help-39

[willow chasm]

you basically did it on your own

I just asked a couple more questions

[inland ingot]

https://tenor.com/view/ibepopimgbottles-meme-my-honest-reaction-gif-16804315594618966703

thanks bro

[willow chasm]

np

you can do .close if you don't have any more problems

[inland ingot]

bet

.close

[pearl pond]

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[inland ingot]

.reopen

[pearl pond]

✅

[inland ingot]

can someone help?

[midnight haven]

I guess they’re talking about the interception pt.

So B and C would be wrong (middle 2)

[inland ingot]

back

[midnight haven]

The y intercept for the negative like is +18

Then the final answer should be the last.

[inland ingot]

bet

[inland ingot]

could you help with the others?

[midnight haven]

Every unit for x changes, the y axis changes the same amount.

[midnight haven]

Well im going for class soon so

Im sure the other helpers will respond:) goodluck

[inland ingot]

thank you

[inland ingot]

alr

can anyone help?

[pearl pond]

@inland ingot Has your question been resolved?

[inland ingot]

no hoe

[versed mica]

what’s this for

[pearl pond]

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[inland ingot]

uhh

[unborn abyss]

it's important to understand that robots are not people

[inland ingot]

I understand

although it is still a hoe

https://tenor.com/view/african-kid-africa-grand-m-eat-smile-gif-17321775

[pearl pond]

@inland ingot Has your question been resolved?

[pearl pond]

@inland ingot Has your question been resolved?

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[visual canyon]

2/3 * (m+6)= -(1/2)

.close

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[proven arch]

.close

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[pearl pond]

@topaz anvil Has your question been resolved?

[tardy bloom]

oh hi

you don't have a vertical asymptote at x=-1

also what is your function f

yeah but that's not on your graph

like your function crosses your vertical asymptote

which shouldn't be happening

also your function doesn't approach your horizontal asymptote

[pearl pond]

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[unkempt olive]

i got 426.6667 a few times and it still says i'm wrong

[dusty flame]

,w integral from 0 to 10 |(x^2/2 + 4x + 6)| dx

[jolly parrot]

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Requested by nuclearinteraction

[dusty flame]

huh

[unkempt olive]

😭

[rose robin]

can you input the fraction instead?

or does it have to be a number

[unkempt olive]

lemme try

IT WORKS

thank u sm HAHA

[rose robin]

nice!

[unkempt olive]

.close

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[rotund nacelle]

I've been struggling with bond premium and it's formula. I do get the basics of it, but the rR is something I struggle with finding. Along when I thought I had figured it out, the smartest in my class saw my answer and told me that my answer (-5,816.89) is wrong, because if it's a negative answer it's automatically wrong. I'm stuck on whether this is true or not.

(I'm using the example questions instead of the real questions because I want to answer them on my own without help, if that makes sense?)

[pearl pond]

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[vague pollen]

stuck at c

[vague pollen]

i think the first step in to sub in (1/3,1/3) into a(x,y) but i dont get why im doing that

[west bluff]

I assume you're taking multivariable calc? Cuz fairly sure this can't be reduced to a single variable function

[vague pollen]

i guess

this topic is multivariable optimisation

[west bluff]

Welp, then guess it all comes down to making the derivative(or all its partials) = 0

[vague pollen]

is that what i did with the fx and fy in part b

[west bluff]

Looks like it, though I haven't heard the term “critical point“ used all that often, I'm gonna assume it refers to stationary points

[valid kettle]

yes

critical points are what we call stationary points as here

[west bluff]

Welp, if you've found all the stationary points of the function, then you need to find a point for which:
1: the partial derivatives are equal to zero for all variables of the function
2: there is no stationary point with a higher function value

If I'm not mistaken, that should be your answer

[vague pollen]

what is

is the point not 1/3,1/3

[west bluff]

I think it actually is, yeah

Since you've only got one stationary point, that must be it

Show it's a maximum regardless

[vague pollen]

my guess is i do this

is that the max

cause its positive?

i subbed and its 2/3

[west bluff]

If your point is correct, then yes, A(1/3,1/3) should be the maximum area

[vague pollen]

but how do you know it is the maximum

[west bluff]

Formally, you've yet to show it, by showing that it's a stationary point(all the derivatives are equal to 0) for which the function increases before reaching it and decreases after, for example

But in most cases, you'll assume the exercise has a solution, so if you only get one stationary point, it will almost certainly be what the exercise is asking you

[vague pollen]

i have the answer

but how do i show it

[west bluff]

It was right then

[west bluff]

Alright

[vague pollen]

im thinking of getting the z with 1/3,1/3 and doing hessian

if the lambda is positive then i guess its shown?

[west bluff]

The function only has a single stationary point, right? So for any x, y lower than 1/3, the function will have a certain growth(be it increasing or decreasing) which will remain unchanged until reaching 1/3, alternatively, it will have another growth for all x,y greater than 1/3, showing that the growth before is increasing and after is decreasing id enough

[west bluff]

But yeah, I think the hessian should work too

[vague pollen]

hmm

do i differentiate the surface area equation to do it?

partially

[west bluff]

Yeah

[vague pollen]

okie ill try now

[west bluff]

For the method I told ya, you need the first derivatives, for the hessian, you need the second ones

[vague pollen]

yes

like fxx fxy right

im not quite sure on how to do your method

[west bluff]

Just forget about it then, the hessian should work too

[vague pollen]

my hessian looks weird i cant sub in 1/3,1/3

[west bluff]

That just means that for any point in the domain the hessian is constant

[vague pollen]

oh

orite

is it cause it only has one stationary point

[west bluff]

I'm actually not quite sure if that's generalizable

But essentially, that means its curvature is constant, it doesn't shift from concave to convex

I.e it has no inflection points

[vague pollen]

uhoh

[vague pollen]

oh like saddles

would this be correct?

[west bluff]

Yeah, if I'm not mistaken

[vague pollen]

negative so max

[west bluff]

Yup

And there you go

[vague pollen]

okie

nice

but i still dont really understand how this will prove the specific coordinate is the max

cause technically i didnt sub anything in

[west bluff]

No other point is a stationary point for the function

Being a stationary point is, while not sufficient, a necessary condition be a maximum or minimum

[vague pollen]

right

[west bluff]

Satisfying that and the curvature being concave IS sufficient to determine that it is a maximum

While all points satisfy the second condition(i.e the function being concave there) no others satisfy the prerequisite

[vague pollen]

wait how did you know the curvature is concave

from the eigenvalue?

[west bluff]

Because all the second order partial derivatives are negative

By taking a partial derivative, you're considering the function to be of n-1 variables, that is, considering one of them to be a variable and the other one a constant

If all those “functions“ are concave, the greater one will be as well

If they had different signs it'd be more problematic though, so it's more of a case based approach

[vague pollen]

i see

[west bluff]

Actually, made a little mistake here, rather than n-1, it reduces the number of variables straight to 1

Regardless of their number, you're considering ALL the other ones as constants

[vague pollen]

thank you so much for guiding me through

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[wind tangle]

I have a data set like this, what is the best way to express it via a graph?

[wind tangle]

I tried with excel and error bars but that only lets me do vertical error bars

Have no idea how to do it in geogebra

So yep

[pseudo oxide]

piecewise functions, probably

frankly i'm not too good with desmos

[wind tangle]

So I should try graphing it in desmos?

[pseudo oxide]

yeah

something like

f(x) = {5.9 < x < 6.1:77183.9524, 7.9 < x < 81750 ... 129139, 0}

this won't do the vertical error bars btw

[wind tangle]

Ill try that!

Thank you

[pseudo oxide]

nw

https://cl.desmos.com/t/how-can-i-graph-a-piecewise-function-in-desmos/4874

[wind tangle]

.close

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[solid drum]

How many 3 digit numbers can be made from the digits 1, 2, · · · , 9, if each digit can at most be repeated
twice?

[solid drum]

This is a probability and statistics questions

[acoustic belfry]

first off define "repeated twice"

do you mean it can appear 2 times or 3 times

[solid drum]

So 992 is allowed

999 is not

[acoustic belfry]

ok

that makes it a lot easier

[solid drum]

idk if you got it

[acoustic belfry]

in this case find the numbers which cannot be made

[solid drum]

can you help me through the complement?

[acoustic belfry]

and then subtract it out of the numbers which can be made

[acoustic belfry]

well you just said this

there are only 8 other numbers which are like that

[solid drum]

oh 111,222,333....

until 9?

[acoustic belfry]

ye

[solid drum]

bruh

[acoustic belfry]

lol

[solid drum]

and what is the total number

to substract from

[acoustic belfry]

well you have 3 digits

each digit can possibly have 9 values

so total is 9x9x9

[solid drum]

9 x 9 x9

[acoustic belfry]

ye

[solid drum]

minus 9

720

[acoustic belfry]

ye

[worthy hare]

yes

[solid drum]

wow

Okay thanks

[acoustic belfry]

nice

[solid drum]

is this solvable without using complement?

[acoustic belfry]

yes

[worthy hare]

9×9×9-9 = 9×9×0 😨

[acoustic belfry]

a lot more work

[worthy hare]

=0 😨😨😨

[acoustic belfry]

lmaoooo

rules of precedence have left the chat

[vast finch]

i mean u can also use how many options u have for each digit

[worthy hare]

it's propaganda

alright jk

[vast finch]

so 9x9x8

[acoustic belfry]

ah yea i thought that at first

but it is not correct

[solid drum]

I have to take each case by itself?

[acoustic belfry]

you need to split into 2 cases

[vast finch]

why not?

[acoustic belfry]

1. first two repeated
2. first two not repeated

[vast finch]

ah fair

[acoustic belfry]

in 1.
you have 9x8x9

in 2.
you have 9x1x8

[vast finch]

its easier if no one can repeat

[acoustic belfry]

add them up for 720

[acoustic belfry]

ye ofc

[vast finch]

then its just 9x8x7

[jolly parrot]

ren

[acoustic belfry]

lol

[vast finch]

fair enough mb frfr

[acoustic belfry]

this was tricky tho i fell for it the first time round

[solid drum]

why 9x1x8

[acoustic belfry]

first digit can hold 9 values, second digit must be equal to first digit, so only 1 value

since you repeated the value, the third digit cannot be the same as the first two, leaving you with 8 options

you end up with 9x1x8

permutations are tricky i know

[solid drum]

Yeah you need to be careful

I'm rushing through them

[acoustic belfry]

yes

[solid drum]

Thanks for the help

Are you a math major?

[acoustic belfry]

not yet

im hoping to become one

[solid drum]

You got this

[vast finch]

im actually 3rd year lmao

[acoustic belfry]

nice

[vast finch]

this is a diagram of the decision tree

[acoustic belfry]

ooo ok

i usually draw boxes or empty lines and fill in how many options there are

[pearl pond]

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[tawny pewter]

how do i do this

[pearl pond]

@tawny pewter Has your question been resolved?

[pearl pond]

@tawny pewter Has your question been resolved?

[zenith cipher]

Do you know rolles theorem?

[pearl pond]

@tawny pewter Has your question been resolved?

[pearl pond]

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[wooden rock]

I need help with a third question.

[spare lark]

C'est quoi qui te bloque ?

[wooden rock]

Bonjour
Comment puis-je etudier les variations de f ?

[spare lark]

Derivation, tableau de signe de la derivee, déduction des variations

[wooden rock]

Mais je n'ai pas étudié la derivation

Mais j'ai suivi une vidéo YouTube l'expliquant

Après les calculs pour f¹(x), que vais-je faire

[spare lark]

f'(x) >= 0

Il faut faire le tableau de signe idéalement

[timid remnant]

soient deux nombres a et b tel que a < b et après on fait la difference entre f(a) et f(b)
toute fonction croissante on a : a < b alors f(a) < f (b)
toute fonction décroissante on a : a < b alors f(a) > f(b)

[spare lark]

Entre a et b oui

[wooden rock]

Hmmm, tu parles arabe ?

[wooden rock]

Mais Ce n'est pas une solution à ma question

[wooden rock]

Peux-tu le dessiner ?

[spare lark]

https://youtu.be/50CByVTP4ig

This goat will do it for me

[spare lark]

Il faudrait résonné avec la limite en +infini de la fonction f à la place de f(b)

Par continuité on a un tel résultat

[spare lark]

On te demande sur 0, +inf

[wooden rock]

Oh ok

@spare lark

[spare lark]

Sur 0,+inf

[wooden rock]

Seulement ?

[spare lark]

C'est la question qui le dit

[wooden rock]

Lisez-le entièrement

[spare lark]

Oui, sur 0,+inf

On fera une phrase pour expliquer les variations de f

[wooden rock]

[0,+ inf [ et ]-inf , 0 ]

[spare lark]

Sur -inf,0

[wooden rock]

Qu'est-ce que c'est

[timid remnant]

il faut deviser le domaine de définition en deux avant que la fonction s'annule et après? non?

et faire le travail deux fois 😅

c'est plus difficile sans utiliser la fonction dérivé

mais c'est le seul choix je pense

[spare lark]

Effectivement c'est une possibilité

[wooden rock]

@spare lark la phrase

[spare lark]

Fais les variatilns pour 0,+inf

[wooden rock]

Comment ?

[spare lark]

Comme il faut en déduire j'ai besoin du résultat

[spare lark]

Ici

[wooden rock]

F(x) : x²+3
F¹(x) : 2x

Et maintenant ?

[spare lark]

Est ce que c'est positif ou négatif quand x appartient à 0, + inf

?

La dérivée

[wooden rock]

Poditif

[spare lark]

Donc que peut on dire de f ?

[wooden rock]

Est croissante sur [0 , +inf [

[spare lark]

Parfait

Et donc pour les variations de f sur -inf, 0 ?

Sachant que f est paire

[wooden rock]

Qu'est-ce que le fait que la fonction soit pair au la variation

[spare lark]

Une des propriétés d'une fonction paire est quelle est symétrique par rapport à un axe vertical

En l'occurrence l'axe des ordonnées

Donc si tu fais le symétrique de f par rapport à y

Tu obtiens quoi comme variation ?

[wooden rock]

Eh bien, alors j'écrirai que f est pair, donc il est décroissant

[spare lark]

Oui

[wooden rock]

Je ne comprends pas encore la relation entre eux, mais je garderai la règle

[spare lark]

Fais un dessin

[wooden rock]

Tableaux de variation ou le graph ?

[spare lark]

Le graph

,w graph x^2

[jolly parrot]

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[spare lark]

,w graph cos(x)

[jolly parrot]

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[spare lark]

,w graph x^2 + 3

[jolly parrot]

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[wooden rock]

Parabole

[spare lark]

Tu vois comme c'est symétrique par rapport à l'axe des y ?

[wooden rock]

Oui

[spare lark]

Les trois que j'ai graph sont paires

Et pour les fonctions impaires

C'est une symétrie centrale

Par rapport à l'origine

,w graph x^3

[jolly parrot]

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[spare lark]

,w graph sin(x)

[jolly parrot]

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[wooden rock]

Il en sera ainsi alors

Quel était le nom de ce smiley ?
Parabole

[spare lark]

,w graph xe^(-x^2)

[jolly parrot]

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[wooden rock]

Forme comme un sourire haha, j'essaye de m'en souvenir comme ça

[spare lark]

Oui

[wooden rock]

Oui, j'écrirai ces choses et je répondrai aux questions pour vous si je ne comprends rien

[spare lark]

Les fonctions impaires peuvent sauver la mise quand il faut calculer une integrale entre deux bornes opposés

[wooden rock]

Tableau de variation de f :

[wooden rock]

Le professeur ne nous a enseigné cette leçon qu'il y a une semaine, donc je n'ai pas bien compris ce qu'elle disait

[spare lark]

Voilà parfait

L'idéal serait de mettre les limites si vous savez les calculer

[wooden rock]

Je n'utilise pas les limites, je ne l'ai pas encore étudié, je pense que nous l'étudierons l'année prochaine ou lors d'un deuxième tour cette année.

,help

[jolly parrot]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[wooden rock]

G(x) = rac(x-4)

Comment vais-je faire leur tableau de variation ?

@spare lark

[wooden rock]

<@&286206848099549185>

Désolé pour la mention

[4, +inf [

[spare lark]

Et donc l'ensemble de derivation ?

[wooden rock]

Positif

Il est Croissante ?

[spare lark]

Oui mais c'est quoi la derivee de g ?

[wooden rock]

G(x) ni pair ni impair

[spare lark]

La derivee pas la parité

g'(x) = ?

[wooden rock]

1

?

[spare lark]

$(\sqrt{u})' = \frac{u'}{2\sqrt{u}}$

[jolly parrot]

YakuBros

[spare lark]

Avec u une fonction dérivable

[spare lark]

Cependant il faut exclure 4 de l'ensemble de definition de g'(x)

[wooden rock]

Pourquoi

[wooden rock]

J'essaie de comprendre ça

[spare lark]

Que ce passe t'il si on applique la formule avec u = x-4 ?

[wooden rock]

Je ne pense pas que g(x) a besoin la derivé

[wooden rock]

Pouvez-vous vous remplacer u par quelque chose que je peux comprendre ?

[spare lark]

En effet

Si deux fonctions ont le même sens de variations, alors la composee de ces deux fonction est croissante

[wooden rock]

Tableau de variation

[spare lark]

Voila

[wooden rock]

Eh bien, je dessinerai pour la première fois le graph

[spare lark]

C'est pas une démonstration,

[spare lark]

Ça c'est une démonstration

[spare lark]

Ou ça

[wooden rock]

Cf.
Je dois la compter comme celle dont j'ai oublié le nom mais x = -b/2a , y = f(-b/2a )

[wooden rock]

Merci ♥️🫡

[spare lark]

Ça c'est le minimum/maximum de ta parabole

Le sommet

[wooden rock]

Je n'en aurai pas besoin ?

[spare lark]

Tu l'as déjà

C'est quand x = 0

Et ça fait f(0) = 3

[wooden rock]

,w graph rac(x-4)

@spare lark

[wooden rock]

Cf : green
Cg : black

@spare lark

[spare lark]

,w graph sqrt{x-4}

[jolly parrot]

Results provided by WolframAlpha

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

Requested by yakubros

[spare lark]

Essaye d'avoir deux, ou trois points pour g

Cf est correct

[wooden rock]

Merci

@spare lark j'ai une question

M.que
Lorsque on calcule delta , il donne -7<0
Bien
Mais pouquoi nous ajoutons le coefficient x² est positif au justification

@spare lark

<@&286206848099549185>

$$-2x + 4 \leq 2(x^2-x + 2)$$

[jolly parrot]

Roman Sama

[wooden rock]

$$0 \leq 2x^2$$

[jolly parrot]

Roman Sama

[spare lark]

Le sens de la parabole

Si il est positif, alors le polynome de second degré est positif, sinon il est négatif

[wooden rock]

$$ backslash frac{- 2x + 4}{x ^ Lambda * 2 - x + 2} =-1$$

[jolly parrot]

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

[wooden rock]

Merci

[wooden rock]

Whattt

$$\ frac{- 2x + 4}{x ^2 - x + 2} =-1$$

[jolly parrot]

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

[wooden rock]

@spare lark

3_b )

[spare lark]

Est ce que f(x) = -1 admet une solution ?

Si f est minorée par -1 et $\exists x \in \mathbb{R} , f(x) = -1$ alors -1 est une valeur minimal de f

[jolly parrot]

YakuBros

[pearl pond]

@wooden rock Has your question been resolved?

[pearl pond]

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[vernal belfry]

Hi, can someone help me with this problem?

[vernal belfry]

The top line is the problem. The paragraph beneath it is my attempt to solve it. I thought I was done and then I remembered negative numbers exist and now I don't know what to do...

How can I change this to work for integers and real numbers, instead of positive reals and natural numbers? And can I still use the axiom of achimedes and the well-ordering principle for it?

[tropic saddle]

if x is negative, consider-x

[vernal belfry]

hm how do I turn a natural number into an interger?

for example what do I do for x = -1 and x = 0?

[tropic saddle]

for x=-1 consider -x=1 and then find a natural number z with z<=1<=z+1

for x=0 you can just write down z=0

[vernal belfry]

thank you! I'll go and try that

[pearl pond]

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[warm depot]

Prove that if A and B are denumerable, then AuB is denumerable

[warm depot]

if A and B are disjoint then this seems easy

what do i do if they arent disjoint

can i just consider A-B which will certainly be denumerable because A is denumerable, but it will also be disjoint with B

define $f : \mathbb{Z^+} \to A$ is a bijection since A is denumerable

[jolly parrot]

syecko

[warm depot]

define $g : \mathbb{Z^+} \to B$ is a bijection since B is denumerable

[jolly parrot]

syecko

[warm depot]

define $h : \mathbb{Z^+} \to A \cup B$

[jolly parrot]

syecko

[warm depot]

[h(i) = f(k), \quad i = 2k-1 \quad h(i) = g(k), \quad i = 2k]

[jolly parrot]

syecko

[warm depot]

<@&286206848099549185>

[pearl pond]

@warm depot Has your question been resolved?

[pearl pond]

@warm depot Has your question been resolved?

[lunar apex]

smth like that yeah

[warm depot]

ok thank you

.close

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[shadow monolith]

I'm trying to understand this guy's probabilistic reasoning. He's trying to show a case where Pr(Y) > Pr(X), given X entails Y & Pr(Y & !X) =! 0?

Then he talks about how it could be that X & Y are equally likely, and this would be a counter-example?

I'm not fully following. What is he saying?

[cyan lily]

Pr(Y and X) = Pr(X) since X entails Y

[shadow monolith]

right, the last bit im not understanding

Pr(X) + Pr(Y & not X) >= Pr(X)

Pr(Y) > Pr(X) unless Pr(Y & not X) = 0

oh sorry nevermind, i understand Pr(X) + Pr(Y & not X) >= Pr(X)

since X entails Y, and Pr(Y) = Pr(X) + Pr(Y & not X)

[cyan lily]

P(X) >= Pr(Y) if and only if Pr(Y and not X) = 0
since Pr(Y) = Pr(X) + Pr(Y and not X)
so
0 >= Pr(Y and not X)
therefore Pr(Y and not X) = 0 is the sole counterexample

yea

kinda a funny post though

[shadow monolith]

okay dope

what about the last thing he said there

"it could be that X and Y are eqaully likely in weird edge cases"

i dont get his point

[cyan lily]

neither do I, i think that point would be covered by the previous statements

[shadow monolith]

how so

[cyan lily]

why not

[shadow monolith]

im not disagreeing

i dont know what you mean

"would be covered by previous statement"

[cyan lily]

since X entails Y, them having equal probability would imply Pr(Y and not X) = 0

[shadow monolith]

how?

[cyan lily]

idk im now confused

[shadow monolith]

im not saying it wouldnt follow

im just not familiar with the probability enough to see

how it does follow

[vestal tapir]

you obviously can't do what he's saying

one circle inside the other

hm

[shadow monolith]

oh i see

if X entails Y, then the Y circle is inside the X circle, but if they are the same probability, then the X circle & the Y circle are the same. which means there is no Y circle that has no X circle. which means Pr(Y & not X) = 0

[vestal tapir]

yeah i see

the difference would be 0 by measure

he's right

they are both right, he's saying there's a weird case but it's almost true

[shadow monolith]

like this case

the sachin varghese

[vestal tapir]

it's like "how much distance i walk from t=0 until t > 5"

and "until t >= 5"

it's the same distance somehow

weird stuff

[shadow monolith]

yeah

if i was to formally write this

if X entails Y, then the Y circle is inside the X circle, but if they are the same probability, then the X circle & the Y circle are the same. which means there is no Y circle that has no X circle. which means Pr(Y & not X) = 0

how would i write this?

like

Pr(X) = Pr(X & Y) = Pr(Y)

therefore, Pr(Y & not X) = 0

what are the intermediate steps

[pearl pond]

@shadow monolith Has your question been resolved?

[shadow monolith]

oh i se

Pr(X) = Pr(X & Y) = Pr(Y)

Pr(Y) = Pr(X) + Pr(Y and not X)

Pr(Y) = Pr(X)

therefore, Pr(Y and not X) = 0

Pr(A) = Pr(A & B) + Pr(A & ~B)
Pr(B) = Pr(A & B)
so: Pr(A) = Pr(B) + Pr(A & ~B)
Pr(A & ~B) > 0
so: Pr(A) > Pr(B)

.close

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[quiet bridge]

how can i make a pie chart with this data

[unkempt yacht]

pie chart is not the correct chart to make this

[heady finch]

Doesn't seem like great data for a pi chart

[quiet bridge]

but that’s the assignment 🥲

[unkempt yacht]

then the assignment sucks

[quiet bridge]

it does

[unkempt yacht]

pie chart is percentage based

i dont see any percentage here

[quiet bridge]

the title is pie chart enzyme activity as percentage

[unkempt yacht]

based on your data, your chart should convey the enzyme activity as time increases

[quiet bridge]

don’t i have to convert them to percentages

[unkempt yacht]

the suitable chart would be bar chart or line chart

[quiet bridge]

i already did a line graph but the next part is pie chart

[unkempt yacht]

that makes absolute zero sense, percentage of what?

[heady finch]

What you could do is separate each section into its temperature and plot the amount of enzyme activity as the proportion of the circle

[unkempt yacht]

they dont even add up to 100

[quiet bridge]

enzyme activity

let me convert to percentages and then i’ll try to make it work

thanks

.close

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[quiet bridge]

.reopen

[pearl pond]

✅

[quiet bridge]

how can i find UQ and LQ

[dawn sand]

,rccw

[heady finch]

25th and 75th percentile so you can most likely count terms

[quiet bridge]

what

i feel so dumb rn

[heady finch]

Well basically

LQ will be the terms in the bottom 25 percent or 1/4 of scores

UQ will be top 25%

So you'll want to group your terms into bottom quarter and top quarter

The cutoff point will be the LQ // UQ

[quiet bridge]

is it 66 and 72

[heady finch]

If it's between two numbers you find their average

If there's 14 terms it would lie at 3.5 and 10.5 since 14/4 = 3.5 which is 25%

So you would take avg of 3 and 4 and avg of 10 and 11

[quiet bridge]

im so confused

i shouldve taken better notes

[heady finch]

It might be diff for how you were taught, that's just how I learnt it

Sorry

You know what LQ and UQ are?

[quiet bridge]

lower quartile and upper quartile

[heady finch]

What do they represent

[quiet bridge]

but how can i divide the data into 4

[heady finch]

You can't exactly

[quiet bridge]

so what do i do

[heady finch]

It falls inbetween 2 numbers

A quarter of 14 is 3.5

Agreed?

[quiet bridge]

14 is the number of terms right

[heady finch]

Yes

[quiet bridge]

okay

[heady finch]

So the bottom quartile would be one quarter of 14

Which is 3.5.

[quiet bridge]

okay

[heady finch]

What you do when you have a quartile between two numbers is you take the average of those two numbers

So for 3.5

You would take the average of term 3 and term 4.

[quiet bridge]

so add them and divide by two

63

[heady finch]

yep

that should be your value for the quartile

at least as I was taught

[quiet bridge]

and it would be 72.5 for uq

?

[heady finch]

yep

[quiet bridge]

okay that makes a lot of sense

thanks so much

[heady finch]

np

[quiet bridge]

like this?

how do i find the iqr

[heady finch]

Iqr is just the range from 25/75

Between the LQ and UQ

So you've already found it

[quiet bridge]

oh what is it

is the lq the second term

.close

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[midnight haven]

doubt in the trig concept used in this step.

(it's a determinant question but not posting the entire thing because it's not relevant to my doubt here)

[heady finch]

what exactly is the issue?

as in is it doubt of that being a correct step or doubt in how they got to that step

[versed mica]

second line to third?

or third to fourth

[midnight haven]

i marked that step with red. The third step. I don't understand how they got to sin(2alpha...
from the 2nd expression

[midnight haven]

2nd line to third

[heady finch]

so step 2 to 3

[versed mica]

$\sin(u-v) = \sin(u)\cos(v) - \sin(v)\cos(u)$

[jolly parrot]

knief

[versed mica]

match the terms

we need cos(v) = 1/sqrt(2) and sin(v) = 1/sqrt(2)

and u = 2 alpha

[versed mica]

from here we get v = pi/4

[midnight haven]

thank you for your help.
I couldn't observe it before

[versed mica]

you’re welcome

[midnight haven]

.close

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[sterile turtle]

for part b

this is my working out

this is the answer

why am i missing the -e?

phh

icic

nvm

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[sharp smelt]

I think contradiction would be the best way to prove this

Let's for the sake of contradiction ,assume $dim(null(ST)) > dim(null(S)) +dim(null(T))$.
\
$null (ST) ={v \mid v \in U \land T(v) \in null(S)}$

[jolly parrot]

A dense set

[pine jay]

I mean if you can use rank nullity this is pretty straightforward (i.e contradiction is not necessary)

A hint would be to work with the restriction of a certain map

[sharp smelt]

If an element is part of the kernel part of T, Then it's a part of the nullity of ST. Thus it follows that $dim(null(ST)) \geq dim(null(T))$

[jolly parrot]

A dense set

[leaden crater]

This seems a bit of a venn diagram of intersecting sets kind of thing as well

[sharp smelt]

yeah

[leaden crater]

If you consider some base and the bases of the kernels of T and S, you can imagine T either maps something into its kernel(then ST is 0) or into the kernel of S or outside of both the kernels of S and T

To be in the kernel of ST you need either of the first two cases, so if a vector v is in kernel of ST then either v is in the kernel of T or it is in the image of T and Tv = u is in the kernel of S, so the dimension of this set is the dimension of the union of those two sets, the first one is null of T, second one is smaller than null of S due to the extra restriction

Mb typed wrong smth in second case

[pine jay]

Don’t even have to think about bases here

[sharp smelt]

Alternatively, All elements in $null(T) \in null(ST))$. We also know that all elements in the nullity of $S$ , map to 0. We thus have $dim(Null(S) + dim(null(T))$. However, there may be some elements in common we remove these by finding $ dim(Null(T) \cap null(S)$. This gives us $dim(null(ST)) = dim(S)+ dim(T)-din(null(S) \cap dim(null(T))$

[jolly parrot]

A dense set

[autumn fossil]

All the elements in nullity of S wont get mapped to 0 by ST

[sharp smelt]

oh yeah, was apprehensove about that

ST=0 in two cases, the first case is v is in the nullity of T, in the second case T(v) is in the nullity of $V$

[jolly parrot]

A dense set

[sharp smelt]

which need not always be true

thus $dim(null(ST)) \leq dim(null(S)) + dim(null(T))$

[jolly parrot]

A dense set

[sharp smelt]

This feels sus

[pine jay]

Just write out what you want to consider

STv = 0

[sharp smelt]

yes

So now either T(v)=0

or $T(v) \in null(T)$

[jolly parrot]

A dense set

[pine jay]

So the image of T restricted to Ker ST is a subset of Ker S

[sharp smelt]

Ah, okay

[pine jay]

I just a hint really, I guess there’s different approaches

[sharp smelt]

So essentially we have $Ker(T)$ + ${v \in ker(S) \mid T(v) \in \ker(S)} = ker(ST)$

[jolly parrot]

A dense set

[sharp smelt]

This works?

[autumn fossil]

Yes, even it's true even if u dont incluse ker T

[sharp smelt]

So This would be my proof: If $v \in ker(T)$, then $S(T(v))= 0$ and $ker(T) \subseteq ker(ST)$. However, $S(T(v))$ can also map to $0$ $ v \in { v \mid T(w) = v \land v \in ker(S)}$. However, depending on the transfomration, not every element has a pre-image, thus $dim(ker(ST)) \leq dim(ker(S)) + dim(ker(T))$

[jolly parrot]

A dense set

[sharp smelt]

This feels sussy

<@&286206848099549185>

[pearl pond]

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[north talon]

how do i approcach this?

i probably dont multiply the terms right

[pearl pond]

@north talon Has your question been resolved?

[north talon]

<@&286206848099549185>

[zenith cipher]

Just a minute

[north talon]

okay

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[midnight haven]

send help

[midnight haven]

<@&286206848099549185>

[acoustic belfry]

complete the square

do yk how to do that?

[midnight haven]

yes

[acoustic belfry]

ok good

do that

[midnight haven]

(x+3) (x-3)

no wait

[acoustic belfry]

...

[midnight haven]

it’s

[acoustic belfry]

1. that is not correct
2. that is not completing the square

[midnight haven]

(x+3)^2 -9

right?

[acoustic belfry]

that is called factorisation and not what we are doing rn

[acoustic belfry]

mmm yes there we go

correct

[midnight haven]

yay

[acoustic belfry]

so now you have g(x) = (x+3)^2 -9

[midnight haven]

yh

[acoustic belfry]

let u be the inverse function

then g(u) = x

[midnight haven]

so replace x with u

[acoustic belfry]

yes

and equate that to x

so (u+3)^2 -9 = x

[midnight haven]

make u the subject

[acoustic belfry]

mhm

oh yea one more thing

cuz you are dealing with (u+3)^2

when you take the square root, be careful about the +/- sign

[midnight haven]

( + or - root x plus 9) -3 =u

[acoustic belfry]

yes

[acoustic belfry]

not to find g^-1, you need to pick one of them

[midnight haven]

which one

[acoustic belfry]

that depends on g, your initial function

the domain of g(x) is x>=-3

[midnight haven]

yh

[acoustic belfry]

so the range of g^-1 (x) is g^-1 (x) >= -3

[midnight haven]

yh

so it’s

-?

[acoustic belfry]

.

what do you need to do to make u >= -3

= not <=

[acoustic belfry]

-(something)-3 is less than -3

[midnight haven]

is that right

[acoustic belfry]

no

take the +ve root

[midnight haven]

okay

[midnight haven]

what about part (b)?

[acoustic belfry]

that was part (b)...

[midnight haven]

oh.

what’s a😭😭

[acoustic belfry]

.

[midnight haven]

on the mark scheme it says y>-3

is that the same thing

[acoustic belfry]

weird

first off it should be >=

so idk why it says > only

and secondly they asked for g^-1 so use that

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

wait mb i read it wrong

[midnight haven]

okay

tysm

[pearl pond]

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[pallid oriole]

the question is find the range and domain of the function f(x)=2-3x, for x>0

and also generally how to find the range and domain of any function?

[pallid oriole]

so i think in this case the range, which is nothing but the output for given values of x is the interval (2, -inf)

domain is all real numbers>0

[pallid oriole]

pls correct me if im wrong. also help with this.

[compact ridge]

the domain is given to you
it's just x > 0 for real x

it's not all real numbers is my point

[pallid oriole]

ok

o

[compact ridge]

yeah so since this function is 1-to-1

[pallid oriole]

k

what

[compact ridge]

you can sub in x = 0 and x = infinity to get y

[compact ridge]

this function has no turning points

[pallid oriole]

ok

u mean parabola

[pallid oriole]

always?

[compact ridge]

so yeah that is nearly correct
(-infinity, 2)

[compact ridge]

only when the function is 1-to-1

[compact ridge]

no, I mean cases when it's not a parabola

[pallid oriole]

oh. so for quadratic?

[pallid oriole]

yes i get it

[pallid oriole]

ans to this?

[compact ridge]

then you would have to look at if the turning point is in the domain

if it is, then the max (concave down) or min (concave up) is the y-coordinate of the vertex
otherwise it's just given by the 2 endpoints

[compact ridge]

there are a lot of types of functions

there's a different method for each type of function, like quadratic, polynomial, exponential, logarithmic, trig etc

[pallid oriole]

oh

[pallid oriole]

oh actually?

how to?

[compact ridge]

ofc

that's why you have to spend time learning this stuff

like at least a semester

if not more

I mean domain and range of these functions is only one part

[pallid oriole]

ok

there is another question

find range for x^2+2

i will try this

[compact ridge]

actually it's easy if you have the graph

I meant like algebraically

anyways

[compact ridge]

great, is there a domain?

[pallid oriole]

i will sketch the graph

[pallid oriole]

yes all real x

[compact ridge]

cool good step

[pallid oriole]

i see turning point

at x=0 y=2

i think it's [2, +inf)

[compact ridge]

cool that's the range

[pallid oriole]

because at x=0 y=2

what

oh okay

[compact ridge]

and the domain is all real x yes

[pallid oriole]

then the question is done.

is there any good yt resource to see how to find range and domain of any funciton

[compact ridge]

https://www.youtube.com/watch?v=KyOQhC8ctxc

this covers most of these functions you will learn at this level of maths

[pallid oriole]

ok

thanks

.close

[pearl pond]

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[pearl pond]

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[chilly terrace]

When to use this formula?

Sv = Kv(1+p/100)
Úv = Sv(1-p/100)
Kv is cost price, Sv is sales price (price with surcharge) and Úv is sale price (price with discount)

[pearl pond]

@chilly terrace Has your question been resolved?

[chilly terrace]

<@&286206848099549185>

<@&286206848099549185>

[obtuse tangle]

You use them if you given p and asked to find cp or sp

and also given cp or sp

[acoustic belfry]

dont spam ping

[chilly terrace]

SHUT UP

[pseudo oxide]

@chilly terrace i have warned you several times in the last few days

be civil to your helpers

they are VOLUNTEERING their time to help YOU

if this happens again, i will be forced to call moderators

[pearl pond]

@chilly terrace Has your question been resolved?

[pearl pond]

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[warm sonnet]

we're asked to find the radius

[warm sonnet]

circle inscribed inside an isosceles

I'm aware that you can drop the height to the base, but how does that help?

[midnight haven]

the sides are given?

[warm sonnet]

nope

just a and b

[midnight haven]

k

[warm sonnet]

this is the answer

[limber glacier]

The sides are tangent to the circle

[warm sonnet]

they get split in two right?

since it's an isosceles

[limber glacier]

Well there is symmetry too

[warm sonnet]

this doesn't make sense

[limber glacier]

Introduce r as a variable, draw some lines joining the center to those points of tangency

And try to use pythagoras theorem or something ig

[warm sonnet]

somemthing like this I guess

h = hypotenuse

[limber glacier]

I don't think u can say that's a/2..I'm not sure

[warm sonnet]

doesn't the circle cut the triangle in half

[limber glacier]

But the line going to side b is also perpendicular

[warm sonnet]

it's an isosceles so

[limber glacier]

And that part is b/2

[warm sonnet]

oh yea

you're right

does it matter that we don't know the hypotenuse?

or can we do something like this

you'd have to introduce another unknown though

[limber glacier]

Well that unknown hypotenuse is also part of the other triangle

U cud probably call that length u previously called a/2, something like x, and use pythagoras again, but there'll still be another triangle u can use pythagoras with one side a - x

And yeah then the red triangle also

[warm sonnet]

I don't think the red triangle helps

[limber glacier]

Well u know the height of that triangle in a and b

[warm sonnet]

wait how?

oh

wait do you just write a pythagoras to find h?

I got something like this from this triangle

but yea it's invalid

something like this?

[limber glacier]

There are 3 triangles

[warm sonnet]

yea

[limber glacier]

U know OC in terms of r and b

[warm sonnet]

OC?

i'll label the triangles

[limber glacier]

Then u also know CD in terms of r and b

And AD = a - CD, so u know AD in terms of r, b, and a

[limber glacier]

Oh wait I'm trying to send u a picture

[warm sonnet]

alright

[limber glacier]

I just realised it's not been sent

Got it?

[warm sonnet]

I see

yea

[limber glacier]

Now when u know AD, u also know OA

In terms of a, b, r

[warm sonnet]

I see

oh so you just find OA

[limber glacier]

Now write pythagorass theorem for ABC, and u habe an equation relating a b r that u can solve for r

[warm sonnet]

and write a pythagoras?

alright i'll give it a go

I got something like this

the equation below is just the pythagora of triangle ABC

isn't AC actually equal to b/2 + x

[pearl pond]

@warm sonnet Has your question been resolved?

[warm sonnet]

<@&286206848099549185>

is there a way to find AD?

I don't think it's x

just doesn't fit the answers in the end

[brave stirrup]

What are you doing

[warm sonnet]

trying to find the radius

[brave stirrup]

Do you have a diameter? Assuming not. What values do you have to work with?

[warm sonnet]

base is b, the sides are a

that is all is given

(it is an isosceles)

[brave stirrup]

So no numbers whatsoever is given???

[warm sonnet]

no, the answer is unknowns anyway so

[brave stirrup]

Lord have mercy I could never 💀

[warm sonnet]

via pythagoras

you could say that the diameter is

$sqrt(x^2 + r^2) + r$

[jolly parrot]

asm

[warm sonnet]

since k = sqrt(x^2 + r^2)

(k is AO)

[brave stirrup]

I'm not sure I'm at this level yet 💀
But atleast I know the hell that's to come ahead 😭

[warm sonnet]

good luck 🙂

[brave stirrup]

You too XD

[warm sonnet]

.close

[pearl pond]

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[pearl pond]

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[knotty tusk]

why is the particular solution At^3+Bt^2+Ct and not just At^2+Bt+C

equation is x'' + x' = t^2 - 2t + 1

[pearl pond]

@knotty tusk Has your question been resolved?

[knotty tusk]

yall are actually so helpless for like my past few questions no one even helps me this server hard fell off

[pearl pond]

@knotty tusk Has your question been resolved?

[knotty tusk]

frick this

.close

[pearl pond]

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[pure glen]

r^10 + 10r^9 + 45r^8 + 120r^7 + 210r^6 + 252r^5 + 210r^4 + 120r^3 + 45r^2 + 10r + 1

[pure glen]

is there any way

to get this down

to root of 2

without having to spend 6 hours

?

[brittle tinsel]

that simplifies to (r+1)^10 by the binomial theorem

[pure glen]

ye ik

[brittle tinsel]

wdym get down to sqrt(2)

[pure glen]

i need it to be in

quadratic formula

like u do with horner

actually

mr. @brittle tinsel

i am in fact

re-tarded

my bad

.close

[pearl pond]

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[junior carbon]

@deep turret