help-42

[unkempt nimbus]

does that mean there are special conditions for that method to be used?

[ivory lion]

not really, as long as its a finite geometric series

if its infinite, then yes, a couple restrictions

[unkempt nimbus]

for example how?

[ivory lion]

what do you mean how

how what

[unkempt nimbus]

the implementation

[ivory lion]

i dont understand your question

[unkempt nimbus]

nvm, i forgot what i wanted to ask

but is there a way to simplify this

[ivory lion]

not really, thats about as simple as it gets

thats why it's the formula

[unkempt nimbus]

okay then, thanks for everything

.close

[calm coral]

Channel closed

Closed by @unkempt nimbus

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[nocturne heron]

Let $\Gamma$ be the parabola with equation $y^2 = 2px$ for $p>0$.

1. If $M(x, 0)$, with $x \geq 0$, show that there exists a unique point $M'(x', 0)$ with $x' > x$ such that the square whose one diagonal is $(MM')$ has its two vertices on $\Gamma$

[potent lotus]

tm

[nocturne heron]

do someone even understand the problem ? 😭

[glad parrot]

tu as trouvé ça où

[nocturne heron]

exo de colle lol

[stoic oyster]

t'as essayé de dessiner ce qui se passe avec cette parabole et ce carré pour commencer ?

[nocturne heron]

oui j’ai essayé de dessiner les points M et M’

[stoic oyster]

ouais donc faut calculer les 2 autres points du carré qui sont censés intersecter la parabole

[nocturne heron]

le carré de côté MM’ ?

[stoic oyster]

de diagonale MM'

c'est pas la même chose

[nocturne heron]

ah

[stoic oyster]

l'autre diagonale elle serait où ?

[nocturne heron]

perpendiculaire à MM’

[stoic oyster]

effectivement c'est un aspect important

[nocturne heron]

pour les côtés du carré je trouve MM’ *sqrt(2)/2 ?

[stoic oyster]

ouais

ça localise pas trop les deux points encore

[nocturne heron]

okay cool

jvais aller manger je reviens après 😅

[stoic oyster]

aight

[stoic oyster]

de quelle manière t'as trouvé ce résultat ?

(ping moi quand t'es de retour)

[calm coral]

@nocturne heron Has your question been resolved?

[nocturne heron]

pythagore

[stoic oyster]

oui donc tu connais les 2 "petits côtés" de ce triangle

[nocturne heron]

oui

[stoic oyster]

c'est suffisant si tu veux trouver les coordonnées des autres points

comment tu trouves P là ?

[nocturne heron]

P c’est le milieu de MM’ à la hauteur MM’/2

$P((x’-x)/2,(x’-x)/2)$

[potent lotus]

tm

[stoic oyster]

almost

[nocturne heron]

[stoic oyster]

(x-x')/2 ça marche pas si tu veux le milieu entre M et M'

[nocturne heron]

ahhhh

oh purée

x’+x/2

[stoic oyster]

tu veux une moyenne

oui

donc P est en (x'+x)/2, (x'-x)/2

et le 4ème point tu passes juste la coordonnée verticale de P en négatif et c'est bon

le 4ème point est pas super utile de toute façon

[nocturne heron]

okay attends je vais faire un meilleur dessin lol

[stoic oyster]

si P est sur la parabole, le 4ème est automatiquement dessus aussi

[nocturne heron]

ça fait un truc comme ça quoi

[stoic oyster]

oui

[nocturne heron]

stylé

comment on montre que M’ est unique ?

[stoic oyster]

bha tu connais P

faut trouver les conditions sous lesquelles P est sur la parabole maintenant

faut que j'aille graille aussi :/

mais ouais faut juste insérer le point dans l'équation de la parabole

[nocturne heron]

t’inquiète je vais essayer de m’en sortir 😭

[stoic oyster]

puis t'essaie de résoudre x' en fonction de x (et p aussi)

[nocturne heron]

oki merci

[glad parrot]

tm qui met une vitesse à toute sa promo

[nocturne heron]

🤣 sans aplatypus j’aurais rien pu faire

[glad parrot]

x = abs(argth(y))

[nocturne heron]

pk ya un arctan qui pop comme ça mdrr

[glad parrot]

Graph la tu verras pourquoi je dis ça

[nocturne heron]

on dirait l’inverse de la fonction

pour déterminer sous quelles conditions P appartient à la parabole, qu’on remplace la composante en x ou en y ça change rien enft ?

oopsie j’ai rien dit

[glad parrot]

Tant que tu l'écris pas sur la belle feuille A4 d'exam c'est fine

[nocturne heron]

t’inquiète c’est un exo de colle de 2008, j’pense qu’il tombera pas dans mes partiels de fac

[glad parrot]

L'archive carrément

[nocturne heron]

ah gars j’suis entrain de me booster la

[glad parrot]

Après les coniques ça s'enseignent plus ici

Ducoup ça donne quoi pour tes conditions sur P ?

[nocturne heron]

attends 😭

relou l’équation du second degré

$x’=x+2p+2\sqrt{p(2x+p)}$

[potent lotus]

tm

[stoic oyster]

looks good

[nocturne heron]

on peut montrer que c’est une bijection j’imagine ?

[stoic oyster]

effectivement

[nocturne heron]

injective ca se voit et pour la surjectivite on résout juste pour x f(x)=y quoi

[stoic oyster]

en fait surjective non

pas sur R+ en tout cas

en x=0 t'as pas x'=0

[nocturne heron]

oopsie

elle est pas bijective alors ?😭

[stoic oyster]

nope

[nocturne heron]

ah rip

bon au pire pas grave on a trouvé la fonction c’est déjà bien

pour la 2), pour montrer que x_n diverge faudrait montrer qu’elle est croissante et pas majorée ?

je dirais même strictement croissante si je ne m’abuse

u_1 > u_0 et après ça se fait par récurrence je pense

[stoic oyster]

oui c'est pas très dur t'as f(x) > x pour tout x de toute façon

[nocturne heron]

yes

[stoic oyster]

le pas majoré est quand même un peu galère comme ça

[nocturne heron]

oui c’est que je me disais lol

[stoic oyster]

mais en vrai il suffit de montrer que x_n converge pas vers une valeur finie

[nocturne heron]

on peut montrer que c’est pas une suite de cauchy?

[stoic oyster]

et ça avec les suites récurrentes il y a de bons théorèmes

points fixes etc...

si la suite converge, elle converge vers un point fixe

[nocturne heron]

yes

[stoic oyster]

si t'as pas de point fixe elle converge pas

[stoic oyster]

mais ça ça te dit direct qu'il y a pas de point fixe

tu vas pas avoir de points avec f(x) = x si f(x) est toujours > x

[nocturne heron]

purée t’es un génie enfaite

moi qui voulait m’embêter a montrer que c’est pas de cauchy 😭😭

[cunning veldt]

hmm c'est pas suffisant de montrer qu'elle converge pas par contre

[nocturne heron]

ah ui

[stoic oyster]

oui c'est possible que je dise n'imp

[nocturne heron]

$u_n=(-1)^n$ contre exemple ?

[potent lotus]

tm

[stoic oyster]

oui mais strict croissante + ne converge pas

[cunning veldt]

toute suite qui ne converge pas mais qui diverge pas vers l'infini est un contre exemple

[stoic oyster]

c'est ça qu'on a

[glad parrot]

Si tu marches tout droit sans t'arrêter oui tu finis à l'infini

[cunning veldt]

oui ça oui

[stoic oyster]

oui je parlais dans le contexte d'une suite str croissante, my bad

[cunning veldt]

mais le problème c'est que si tu forces les choses comme ça au final montrer pas convergente ça revient à montrer pas majorée, t'as le même pb d'origine

[glad parrot]

Aah Avec le point fixe ça ma l'air quand même plus simple non ?

[cunning veldt]

lequel des théorèmes du point fixe ?

[glad parrot]

Si ya pas de point fixe ça converge pas

Celui que aplatypus a mentionné plutôt d'ailleurs

[cunning veldt]

en effet c'est p-ê moi qui suis bête là

[stoic oyster]

ce genre de théorème

[cunning veldt]

oui ok ça c'est vrai

[nocturne heron]

bon c’est un peu chaud de montrer que c’est pas de cauchy

[glad parrot]

Sob

[nocturne heron]

j’ai $u_{n+s}-u_n= \sum_{k=n}^{n+s-1} (2p+2 \sqrt{p(2u_k+p)})$

quoique

du coup $u_{n+s}-u_n \geq \sum_{k=n}^{n+s-1} (2p+2 \sqrt{p(2+p)})$

ah mais purée les indices se mélangent

[potent lotus]

tm

[cunning veldt]

tu t'embêtes bcp non ?

[potent lotus]

tm

[glad parrot]

Mais que fais tu la ?

[nocturne heron]

bah j’essaye de montrer que c’est pas de cauchy 😢

[glad parrot]

J'pense qu'il faut pas attraper la veste de cauchy pour cette question

[cunning veldt]

là avec ce que t'as fait tm tu peux conclure p est une constante

[nocturne heron]

bah oui justement, tu peux même dire que $u_{n+s}-u_n \geq \sum_{k=n}^{n+s-1} 2p$ pour te faciliter la tâche

[potent lotus]

tm

[nocturne heron]

enfin je pense

[cunning veldt]

on est d'accord

t'avais quoi comme x_0 par contre

[nocturne heron]

plus grand ou égal a 0

[cunning veldt]

ouais donc tu peux juste dire que le truc de droite c'est à peu près 2ps donc pas dur de trouver comment rendre u_N > M rigoureusement

[nocturne heron]

yes

bon maintenant faut trouver un équivalent de u_n 😥

$u_{n+1}=u_n +2p +2 \sqrt{p(2u_n+p)}$

[potent lotus]

tm

[nocturne heron]

j’ai pensé à faire un DL de la racine carrée maisss c’est pas sous la bonne forme

[cunning veldt]

hmm on dirait qq chose équivalent à un polynôme de degré 2

parce que tu peux écrire un+1-un

et sommer sur n

[nocturne heron]

pk sommer ?

[cunning veldt]

pcq ça redonne un+1 à gauche

t'as un+1 = somme des 2p+2sqrt(p(2un+p))

[nocturne heron]

oui?

[cunning veldt]

somme des 2p c'est c'est 2p*nombre de terme

et somme des 2sqrt(2pun)

c'est grosso modo du 4/3 2pun * sqrt(2pun)

[nocturne heron]

j’ai $u_{n+1} \approx 2\sqrt{2p} \sqrt{u_n}$

[potent lotus]

tm

[cunning veldt]

t'as un un qui a disparu là nn

[nocturne heron]

ah ouais bien vu 😭

att

[cunning veldt]

non un a + en + de poids dans 2sqrt(p(2un+p))

ça finit par être largement plus important que 2p

[nocturne heron]

$u_{n+1} \approx u_n +2p+2 \sqrt{2u_n p}$

[potent lotus]

tm

[cunning veldt]

ça oui

[nocturne heron]

même sans 2p c’est bon

[cunning veldt]

non

tu peux un dév asymp à deux termes

le 2p va sûrement donner le 2e

[nocturne heron]

ah

[cunning veldt]

en ordre de grandeur

c'est le terme de moindre importance par rapport à la racine de un

[nocturne heron]

et maintenant on cherche $u_n \approx K n^{\alpha}$

[potent lotus]

tm

[nocturne heron]

en remplaçant dans l’équivalence ?

[cunning veldt]

moi la manière dont je vois le truc là c'est un+1 - un = 2p + 2sqrt(2pun), donc c'est à peu près l'équa diff y' = 2p+2sqrt(2py)

notre équivalent asymptotique devrait être à peu près une sol à cette éq diff en fait

[nocturne heron]

ouais je vois

mais on résout pas l’équation diff quand même ?

psk moi je sais pas faire lol

[cunning veldt]

bah elle est pas très dure là

[nocturne heron]

uh mais c’est pas linéaire ça

ahhhh ok nn je vois

t’intègre y’/sqrt(y) ?

[cunning veldt]

oui

mais l'idée c'est pas juste de résoudre pour résoudre stv, pcq il faudra ajuster les constantes à la fin pour avoir un équivalent exact de toute façon

l'idée c'est plutôt de reconnaitre les termes dominant par ordre

un on dirait que le terme dominant va être du 2pn² quand même

[nocturne heron]

a gauche ça fait du $2 \sqrt{u_n}$ mais a droite jsp 😭

[potent lotus]

tm

[cunning veldt]

comme je disais tout à l'heure ça ressemble vachement à un polynôme de degré 2

parce que si t'as une dérivée qui donne une racine du terme d'avant par réc

ça te fait penser au fait que racine carré de deg 2 donne du deg 1

[nocturne heron]

aaah ouais

[cunning veldt]

et intégrer redonne du deg 2

en soi

[nocturne heron]

ouais jvois

[cunning veldt]

si tu tentes un = an²+bn

+c

la constante devrait finir négligeable

ce qu'on cherche c'est du an²+bn

[nocturne heron]

je remplace dedans ?

[cunning veldt]

j'ai pas envie de te faire faire un truc potentiellement inutile mais bon moi je serais tentée de le faire

[nocturne heron]

moi aussi j’suis tenté j’avoue

[cunning veldt]

t'as un+1 = un + 2p + 2sqrt(p(2un+p))

a(n+1)²+b(n+1) = an²+bn + 2p + 2sqrt(...)

2an+a+b = 2p + 2sqrt(...)

ouais bon m'en sortir joliment c'est mort faut remplacer dans la racine aussi pour comparer

[nocturne heron]

jvais tenter sur mon tableau attends

[cunning veldt]

sqrt(p(2un+p))

je crois que ça donne un truc joli

on dirait que a = 2p

[nocturne heron]

pas trop

$2an+a+b=2p+2 \sqrt{2p(an^2 +bn+c)}$

[potent lotus]

tm

[cunning veldt]

regarde ce qu'il se passe si a = 2p

le coeff devant n² à droite c'est (2p)²

donc 2p, passé à la racine

t'as du 2pn à droite

multiplié par le 2 de la racine

donc du 4pn

et à gauche t'as du 2 * 2pn = 4pn

et le terme qui reste constant sans être tiré par le polynôme, à droite, c'est le 2p

donc b = 2p

si tu remplaces avec a = b = 2p

je pense ça donne une égalité

[nocturne heron]

j’vais essayer

[cunning veldt]

ça donne 4pn+2p+2p à gauche donc 4p(n+1)

[nocturne heron]

oui

[cunning veldt]

et à droite 2p + sqrt(4p²n²+4p²n) = 2p+4psqrt(n(n+1))

ça colle

[nocturne heron]

euh attends

[cunning veldt]

au choix de c près

donc constante osef

[nocturne heron]

comment le 4 il va devant la racine

[cunning veldt]

je sors le (2p)² et y a un 2 devant la racine à la base

[nocturne heron]

ahh oui mb

ouais nan ça a l’air de marcher

du coup $u_n \approx 2pn^2+2pn+K$ ?

[potent lotus]

tm

[cunning veldt]

le K est complètement inutile dans un dev asymptotique où y a du n

t'as juste qu'en l'infini un c'est à peu près 2pn²+2pn

[nocturne heron]

yes

bon bah $u_n \approx 2pn^2 +2pn$ alors

[potent lotus]

tm

[cunning veldt]

et l'idée d'en faire une éq diff discrète, c'était d'avoir une fonction qui donne à peu près le 2pn²

de sorte à trouver une idée d'équivalent

ça marche pas toujours mais parfois faut penser à cette analogie continu <=> discret

un+1 - un en soi c'est une dérivée pour un, et sommer sur n c'est un peu comme intégrer

et retrouver une primitive

[nocturne heron]

ouais les suites récurrentes $u_{n+2}=u_{n+1}+u_n$ je les voyais un peu comme du $y’’=y’+y$

[potent lotus]

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

[cunning veldt]

tu penses juste à la méthode des éq linéaires

non là l'idée marche pas t'as pas la bonne analogie

moi ce que je te dis comme analogie

[nocturne heron]

ah

[cunning veldt]

c'est que par ex si t'as un+1 = un + n, tu sens bien que tu sommes les entiers et que t'es censé retrouver le classique n(n+1)/2
mais pourquoi c'est du carré, en ordre de grandeur ?

l'analogie de la dérivée pour une suite c'est un+1 - un

[nocturne heron]

oui

[cunning veldt]

une différence sur le plus petit intervalle possible (1)

donc t'as un+1 - un = n

et tu sommes sur n

et là tu te dis ah oui en fait c'est le même processus qu'intégrer

[nocturne heron]

ouais c’est une somme télescopique à gauche quoi

[cunning veldt]

oui mais ce qui est important à comprendre c'est que ça donne une bonne analogie de dérivée/intégrer

parce que quand tu dérives tu regardes une diff sur un intervalle infiniment petit

et intégrer au sens de Riemann c'est sommer ces diff par intervalle

là on a une suite où le pas est forcément de 1

et sommer téléscope joliment en montant d'un ordre de grandeur

pcq tu sommes sur ces pas

[nocturne heron]

ui

[cunning veldt]

la somme sur les un+1 - un c'est ce que t'as de "plus proche" du concept d'intégration

en un sens

si je te disais un+1 = un + ln(n) par ex, mtn t'aurais vachement envie de me rép ouais un c'est à peu près nln(n)-n quoi

[nocturne heron]

ouep

c’est une manière de résoudre des equa diff de passer du cas continu au cas discret ?

[cunning veldt]

donc pcq t'as une approx assez grossière, vu que tu discrétises par la force, tu trouves seulement une fonction continue sur R qui est une approximation sale

ça va donner un bon équivalent pcq forcément si t'as une relation différentielle

t'as le bon comportement asymptotique

par contre localement c'est loin de forcément donner de bonnes approx

passer de l'un à l'autre ça reste une belle perte d'information

[nocturne heron]

yes j’imagine

[cunning veldt]

et une autre manière de faire quand tu vois ces suites

pour poursuivre l'analogie avec les intégrales

c'est que si tu fais des substitutions ça peut rendre le terme dominant + clair

ce qui peut être pratique pour des raisons calculatoires

mtn que tu sais que le terme dominant c'est du 2pn², tu te doutes que si t'avais posé vn = sqrt(p(2un+p))

ça aurait pu donner une relation de réc en vn qui nous prend moins la tête

potentiellement

et après tu remplaces vn par l'expression en un une fois que c'est "assez joli"

[nocturne heron]

a tester, je vois pas de tête 😅

[cunning veldt]

vn = sqrt(p(2un+p)) et vn+1 idem avec un+1

[nocturne heron]

oui

[cunning veldt]

donc si tu remplaces dans la relation de réc t'as la même chose sauf que t'as une différence de racine qui donne une meilleure idée du comportement

pcq du coup t'as vn qui est à peu près du 2sqrt(p)n

[nocturne heron]

aaaaa ui ok

[cunning veldt]

et un devrait avoir le carré de ça/2

[nocturne heron]

pas mal

[cunning veldt]

dans son expression

vu qu'on a posé vn = racine blabla

tu retrouves essentiellement la même idée qui est que dans certaines manières d'exprimer les choses, c'est très clair quel terme domine

et si t'as un premier terme du dév asymp

tu peux toujours faire suite - premier terme de dév

et essayer à nouveau de dév

pour trouver un 2e terme

en bricolant tu finis toujours par t'y retrouver, mais idéalement tu trouves la forme du truc assez tôt

[nocturne heron]

ah oui j’avais fait ça en prépa, c’est pour trouver un development asymptotique plus précis nan?

[cunning veldt]

bah pour trouver le 2e terme quoi

c'est un qui est "plus fin" que le premier et qui domine avant le 3e

donc si tu fais expression - premier terme du dev

ça doit se comporter comme second terme du dev + o(second terme)

[nocturne heron]

ui jvois

c’était un peu chiant a faire d’après mes souvenirs

[cunning veldt]

en général c'est pénible oui

mais deviner la forme en cours de route ou faire de bonnes analogies, ça arrive souvent à mi chemin

[nocturne heron]

oki ça marche

[cunning veldt]

t'aurais eu besoin de raisonner comme ça avec les éq diff discrètes, pcq le premier terme d'expansion aurait été bon je crois

par contre le 2e aurait visé un peu à côté et y aurait fallu revenir à un - premier terme je pense

[nocturne heron]

equa diff discrètes c’est les u_n+1 -u_n c’est ça ?

[cunning veldt]

ouais l'idée que sommer ça c'est intégrer etc
tu te retrouves avec l'analogie continue qui est y' = 2p + 2sqrt(p(2y+p))

et tu retrouverais le y = 2px² + o(x²)

mais en terme d'après t'aurais pas du 2px

le fait de continuiser crée une différence "trop grave"

il faudrait ensuite regarder y - 2px²

pour se rendre compte que ça se comporte en 2px

pcq c'est un terme linéaire qui vient de la structure même de la suite, sans se soucier du fait que c'est discret on le retrouve pas naturellement

l'idée de l'éq diff permet de deviner le 2pn² ce qui est déjà pas mal quand même

[nocturne heron]

mon cerveau il suit plus 😭😭

il surchauffe

[cunning veldt]

bah l'idée c'est juste si tu vois u comme une fonction
un+1-un c'est censé être un peu comme du u'

donc t'as une éq diff que tu peux traiter comme tes éq diff de fonctions dérivables

et l'approximation est plus ou moins précise

[nocturne heron]

ui

fin bon la j’pense jvais aller dormir 😭

[cunning veldt]

bn

[nocturne heron]

en tt cas merci de m’avoir accompagné durant ce long voyage les amis 🫶

préparez vous pour l’exo de demain ça va être chaud

ciao ciao 👋

.close

[calm coral]

Channel closed

Closed by @nocturne heron

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[normal spindle]

How can I solve this?

$y(-312+k)=-119790$

[potent lotus]

UCYT5040

[normal spindle]

I tried using the table of values to find a valid y & k pair here

i found y=495 and k=70 but thats incorrect cuz then x=6606 and plugging that into the second equation gives 146952 (which looks similar to the correct answer but is incorrect)

and it is really annoying to be looking at values in that table where x>100 because i have to manually set interval for each 100 values of x

[stoic oyster]

what table are you talking about

[normal spindle]

from the grapher app

btw i should clarify the x and y on this table do not correlate to the x and y in the problem, x in the table maybe = y and y in the table maybe = (-312+k)

[lean hinge]

table is not the place to start

[normal spindle]

what else could i do?

[lean hinge]

hold om

[stoic oyster]

it's not super clear how you even end up w/ that in the first place

[lean hinge]

it appears as if they got rid of x by adding equations

[stoic oyster]

yea that's what I assume

[normal spindle]

if theres a better way than that equation lmk but i saw a problem like this before and i was able to do that and the table to solve

[lean hinge]

factorize -119790 into factor pairs

since y has to be positive, -312 + k has to be negative -> k is less than 312

since you need pos*neg for a negative -119790

[stoic oyster]

like you say your result is slightly wrong, maybe you typo'd somewhere while getting that equation, it's impossible to check

[normal spindle]

my equation is right, i know because i just found the answer

x=6606

y=495

k=70 (correct)

and i found it using the table again but deeper down in the table where (-312+k)=-242

but still this was an annoying way to solve

[lean hinge]

find factor pairs man

[normal spindle]

isnt factorizing pretty similar to just using this table (jeez i am saying table a lot)

how can i quickly factorize

[lean hinge]

find prime numbers

[normal spindle]

im not sure i understand

[lean hinge]

for each prime number, it can include any number up to its exponent or it could include none of the prime

from the prime numbers you can get each factor

also I believe your answer is wrong

[normal spindle]

i have the answer key...

[lean hinge]

oops

I thought k=5

[normal spindle]

wait it does

🤦‍♂️

[lean hinge]

check your equation

[normal spindle]

ok i will

when i saw 70 in the key i though i was right

i didnt realize that was q8 not q10

anyways thank you for your help

i think i understand how to do it now

.close

[calm coral]

Channel closed

Closed by @normal spindle

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[timid cape]

How should I approach these?

[kind axle]

the first question is incomplete

the 2nd one is simple @timid cape
you need to remember a^x b^x = (ab)^x

[lean hinge]

how is log base 5 equal to p

that doesn't make sense

log base 5 of what

[timid cape]

Is this right. Idk if you can understand my hand writing

[timid cape]

I was so confused😭

[lean hinge]

my neck is getting sore

for #3, just take ln of both sides instead of log base 10

also that's right if you aren't supposed to approximate the value

[timid cape]

Is it bc i flipped the picture, sorry💀

[timid cape]

[lean hinge]

is math different in the UK or something, I have never seen log base 5 as a standalone variable

[spice portal]

im tweakingover this too lol

[lean hinge]

assuming UK because of the word marks

[spice portal]

i was looking at the problem trying to figure it out and i was googling log base no argument??????

[lean hinge]

this is beyond me

[tacit moat]

they probably meant $\log_5(7) = p$ bc you need that to solve the problem

[potent lotus]

Sepdron

[kind axle]

take log base 5 instead of 10

[calm coral]

@timid cape Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold shale]

for the root test or ratio test, if it converges

[untold shale]

like example L = 1/2, which is <1

would this be the value the function converges to ?

or is it just to show that it does converge but not necessarily the value of L

[potent lotus]

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

Mar 14, 2025, 2:35 AM

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

Mar 14, 2025, 2:35 AM

[short salmon]

,rotate

there we go

[untold shale]

what

answer my question

[short salmon]

oh wait im an idiot

sorry

[untold shale]

its all good

[short salmon]

I thought that old question was active

[untold shale]

im just kiddin around

[short salmon]

let me take a look

[short salmon]

no

[short salmon]

exactly

[untold shale]

type shit

you trollin

[short salmon]

consider the series of partial sums of $\frac{1}{2^n}$

[potent lotus]

00100000

[short salmon]

we can start n=1

[untold shale]

okay

[short salmon]

clearly, this converges to 1, however, apply either the root or ratio test to it

and you get 0

[untold shale]

mann

[short salmon]

why would i troll 💀

[untold shale]

hang on

[untold shale]

im suspicious

[short salmon]

:(

[untold shale]

i believe you

[hidden estuary]

Btw

It works

[untold shale]

lies

[hidden estuary]

I just forgot to add 3375

[untold shale]

show me

[short salmon]

you can check the example I gave to verify :^)

[short salmon]

apply the tests to this

[untold shale]

how is it 0

what

[hidden estuary]

,calc 3375-10205/3

[potent lotus]

Result:

-26.666666666667

[hidden estuary]

I am so happy right now

[short salmon]

sorry, root test gives 1/2

my bad

[untold shale]

good job

[short salmon]

but still, $\frac 12 \neq 1$

[potent lotus]

00100000

[short salmon]

soooo...

the point still stands

[untold shale]

my prof said that the nth root of a constant^n is 1

so wouldnt be 1

[short salmon]

well

think about it

if I'm multiplying x n times, I get x^n

now, if I'm taking the nth root of x^n

well, what do I multiply n times to get x^n?

x

so, the nth root is x

[untold shale]

yes but this is a constant

idk this is just something he said

[short salmon]

yes, x is a constant here

[untold shale]

oh

okay i see sorry

[short salmon]

you can take $x\in\mathbb{R}$

[potent lotus]

00100000

[short salmon]

np np

[short salmon]

I don't think your professor said this

because also

0 is a constant

how could the nth root of $0^n=0$ be 1?

[potent lotus]

00100000

[untold shale]

he did cause in his work he has nrt(3^n) and then said its equal to 1

i really dont know why

[short salmon]

what's nrt() denote?

[untold shale]

nth root

[short salmon]

oh, I see. well, it seems that it's a mistake in his work

haha, for some reason, it reminds me of this amazing anecdote

In the early 1980s, I managed a computer retail store. Several of my employees were college students. One bright your man was having difficulty with his Freshman college algebra class. I tutored him and he did very well, but invariably, he would say, "the professor worked through this problem on the board, and it was nothing like this. I sure hope we got the correct answer."
I accompanied him to class one morning and discovered the source of his frustration. The professor was from the music department, and didn't normally teach college algebra --- he had been pressed into duty when over enrollment forced the class to be split.

During the class, he picked a problem from the assignment to work out on the board. Very early in the problem, he made an error. I don't recall the specifics, but I'm sure it was one of the many typical algebra errors you list.

Because of the error, he eventually reached a point from which he could no longer proceed. Rather than admitting an error and going to work to find it, he paused staring at the board for several seconds, then turned to the class and said, "...and the rest, young people, should be obvious."

cuz it's a very very simple mistake

[untold shale]

lol

hang on

lemme see if its in the class notes

[short salmon]

to be honest, I am curious about your class notes too

I do want to see if it actually says that

[untold shale]

he hasnt uploaded the ones from this week

but i wrote down the question and his work

gimme a sec

okay wait

so tad error on my part

but essentially the same

its just the nth root of a constant

so nrt(c) = 1

[short salmon]

well that's not true either

take n=2

c=4

then, $c^{\frac 1n}=2$

[potent lotus]

00100000

[short salmon]

not 1, right?

I think what you mean is $\lim_{n\rightarrow\infty}c^{\frac 1n}=1$ for $c>1$

[untold shale]

yeah ive really no clue

thats just what my prof said

[short salmon]

because this is true

I think you may be misunderstanding...

[untold shale]

so ive gotta use this on my quizzes lol

[untold shale]

nah

[short salmon]

well, if you say so

[untold shale]

maybe its wrong

be regardless i have to use this info

to get points on his quizzes

cause i wrote in my notes that the nth root of both n and a constant is equal to 1

[potent lotus]

00100000

[short salmon]

but you should be aware that it is not generally true that the nth root of a constant is 1

[untold shale]

nah

[untold shale]

i figured

because when he said it in class i was confused

but im assuming that as it reaches infinity

it might as well be 0

*1

[short salmon]

nor that the nth root of $c^n=1$

[potent lotus]

00100000

[short salmon]

yes, that's what I'm saying with

[short salmon]

^

[untold shale]

when he uploads his notes ill show you

cause i get what your saying

it makes more sense to me than what he says

[short salmon]

well, that does tend to be the case when you're saying that he is making false statements

💀

[short salmon]

this can be strengthened slightly to $c\geq 1$

[potent lotus]

00100000

[untold shale]

i dont think this is the first time

he has done this

but its okay

anyway back to the og question

so the value of L isnt necessarily the value the series converges to ?

[short salmon]

that's unfortunate. i'm used to my professors being quite accurate

[untold shale]

dont get me wrong, he isnt just spewing misinfo

but he does make mistakes sometimes, which is normal

it could also be error on my part

so 🤷‍♀️

[short salmon]

it was kinda cool catching this tiny mistake in my ODE prof's notes lol (on picard iterators)

[untold shale]

its a trivial detail so it isnt bothering me

[untold shale]

i am confused

😭

[short salmon]

yes. again, consider the sequence of partial sums of $\frac{1}{2^n}$

[untold shale]

1/2^n

[short salmon]

oh that has nothing to do with ur stuff, it's just a funny mistake I caught

[short salmon]

oops

[potent lotus]

00100000

[untold shale]

what is the mistake tho

[untold shale]

it converges to 0 but the root test

1/2

[short salmon]

funnily enough, it also applies to the sequence of partial sums of $\frac{1}{2n}$

[potent lotus]

00100000

[short salmon]

so I didn't even write a false statement

[untold shale]

yes, but you wouldnt do root test for 1/2n

right

[short salmon]

well, you get an L-value from the ratio test of this

which is 1

yet, this series doesn't converge

so..............

the L value isn't the value it converges to!

now we got 2 counterexamples :^)

[untold shale]

okay i see

im glad i asked the question cause at first i thought it was dumb

bahahaha

i was actually expecting that they are the same value

[short salmon]

no it's fine

hopefully I was able to clear it up for u 👍

[untold shale]

yeppers, you did

thank you

i appreciate your help

.close

[calm coral]

Channel closed

Closed by @untold shale

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged knoll]

Okay, I've got this bugger of an initial value problem that I have to solve via Laplace transform -

z'' + 5z'+6z = e^(-5t) u(t-1), y(0)=2, y'(0)= -5

[winged knoll]

Here's what I have so far -

I've taken the Laplace transform of the right side of the equation to get (e^-s * e^-5)/(s+5)

and the left side is in the form
(s+2)(s+3)Z -2s-5

So I move those factors over to the right side to get

(2s-5)/(s+2)(s+3) + (e^-s * e^-5)/(s+2)(s+3)(s+5)

Then I do partial fraction differentiation

On the left side both A and B are 1, so we rewrite the first term as

1/(s+2) + 1/(s+3)

As for the right side I'm a bit stuck

How do I partial fraction differentiate something with an exponential in the numerator?

[cunning veldt]

what's u ?

[winged knoll]

That's the other thing I'm stuck on

[cunning veldt]

z is the function you're trying to solve this for right ? and I assume you meant z(0) = 2 etc

[winged knoll]

yes

variables get confusing

[cunning veldt]

then what is the u(t-1) ?

[winged knoll]

Also it seems that looking at my textbook, u(t) is the unit step function

so it's 0 if t is negative and 1 if it isn't

[cunning veldt]

you forgot to apply the laplace transform on u(t-1)

[winged knoll]

how do i apply it there

Cause it's discontinuous

[cunning veldt]

the laplace transform of u(t-1) is e^-s /s

so your transform was correct

but then it's your question that I don't get, where did you stuck ?

[winged knoll]

The partial fraction expansion of (e^-s * 1^-5)/(s+2)(s+3)(s+5)

How do I do the expansion with that exponential in the numerator

?

[cunning veldt]

I mean you just have exponential * rational fraction

why does the exponential matter exactly ?

[winged knoll]

oh right

so it's just 1/the denominator and I can multiply the exponential back later?

[cunning veldt]

yeah

[winged knoll]

Okay, so now I have (e^-s * e^-5) ([-1/2(s+3)]+1/6[s+5] +1/3[s+2]) + 1/(s+3) + 1/(s+2)

Now I have to somehow take the inverse laplace transform of this monstrosity

The two at the end are easy enough, e^-2t and e^-3t

But the part multiplied by the exponential is leaving me stuck again

[cunning veldt]

that's because it gives back some u

unit step function

[winged knoll]

Yeah but how much

I honestly dont know where to begin approaching this

[cunning veldt]

it should make u(t-1) reappear

[winged knoll]

Well we can redistribute, so L^-1(e^-5 e^-s[-1/2(s+3)])

Take out the constant, L^-1(-1/2 e^-5[e^-s/s+3])

But now what

[cunning veldt]

left column time domain, right: s domain

https://en.wikipedia.org/wiki/Laplace_transform

[winged knoll]

so e^-s / (s+3) can be rewritten as e^-3(t-1) * u(t-1)?

e^(-3t+3) * u(t-1)?

[cunning veldt]

you forgot the e^-5
so let's say you want to find the laplace inverse of the e^(-s-5)/3(s+2)
it's 1/3 of the inverse of e^(-s-5) * 1/(s+2)

you need both frequency shifting and time shifting

to get 1/3 e^(-2(t-1)) * u(t-1)

[winged knoll]

Oh i need the e^-s-5 is the thing I'm missing?

[cunning veldt]

and like you can generalise for the other terms, laplace inverse of e^(-s-5) * 1/(s+k) gets inverted in e^(-k(t-1)) * u(t-1)

the - before the s is the time shift

[winged knoll]

So in the case of -1/2 e^(-s-5) 1/(s+3) it would be

-1/2 e^(-3(t-1)-5) * u(t-1)

or -1/2 e^(-3t-8) u(t-1)?

[cunning veldt]

e^(-s-5) 1/(s+3) gets inverted into e^(-3(t-1)) * u(t-1) * e^-5

so yeah

wait no -3*-1 = 3
3-5 = -2

not -8

be careful with signs

[winged knoll]

That worked!

Looks like it was correct

What a mouthful lol, I spent an hour on this question

[cunning veldt]

yeah laplace transforms are a pain to calculate, you need to look at a sheet of common transforms

so that you can transform/invert quickly

[winged knoll]

This is actually a huge help

Makes me understand Laplace transforms in general better

Tysm

[calm coral]

@winged knoll Has your question been resolved?

Channel closed

Closed by @winged knoll

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[outer sedge]

find the line of intersection resulting from the planes z = 1/2 and y = 0

[winter elbow]

Wat hav u tri?

[outer sedge]

graphing it

[winter elbow]

Well, you can find the line with some algebraic concept

[outer sedge]

how?

there's a z value and then a y value

[winter elbow]

There are two requirements for any points to be on the line

[outer sedge]

this is a 3D line i think

[winter elbow]

First, its y value must be 0 and their z value must be 1/2

[winter elbow]

Yes it is

[winter elbow]

And there is no restriction on the x value

This is everything you need.

[outer sedge]

why do u have a matrix

[winter elbow]

That’s my linear algebra hw, ignore it

[outer sedge]

the line is parallel to the x axis?

[winter elbow]

No, the direction vector of x axis is (t, 0, 0), where t is a real number. On the other hand, the vector which is parallel to the line’s direction vector is (t, 0, 1/2)

[outer sedge]

i have a bit of problems graphing in 3D

how do u get the x, y, and z value of a point

[winter elbow]

Graphing makes things more complicated since you have to be creative

[outer sedge]

what if I have this point

for example

[winter elbow]

Ok

[outer sedge]

how do i get the (x, y, z) for this point

[winter elbow]

It’s hard to determine with naked eyes

[outer sedge]

how do we get the y-value

the answer is (1, 0, 2)

.close

[calm coral]

Channel closed

Closed by @outer sedge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[echo lance]

(x+y+z)(-x+y+z)(x-y+z)(x+y-z) x being square root of 15 , y being square root of 14 and x being e square root of 13 find th e value of ths equation

[echo lance]

i made it so x+y = a

so then made it (a-z)(-x+y+z)(x-y+z)

but is there any way to simplify this

[velvet osprey]

i mean it looks suspiciously similar to heron's formula

[echo lance]

huh

[velvet osprey]

wait, what's z?

first you say that $x = \sqrt{15}$ and then you say $x = e \cdot \sqrt{13}$ but you don't say what $z$ is

[potent lotus]

Ann

[velvet osprey]

x being square root of 15 , y being square root of 14 and x being e square root of 13

[echo lance]

oh i meant to say z = square root of 13

[velvet osprey]

also sqrt()

x = sqrt(15), y = sqrt(14) and z = sqrt(13)

[echo lance]

ye so i was able to js kidda simplify it into (a-z)(-x+y+z)(x-y+z)

[velvet osprey]

also, you meant expression and not equation

[echo lance]

a being x + y

[echo lance]

😭

[velvet osprey]

anyway

(x+y+z)(-x+y+z)(x-y+z)(x+y-z) = (x+y-z)(x+y+z) * (z-x+y)(z+x-y)

= [(x+y)^2 - z^2][z^2 - (x-y)^2]

your a will not do much good here unless you like confusing yourself with myriads of variables

[velvet osprey]

1. do you understand how this was obtained?
2. do you see how to continue?

[echo lance]

i understand the
[(x+y)^2 - z^2]

[velvet osprey]

!occupied

[calm coral]

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

[echo lance]

rn im trying to figue out the second oje

one

[velvet osprey]

the second one is the same difference of squares idea

[echo lance]

uhhh

lemme js write this down rq

[echo lance]

uhhh im js getting (z+x-y)(z-x+y) for the second one ,-.

[velvet osprey]

what do you mean

do you mean that you find yourself faced with a mental wall?

[echo lance]

yes

i understand the first one cus of the formula a^2 - b^2 = (a+b)(a-b)

but then the second one then idk

[velvet osprey]

$(z + (x-y))(z - (x-y)) = z^2 - (x-y)^2$

[potent lotus]

Ann

[velvet osprey]

does this run you into a wall still or do you get it

[echo lance]

ohh wait i understand it now

tysm for your help.

.close

[calm coral]

Channel closed

Closed by @echo lance

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warm pewter]

How does a negative reaction force tell that a uniform rod is tilting

[unborn cove]

hello

@practicing._.dhamma is not a bot. He was just kidding.

[calm coral]

@warm pewter Has your question been resolved?

[calm coral]

@warm pewter Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Please don't occupy multiple help channels.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[echo lance]

I js decided to manually calculate this 💀 but ik this is prob not even close to the correct anwser

[echo lance]

In the solution it made it so like 4(sqrt 3 - 2(sqrt2)) = (4sqrt (sqrt 2 - 1)²)

And I don't understand shit

Anwser is 3+sqrt2 tho

[velvet osprey]

what did i tell you about resting just an hour or so ago

[echo lance]

Yea and I did 💀

[velvet osprey]

you're up doing math again 💀 ☠️

[echo lance]

It's now js a habit atp

[velvet osprey]

you can and should break it

[echo lance]

How

[velvet osprey]

set yourself a strict rule: from now until tomorrow morning you do NO MATH.

[echo lance]

Uhhhhhhh ._.

But then I'm bored

: P

[velvet osprey]

play video games, read a book, eat something, watch YouTube

anything but constant math math math math math math math math all fucking day

there's more to life than constant cramming for math competitions

[echo lance]

😭 lmao

[echo lance]

I mean yea-

But math is literally a hobby of mine so

It's js to fill up time

💀

[velvet osprey]

get other hobbies to supplement it

im not saying to drop math outright but like literally get a second or even third hobby

[echo lance]

TvT

👍

.close

[calm coral]

Channel closed

Closed by @echo lance

Use .reopen if this was a mistake.

[echo lance]

TvT

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[indigo cloak]

In a parallellogram ABCD is A(−1, 1, 1), B(3, 3, 4) og D(1, 3, 2).

Find the coordinate C.

How can we do this correctly, especially with 3 dimensions? From what I know of a parallellogram is that 2 sides must be the same, but how can we determine which ones should be the same?

[frosty spruce]

use the property of a parallelogram

parallelogram has 2 pairs of opposite sides that are parallel and equal

(though you only need one pair for this problem)

[calm coral]

@indigo cloak Has your question been resolved?

[indigo cloak]

I drew it in the yx plane like this, is this kind of what you meant?

[indigo cloak]

.

[frosty spruce]

you messed up the order of points

[indigo cloak]

Woops, like this I assume?

[frosty spruce]

no????

[indigo cloak]

I'm not following, what did you mean instead?

[frosty spruce]

it's just as simple as ABCD

in the first one, you drew ACBD
in the 2nd one, you drew ADBC

[indigo cloak]

Could you show how it would look in the coordinate system?

[frosty spruce]

idk, does it really matter? it's still a normal parallelogram with the only difference is that the point has numbers

yes, the only difference is that they have numbers

otherwise it doesnt matter how it would look like in Oxyz

[indigo cloak]

So it doesnt matter however we draw it?

In the coordinate system

[frosty spruce]

yea, as long as it follows the standard when you draw the shape normally

which in this case, the only thing matters is how the points are arranged

[indigo cloak]

Will the points always look like that or could you also draw it like this

[frosty spruce]

that works too

as long as it's ABCD and not something else

i can rotate it 90 degree and it still is ABCD

[indigo cloak]

I'm mainly asking as we could find the coordinates by saying that AC = BD

AC = [x-Ax,y-Ay,z-Az]

[frosty spruce]

oh wait no

sorry

that doesnt work

that one is ABDC

i thought reading name of a shape was taught from like, primary school?

[austere tartan]

The letters have to be consecutive. Start at a point and go either clockwise or anti clockwise

[indigo cloak]

Ah right, that makse sense

[indigo cloak]

not sure what you're trying to say with that lol

[frosty spruce]

"lol"

ok let's just get back to the original problem

so, now you know how to read name correctly

now u use the fact that the parallelogram has a pair of opposite sides that are equal and parallel

[indigo cloak]

Sure, so AC = BD

[frosty spruce]

for example, i can choose AB = DC and AB || DC

no, those are diagonals, and the only thing you can use from them is their sides are equal

i mean you can go this route but you'll get stuck very quick

for like, 5 seconds later

wait

have you learned vectors?

[indigo cloak]

yeah

[frosty spruce]

perfect

[indigo cloak]

this chapter is about vectors in space

[frosty spruce]

i mean, what would u use for this problem other than vectors lol

[indigo cloak]

AB = [4,2,3]
CD = [1-x, 3-y, 2-z]

correct?

[frosty spruce]

yep

[indigo cloak]

So we said that AB = CD, right?

[frosty spruce]

wrong

[indigo cloak]

DC*

[frosty spruce]

yep

[indigo cloak]

Meaning
4 = 1-x
2 = 3-y
3 = 2-z

correct?

[frosty spruce]

if you did DC, then DC = [x-1,y-3,z-2]

[indigo cloak]

Yeah I meant DC is that, I switched letters

Meaning that point is C(-3, 1, -1)?

If we solve for x,y,z

[frosty spruce]

note that you did AB = DC, so 4 = x-1 and so on

got a lil messed up there

[indigo cloak]

Yeah alright so C(5,5,5)

[frosty spruce]

ye.

[indigo cloak]

So question

Couldn't we also say that AD = BC, without any complications?

Considering I just did the math and seem to get the same answer

[frosty spruce]

that works the same way

[indigo cloak]

So main takeaway is that the numbers go either clockwise or counter clockwise

Is this for each figure, also for example trapezoid?

[frosty spruce]

suppose i have a trapezoid like this, yes, that is, supposing the problem gives the ratio of AB and CD

[indigo cloak]

Same goes for rectangle etc I assume

In the same sort of problem sets

Yeah alright

Thanks for the help

.close

[calm coral]

Channel closed

Closed by @indigo cloak

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[rare inlet]

Can someone please help me simplify this

[rare inlet]

Idk how to start

[dull wagon]

,rotate

[dull wagon]

determine whether the insides of the abs vals are positive or negative to rid yourself of absolute value bars

[rare inlet]

[(3√ 2 - 2√3)/6 - (4√ 2 - 3√ 3)/4 + (2√ 3 -√ 2)/12] / (3√3/8)

[hushed estuary]

LCM

[dull wagon]

the denom of the main fraction 3√3/8 should be in ()

then lowest common denominator

also for that third fraction you should have 2sqrt(3) - sqrt(2) not 2sqrt(3) - 2

[hushed estuary]

The LHS becomes (6root(2) - 4root(3)-12root(2)+9root(3)+root(2)-2root(3))/12

[velvet osprey]

don't do your problem in discord messages

[rare inlet]

Is this a rule in this server?

[velvet osprey]

no

it's advice i am giving you

[dull wagon]

ideally you should be doing this on paper

[rare inlet]

I am

[dull wagon]

take a photo if you are capable of doing so

[rare inlet]

I can

[velvet osprey]

also strongly suggest breaking the problem into pieces and not trying to drag all of it with you in every single line

as another measure to reduce the number of places where you might make a mistake

[dull wagon]

tried to do too many things at once and made a mistake
slow down a bit for now

[rare inlet]

They teach us to do it like this and idk how to do this easier

if my teacher would see me not multi-tasking she'd be so mad lol

[dull wagon]

take things slow,
focus on getting things right first, that is your #1 priority
then you can combine smaller steps with more experience

[rare inlet]

AHHH

i saw

lemme fix

[dull wagon]

separate the steps for
multiplication to get a lcd
combining your fractions

until you stop making these kinds of mistakes

[rare inlet]

or could have simplified 12 with 8 i just see it and made things simpler

but this works too right?

[dull wagon]

your final result should be simplified as much as possible

you will lose marks otherwise

[rare inlet]

well we can still get there even if I did not simplify that part?

[dull wagon]

yes, it'll just be inefficient

[rare inlet]

I see

ok imma simplify that part

[dull wagon]

$$\frac{123456789}{987654321} \times 987654321 = 123456789$$
without having to go through:
$$\frac{609663155563176345}{987654321}$$

[potent lotus]

ℝαμOmeganato5

[rare inlet]

got stuck here

something must be wrong

bec it would give me 0 which is not the final answer

yea nvm i see it

Ig i am a bit tired😭

now what

[dull wagon]

now ur done

assuming there aren't any mistakes i didn't catch

[rare inlet]

this is not the final answer the book mentions

actually

wait

14(3-root6)/27 is what we are looking for

[dull wagon]

factor out 14 then

but there's no real need

[rare inlet]

we still won't get the -root 6

oh

i see it

but how about the 42

ok yea

got it

[rare inlet]

so i won't lose marks for not factorizing

[dull wagon]

yes

unless they told you they want the answer in a specific format

[rare inlet]

got itt

tysm

.close

[calm coral]

Channel closed

Closed by @rare inlet

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tall tartan]

how do i solve after this?

[velvet osprey]

work out 91^2 + 1 by hand

see how it factorizes

[tall tartan]

so i do that

then use prime factorization?

[brisk pelican]

That's so gross

[velvet osprey]

well the thing's gonna be divisible by 2 at least

[tall tartan]

yeah

it is

thanks i understand it now

.close

[calm coral]

Channel closed

Closed by @tall tartan

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[remote mural]

where did w get the 2 next to ln from

2nd line

[dull wagon]

chain rule

[remote mural]

what

[dull wagon]

(original 4 on the numerator) / (derivative of 1 + 2x)

[remote mural]

oh

okay

then what

wait

woahh

[dull wagon]

that's where the 2 comes from

[remote mural]

derivative of 1+2x is 2

ohh

4/2