help-39

[zealous bluff]

f(-x+2) as a function of -x+2 would peak at -4 too

so it peaks at -x+2=-4

i.e. x=6

i.e. it peaks at (6,4)

[vernal tinsel]

ok so i got it wrong 🔥🔥🔥🔥

lovely

[dire tapir]

btw I just realised it's turkish

[vernal tinsel]

can u guys check the trig one as well 💔

[vernal tinsel]

yes

[zealous bluff]

can you translate that one as well

[zealous bluff]

you got the x value correct

[vernal tinsel]

so the diagram is a square and (ac) is uhhhh the point where it splits evenly at 45/45° idk what it is in english 💔

im so helpful

[zealous bluff]

oh wait i get it

[vernal tinsel]

de is the same length as ec the angle EFC is alpha(i dont have that symbol) and tan(a) is 2/5

so its asking for ac over af

[zealous bluff]

ok so

let me think

[vernal tinsel]

i wanna go back to the days where we was counting apples

[dire tapir]

i'll attempt the question then and see if i get same answer

[vernal tinsel]

okay

we were 6 people in the classroom taking it and all six of us guessed c 😭

[zealous bluff]

lol

[sterile python]

Huh its 8

[zealous bluff]

if (\tan\alpha=\frac25), then we can calculate (\sin\alpha)

[jolly parrot]

Richard Liu

[sterile python]

CEF = 135

[zealous bluff]

i mean we can calculate sin alpha whether tan alpha is 2/5 or not

[dire tapir]

okay yeah it's 8

[sterile python]

EC/sin a = FC/sin(135+a)

[zealous bluff]

right yeah

[sterile python]

If let a be the side of square so let EC half a

[zealous bluff]

then that trivialises the problem

[sterile python]

From here got FC and then solve from now on

[sterile python]

You guys chose the wrong one😭

[vernal tinsel]

FAAH

welp

guess were all idiots ❤️❤️❤️

i mean I could only do like fifteen of the 30 questions

[zealous bluff]

what test was this

[vernal tinsel]

bilgi sarmal 11. sinif deneme 3

💔

fuh bilgi sarmal

welp im closing the channel guys thanks for telling me im wrong 🔥

appreciate the help 😝

.close

[pearl pond]

Channel closed

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[sterile python]

No youre good!

[pearl pond]

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[runic hazel]

@strange hill

[strange hill]

yea

[chrome plank]

!da2a

[pearl pond]

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

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

[ruby cargo]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[runic hazel]

Ok

[strange hill]

sorry he is new here

[ruby cargo]

It's fine it's a PSA

[strange hill]

alr

[runic hazel]

Sorry for mistake

[crystal dew]

so what's your question?

[west sail]

do you have a question OP?

[strange hill]

no he is just new

so i am telling him how to ask quetions

[west sail]

then are you guys done with the channel or would he like to ask a question?

[strange hill]

yea we are done sorry for intruption

[west sail]

then would he like to learn to close the channel?

[strange hill]

ha uh one sec then

@runic hazel

@runic hazel

[runic hazel]

.close

[pearl pond]

Channel closed

Closed by @runic hazel

Use .reopen if this was a mistake.

[warped violet]

like watching a baby take their first steps 😍

[strange hill]

bro wtf lol

[warped violet]

just a joke about learning how to use these channels

[runic hazel]

hey shutup

[strange hill]

yea i know

[warped violet]

I didn't mean it in a rude way

Sorry

[runic hazel]

okk

[strange hill]

bro he is joking

[outer canyon]

tf

[runic hazel]

ya

[warped violet]

I worded it poorly yeah

[strange hill]

nono he just types this way

[strange hill]

dont get offended he just types this way

[runic hazel]

sry foe me also

[outer canyon]

alakh daddy

[strange hill]

bro dont talk like this here its for doubts only channel

[outer canyon]

ok mb

[runic hazel]

yes dont talk like that

[outer canyon]

..

[strange hill]

its alr

[strange hill]

srry he is that way

[midnight haven]

https://tenor.com/view/kamai-gif-13475771597659100799

[outer canyon]

jeetards take over

[crystal dew]

!redir pls

[pearl pond]

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

[midnight haven]

why is the bot,

taking time

[pearl pond]

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[jagged mountain]

can anyone explain how to get α

[glass meadow]

Let x be the side length of the square, then use trigonometry to find other lengths and angles until you get to alpha

[jagged mountain]

sorry, how is that

[glass meadow]

How is what

[proper nova]

im thinking this way @jagged mountain

do you notice that AD || BC

this is kinda impossible without trig

even with angle chasing

[wooden merlin]

Try making a triangle beside BF, which is identical to ABF.

[proper nova]

ooh

what for?

[wooden merlin]

Do you know angle BAF?

[trail otter]

It’s 40

[wooden merlin]

@proper nova Do you get this?

[trail otter]

EAD is 5 and AED is 85

[proper nova]

you meant like with the triangle you made, you have an isosceles

[wooden merlin]

yes

[pearl pond]

@jagged mountain Has your question been resolved?

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[warped kite]

how do without calc

[worldly jewel]

Use the identity sin(2x) = 2 sin x cos x

[warped kite]

im supposed to compute it without identity

[worldly jewel]

Huh?

Maybe you should state what is allowed then

[warped kite]

mb

[worldly jewel]

If neither calculators nor algebraic manipulation are allowed, it is hard to see what is allowed

[warped kite]

even if i did that way how am i meant to know what sin(2) is

[worldly jewel]

you could use a series expansion for sine

[warped kite]

im not at that level yet

[worldly jewel]

maybe you should just state what you are allowed to do then

you are allowed to use "6️⃣7️⃣"?

what does that mean

[warped kite]

rizz

.close

[pearl pond]

Channel closed

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[proper nova]

there's no real definition for this, it's just a meme in the 2025s

[worldly jewel]

Are you in cahoots with the other guy #help-23 message

[eager jewel]

why are so many clowns entering the help channels bruh

[warped kite]

calm down buddy its not that deep

[spiral pivot]

You can calculate an approximation for it using the taylor series expansion of sine, but there's no easy (or really possible) way of getting a closed form for this value (where 2 is in radians) without appealing to the sine function itself.

if 2 is in degrees, then you can use some utterly horrible trigonometry to get a result in radicals

[pearl pond]

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[outer crystal]

can someone explain how prof leonard does this translation

<@&286206848099549185>

[warped violet]

!15mins

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[west sapphire]

by translation, do you mean what he is labeling as w and v?

[outer crystal]

i meant the process of making w a position vector

[west sapphire]

he wants to keep the same length and orientation but shift the "tail" to the origin, right?

you do that by subtracting the tail's coordinates (1,2) from both the tail and the "head"

[outer crystal]

i dont get it

[west sapphire]

do you agree that if you subtract (1,2) from both endpoints then you still have the same vector, just relocated so that it starts at the origin?

[outer crystal]

oh so basically we subtract (1,2) from the tail to make it (0,0) and then we subtract (1,2) from (6,4)

[pearl pond]

@outer crystal Has your question been resolved?

[pearl pond]

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[jolly meadow]

hey how would i go about solving x^4 - x^3 -(a+2)x^2 +bx +4 :(x^2 - 2) im completely lost i know about long division and synthetic division but how does one handle the coeffiecients? I also have to find a and b so that the remainder is 5x+6 but im a bit lost 🙏

[tropic saddle]

lets call the original polynomial f, then f(x)=q(x)*(x^2-2)+5x+6 for some polynomial q(x)

now plug in some values for x

[jolly meadow]

i think im slow

[shell brook]

Nah you're good

[jolly meadow]

so the quotient should be something with coeffiecients? and how do i figure out a and b so the remainder is 5x+6 😭

[tropic saddle]

you do not have to figure out what the quotient is

you do not have to do the polynomial division

the result will be some polynomial q(x)

[tropic saddle]

which satisfies this equation

[verbal whale]

Mmh but there's an "also" before "find a and b"

[tropic saddle]

either you can find it afterwards. or its xy and the question is just asking for a,b

[verbal whale]

!xyp

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[jolly meadow]

for polynomial P(x) = x^4 -x^3 -(a+2)x^2 +bx +4 A. P(x):(x^2-2) B. find a and b so the remainder is R(x) = 5x+6 C. for a=-1 and b=7 write the division identity

wow this formatting is dogshit

[verbal whale]

Yeah that's why we usually ask for a picture or a screenshot

[jolly meadow]

but its in greek 💔

[verbal whale]

Send it anyway

[jolly meadow]

If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

lol

[verbal whale]

Indeed 🫠

[jolly meadow]

might

wanna rotate this one 💔

[tropic saddle]

,rotate

[jolly meadow]

WOW

good shit

[tropic saddle]

thats not the correct problem

[jolly meadow]

omg

,rotate

[jolly meadow]

monkey see monkey do

[tropic saddle]

hmm well unlucky then you do have to do the division

just imagine instead of the alpha and beta you had the numbers 1230741 and 107513. do the calculations like you usually would and dont simplify anything

[pearl pond]

@jolly meadow Has your question been resolved?

[jolly meadow]

How's this

if this is correct do i then just substitute in 5x+6 and solve the equation?

[eager jewel]

<@&268886789983436800>

[jolly meadow]

pretty but doesnt help

[tropic saddle]

ok I'm not sure if you planned that +2x^2 or whether your sign errors just cancelled out

but if you did plan that, I would recommend to stop doing it that way

[jolly meadow]

honestly idk

but is the answer correct because it looks really sketchy

and i have no clue on how to continue with B. 💔

[tropic saddle]

well the x^2-x-a is correct

god the way you do your signs is fucking me up

[jolly meadow]

same

idk how to describe this in words but i was taught to multiply the quotient with the divisor and then reverse the sign

[tropic saddle]

imo it is much better to separate those steps

first just multiply the terms

in step one that gives you x^4-2x^2

and then put -(.....)

anyway, I have to go

[jolly meadow]

see u 🙏

Okay I think im just left with this bs B. question now

gyatt β=7 α=-1

.close

[pearl pond]

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[pearl pond]

Available help channel!

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[crude aspen]

i dont understand what i did wrong can anyone help (calc 2)

[dense jasper]

Would you like to show how you obtained these simplifications?

We're not mind readers, so you need to show us what you did

In general, try to give all of the possible context since it also saves time from explaining things unnecessarily

[crude aspen]

.close

ty

[pearl pond]

Channel closed

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[pearl pond]

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[brave halo]

Hello, where Is my mistake in thinking?

[brave halo]

this is the original

why is the red bordered this ?

we did butterfly

[sharp vigil]

there was already a common factor of (n+1) in both denominators so no need to multiply it to both

[brave halo]

wym

dont you just mutiply like butterfly

so the diagonlaes

[glass meadow]

,, \frac{n}{n+1} = \frac{n(n+2)}{(n+1)(n+2)}

[jolly parrot]

Nel

[glass meadow]

Now add the other term

[brave halo]

why n (n+2) where does that come from

[glass meadow]

Multiplying by (n+2)/(n+2)...

[brave halo]

what

I cant follow

Im sadly to exhausted to ask precise questions

[glass meadow]

,, \frac{n}{n+1} = \frac{n}{n+1} \times 1 = \frac{n}{n+1} \times \frac{n+2}{n+2} = \frac{n(n+2)}{(n+1)(n+2)}

[jolly parrot]

Nel

[brave halo]

why does it look to different to

[glass meadow]

I'm only showing you the first term

Then you add the second

[brave halo]

Did I addition wrong?

How do you add?

by multiplying and then addition?

[glass meadow]

Are you asking me how to add fractions?

[brave halo]

yes

how you added them

[glass meadow]

Me? I didn't add anything

[brave halo]

wym 😭

[glass meadow]

All I did was multiply by 1, I don't know why that seems so complicated for you

[brave halo]

why did you cut the n?

[glass meadow]

Cut?

[brave halo]

on the left?

on the right i mean

[glass meadow]

I don't understand

[brave halo]

why did you mutliply by 1

i mutliplied the left down with the right up

and the left up with right down

[unborn abyss]

the butterfly method is a shortcut but if you're at this level of math you should be comfortable with these sorts of manipulations

[midnight haven]

What math grade is this

[glass meadow]

I think you've been relying on that "butterfly method" a little too much and now you don't understand how to manipulate fractions at all

[brave halo]

whats the normal wy

[glass meadow]

To add fractions? Rewrite them so they have the same denominator, then add the numerators

[midnight haven]

Is this recurrence

[brave halo]

induction

[brave halo]

oh

[midnight haven]

Oh ok

What do you need to prove

[brave halo]

i think it was every number n must be bigger/equal then n subscript 0

[midnight haven]

Hm

Can you show me the question i would like to try

[brave halo]

What you want me to translate

[midnight haven]

No no need

[brave halo]

?

Vergessen wie Addition von Brüchen geht

[brave halo]

Spezifisch

[brave halo]

This

Schwierig bei so vielen Klammern

oder?

und variablen

Was ist der gleiche nenner bei solchen Zahlen? Verstehe ja sowas wie 5 und 9 aber bei n + 1

[midnight haven]

So you need to only prove that the sum equals to that n/n+1 right

[brave halo]

oh

Also sprich selbes ergebnis

produkt

und dann?

much ich nicht die werte des linken dann erheben?

multiplizieren mit den selben hinzugefügten werten

ich habe unten links mit unten rechts erweitert. ....

ka was mache ich denn nach dem nenner ausgleich

mit (n+2)

dachte ich jz

und was passier tdann

addieren? einfach so?

addieren der zähler i mean

abschreiben?

und die zähler müssse aufgrund der erweiterung nicht verändert werden?

also alles was ich von unten rechts nehme auch zu oben links hinzufügen oder, multiplieren mit allem was oben links steht

[brave halo]

bruh

irgendiwe hänge ich da, ka was genau nochmla

ja

checks immernoch nicht 😭

da passieren doch garnicht so viele schritte. Was wäre dein aller erster schritt

so du siehts das und dann?

[dark hound]

have you heard of method of differences

[brave halo]

checke immernoch nciht warum da n + 2 / n+ 2 steht

[brave halo]

warum direkt n+ 2 / n + 2

und warum mal

ka

aber der nenner ist doch n +2 warum ändere ich dann den zähler

oh

fast

ich mache also dieses komische n + 2 / n+ 2 um den nenner zu gleichen

warum ist dann schon im nächsten schritt n(n+2) im nenner

was ist da passiert

oh

irgednwie dachte ich wir streiche das weg

also das einzige was man macht ist diese hinzuzufügen

immer?

und dann gehts normal weiter

mhh eigentlich ist nur der erste schritt mit aufwand verbunden

also erfinde ich einfach diese konstrukt?

[brave halo]

muss ich nur 1 konstrukt erzeugen oder mehrer

[plush bramble]

,rotate

[pearl pond]

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).

[dark hound]

nah i don’t need help, im trina help

[brave halo]

ok, also nehme ich ab sofort immer als vorschritt wenn die nenner ungelich sind

das multiplizieren einen bruchs mit

linke seite * erweitertem fehlenden teil \erweitertem fehlenden teil

klingt schonmal besser . danke

wo ist der unterschied zu dem hier?

würde niucht 7/8 * 10 ausreichen?

7/8 * 10 / 10

warum reicht es bei dem n/ n+1 * n +2 / n+ 2 zu schrieben und nicht

mehr?

ok?

[brave halo]

im sry I dont know atm

[midnight haven]

Its ok

[pearl pond]

@brave halo Has your question been resolved?

[dark hound]

yi check this out this another way of looking at it

[pearl pond]

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[unkempt yacht]

i don't quite understand the fourth sentence of the definition

oh my god the minus sign is the en dash not the actual minus sign

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

So here I'll want the expectation on 30 rolls, I suppose, right

[left steppe]

yes

[cinder flower]

what is the section name in the book this is from?

[sharp smelt]

Inferences Based on the MLE

[sharp smelt]

wait, won't I want something to do with 2 pops

[left steppe]

yes

what’s the expected number of times for two pips showing up for 30 rolls

[sharp smelt]

right

my bad

oops

It would be $2 \sum_{i=1}^{30} P(x=i)$ no?

[jolly parrot]

wai

[sharp smelt]

where x is the number of times 2 occurs

[brave sluice]

maybe you should calculate the chance of it happening 10 or more times

i think that was how you do a p-test

[sharp smelt]

$1 - \sum_{i=0}^{10} \left(\frac{1}{6}\right)^i \left(\frac{5}{6}\right)^{10-i}$

[jolly parrot]

wai

[brave sluice]

shouldn't it be 30-i?

[sharp smelt]

oops, yes

though shoul;dn

shouldn't we find the expectation

[brave sluice]

ok

[sharp smelt]

of 2 pips

[brave sluice]

yeah maybe

does the book talk about p-tests or?

is there a different technique you should be using?

[sharp smelt]

$2 \sum_{i=1}^{30} \binom{30}{n} \left( \frac{1}{6} \right)^n \left(\frac{5}{6} \right)^{30-n}=2$

[sharp smelt]

idt it does so far

as the bias is 0, it is unbiased

[jolly parrot]

wai

[brave sluice]

i got a p-value of about 0.02

so i'm thinking it's biased

[sharp smelt]

hmm, but this is the expectation of 2 pips no?

[sharp smelt]

this is just wrong

[brave sluice]

the expected number of times to roll a 2?

[sharp smelt]

just realised

[brave sluice]

out of 30 rolls?

[sharp smelt]

nah, it would be 5

expecation would be 5

as all outcomes are equally likely

🤦

[brave sluice]

yeah 5 i think

you need to add a factor of n inside your sum

of i, i mean

not n

or, oh

yeah whatever the index variable is

[sharp smelt]

yup

got it

thanks

.close

[pearl pond]

Channel closed

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[pearl pond]

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[carmine schooner]

looking for proof verification. unsure if there is some edge case i haven't considered for the construction of R\setminus E as the union of open intervals

[carmine schooner]

this is the general idea

actually, i suppose i've forgotten the case where I_k is unbounded

[worldly jewel]

you might want to explain why the intervals are disjoint

[carmine schooner]

ah true, i left that out

does making the amendment "observe that as each x in R\setminus E is contained in exactly one I_x" suffice?

(otherwise x is contained in some other interval (c,d) but either a_x < c or a_x > c contradicting our construction of a_x)

yeah that works i think, thank you

[worldly jewel]

sounds correct

[carmine schooner]

.close

[pearl pond]

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[shadow jay]

prove that the diagonals of a parallelogram bisect each other through vectors

[pearl pond]

@shadow jay Has your question been resolved?

[pearl pond]

@shadow jay Has your question been resolved?

[desert wagon]

this settlement was promised to me 3000 years ago...

[pearl pond]

@shadow jay Has your question been resolved?

[timber cape]

whar

[dire tapir]

first thing I'd do is draw a diagram of a parallelogram ABCD, and then label the points of as A, B, C, D.

then draw the diagnols

just to help visualise it

[pearl pond]

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[quick venture]

Why isn't this -3y/x?

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[quick venture]

,rotate

[quick venture]

It's to solve the above differential equation

Using an integrating factor

[summer imp]

They aren’t equal

[summer imp]

p is just not identified correctly in this

p is what is attached to the y(x) term

[quick venture]

Thanks

.close

[pearl pond]

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[pearl pond]

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[royal galleon]

For number 19 if we use the fundamental theorem of calculus don't we get f'(x)=0. But then if we accept what they have for f'(x) and find f''(x) we have that it is equal to int 0 to 1 g(t)dt =0 but then we have 2=0

[sterile python]

Is the question asking for f'1 and f''1?

[royal galleon]

First it wants me to prove that f' is equal to some expression then to calculate f'' at 1 and f''' at 1

[sterile python]

Can you zoom 19 bigger i cant see well its blurred

[summer imp]

You don't get f'(x) = 0 because you get extra terms when you differentiate inside the integral

It's clearer if you expand the (x-t)^2 I think

$\int_0^x (x-t)^2 g(t) \dd{t} = \int_0^x (x^2 - 2xt + t^2) g(t) \dd{t}$

[jolly parrot]

Azyrashacorki

[summer imp]

Now you can split that integral into three integrals, and note that x is constant with respect to the integrals, so you can bring the x's out

The rest is product rule + FTC

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

Ok so two questions. I understand that once we choose an x it is a constant but untill then doesn't isnt it a variable since it can be any constant. Also why can't we apply the fundamental theorem directly?

[sterile python]

I solve like this

,rccw

[royal galleon]

I still don't get how we can treat x as a constant. Is it because though it can be any constant it is still a constant? Also I don't understand why we can't directly apply the first fundamental theorem

[sterile python]

Well i think x here is for domain set, while t actually change to calculate area in integral

And for instant if you have integral of 5t from 0 to x right? If you put 5 in x, it will be 0 to 5 and wont change from the start until you finish integrating

[royal galleon]

What about not being able to apply the first fundamental theorem directly?

[sterile python]

You already using it on t?

[royal galleon]

But applying directly to the original function

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

I don't understand why it doesn't lead to (x-x)^2g(x)

[pearl pond]

@royal galleon Has your question been resolved?

[pearl pond]

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[green aurora]

can someone explain this?

[pearl pond]

@green aurora Has your question been resolved?

[plush bramble]

don't troll help channels

[green aurora]

pls help

[plush bramble]

X is a function that elements in a set Omega to another set T

[green aurora]

yes

[plush bramble]

X^-1 is the inverse map of the image

[green aurora]

but its a function?

[plush bramble]

yes it's a function

[green aurora]

<@&268886789983436800>

[plush bramble]

<@&268886789983436800> i tried

[green aurora]

because if I consider the roll of a die I find that the inverse does not exist

[plush bramble]

a little more details on the subtlety https://math.stackexchange.com/a/1755385

[proper nova]

OP good job on calling the mods

[plush bramble]

oh you were asking about the inverse

no the inverse is not a function in general

[green aurora]

yes

[plush bramble]

same thing that not all functions are invertible

like $y = 3$ is not invertible

[jolly parrot]

riemann

[cinder bough]

Gone

[green aurora]

...

[cinder bough]

Sorry you had to see that

[green aurora]

👍

[green aurora]

But I can still go back from codomain to domain , even if it's not invertible, right? I mean, I can always apply X^_1

like omega={1,2,3,4,5,6} and T={0,1} even=0, odd=1 I can still take an s of T and apply X^-1 even though in this case it is not invertible

[tender crest]

yes, the inverse mapping of every function of a set exists and is defined as $f^{-1}(S) = {x \in T: f(x) \in S$, but the inverse of the individual values may not exist

[jolly parrot]

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

[tender crest]

(if $f:T \to S$

[jolly parrot]

Ada Conway

[green aurora]

but if I take my example and s=0, I find that it doesn't return a single value, but more

[tender crest]

Yes, so the inverse function does not exist on single values

but the inverse mapping of a set is well defined

the inverse mapping of S is just all the members of omega whose image is in T

[green aurora]

so in probability we are not talking about an invertible function

[tender crest]

Generally no unless there is only 1 value which the input to a function could take, and the function is onto and one-to-one

[green aurora]

bayes

thanks, can i ask a question on bayes

[tender crest]

yeah!

[green aurora]

why the vertical stick is P(A|B_1)?

[tender crest]

Well what does P(A|B_1) mean?

[green aurora]

i wanna prove that P(A|B_1)=P(A and B_1)/P(B_1)

[green aurora]

the probability of A knowing that B_1 happened

[tender crest]

Right, so given that B_1 happened, your probability space collapses into B1

If the entire world is B_1, A intersect B_1 is the only way that A can happen

[green aurora]

yes

[tender crest]

Does that help at all?

[green aurora]

why?

but where comes from P(A|B_1)

[tender crest]

The width of the rectangle is P(B_1), and the area of the shaded region is P(A intersect B), so the width of the side of the rectangle comes from P(A intersect B) / P(B_1)

by the area of a rectangle

[green aurora]

yes, but why the vertical side is P(A|B_1)

[tender crest]

by the thing you're trying to prove, or are you trying to use this to prove it?

[green aurora]

i wanna prove that P(A|B_1)P(B_1)=P(A and B_1)

I don't understand how to find the components in the graph

i didnt the graph in the pic

[tender crest]

yeah I think venn diagrams are easier to show this

but we know that B_1 happened with a probability of 1, and we want to find out if A will also happen. So our denominator is P(B_1) and our numerator has to be P(A intersect B_1)

[green aurora]

okay!

[tender crest]

because A and B have to happen, but they happen in the space of B_1 rather than the full probability space

[green aurora]

mmm

[pearl pond]

@green aurora Has your question been resolved?

[green aurora]

<@&286206848099549185>

[balmy monolith]

why r we still at the area based explaination if u dont understand it ;-;

ngl u should get a algebra kinda proof and it will be clear afterward to understand this area one

but eghh my bad not i am really being helpful here...

[pearl pond]

@green aurora Has your question been resolved?

[pearl pond]

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[torn flicker]

hmm what have you tried

[stray tulip]

All I know is what’s given

Which is 15 = AC

Ac = ef = fc would mean all of them are 15 I believe

[torn flicker]

so 5 cm 5 cm 5cm

[woven falcon]

Oh wait lemme try this, might take some time tho if u don’t mind

[stray tulip]

Ok

[nocturne grail]

AC=15 AND AE=EF=FC

AC != EF or FC.

in fact, AC contains EF FC and AE so logically it must be bigger than each one of them.

(as long as we're talking about lengths)

[stray tulip]

Ac is the diameter

So I assume 15/3 =5 cm

[nocturne grail]

well yeah you dont have to "assume"

AC is split into AE and EF and FC

which are three equal parts

so AC is split into three equal parts and thus its 5 + 5 + 5 (15).

you're correct tho!

I need to leave hope you find help soon

[woven falcon]

U can use coordinate geometry

Assume centre of circle to be at origin

And then assign coordinates according to measurements given in question

This way it becomes much more easier to use points of trisection E n F

I’m getting the answer as 3 . Root15

[stray tulip]

Ok

What next

[pearl pond]

@stray tulip Has your question been resolved?

[woven falcon]

Try it

I could send u the solution

But it’s better if u try it

Just draw the cartesian plane, ull see ehat to do next

[stray tulip]

What’s coordinate geometry

[woven falcon]

which grade r u in??

for us they did it in 9th grade

[stray tulip]

12

[woven falcon]

wait, they must have taught u coordinate geometry by now then

do uk what a cartesian plane is??

[stray tulip]

Do you mean graph

[woven falcon]

yeah that

ok wait lemme send u the working n try explaining it from that

so first thing we did here was take centre of circle at origin

and assign the coordinates

as we know DE is perpendicular to AC we can assign "y" coordinate to D too

now we know origin to D is radiues

so we apply distance formula

to find value of x coordinate of D

once we find D

we can apply distance formula on points C nd D to find CD

[stray tulip]

So what would CD be

[woven falcon]

3 root 15

but did u understand how we got that??

[stray tulip]

Wouldn’t it be 5 radical 5

[woven falcon]

radical??

[stray tulip]

Or square root

[pearl pond]

@stray tulip Has your question been resolved?

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[floral river]

Hello! I don't have a specific question, but could someone pls explain the conditions under which I can apply L'Hopital's Rule?

[west sapphire]

isn't it stated with hypotheses in your book or lecture notes or whatever?

[floral river]

I am still doing highschool maths. It was just a trick my teacher taught me, and I thought it would be useful

[smoky gull]

L’Hopitals rule can only be applied when both, the Dr and Nr are tending to 0 or infinity when the limit is applied

[floral river]

okok ty :D

[smoky gull]

So you get smth like 0/0 or inf/inf form

[floral river]

and I can differntiate the numerator and denominator over and over until I get an answer?

[smoky gull]

Note, you cant apply it if you get inf/0 form

[tame topaz]

note if you get something like 0 * inf you can rewrite it to take on the form 0/0 or inf/inf

[tame topaz]

if lhopital applies again yes

[smoky gull]

Yes but after diff once, you have to check again if its applicable

[floral river]

ok cool!

[smoky gull]

if it is, then sure go on

[floral river]

thanks guys

So just confirming, its ONLY when its 0/0 or infty/infty

[tame topaz]

^

[floral river]

ok cool ty 😎

[tame topaz]

for example think of $\lim_{x \to 0^+} x \ln(x)$

[jolly parrot]

Altanis

[tame topaz]

it takes the form $0 \times \infty$, but you can rewrite it to be $\frac{x}{\frac{1}{\ln(x)}}$ so it takes on the form $0/0$

[jolly parrot]

Altanis

[tame topaz]

then u can apply lhopital

[floral river]

mmmmm

thank uuu

[tame topaz]

np

[pearl pond]

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[short loom]

Hello! i have no clue how to deal with the 3-x thing

[short loom]

rahh

[warped violet]

What is the objective

[short loom]

uhh simplify

i suppose

yea :D

basic algebra

[warped violet]

Right

What do you mean by "no clue how to deal with the 3-x 'thing'"

[short loom]

well you normally solve these equations

by making the denominators the same

but idk how to convert 3-x to x-3 correctly (so i can mulitply entire fraction w/ x+3)

and keeping the equation consistant

[warped violet]

Let me start by saying that this is not an equation

It is an expression

And we can not solve expressions

[short loom]

i am expressing tiredness

oh ok

the expression mb chief

i learned smth new today 👍

[warped violet]

No problem

It's a very important distinction

[short loom]

gotchu

[warped violet]

And I want to make sure people understand it

Anyway

We solve equations

But here we have an expression

And among other things, we can simplify it

[short loom]

indeed

[warped violet]

And as you said that is what we are supposed to do

And yes

Since both terms are a fraction

We will usually try to combine them

[warped violet]

And we do that by finding the least common denominator

[warped violet]

Yes

This is a very simple step 🙂

Any ideas at all?

[short loom]

eager to hear how to do it

ummm i think you

multiply (3-x) by -1 and (x-2) by -1 but do you touch the left fraction at all

or no

i put it into math way and it did some weirdd shenanigans w/ the left fraction

[warped violet]

On the left fraction

You got the denominator to (x-3) * (x+3) right

[short loom]

yea

[warped violet]

Okay

And on the right

With one step (2) we can turn 3-x into x-3

3-x is also an expression by itself

And we can factor things in an expression

Can we factor something in 3-x

[short loom]

-1 (x+3)

i think?

[warped violet]

Well be careful

[short loom]

wait im bad

[warped violet]

If we multiply that we get -x-3

[short loom]

oops

[warped violet]

Or -3-x

[short loom]

-1 (x-3)

[warped violet]

Yes!

[short loom]

gang its so late im forgot the minus sign mb

[warped violet]

So $\frac{4-x^2}{(x-3)(x+3)} - \frac{x-2}{-(x-3)}$

[jolly parrot]

USS-Enterprise

[warped violet]

Now can we "move" the minus in the denominator in the second fraction anywhere

[short loom]

ohh

so you can make it

$\frac{4-x^2}{(x-3)(x+3)} + \frac{x-2}{(x-3)}$

[jolly parrot]

Azue

[warped violet]

Yes exactly

[short loom]

that makes sense

[warped violet]

Because $-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}$

[jolly parrot]

USS-Enterprise

[warped violet]

Just multiplying a fraction by (-1) right

[short loom]

right that makse sense

[warped violet]

Right

[short loom]

ok im pretty sure i can do the rest of the problem now

[warped violet]

Now from here you should be able to continue

Yeah

[short loom]

i js had trouble on the negative factoring thing

[warped violet]

No problem

Hope it's clear now

[short loom]

tysm :D good teaching by u

[warped violet]

Always think of factoring

🙂

[short loom]

got it

[warped violet]

Good luck!

[short loom]

thanks alot

cyaa

[warped violet]

No problem

[short loom]

!close

[warped violet]

!done

[pearl pond]

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

[short loom]

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

uh, confused

[vital estuary]

with what part

[sharp smelt]

(a)

I suppose I could show R / [lcm(a,b) ]is a field

[rustic gate]

why should that be true

[sharp smelt]

R/I is a field iff I is a maximal ideal is true

[rustic gate]

but why should (lcm(a, b)) be maximal

[sharp smelt]

oh, that need not be maximal

right

[rustic gate]

its contained in (a) ∩ (b), so its definitely not maximal generally

[sharp smelt]

oh

I can show (lcm(a,b)) is contained in (a) cap (b)

and then any princiap ideal in the intersection msut eb too

[rustic gate]

thats usually false

(1) is not contained in (2) ∩ (3) = (6)

[sharp smelt]

I meant lcm

[rustic gate]

ok

[vital estuary]

first off do you know the objective of the problem

[rustic gate]

start by showing this then

[sharp smelt]

it's leading towards (c) yes

hmm\

could I get some help

I'd normally be able to do this, but I'm panicking rn :(

[vital estuary]

how much sleep have you had

[sharp smelt]

~8/9 hours

[vital estuary]

nice

[sharp smelt]

no point in working while strssed

i'll be back

.close

[pearl pond]

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[shell brook]

just chill mate

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[sharp smelt]

$bc=qa$;Then $c=pa+r$. So $r= \frac{qa}{b}-pa \implies r= a(\frac{q}{b} - p)$. Which means $r \mid a$ which is a contradiction

[jolly parrot]

wai

[junior pecan]

has anyone of u tried euler circle?

[sharp smelt]

but this feels sus

[pearl pond]

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).

[sharp smelt]

I mean I need to say tha here b|q, but that feels wrong too

[dense jasper]

Does $q/b$ necessarily exist?

[jolly parrot]

Civil Service Pigeon

[sharp smelt]

no

so this is invalid, right

[dense jasper]

yeah

[sharp smelt]

hmm, then how do I do it

[dense jasper]

Your overall idea of using remainders is fine

||Bezout's identity||

[sharp smelt]

ofcourse

[junior pecan]

can anyone help me out

[sharp smelt]

#❓how-to-get-help

[junior pecan]

i got in a mathematical thinkings class

[junior pecan]

oh okay

then what is this channel for?

[sharp smelt]

hmm,

[junior pecan]

oh im sorry i thought this was the general chat mb

[sharp smelt]

Oh, I had another idea, if I may shoot it by you

let bc=qa. c=pa+r so bc=bpa+br, so qa-abp=br, then a(q-bp)=br. This would mean a|br which is only possible if r=0?

if this doens't work I'll resort to bezout's

[dense jasper]

a|br which is only possible if r=0
Take a=6, b=2, c=3 in the integers

The coprimality is important

[sharp smelt]

ya but here a and b aren't co prime

it can't be that a \mid r as r<a and a \nmid b as they are co-prime

[dense jasper]

Ok now you're using the $(a,b)=1$ to say $a \mid br \implies a \mid r$?

[jolly parrot]

Civil Service Pigeon

[dense jasper]

I mean this follows from bezout anyway so

[sharp smelt]

Oh, right

okay, so this works

[dense jasper]

I'd be more explicit at the end, but yes.

[sharp smelt]

cool

thanks!

[dense jasper]

[sharp smelt]

.close

[pearl pond]

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[pearl pond]

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[solar vault]

Can anyone help with this, I don't understand how to get the correct answer

[solar vault]

I can read the box plot but dunno how to get the number. I thought 75% is just a quartile 3

"more than" is probably also a key to solve it

Or I should think about the remaining 25%?

[vivid thistle]

Well 25% of beaches have less than 8 surfers right?

[solar vault]

Yea, but that also implies that 75% of beaches has more than 8 surfers

Wait, is that the answer 😭

[unborn abyss]

yes

[solar vault]

Wtf 😄

Thank you

[dense crater]

Hii

[solar vault]

I'm pretty sure that if "more than" becomes "less than" the answer will be reversed or something

[dense crater]

What are you solving

[solar vault]

Logic puzzles

.quit

.close

[pearl pond]

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[vestal tapir]

@silk whale

[shell brook]

so real

[midnight haven]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[midnight haven]

also wtf is this

[vestal tapir]

i forgot to check if they have dms open

.close

[pearl pond]

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[pearl pond]

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[silk whale]

Thank you frowny I just saw this

[pearl pond]

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

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

[silk whale]

.close

[pearl pond]

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[light jungle]

hi

[shell brook]

hello

[light jungle]

no

[pearl pond]

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[sharp smelt]

Let's consider the partition $P_n= {0} \cup \left{\frac{1}{n}\right}_{i=1}^{n}$

[jolly parrot]

wai

[sharp smelt]

The $L(P_n,f)= \sum_{i=2}^{n} \frac{i-1}{i+1}$. Which is not bounded above. Thus the integral doesn't exist?

[jolly parrot]

wai

[autumn fossil]

uh, how exactly did you compute that?

doesnt look quite right

[tropic saddle]

its piecewise constant

[sharp smelt]

on the interval [1/n+1,1/n] the minimum of the function is at 1/n and is n/n+2

[tropic saddle]

surely you have a lemma for that

[sharp smelt]

oh, right

[sharp smelt]

yea, didn't realise

it has countably many discontinuities and is thus integrable

.close

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[autumn fossil]

are you supposed to evaluate it?

[sharp smelt]

idts

[pearl pond]

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[daring bay]

this fucntion derived

[daring bay]

i dont understand

why isnt it 9/6b x^2

atleast the first part

[summer imp]

What is the derivative of x^3?

[daring bay]

3x

^2

[summer imp]

No

Yes

[daring bay]

but here bro why dont we do the whole fraction

[summer imp]

3/(2b) is a constant

$\left(\frac{3}{2b}x^3\right)' = \frac{3}{2b} (x^3)'$

[jolly parrot]

Azyrashacorki

[daring bay]

so if we had 1/2x^3

bro this doesnt make sense to me

why is it still 2b

b is just a number

so it should get multiplicated

[light helm]

that's not how fractions work

you multiply to the numerator

not both numerator and denominator

as then you'd be effectively multiplying by 1

you're implying that
$$\frac 11 \times 2$$ and
$$\frac 11 \times \frac 22$$
will both result in $\frac 22$

[jolly parrot]

ραμOmeganato5

[daring bay]

but if we have 1/2x^2

we get 1/4x

[summer imp]

No

[daring bay]

no

i mean 1

so

whatever ill jsut memorize it

this rule

[summer imp]

$\frac{1}{2} x^2$ differentiates to $\frac{1}{2} (2x) = x$

[jolly parrot]

Azyrashacorki

[light helm]

$a \times \frac bc = \frac{ab}{c}$

[jolly parrot]

ραμOmeganato5

[daring bay]

also if we have 3b/2 x^2

[light helm]

something that should've been drilled well before calculus
applying power rule wasn't really the issue here

[daring bay]

then we do multiply the 3b or do we also dont

[proper nova]

agreed

[light helm]

you multiply as per rules relating to multiplication of fractions

[daring bay]

yeah i never learned all these ffraction rules

properly

[summer imp]

Are you learning calculus from a course or by yourself?

[daring bay]

im in high school

😭

[proper nova]

🥀

multiplication rules should be covered in high school right? or am i wrong

[daring bay]

we did those in like 10th grade i never properly learned

[light helm]

a review of fractions is highly recommended

[summer imp]

Like omeganato said those are things you need to be comfortable with from the get-go because then your understanding of calculus is thwarted by lacks in algebra and arithmetic.

[proper nova]

crash course

[daring bay]

i literally got an exam tomorrow

[proper nova]

$\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$

[jolly parrot]

1 divided by 0 equals Infinity

[daring bay]

i know this stuff

it was just when deriviating

[light helm]

normal algebra rules still apply

applying power rule wasn't really the issue here
you didn't seem to have an issue with that

[daring bay]

hold up

wait

im calcualting something

[light helm]

after just differentiating, you'll have
$$\frac{3}{2b} \cdot 3x^2$$
then the rest is algebra

[jolly parrot]

ραμOmeganato5

[proper nova]

oh with respect to $x$ ig

[jolly parrot]

1 divided by 0 equals Infinity

[daring bay]

oh my god

look

3/2b * 3/1

[proper nova]

?

[daring bay]

wait

no

i mean yes

look

9/2b

[light helm]

do that if it helps you understand

[daring bay]

logical

it helps me yes

but its time consuming

especially

[light helm]

just one more step, better than getting it wrong

[proper nova]

if you are used to algebra alr, then the algebra steps takes like 1-2 minutes long

[daring bay]

also

square root of 4/9b^2

do i square root the 4/9 too

or just the b^2

which results in b

[proper nova]

$\sqrt{\frac{4}{9b^2}}$?

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

imagine it as $\sqrt{\frac{4}{9}} \cdot \sqrt{\frac1{b^2}}$

[jolly parrot]

1 divided by 0 equals Infinity

[daring bay]

no this

[proper nova]

square root of 4/9b^2
🥀

[proper nova]

the big root that you had here?