
ok
glacial hedge
because there are only n possible rest from division by n ... and there is 2^n s...
alright, so we have 2^n - 1 subsets
since we look at divisibility by n, it is sufficient to look at the remainder of the division by n
frosty linden
yes
glacial hedge
we want to find a set such that it is 0
frosty linden
yes
glacial hedge
since we look at divisibility by n, it is sufficient to look at the remainder of...
which we call the value mod n btw
mod for modulo
frosty linden
yes
glacial hedge
this means these 2^n - 1 subsets take values in [0, n-1]
frosty linden
ok
glacial hedge
then by the pigeonhole principle, two of them are equal
frosty linden
?
glacial hedge
this means these 2^n - 1 subsets take values in [0, n-1]
2^n - 1 > n-1
you do know the pigeonhole principle right ?
frosty linden
ok can we say also that there exists the subset n-1 +1 which is divisible by n also
glacial hedge
you don't have a guarantee that every number is in it
frosty linden
yes this exercise is an introduction to it
frosty linden
you don't have a guarantee that every number is in it
wdym
glacial hedge
you don't have a guarantee that every number is in it
it might be n-1 3s and one 4. In that case you can't do (n-1) + 1 because you don't have these numbers
frosty linden
can't I make a subset which is (1;n-1)
frosty linden
yes but this set exists
glacial hedge
doesn't mean it's a subset though
frosty linden
?
glacial hedge
it might be n-1 3s and one 4. In that case you can't do (n-1) + 1 because you do...
if you have {3, ..., 3, 4}, you can't extract {1, n-1}
frosty linden
oh yes mb I was thinking like (1,2,3,4) like sort mb it's me
glacial hedge
the numbers are random
frosty linden
yes
glacial hedge
then by the pigeonhole principle, two of them are equal
this still applies though
since you have 2^n - 1 values in [0, n-1] which are n "boxes"
frosty linden
no I don't get it sorry
glacial hedge
what does the pigeonhole principle say ?
frosty linden
if n balls are placed in k drawers, one drawer contain at least n/k balls
frosty linden
yes
glacial hedge
so in particular, if n > k, one drawer contains 2 balls
this is the version we're using here
2^n-1 > n
we have more balls than drawers
frosty linden
what is n equal sorry
glacial hedge
Let n integers a1,a2 ..., an be given. Prove that there exists a subset of {a1, ...
n is defined at the very start
frosty linden
oh yes i was confused with the subset
like it says more that there is more subsets than values no ?
so more drawer than balls ?
glacial hedge
the opposite
the subsets are balls, the values are drawers
we're mapping the each subset to its sum mod n, which is a value in [0, n-1]
and there's 2^n - 1 subsets
frosty linden
ok
glacial hedge
then by the pigeonhole principle, two of them are equal
so is this clear ?
frosty linden
no
glacial hedge
you don't see how the principle applies ?
frosty linden
yes
glacial hedge
there's 2 subsets that have to end up in the same drawer
frosty linden
?
glacial hedge
because 2^n - 1 > n so there's more balls than drawers
frosty linden
we ranged subset in a number ?
glacial hedge
I don't understand that question
each subset is mapped to a number, its sum modulo n
the S(A) the others were talking about
frosty linden
if balls are subset and number drawer we range subset in a number
glacial hedge
are you french ?
frosty linden
yes
glacial hedge
ranger se traduit pas par range
glacial hedge
bon anglais pour un lycéen quand même
glacial hedge
ranger se traduit pas par range
store, put in, arrange, place, but not range
frosty linden
oh yes arrange
glacial hedge
a range c'est plus une portée, du style "ranging from 1 to n" pour dire dans [1, n]
frosty linden
oui en plus je sentais que c'était pas le bon terme, on va mettre cela sur le coup de la fatigue
tu es français aussi ?
glacial hedge
1 seule faute gênant la compréhension à cet âge là c'est irréprochable en soi
yes
par contre niveau grammaire, fais attention à tes singuliers-pluriels
et le s à la 3e personne du singulier
frosty linden
ouais là dessus je sais que je fais jamais d'efforts
glacial hedge
t'es en première ?
frosty linden
terminale cette année
glacial hedge
en soi c'est quand même un bon niveau d'anglais
à partir du moment ou t'arrive à communiquer efficacement, c'est bien au dessus de la moyenne je dirais
frosty linden
moyen mais on arrive un peu à se faire comprendre (enfin des fois)
glacial hedge
what is n equal sorry
"equal to" notamment
to equal something
to be equal to something
pas d'entre deux
frosty linden
oui quand j'écris je me permets plus de fautes et je relis pas trop en me disant que la personne comprendra
glacial hedge
mais anyways, c'est un server de maths
frosty linden
english is shit btw
glacial hedge
niveau conjugaisons c'est quand même beaucoup plus simple
à parler faut l'écouter pour l'apprendre c'est sur
et niveau quel petit mot va avec quel verbe c'est pas toujours évident, mais à force on finit par en avoir une intuition
frosty linden
oui je parle juste dans ce serveur et lis quelques bouquin en anglais et le tour est joué
frosty linden
j'y arrive assez bien sans avoir un grand vocabulaire
par déduction on comprend assez bien
glacial hedge
en vrai youtube est très utile pour, car le vocab est moindre et tu vois plus d'accents et de prononciations
c'est comme ça que je l'ai appris
frosty linden
oui je regarde pas mal en anglais(toute façon les contenues si tu veux en avoir des biens ya que ça)
glacial hedge
et ça donne accès à plein de contenu de maths très qualitatif, comme 3blue1brown
ou mathologer
frosty linden
je l'aime bien ce type, il est marrant
glacial hedge
et il vont assez loin dans ce qu'ils peuvent aborder
frosty linden
c'est ça qui rend la chose meilleur
glacial hedge
une des masterclass de mathologer était tombé à Centrale en 2007 je crois. Formule de sommation d'Euler Maclaurin
frosty linden
le type explique plutot bien je trouve n'ayant pas un énorme niveau j'arrive à suivre ces vidéos, en me renseignant aussi de mon coté il aborde plutot bien le coté et est très pédagogue
je ne t'es pas demandé mais tu es de quelle année toi alors ?
frosty linden
quelle école
glacial hedge
louis le grand
frosty linden
ah d'accord une bonne école alors, mais alors tu vas passer les concours bientôt
glacial hedge
3 semaines
frosty linden
ENS et polytechnique c'est ça, ou il y en a d'autres
glacial hedge
cette semaine là oui
frosty linden
va réviser qu'est ce que tu fou là 
glacial hedge
lol
frosty linden
à aider un vieux clampin
frosty linden
tu es de 2005
glacial hedge
2004
frosty linden
ou 2004
glacial hedge
2^11 - 42 - 2
frosty linden
2^11 - 42 - 1
glacial hedge
2^11 - 42 - 2
t'a lu les Douglas Adams ?
frosty linden
non
attends
oui c'est le livre de science fiction qui parle du nombre 42 justement
glacial hedge
affirmatif
frosty linden
d'accord je vois lequel c'est oui, non pas encore lu mais il me tentait bien après en avoir entendu parler par ci par là
glacial hedge
l'humour est très britannique
frosty linden
il faudra que je m'y tente alors
glacial hedge
lit aussi 1984 un de ces 4. Un classique qui ne doit pas être inconnu
frosty linden
oui déjà lu, en anglais d'ailleurs très bon livre je suis friant des dystopies
glacial hedge
je l'avais en 3e car c'était un livre de français
mais j'avais lu animal farm car ma prof d'anglais me l'avais prêté
frosty linden
oui pas mal aussi, j'aime bien georges orwell enfin je n'est lu que ceux-ci
glacial hedge
si tu compte aller en prépa lis des livres de philo aussi
ça sauve des vies aux oraux de français
car actuellement je suis en PLS personnellement
frosty linden
j'ai lu que aldous huxlay et schopenhauer l'art d'avoir toujours raison
pourquoi tu dois faire des oraux de français
frosty linden
oui sartre aussi j'aime bien
glacial hedge
pourquoi tu dois faire des oraux de français
les programmes sont faits par des littéraires hein
le pire c'est les coefs
regarde le poids du français à Centrale
frosty linden
je connais pas trop c'est beaucoup j'imagine
tu as une épreuve écrite de français aussi ?
glacial hedge
17 % aux écrits
contre 22% pour la physique et 34% pour les maths
frosty linden
ah oui c'est énorme
glacial hedge
mais bon
Retournons en à nous tiroirs et nos parties
btw en terminale tu devrais retenir que |P(A)| = 2^|A|
frosty linden
ok
glacial hedge
des fois que ça serve en dénombrement
frosty linden
oui bien sûr mais je voyais pas pourquoi, car dans la correction il ne nous parlait pas de ça
glacial hedge
depuis le temps, est-ce que c'est clair qu'il y a deux parties A et B telles que S(A) = S(B) ?
frosty linden
44²+46²-42 = 2*2005 ok c'est bon j'ai réussi à caser mon 42
non je l'ai pas celle là comment peut on affirmer qu'il existe un S(A) = S(B)
glacial hedge
S va envoie 2^n - 1 parties sur les mêmes n valeurs
2^n - 1 > n donc S n'est pas injective
il y a forcément deux ensembles A et B qui sont envoyés sur la même valeur
t'a pas assez de valeurs possibles pour qu'ils soient tous distincts
frosty linden
est ce que (3,2) = (2,3) ou c'est considéré comme un autre ensemble
glacial hedge
on considère des ensembles d'indices
de sorte à en avoir 2^n - 1
donc {2, 3} = {3, 2} correspond à (a2, a3) (que je ne mets pas comme ensemble car à priori ils ne sont pas distincts)
frosty linden
d'accord
glacial hedge
depuis le temps, est-ce que c'est clair qu'il y a deux parties A et B telles que...
donc?
frosty linden
ils sont congrues pareil mod n
frosty linden
oui
glacial hedge
on est enfin arrivé à la conclusion d'everg
frosty linden
donc si S(A)=S(B) alors on aura un sous ensemble divisible par n ?
glacial hedge
attends depuis le temps il faut que je retrouve comment on finit l'exo
frosty linden
glacial hedge
mais en attendant, j'affirme qu'on peut prendre A et B disjoints
i.e. d'intersection nulle
réfléchit à pourquoi
glacial hedge
on peut trouver deux parties A, B tels que S(A) = S(B) et A inter B = vide
glacial hedge
la démo d'everg marche pas à priori
oh well screw it
on considère Si = {a1, ..., ai}
glacial hedge
la démo d'everg marche pas à priori
terrible désillusion
glacial hedge
on considère Si = {a1, ..., ai}
ça fait n sommes
si l'une d'elle est nulle, CQFD
sinon elles prennent des valeurs dans [1, n-1]
frosty linden
une somme ne peut être nulle
glacial hedge
nulle mod n
glacial hedge
si l'une d'elle est nulle, CQFD
c'est juste une disjonction de cas
frosty linden
une disjonction de cas, comment faire ici
glacial hedge
si l'une d'elle est nulle, CQFD
soit on directement une solution
frosty linden
et puis j'ai l'impression qu'on s'écarte du principe de dirichlet en plus non ?
glacial hedge
sinon elles prennent des valeurs dans [1, n-1]
soit 0 n'est pas une somme atteinte par les Si, auquel cas elles sont dans [1, n-1]
celle là elle marche tkt
glacial hedge
soit 0 n'est pas une somme atteinte par les Si, auquel cas elles sont dans [1, n...
ok ?
frosty linden
ok
glacial hedge
principe des tiroirs
n valeurs, n-1 tiroirs
juste ce qu'il faut
d'où i et j tels que Si et Sj aient même somme
frosty linden
oui
glacial hedge
essaie de conclure
frosty linden
non même avec ça j'arrive pas à comprendre comment faire, donc on a des ensembles qui ont au moins 1 valeurs voire plus et 2 ensembles ayant la même somme
frosty linden
qu'est ce que les Sk déjà j'ai surement déjà mal compris ce passage, c'est l'ensemble (a1,a2,...ak)
frosty linden
oui d'accord
mais avec ça je comprends pas comment conclure, ça nous sert à quoi de savoir ça
glacial hedge
donc a1 + ... + ai = a1 + ... + aj mod n
ai+1 + ... + aj = 0
en soustrayant la somme jusqu'à ai de chaque coté
frosty linden
d'accord oui mais là on dit que la différence de deux sommes égale fait 0 donc divisible par n ça va de soit mais ça ne montre pas que la somme même de l'ensemble est divisible par n ?
glacial hedge
A = (ai+1, ..., aj) convient
frosty linden
A = 0 non car tout les termes s'annule
frosty linden
oui mais on pose S(A) = S(B) et ensuite on dit que l'ensemble A = S(A)-S(B) mais cela vaut 0
frosty linden
mais si ça vaut 0 alors c'est l'ensemble vide ?
glacial hedge
oui mais on pose S(A) = S(B) et ensuite on dit que l'ensemble A = S(A)-S(B) mais...
attention aux notations là
glacial hedge
mais si ça vaut 0 alors c'est l'ensemble vide ?
justement tu compare des nombres et des ensembles là
frosty linden
mais la somme de l'ensemble est la même donc si j'associe chaque termes de deux ensembles on aura 0 donc un ensemble vide non ?
glacial hedge
A = (ai+1, ..., aj) convient
A est non vide car j > i (spdg)
mais la somme de ses éléments vaut 0 mod n
frosty linden
mais supposons alors que dans l'ensemble B il y ait un ak > ai alors quand on va faire ai-ak on va avoir un nombre négatif ?
glacial hedge
mod n
tout est dans [0, n-1]
tous les calculs sont faits mod n
potentiellement ak = 10^k, mais mod n c'est pas gênant
frosty linden
oui mais j'ai l'impression que dans tous les cas on se retrouvera pas avec une somme qui vaut kn et donc divisible par n, où k un entier positif mais plutot une somme qui est égale à 0 soit un ensemble vide
glacial hedge
tous les calculs sont faits mod n depuis le début
glacial hedge
sinon elles prennent des valeurs dans [1, n-1]
ça c'est les valeurs prises par les sommes mod n
frosty linden
ah donc S(A) = S(B) mod n
glacial hedge
oui
frosty linden
plutot que juste S(A) = S(B)
glacial hedge
il était sous-entendu qu'on considérait les nombres mod n
(on peut définir rigoureusement le concept de "nombre mod n" mais ça dépasse le cadre du lycée)
frosty linden
vraiment plus dur ou pas ?
glacial hedge
relation d'équivalence -> ensemble quotient -> Z / nZ
à priori ça passe depuis la terminale
les ensembles quotients c'est un des premiers trucs qu'on voit en MPSI
Z/nZ en est pas un exemple bien compliqué
frosty linden
bon je tacherai de regarder alors, oui j'en ai déjà entendu parler, enfin vue un peu
glacial hedge
il y a d'ailleurs des algorithmes de récupération de données (pour la transmission de signaux légèrement corrompus par le voyage) qui se base sur cette construction, mais en allant beaucoup plus loin
mais là c'est plus accessible à un lycéen
(cf l'arithmétique dans F_256 [X], où F_256 est le corps (unique à un isomorphisme près) à 256 éléments, qui se construit, il me semble, comme ensemble quotient à partir de polynomes de Z/2Z [X])
glacial hedge
glacial hedge
relation d'équivalence -> ensemble quotient -> Z / nZ
tu trouveras les 2 premiers dans le chapitre 2
frosty linden
oh oui merci je vais y jeter un oeil
frosty linden
j'avais déjà trouver quelques ressources
glacial hedge
je te conseille de les télécharger avant la fin de l'année si t'es intéressé
frosty linden
oui je comptais commencer le programme enfin faire ce que je pouvais maintenant de toute façon
glacial hedge
Z/nZ techniquement est dans le chapitre 6
mais à par la notion de groupe en soi le saut se fait bien
frosty linden
et donc là ya tout les chapitres de MPSI
glacial hedge
et donc là ya tout les chapitres de MPSI
il en fait 21 ou 22
il les met au fur et à mesure
d'où le fait de les télécharger avant la fin de l'année, avant qu'il les enlève
ainsi que les TD, DM, DS si t'es intéressé
frosty linden
ah d'accord donc là les MPSI ont fait 14 chapitres sur les 21
glacial hedge
la majorité sont pas bloqués derrière le mot de passe
frosty linden
ok parfait oui je vais télécharger tout ça et essayer de bosser dessus
ah merde ils mettent des mots de passes pour certains cours
glacial hedge
si tu veux un défi regarde le DM 1
le poly de transition de LLG est une mine d'or aussi
frosty linden
pour faire les DM il faut que je suive déjà certains chapitres
glacial hedge
lire le chapitre 1 ne te fera que du bien
et après le DM 1 c'est juste du lycée ++
comme le concours général mais en plus dur quoi
frosty linden
le poly de transition de LLG est une mine d'or aussi
tu les as aussi ?
frosty linden
comme le concours général mais en plus dur quoi
ah oui donc bien énervé déjà
glacial hedge
https://www.cpge-paradise.com/pdf2/Poly_transition_Tosel_New.PDF
Suffit de le chercher
Tosel qui est le prof des MP*1 depuis 31 ans btw
glacial hedge
ah oui donc bien énervé déjà
ce serait très beau d'en faire un tiers en 4 heures à ton niveau
en vrai les 4 premières questions sont juste de l'échauffement
frosty linden
d'accord maintenant que j'ai mon temps de référence plus qu'à charbonner, est ce finissable
glacial hedge
ce serait très beau d'en faire un tiers en 4 heures à ton niveau
tous les DS durent 4h. Jamais personne les finit, à par peut-être le majorant
j'avais eu 13/70 à ce DS
16e/42 en classement
un début d'année mémorable
mais après tkt ça devient plus normal, du style 20 ou 30 sur 50
glacial hedge
de mémoire 1-4, 6,7, 9, 10, 11, 13, 14, 18
je dirais
la beauté du premier DS, c'est que personne a encore appris à rédiger
glacial hedge
comme tout le monde
frosty linden
nan pire
glacial hedge
regarde ses corrigés. ça vaut te montrer des choses
frosty linden
tu ne peux l'imaginer
frosty linden
d'accord je tacherais de le faire alors mais pas sûr que je réussisse à faire déjà un tier
glacial hedge
et puis il y a penser avoir fait et avoir bien fait. Deux choses très différentes
d'où, pour toi, l'intérêt de l'autocorrection
frosty linden
oui c'est sur ce n'est pas la même chose
peut etre devrais je faire le poly de transition avant
glacial hedge
espère pas le finir
not a chance
561 exercices, certains pouvant prendre plus de 2h
frosty linden
mais alors à quoi il me sert si je le finis pas, je dois en faire que quelques un ?
d'ailleurs à t'on la correction des exercices
glacial hedge
ça doit être quelque part en ligne
frosty linden
ok parfait, au pire je demanderais de l'aide sur ce serveur
glacial hedge
apparemment ce site existe
https://cpge-paradise.com/sites.php
donc fais toi plaisir
frosty linden
oh oui quel bonheur
glacial hedge
d'ailleurs à t'on la correction des exercices
je suis pas sur qu'ils aient finis de rédiger le poly de correction lol
frosty linden
un bonheur inefable
glacial hedge
mais un certain nombre d'entres eux sont corrigés sur place
(dans le poly hein, pas en prépa)
frosty linden
d'accord nickel ce fameux tosel ma l'air d'être un sacré matheux
glacial hedge
Tosel qui est le prof des MP*1 depuis 31 ans btw
ah bon
frosty linden
bon je vais close ce channel je pense, pourrais je te poser des questions en privée si besoin (enfin quand t'auras fini tes concours)
glacial hedge
sure
rose wedge
.reopen
How do I use the ratio or root test to show this is convergent or divergent? $$\sum_{n=1}^{\infty}\frac{e^{n}+5}{ne^{n}+5}$$
marsh citrusBOT
How do I use the ratio or root test to show this is convergent or divergent? $$\...
elfin berryBOT
donut
rose wedge
I setup $$a{n}=\frac{e^{n}+5}{ne^{n}+5}$$ $$a_{n+1}=\frac{e^{n+1}+5}{\left(n+1\right)e^{n+1}+5}=\frac{e^{n}e^{1}+5}{\left(n+1\right)e^{n}e^{1}+5}$$
elfin berryBOT
donut
rose wedge
Nothing seems to come out of it
I see no cancellations
That is for
$$\lim_{n \to \infty}{\left|\frac{e^{n}e^{1}+5}{\left(n+1\right)e^{n}e^{1}+5}\cdot\frac{ne^{n}+5}{e^{n}+5}\right|}$$
elfin berryBOT
donut
marsh citrusBOT
@rose wedge Has your question been resolved?
oak void
marsh citrusBOT
oak void
i don't get how they got the 1st line on the second picture
why did they times it by 1/2 at the end
proper zodiac
dx can be replaced by 1/2 du
marsh citrusBOT
marsh citrusBOT
Available help channel!
graceful herald
marsh citrusBOT
tight furnace
what's the period of sin(x)
desert dirge
sin(x) would have a period of 2pi, but doing x/4 stretches sin(x) in the x direction by a factor of 4, so the period becomes 4*2pi=8pi
median anvil
does this just = does not exist because of the denominator I’ve used synthetic division for the numerator but the bottom doesn’t factor
marsh citrusBOT
does this just = does not exist because of the denominator I’ve used synthetic d...
elfin berryBOT
Stephen
cobalt sentinel
The -2 isn’t the coefficient to the leading term in the denominator
mighty snow
does this just = does not exist because of the denominator I’ve used synthetic d...
2?
cobalt sentinel
Therefore the denominator can be factored
mighty snow
-12
rugged cobalt
stop just saying numbers 💀
median anvil
Therefore the denominator can be factored
but how do I factor it out I did the top with synthetic division so far the numerator is (x+2) (x^2-2x+4) and the denominator so far I just got (x+2)
-12 is an option 🤷♂️
just don’t know how yet
rugged cobalt
(x+2)(x+1)=x^2+3x+2
median anvil
hows 2 x 1 = the 3 though
mighty snow
just don’t know how yet
factor
mighty snow
crimson dawn
Can someone describe what combinations and permutations are and perhaps help me figure out the steps for some practice problems?
marsh citrusBOT
Can someone describe what combinations and permutations are and perhaps help me ...
flat raft
crimson dawn
https://www.khanacademy.org/math/statistics-probability/counting-permutations-an...
tysm!
.close
cedar isle
heyo, i was doing one of my study sheets for a test i have and im pretty sure i went wrong somehwere
marsh citrusBOT
heyo, i was doing one of my study sheets for a test i have and im pretty sure i ...
still parrot
well whats the original question and what are you trying to solve
cedar isle
i mightve messed up finding slope or y-int
@still parrot any thoughts?
<@&286206848099549185> anybody else know what i could have done wrong or if i did at all
marsh citrusBOT
@cedar isle Has your question been resolved?
marsh citrusBOT
Available help channel!
cerulean oxide
I need someone who is quite familiar with the mathmatical program "Maple"
cerulean oxide
If so please respond
thin hinge
What do you need help with
cerulean oxide
I have set up
Now, I need to get maple to make an approximation with Eulers method
Im not sure if i first need to set up a differential equation for this set curve or what
marsh citrusBOT
@cerulean oxide Has your question been resolved?
marsh citrusBOT
@cerulean oxide Has your question been resolved?
odd sparrow
marsh citrusBOT
odd sparrow
What exactly is this trying to say?
Everything is included but PnQnR?
or everything isnt included?
marsh citrusBOT
Show your work, and if possible, explain where you are stuck.
odd sparrow
Highlited is the answer, im trying to determine if the highlighted means its included or not included.
burnt garden
I'd say the highlighted part is included.
burnt garden
Everything **is** included but PnQnR?
This.
odd sparrow
Just trying to figure out why that is the case
Would PUQUR just be the full 3 circles?
burnt garden
That's right.
In this case, it's everything that's not P, Q or R. So the only thing the three have in common must be what's being excluded.
odd sparrow
Right...
How are you meant to wrap your head around this stuff?
No matter how much practice I do, I just can't get confident with all of this.
Simple stuff with 2 circles is not as bad
but once it gets to 3 circles, I really do not feel confident.
For some reason, im not really understanding why thats the case
im ending up just route learning the different options.
burnt garden
For some reason, im not really understanding why thats the case
Do you mean in this particular case you sent or in general when it's three circles?
I'm definitely no expert, and perhaps someone does have a way of making this easier, but I'd say just divide the problem into several chunks. If you don't feel confident enough when it's three circles, then just go one by one following the order. In this case, maybe get a piece of paper and draw the diagram with no color. Then just start highlighting everything other than P; afterwards, everything other than Q; and then the same for R. You'll reach that solution.
I don't know if that's helpful at all, hopefully it is.
odd sparrow
I'm definitely no expert, and perhaps someone does have a way of making this eas...
Ahh okay no this is helpful
I do understand this way. Thank you.
burnt garden
No problemo. Good luck, man.
odd sparrow
.close
toxic pagoda
Hey it's french but in the board, there are the salaries and i need to calculate the median
marsh citrusBOT
Hey it's french but in the board, there are the salaries and i need to calculate...
toxic pagoda
Effectif = Cumultative
Effectif cumulé croissant: Cumulative increasing number (i think thats the translation)
Salaire: salary
.close
stark ice
I need subtle hints into what I am not understanding on this problem: #help-19 message
marsh citrusBOT
I need subtle hints into what I am not understanding on this problem: https://di...
stark ice
I've done every sinusoidal graph the same way in the past. I don't see why it would be different now.
I need guidance, not the solution.
flat grail
stark ice
That's my answer and I've graphed it on Desmos website and everything checks out, the highs and lows are spot on.
stark ice
The website says no, so I don't know what the problem is. Unless, it's looking for something else. Ahhhhh
flat grail
That's my answer and I've graphed it on Desmos website and everything checks out...
I pasted it here because it was a link In a link
stark ice
I think you just nailed it. My answer is right but let me undo the distributive property and see if the website likes it better.
No, that didn't work.
flat grail
What did you try enter?
stark ice
14.5cos(2pi/365(t-206)) + 63.5
opaque wolf
let me solve it myself
stark ice
But don't blow the answer for me. I just need a nudge in the right direction.
I know it's gotta be so close because I've done enough of these to understand them. But maybe I am missing a small thing.
stark ice
I think it's a tricky wording. Notice in the last sentence how they put in bold lettering the word "years"?
I'm not sure though.
opaque wolf
,w 78-63.5
elfin berryBOT
stark ice
One guy said my problem was inside cosine but I'm not sure he was right.
I still think it's correct.
Worst case scenario is the website has the wrong answer in the database. But that's a small chance.
stark ice
Are you showing me the way it should be?
I figured it out, this was a very tricky problem. They require you using 2 different units in the cosine function.
I've never been given one like that before iirc.
.close
odd sparrow
marsh citrusBOT
odd sparrow
In the case of this question, is it assumed they are replaced?
I originally thought, if it was chosen at random. The final result will show 3 buttons taken out.
I based my probabilities on that only to find out that their answer was recorded when the buttons are replaced.
Is there any hint in this question that tells you that?
gusty flare
it’s honestly just a badly worded question - it should specify replacement or not
however since they do not give you an exact number of buttons but rather a ratio, that would imply that there is replacement
if there was no replacement, you would get different answers based on the total number of buttons
odd sparrow
however since they do not give you an exact number of buttons but rather a ratio...
Right okay, but say they have 3 red buttons and 2 white buttons, you would assume it is not replacement?
gusty flare
there would really be no way to tell and it would be a bad question
late geode
if a fixed amount was given assume no replacement
you could maybe introduce something to represent the total number of buttons
marsh citrusBOT
marsh citrusBOT
Available help channel!
languid elm
How do I start this?
marsh citrusBOT
How do I start this?
languid elm
dim geyser
Conservation of linear momentum
and given the fact that the collisions will be perfectly elastic
knotty trellis
Let group be defined by following axioms:
a*b is element of the group for any a and b (closure)
a*a^(-1)=e (inverse)
a*e=a (identity)
(a*b)*c=a*(b*c) (associativity)
Theorems I gotta prove:
(a*b)*c=a*b*c
a*e=e*a
a*a^(-1)=a^(-1)*a
marsh citrusBOT
Let group be defined by following axioms:
a\*b is element of the group for any a...
twilit grove
!status
marsh citrusBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
knotty trellis
```
What step are you on?
1. I don't know where to begin
2. I have begun but got...
1 I guess
glass silo
chartbit is here to suffer
Looool I was only here to ping you 
hushed egret
knotty trellis
(a*b)*c=a*b*c would this be even provable?
hushed egret
depends on what you mean by a*b*c
a priori * : G x G -> G is only a binary operation
writing a*b*c is ambiguous
knotty trellis
In the video I've seen they used a○b○c as a notation, they started with same set of axiom as I specified above, but they also used a○e=e○a, and they were able to prove all theorems except for the (a○b)○c=a○b○c
hushed egret
well you drop the brackets because its value is unambiguous due to associativity
without associativity the value would depend on the order of evaluation, so dropping brackets would make it ambiguous
glass silo
Cheaters to use a * e = e * a 
knotty trellis
How would I define associativity then? they defined it as (a*b)*c=a*(b*c), does this definition imply also the (a*b)*c=a*b*c
hushed egret
it's not really an implication
it's a notational issue
the fact that you have associativity means a*b*c has a unique value
no matter how you decide to evaluate it
so the brackets are not needed
hushed egret
i think i proved this last time after proving left and right inverses exist and are the same
knotty trellis
Assuming that the left and right inverse exist and are the same,
a*a^(-1)=a^(-1)*a
a*a*a^(-1)=a*a^(-1)*a
a*e=e*a
So proving one would also prove the other
hushed egret
proving left and right inverses exist and are the same proved the identity acts on both sides
but not the other way round as you've written it
knotty trellis
oh I see
anyways, how would the left and right inverses exist and are the same be proved?
hushed egret
well
try shoving a bunch of a and a^-1 and (a^-1)^-1s together
if i tell you the calculation it'd give it away immediately
knotty trellis
Oh I didnt try the (a^(-1))^(-1)
marsh citrusBOT
@knotty trellis Has your question been resolved?
knotty trellis
well I found out that if (a^(-1))^(-1)=a, then a*a^(-1)=a^(-1)*a, but then I got stuck
I gtg, I will return to this later
.close
marsh citrusBOT
knotty trellis
thx for help snow
marsh citrusBOT
Available help channel!
analog sun
I'm integrating a problem, how can I verify if my values for A,B,C and D are correct?
marsh citrusBOT
I'm integrating a problem, how can I verify if my values for A,B,C and D are cor...
analog sun
open cargo
I'm integrating a problem, how can I verify if my values for A,B,C and D are cor...
Substituting the values you got for A B C D into the decomposed partial fractions and see if you get the original fraction
analog sun
Substituting the values you got for A B C D into the decomposed partial fraction...
This one?
analog sun
okay
analog sun
You should get the same thing back if it’s correct
Fractions aren't matching. Got something wrong.
Thanks @open cargo 👍
.close
serene spire
how do i solve this integral problems?
marsh citrusBOT
how do i solve this integral problems?
marsh peak
Consider substituting for 9 - x^2, works out really well
serene spire
eh?
still temple
how?
marsh peak
You will see
still temple
it looks like a IBP
serene spire
ikr?
marsh peak
It looked like to me as well
But after sub it becomes -1/2 * (9 - t)sqrt(t)
Which can be done with simple power rule for integration
marsh peak
\begin{gather*}
t = 9 - x^2 \
\dd{t} = -2x\dd{x} \
\int{x^3\sqrt{9-x^2}}\dd{x} = -\frac12\int{x^2\sqrt{9-x^2}}\cdot{(-2x\dd{x})} = \ = -\frac12\int(9-t)\sqrt{t}\dd{t}
\end{gather*}
elfin berryBOT
A Lonely Bean
still temple
what does gather mean
marsh peak
Use that when you just wanna have a collection of equations centralised without alignment (I don't know why mine isn't centralised but whatever)
marsh peak
Right
still temple
that was a good substitution
marsh peak
Yup
marsh citrusBOT
marsh citrusBOT
Available help channel!
hearty rune
marsh citrusBOT
hearty rune
The question is how can i find the function of price elaticity
<@&286206848099549185>
i think the answer is diffrence in x / difference in p -4
marsh citrusBOT
@hearty rune Has your question been resolved?
hearty rune
.close
soft tangle
Hello can I get some help on this problem?
soft tangle
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
torn gull
marsh citrusBOT
torn gull
How the fuck does this question make any sense
i understand that 6 parts must equal 12 counters
so 1 part is two counters
the counters don't change so 12 + 10 + 14 = 36 counters
counters dont change so there are 36 counters in the original ratio
5+4+3 = 12
36/12 = 3
5 x 3 = 15
so isla started with 15 counters
.close
rugged cobalt
elp
rugged cobalt
so
ik cos of angle A is (-3/5)
so i wrote
using cosine rule
$a^2 = 7^2 + x^2 -2(7)(x) (-3/5)$
elfin berryBOT
rio.
elfin berryBOT
rio.
elfin berryBOT
rio.
elfin berryBOT
rio.
marsh citrusBOT
@rugged cobalt Has your question been resolved?
elfin berryBOT
rio.
rugged cobalt
then quadratic forula doesnt work cus i get a neagtive sqrt
hi rio, are you still there
yes
helper: i have a solution
me: tyhanks what is it
helper: when you took the 5 over to the other side it was a mistake as it wasnt everything overr 5 so you would need to multipl,y all other terms on the right by 5 also
me: ahh yes thankyou, i have now done it corrctly and hvae came out wiht the correct answer
me: thank you so much helper!!!!
helper: no problem any time!
marsh citrusBOT
@rugged cobalt Has your question been resolved?
opaque wolf
@rugged cobalt u good
rugged cobalt
yeah u
short wadi
which theorm tells us this last line?
marsh citrusBOT
which theorm tells us this last line?
proper zodiac
minor mural
Stuck on how to start this question
marsh citrusBOT
Stuck on how to start this question
@minor mural Has your question been resolved?
minor mural
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
plush delta
How do I get the first derivitve equal to zero
marsh citrusBOT
How do I get the first derivitve equal to zero
plush delta
I got 2x^-1/3 - 2
I set it equal to zero and divde by 2 to get x^-1/3 =1
then I multuiply by x^-1/3 to get rid of the cube root and cubed it to get x= 1
Is this right?
crisp lion
seems so for real x
plush delta
so the crit is 1?
crisp lion
by crit do you mean a local extremum of this y(x) function? if so, then yes
plush delta
I think so its the point where it is increasing or decreasing
Im trying to get the absolute mins and maxes on the graph
of the interval [-1,1]
<@&286206848099549185>
.close
red meadow
marsh citrusBOT
red meadow
Can someone help me understand the last paragraph? I don’t understand the concept of taking the mean on both sides. This whole example is confusing me. Why do we need to model this? We can just look at our savings and see if we have more or less than 100 000 SEK.
I don’t understand why they have complicated it. Why do we need a logistic regression model to see how much we got in our savings
rugged cobalt
if you are only told how much income a household makes
you do not necessarily know how much they save
so that's what the model is for
and what do you mean by not understanding taking the mean of both sides
like why in this scenario that is the mean?
marsh citrusBOT
@red meadow Has your question been resolved?
runic parcel
Equation 2x²+px+3=0 has roots α,β.
Express (α-β)² in terms of p
marsh citrusBOT
Equation 2x²+px+3=0 has roots α,β.
Express (α-β)² in terms of p
runic parcel
I got that αβ=3/2
and α+b=-p/2
sweet pawn
why don't we start directly from the quadratic formula
a = U - V
b = U + V
a - b = (U - V) - (U + V) = -2V
(a-b)^2 = 4V^2
where U is the axis of symmetry and V is the distance from the axis of symmetry to each root
we have that V = sqrt(p^2 - 4(2)(3))/(4)
marsh citrusBOT
@runic parcel Has your question been resolved?
hollow dock
marsh citrusBOT
hollow dock
So im a bit confused with those x and y equations
Because when r is equal to 2, it doesn’t work
Lets say r is 2
And theta(the angle) is equal to 30 degrees
Now we have a 30 60 90 triangle
So
The shortest side would be 1
And the other side would be sqrt of 3
But
We have to multiply that sqrt of 3 by the r
Which is 2
So we get 2*sqrt(3)
Which is not the correct value
….?
Anyone?
still temple
Just because a triangle is 30 60 90 doesnt mean its shortest side is 1
it just means that if the shortest side is x, the other leg is sqrt(3)*x and the hypotenuse is 2x
desert dirge
if r was 2 and the angle 30, then the opposite would be 2sin(30)=1, and the adjacent would be 2cos(30) which is just sqrt(3) anyway
marsh citrusBOT
@hollow dock Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
plush delta
How do I get the derivitive of xSQRT(x+3)?
marsh citrusBOT
How do I get the derivitive of xSQRT(x+3)?
elfin berryBOT
AustinU
plush delta
yeahh that
stark trail
looks like you should use the product rule because you have the product of two functions (x and sqrt(x+3))
stark trail
looks like you did the derivative of sqrt(x+3) incorrectly
be more careful with your power rule
and it will probably sort out
plush delta
ohh okay Ill try rq
I have SQRT(x+3)+(X)(1/2SQRT(X+3)) now
ohh its right but its not simplyfied can you show me how to do that rq>
<@&286206848099549185>
jaunty citrus
plush delta
ohh alr thanks tho
plush delta
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
north owl
marsh citrusBOT
north owl
Hi i need help with this question, is the limit of cos^2(1/x^2) as x approaches 0 = 1? or is my working wrong
tight furnace
It is not 1
-1 ≤ cos^2(1/x) ≤ 1 is correct
but you can't square all sides of the inequality
just because a≤b≤c doesn't mean a^2≤b^2≤c^2
north owl
ahh this makes sense to me... i can imagine especially if its like -1<0.1<2
but how do i justify my working then?
tight furnace
How did 1/x^2 get inside the cosine?
north owl
thats for 2b, after differentiating the equation of the question
marsh citrusBOT
tight furnace
np
marsh citrusBOT
Available help channel!
soft tangle
Can you help me with his problem?
marsh citrusBOT
Can you help me with his problem?
soft tangle
desert dirge
what is cot(t)?
lean flame
tan is tangent, cot is...
soft tangle
Yeah at the end
desert dirge
first, you didnt need to separate the cos^2 (t) into cos(t)cos(t), the other problem is when you made the common denominator
it would be sin^2(t)/sin(t)
not 2sin(t)/sin(t)
spark otter
It's just that the question isn't asking you for the final answer, it's asking you for an intermediate result
spark otter
#help-20 message we did
soft tangle
That’s multiplication
Sin(t)^2 is sin(t) x sin(t)
@desert dirge yo can I get your help on finishing this
spark otter
The website is helping you with a solution step by step, so start back to cot(t) = cos(t)/sin(t)
And it should tell you exactly what to put in the rectangle
soft tangle
What
We did verify it
But never did that step
Alright then so what do I do next?
@spark otter so what next???
Bro are you gonna help or???
What?
spark otter
Sorry I'm helping others at the same time
soft tangle
Alright drop me i’ll find someone else it’s all good
It’s taking me too long to figure this out
spark otter
What we did was good, we need to put everything under the same denominator
Here's the denominator is gonna be sin(t)
soft tangle
It needs to be sin(t) + sin(t)
spark otter
You mean sin(t) × sin(t) right?
spark otter
Look, you wrote the answer on the first line
You literally wrote "cos(t)cot(t) + sin(t) = cos(t)×cos(t)/sin(t) + sin(t)"
soft tangle
They aren’t under the same fraction💀💀💀
spark otter
They aren’t under the same fraction💀💀💀
Well, look at what they want you to do
spark otter
Here
spark otter
They're asking you "cos(t)cot(t) + sin(t) = cos(t)×________/sin(t) + sin(t)" and you wrote "cos(t)cot(t) + sin(t) = cos(t)×cos(t)/sin(t) + sin(t)"
So what do you put in the ___?
spark otter
It's not asking about the final answer
soft tangle
Unless there is an input error
I get what they’re askingg
But it’s not being accepted
spark otter
What did you put in?
spark otter
They're asking you "cos(t)cot(t) + sin(t) = cos(t)×________/sin(t) + sin(t)" and...
Can't you see you need to input cos(t) in the rectangle
spark otter
Don't worry
spark otter
We did all of this before didn't we?
Well now, you can use what we did here right?
soft tangle
Bro I don’t have the work💀
spark otter
Don't worry, I kept what you sent on the previous help channel