help-4

[mild cipher]

u wanna talk

eighty

u ther

@stoic adder

[stoic adder]

What do you require?

[mild cipher]

nth

[stoic adder]

Also this is a help channel -- if you need something, I advise opening another one, or dming me

use .close

[mild cipher]

.close

[vale dock]

Channel closed

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Available help channel!

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Remember:
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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[dense basin]

Given a fixed circle (O) with center O, fixed segment BC and moving point A on the large arc of BC on (O). A fixed circle (Oa) passes through BC and intersects AB, AC at F, E. BE intersects CF at P, AP intersects BC at L. K is the Miquel point of quadrilateral BCEF.AG. Prove that KL passes through a fixed point.

[dense basin]

I have provef K,P,Oa and KP bisects angle FKA

[vale dock]

@dense basin Has your question been resolved?

[jovial edge]

your best luck is waiting for someone who has done olympiad willing to help

<@&286206848099549185> someone should help this guys

[vale dock]

@dense basin Has your question been resolved?

[fading trail]

such a difficult geometric problem

i need to use geogebra to draw the graph first i suppose

[vale dock]

@dense basin Has your question been resolved?

[dense basin]

<@&286206848099549185>

[fading trail]

i just drawed the graph()

[honest stone]

oh lord this hurts every single atom in my body

[jovial edge]

-# that's nightmare

[fading trail]

There may be CMO difficulty.

I'm not sure

[jovial edge]

-# It's highschool entrance exam

[fading trail]

this?

is

[dense basin]

not exactly

the entrance exam is a specific case

where o_a is the midpoint of BC

this is generalized to O_a fixed passing through BC

[dense basin]

not really

its about less than VMO

(vietnamese MO)

[fading trail]

Then I feel that there may also be questions for the second test of the high school mathematics league.

[dense basin]

i have a feeling that menelaus and trigbash may work but its too hard

[dense basin]

what league is this

[jovial edge]

Ceva and Menelaus might work but I'm not gonna bang my head to solve it

not worth it

[fading trail]

High School Mathematics League in China

[dense basin]

this question might be a bit easier

cuz chinese geo is harder than vietnamese geo

[fading trail]

emm

you mean cmo?

[fading trail]

This competition is similar to the selection of CMOs.

[dense basin]

oh

CMO is much harder than VMO

[fading trail]

that's maybe true

[jovial edge]

you guys can chat here #competition-math

[fading trail]

I think the premise of using menelous is that there is a proportional relationship of edges.

[jovial edge]

other helper may not see your original problem

[fading trail]

I was just a junior high school student, and I didn't think I was very good at geometry.

[vale dock]

@dense basin Has your question been resolved?

[dense basin]

<@&286206848099549185>

[solemn widget]

Wow

i think essentially what we need to do is to prove K , L and the fixed point to be collinear

hence the first step to the solution should be defining what the fixed point is

it comes as a surprise when you know this is high school entrance examination question but the school is a specialized school for gifted students

[dense basin]

thats the problem, i cant pinpoint the existance of the point

[vale dock]

@dense basin Has your question been resolved?

[dense basin]

.close

[vale dock]

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[vale dock]

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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>.
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• 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.

[novel owl]

Could someone explain polynomials but like in a way where its actuslly understandable

[novel owl]

Its just that the whole thing is confusing. From leading term and coefficient to the end behavior

It might not be confusing but im just tired rn but my exam is close by do I csnt really go sleep. I want a way to understand it so that I csn understand it even whilst tired

[jovial edge]

there are lots of properties about polynomial, do you have any example on the part you don't understand?

[novel owl]

Not a specific example, no. Just the whole thing confused me(mainly because im kind of tired right now) so I kind of want it explained in a way thats easy to understand right noe

[rocky lotus]

brandon

[novel owl]

Ah so the power or whatever it was called is basically the leading term

[rocky lotus]

brandon

[novel owl]

If negative then down. If not then up. But what if the coefficient is x?

Got it

[rocky lotus]

brandon

[jovial edge]

Let take $4x^2$ the 4 we call coefficient and x we call variable

[rocky lotus]

Alexis_Fx

[jovial edge]

So for $ax^n$ a is coefficient

[rocky lotus]

Alexis_Fx

[novel owl]

Ah got it kinda.

I understand it

[rocky lotus]

brandon

[jovial edge]

For polynomials $P=5x^4+2x^2+4$ you can see $x^5$ has the highest exponent so we call coefficient of $x^5$ which is 5 leading coefficient

[rocky lotus]

Alexis_Fx

[novel owl]

What's that v next to the - infinity

[novel owl]

Wait you wrote 5x⁴ why would you say the x is x⁵

Oh

[jovial edge]

Oh sorry

Mb

[novel owl]

Oh

[jovial edge]

Yeah

[novel owl]

Is the end behaviour always a +infinity or a -infinity? Or can It be a number sometimes

[jovial edge]

Yeah, then x goes to ininity then the polynomials also goes to infinity

It can't be number

[novel owl]

Ah ok

[jovial edge]

It's true for polynomials but not for every function

[novel owl]

Again, the subject isn't at all complicated. Its my lack of sleep that makes it more challenging than it should be.

[novel owl]

Right polynomials was what I was worried about.

Anything else im missing?

[jovial edge]

Uhh, roots maybe?

[novel owl]

Ight sure

[jovial scarab]

i dont think you need to understand limits beyond end behavior if youre just learning about polynomials

[novel owl]

Oh quadratic formula i got it lowkey. My teacher said that I should just use mode 5, 3 in my calculator then ill get the answer

[jovial scarab]

what class is this for mantis?

as in what math class

its not calc/precalc is it

[novel owl]

Im studying pre calculus for my uni entrance exam

[jovial scarab]

ohh ok

[novel owl]

I need to score 60% or higher

And im not gonna lie, its tough

[jovial scarab]

when is your exam?

if you have time to learn it would be good to learn the factor theorem/remainder theorem/complex roots in general but I'm not sure how much time you have

[jovial edge]

pre calculus is not that tough , try hard one and you probably pass it

[novel owl]

In 8 hours from now

[jovial scarab]

oh 😭

[novel owl]

Its 4 am for me

Im not sleeping

[jovial scarab]

okay but i agree with alexis and the boundary doesn't seem too high

you got thiss

maybe sleep for a little bit :(

[jovial edge]

Yeah get sleep

[novel owl]

Appreciate you queen/king

[novel owl]

Im too stressed

[jovial scarab]

try doing some khan academy precalc /algebra 2stuff

[jovial edge]

Maybe try to relax a bit

[novel owl]

Trust me, sleep is what im begging myself for right now. But im too stressed to sleep right now

[jovial scarab]

itll give you quick practice and get more used to questions

[novel owl]

Sitting on a couch, listening to music, writing on my notebook

[jovial edge]

Wait, you said now is 4am and 8hour? You take the exam at 12pm?

[novel owl]

Yeah

Well i gotta be there at 12

Its 1

So thats when ill stop studying

Cuz the srive there is a bit long

Especially with the traffic

[novel owl]

Drive*

[jovial edge]

Better sleep at least 4-5 hours

[novel owl]

Mate I got 14 chapters😭

Im not sleeping

[jovial edge]

Then you can watch Khan Academy and practice a bit

[novel owl]

Perchance.

[jovial edge]

You shouldn't review all 14 chapters in 8 hours or you're gonna get exhausted

[novel owl]

I already am

God bless monster energy drink

[jovial edge]

You don't want to exhausted when doing exam

[novel owl]

Oh yeah

[jovial scarab]

yeahh and tbh if you just need a 60% you might want to prioritize a smaller spread of topics you can master more readily

[novel owl]

God bless coffee

[jovial scarab]

i doo

idk a ton about the more advanced residue stuff but i know the basics

is that help-13?

[jovial edge]

alright you better not use any energy drink or coffee until 1 hour before your exam

[novel owl]

Ight

Ill 100% come back for more help

As much as I trust myself, I trust others a lot more

Oh ypu helped a LOT

You and @jovial edge

[jovial edge]

You can DMs me if you want, that exam seems important

[novel owl]

Bless yakl bruh 🙏❤️

[novel owl]

If i score less than 60%, they'll make me learn pre calculus and business math. If i score 60 or higher, ill only learn business math. Thats why im so focused on it

[jovial edge]

Okay sure you can DM me about your chapter I will try to review it for you

[novel owl]

Bless your heart gang

[jovial edge]

No I'm going to have uni exam so I know what's feel like

Just want to help

A bit

[vale dock]

@novel owl Has your question been resolved?

[novel owl]

If i say yes will the channel close

Cause I kinda need it

[oak crane]

react with ❌ if u want to keep ur channel open

otherwise it'll time out and then it'll close the channel

[novel owl]

Ok thanks

Kinda still need this channel

[oak crane]

whats the question here btw?

[novel owl]

Me wanting to know about polynomials

[oak crane]

u should probably watch a video for it

[novel owl]

But thanks to @tawdry lintel and @jovial edge I kinda got the hang of it

[oak crane]

unless u have a specific problem

[jovial edge]

you should practice and then post any problem you get stuck on to get help with

That's much better

[oak crane]

u can always open a new help channel at any time when u have another problem

but to answer ur problem

Me wanting to know about polynomials
a polynomial is a sum of terms with variables raised to whole-number powers 🙂

u r familiar with this right?

source: https://www.geeksforgeeks.org/maths/polynomial-functions/

[vale dock]

@novel owl Has your question been resolved?

[vale dock]

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[mossy inlet]

can someone explain why I got this wrong??

[ruby sleet]

did you see the minus sign on the first 5?

otherwise you have to show your work as I can't guess

[mossy inlet]

OH

yes im so sorry I didnt see that

thank u tho

[vale dock]

@mossy inlet Has your question been resolved?

[terse plinth]

need some help

im familiar with inverse functions but not as much when its just a set of points

[violet ingot]

Well when the y is 4, what is the x?

[terse plinth]

also how do you insert latex into the channel

[terse plinth]

is that it?

[violet ingot]

yeah the idea of inverse function is that the y becomes x and the x becomes y

[terse plinth]

alr thanks

[violet ingot]

np 🙂

anything else you needed help on?

[terse plinth]

yea whats the latex command

to make a latex message

[violet ingot]

um I'm not sure tho there might be smtg in the latex help channel

[terse plinth]

theres a latex help channel?

[violet ingot]

$\frac{a}{b}$

[rocky lotus]

Archith

[violet ingot]

yeye in this server

[terse plinth]

oh you just put $$ at the beginning and end

[violet ingot]

that's what it looks like

anything other than that?

I'll close the channel then

.close

.close

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Remember:
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[gleaming niche]

Hey folks!

I have an arc, an origin of said arc, a radius of the arc, and a segment separate from the arc.
I'd like to find out how I'd find the longest vector (starting from the origin) intersecting the aforementioned segment whose magnitude does not exceed the arc radius, if any.
Any ideas?
I will continue to search for a solution in the meantime.

[gleaming niche]

if it helps, the segment marked in blue will always be either entirely vertical or entirely horizontal

working in 2D

the arc can face any direction though

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@gleaming niche Has your question been resolved?

[gleaming niche]

.close

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[tame bobcat]

can i simply use ||legendres three square theorem and show that n =/= 4^a(8n+7)||

[tame bobcat]

i literally cannot find an answer key anywhere

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[mental robin]

When dividing 𝑥+3/x^2+5x+10 What is the remainder? Question 6Select one: a. -4 b. 14 c. 4 d. 10

[mental robin]

i dont understand this

[hot tide]

$\frac{x+3}{x^2 + 5x + 10}$

[rocky lotus]

artemetra

[hot tide]

just to see it better

[mental robin]

ye

[ebon glade]

are you sure thats the correct way around?

[tidal swift]

but this is a proper fraction

[mental robin]

yes

[tidal swift]

the remainder is the numerator

[mental robin]

idk

[ebon glade]

!original

[vale dock]

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.

[mental robin]

better pic

[tidal swift]

that's the other way round

also is this graded

[short solar]

thats (x^2+5x+10)/(x+3)

[mental robin]

ye but u can take it as many times as you want

im just rlly stuck

[mental robin]

ohhh ye im rlly dumb..

kk ty

i dont need help

have a good day

.close

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[torn crescent]

This is less a pure maths question but I want to see why the math isnt mathing, I have the following circuits and values, my understanding is that developed means supplied, therefore all my negative values however it seems my answer is wrong?

[hearty belfry]

probably better to ask this in the physics discord. See #old-network

or maybe someone here who knows can help you

[torn crescent]

thats what I was hoping because its purely a math not adding up

[hearty belfry]

you can ping helpers after 15 minutes if you don't get help

[torn crescent]

ik

thank you tho

[vale dock]

@torn crescent Has your question been resolved?

[torn crescent]

<@&286206848099549185>

[gaunt whale]

Is this correct?

[viral idol]

uh what?

where's your channel?

[torn crescent]

!occupied

[vale dock]

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).

[viral idol]

go get another channel and i'll help you

[vale dock]

@torn crescent Has your question been resolved?

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[fossil folio]

Would anyone help me solve number 11? I haven’t try to work with isomorphism between functions and I guess I don’t get the fundamentals to know where to start as compared to sets.

[ebon glade]

the question isnt clear

are you supposed to show those are homomorphisms?

[fossil folio]

Isomorphism

[ebon glade]

none of those are isomorphisms

[fossil folio]

Oh how would you know so quickly?

Just like what’s the train of thought there

[ebon glade]

isomorphisms are bijective homomorphisms.

those maps are clearly not bijective

well ok I havent looked at all of them

and I dont actually know what F is, I am just assuming F is suitable

[fossil folio]

I’m just looking at 11 for now

F is a set of the functions mapping R to R

That have derivatives of all orders

[ebon glade]

11 is asking whether phi(f+g)=phi(f)+phi(g)

aka, (f+g)'=f'+g'

[fossil folio]

Ohhh I see.

Took me a second to get it.

My guess is not but I’m not sure why, it just feels instinctively. I’m unsure how to work the math to check though, since I can’t think of an example

The key would be to find two functions with distinct structural properties?

[ebon glade]

not sure what you mean

this either is obvious and you are allowed to use that derivatives behave that way

or you have to go back to the limit def of derivatives to prove it

[fossil folio]

Well is obvious for simple functions like x^2+ x. That would lead me forward saying that (f+g)’ = f’ + g’
But I’m unsure about more complex functions

[ebon glade]

do this then

how did you arrive at group theory without knowing this...

[fossil folio]

Just picked up a book and read through the first 3 chapters till arriving here. It may be obvious I just don’t see it.

[ebon glade]

but you took a calc course or something, right?

this is like the first thing you learn after learning about derivatives

[fossil folio]

Yeah it just that it’s been time. I know the definitions okey? I just can’t tie it together. But I’m getting ur comment.

Okey quick search cleared all my doubts. I’m sorry you were right it is pretty damn obvious my bad.

[fossil folio]

How did u arrive at this again though?

[ebon glade]

plugging in the definition of phi

[fossil folio]

Oh I see thank you. I’ll try the other ones now. Thank for you help and sorry for the bother/ stress lol.

.close

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[pure kiln]

That is the end of a tedious proof of a simple looking theorem about projective geometry, but I am just losing time on it (a few hours already) because I can't fin the answer

Thank you 🙏🏻

[vale dock]

@pure kiln Has your question been resolved?

[pure kiln]

<@&286206848099549185>

[wraith heart]

did you try to find a counterexample

If $\lambda_1 = 1$, $P^\top P = I_3$ has only a few solutions

[rocky lotus]

riemann

[distant pulsar]

it is wrong

[pure kiln]

Oh, what is a counterexample ?

[distant pulsar]

$P = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 \end{pmatrix}$

[rocky lotus]

Mqnic_

[distant pulsar]

lambda_1 = 1, lambda_2 = w

[pure kiln]

wtf

Ok my statements were not complete, I will write it again

[pure kiln]

And what I wrote here is evrything that I deducted, supposing det(P) =1

[vale dock]

@pure kiln Has your question been resolved?

[wraith heart]

sounds like you're making up a problem as you go

if you have a proof for your problem, can you just show it

[pure kiln]

I have a big chunk of proofbut it's in French

And it would be too long to rewrite in english

[spring jackal]

there are a couple of french speakers here, you can post if you'd like and maybe they can help

[pure kiln]

Where B_1, ..., B_6 are the matrices defined by these quadratic forms

(They have the property that Tr(B_i.inverse()B_j) = 0 for i and j different)

(and so are linearly independant)

(omega is a cube root of untiy)

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@pure kiln Has your question been resolved?

[vale dock]

@pure kiln Has your question been resolved?

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[torn bay]

no idea how to go about doing this (the first part is easy i'm on abt the second one)

[normal hollow]

What have u tried

[torn bay]

can i just write pi as n

so nrl for the formula of curved surface area

and i think to grt the radius

you do pythagoras with

x as hypotenuse

so x^2 - h^2 = radius^2

but

that just feels and looks wrong

[normal hollow]

The arc PQ,s=r(theta)=120x

And the circumference of the base of cone is PQ
So these two r equal

Finding a relation between x and r and substituting in V=3A,might help

[torn bay]

so r is equal to the arc length?

[normal hollow]

Y do u want to write it as n

[torn bay]

just easier cz it looks like pi

[normal hollow]

No circumference =2(pi)r

[normal hollow]

I see

Do u have any progress on it rn?

[torn bay]

ok i need to go sorry i'll try work it out on my own later i think i've made some progress

bye

[normal hollow]

The s=r(theta) is useful

Bye

[torn bay]

.close

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[lapis widget]

how can i solve this quickly with desmos?

[lapis widget]

<@&286206848099549185>

[inner bloom]

enter the equations into desmos and use a slider for c

[lapis widget]

i tried that but its very inefficient

.close

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[obtuse herald]

I'm having problems finding the last two eigenvectors. Could someone check my work?

[obtuse herald]

-1

arithmetic error

anything else?

[distant galleon]

No

[obtuse herald]

ok. i kept having an issue with the calculator rounding during row reduction.

i'll have to look into that

thanks

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[oak crane]

how do I get started with this?

[inner abyss]

looks like laplacian expansion along the first column might be helpful for the induction

[oak crane]

oh yeah thats a good start lemme try it

ok i think I can continue the rest, thanks

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[silver whale]

Can u help me with this question

Q.7

[jovial edge]

,rccw

[flint kindle]

1+a+a^2…+a^k-1 = a^k-1/(a-1)

Or actually

Take the modulus of zk

[river shale]

Do you have any idea in advance?

[flint kindle]

|zk|=|1+a+…+a^k-1|<= 1+|a|+…+|a^k-1|= (|a|^k-1)(|a|-1)

Ah perhaps rearrange zk

Zk-(1+a)=a^2+…+a^k-1, |zk-(1+a)=|a^2+…+a^k-1|<= |a^2|+…+|a^k-1|

And then bound it more using gp sum

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[distant galleon]

<@&268886789983436800>

[gleaming kindle]

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[north scarab]

for the first question, I wrote x^3/3

[north scarab]

but the answer was

is my answer considered correct?

[next vortex]

you wrote down an indefinite integral to the function y = x^2

definite integrals are the ones with bounds, and correspond to areas under graphs

[north scarab]

but with x^3/3, I didnt include the integral symbol

[frozen ledge]

youd need to apply FTC

[river shale]

Just x^3 /3 itself does not work

[north scarab]

why not

[frozen ledge]

it’s just an anti derivative

it’s a function

[river shale]

You cannot omit the bound

[north scarab]

I tried understanding it, i understanded just few parts of it

I wanna understand ftc (1)

[north scarab]

yeah but

the proof

[frozen ledge]

do you know anything about derivatives formally?

[north scarab]

yes i do

[frozen ledge]

consider the difference quotient and apply rules for splitting integrals

the intuitive explanation is probably best explained with an analogous scenario involving the surface area of a sphere and the rate of change of a spheres volume

think of the rate at which a spheres volume increases to be proportional to the surface area of the sphere since adding on layers of paint to the sphere is an increase in the spheres volume

you can form a similar idea with integrals in this situation

[north scarab]

wait what are you explaining?

[frozen ledge]

you asked about FTC

[north scarab]

oh yes

[frozen ledge]

do you see how to connect this idea to integrals

think of f(x) as being the layer of paint

instead of adding on to 3 dimensional volume for a sphere you’re adding on area

f(x) is the height instead of the surface area

[ashen prawn]

what a brilliant one

wow

[north scarab]

wait do u know what f(t) is

this is part of understanding ftc

is f(t) the graph?

or the y axis

[frozen ledge]

seems to be the y axis

to try and say y = f(t)

[north scarab]

and f(x) is the graph?

[frozen ledge]

value of f at x

they drew the point x in

[north scarab]

yes

why did they say f(t) be a continuous function

arent they referring to the graph?

[frozen ledge]

wdym

[north scarab]

Let f(t) be a continuous function on the interval [a,b]

[frozen ledge]

continuity implies integrability and is used in the proof

[north scarab]

where is a?

[frozen ledge]

idk what you mean by this though

[north scarab]

yea forget that

[frozen ledge]

here it appears to be at 0

but it doesn’t really matter, just some fixed value

[north scarab]

you said that f(t) is the y axis, but then why are they refering to the interval a,b?

how can a y axis be a continuous function?

[frozen ledge]

f(t) isn’t the y axis.

they wrote f(t) to denote that y = f(t)

the y axis is the line t = 0

[north scarab]

wdym

[frozen ledge]

they’re just saying f takes values along the vertical axis

this shouldn’t be new to you

you’ve never seen someone say y = x^2 or whatever

it’s poor notation i agree

but it shouldn’t be new

[north scarab]

alright ill come back to this

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[modest jolt]

Help

[modest jolt]

Why is the median important

Or mode

Whats the usefulness of using mode

[tawny moon]

mode finds the most frequent value

[hybrid hound]

Median is useful for when a set might be skewed for a nicer average than mean

[tawny moon]

median finds the center value regardless of outliers

[modest jolt]

What outliers

[tawny moon]

let's say you have a set of data {1, 2, 3, 4, 5, 10000}

[modest jolt]

What do you mean by skewed

[tawny moon]

that 10000 is an outlier

[modest jolt]

Yes

[modest jolt]

Why

[tawny moon]

because it's so far off from the rest of the values

[hybrid hound]

A set with outliers or a tail pushing the mean away from the medial

[modest jolt]

What about {1,2,3,5}

[modest jolt]

Ohh

[tawny moon]

this is still fine
honestly, to determine outliers, you need to draw a box-and-whiskers graph

[modest jolt]

Haven't learnt that

[tawny moon]

but that set of data you provided shouldn't have an outlier

[modest jolt]

What about mode?

Why is it important to calculate the mode

[tawny moon]

if you have data where one or two variables happen a lot or you want to find the most frequent event, that's when you use the mode

suppose you have a school lunch program that offers 3 types of meals and you want to find the most popular type

and this is your set of data for some students:
{3, 3, 3, 1, 3, 2, 2, 3, 3, 3, 1, 2, 3, 3, 1, 1, 2, 2, 3, 3}

[modest jolt]

What if there's 2 mode

[tawny moon]

then two or more events are equally popular and both of them become the mode

[hybrid hound]

I remember a video from a while ago where the graph of test scores had 2 peaks(essentially modes) which showed that cheating went down

[modest jolt]

Oh okay

[modest jolt]

EH

[tawny moon]

btw the mode of this set is 3, if you can't tell

[modest jolt]

Kkai

Okay

Bye now

[tawny moon]

!done

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[jovial scarab]

i wanna share a fun fact but its not at the correct level for this but i will share it anyway

the mean, median and mode can all be thought of as numbers that minimize specific error functions over a dataset

the mean minimizes the squared error, which you get by subtracting a fixed number from each element of the set, squaring the differences and adding them together

the median minimizes the absolute error, which you get by subtracting a fixed number from each element of the set, taking the absolute value of the differences and adding them together

the mode minimizes the discrete error, which you get by taking a fixed number and counting the number of elements in the set that are different from that fixed number

in that sense you can come up with lots of statistics for datasets by picking some sort of error function and minimizing it, and there's contexts where these statistics make sense depending on what notion of error makes sense for you

[tawny moon]

i'm saving this, this is very good explanation

thank you! (even though i'm probably not your target hahah)

[jovial scarab]

no im glad this was useful!!

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[rustic knoll]

What program/tool(s) can I use for results like this?

[rustic knoll]

Like how can I have a computer turn my expressions into sum of squares

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[visual saddle]

What is the sum of the real solutions of the equation?
$t = 1+ (t^2-1)/(1+sqrt(t^4 - 7t^2 + 16))

[visual saddle]

$ t = 1+ (t^2-1)/(1+sqrt(t^4 - 7t^2 + 16))

[candid bolt]

you gotta end it with a dollar sign

[visual saddle]

$t = 1+ (t^2-1)/(1+sqrt(t^4 - 7t^2 + 16))$

[rocky lotus]

arachi

[stark wedge]

if you're gonna do latex, do it properly

$t = 1 + \frac{t^2-1}{1+\sqrt{t^4-7t^2+16}}$

[rocky lotus]

Ann

[visual saddle]

Yea thanks

[stark wedge]

progress?

[visual saddle]

Don't know how to start

My first thought was to put everything on the same denominator and equal the fraction to zero

But the fraction I get is very complicated

[normal hollow]

Taking 1 to lhs, u can eliminate (t-1)

[stark wedge]

subtract 1 from both sides and note t=1 is a solution, then divide by (t-1) to look for others

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[celest bay]

i have no idea how to proceed

[stark wedge]

consider the function f(x) = (a+x)(b+x)(c+x)(d+x)

this is a polynomial function of degree 4 and it also satisfies f(1)=1, f(2)=2, f(3)=3, f(4)=4

does that give you any ideas?

[celest bay]

uh

f(x)-x=k(a+x)(b+x)(c+x)(d+x)

put 5 in there

and maybe smth will happen

wait

no

f(x)-x=k(x-a)(x-b)(x-c)(x-d)

will have roots -1 -2 -3 -4

k(a+5)(b+5)(c+5)(d+5)=f(-5)+5

but k and f(-5) are unknown

@stark wedge

u there?

[jade ivy]

Subtracting x from f is only going to modify the coefficient of x in the polynomial, you are not going to get a "k" like that.

[stark wedge]

not quite.

a, b, c, d are given in the problem. you can't just assign them new meanings out of the blue.

f(x) - x has 1, 2, 3 and 4 as roots.

and it is also monic.

thus $f(x) - x = (x-1)(x-2)(x-3)(x-4)$

[rocky lotus]

Ann

[stark wedge]

and the thing we're asked for is f(5)

not f(-5)

[celest bay]

yeah ik i messed that up

[celest bay]

k is the leading coefficient

[celest bay]

how about this

this should do it right

[celest bay]

(a+5)(b+5)(c+5)(d+5) is what we are looking for

f(-5) takes us there

[stark wedge]

no

consider the function f(x) = (a+x)(b+x)(c+x)(d+x)

it is f(5) that we want, not f(-5) but f(+5)

we want f(+5)

[celest bay]

i didnt

i factorised the polynomial

[lyric sundial]

You did...

[stark wedge]

then this

f(x)-x=k(a+x)(b+x)(c+x)(d+x)
is wrong

[jade ivy]

You were already given a very very large hint by Ann to the point of basically giving the solution away. You should think for a bit longer on why what they said is helpful before blurting the first thing that comes to mind.

[celest bay]

just like ax^2+bx+c can be written as a(x-root1)(x-root2)

where a and b are roots

[stark wedge]

variable name conflict.

you are reusing variables again.

[celest bay]

?

that was an example

[stark wedge]

an example of the very same mistake you keep making lol

[celest bay]

oh cmon

[stark wedge]

you're using a and b in two different meanings

first they're the leading and linear coeff of the quadratic

then they're somehow the roots?

don't do this.

[celest bay]

ok i messed up

but what i did with the original question

isnt wrong is it

[stark wedge]

f(x)-x=k(a+x)(b+x)(c+x)(d+x)
this?

[celest bay]

this

[stark wedge]

is that a yes or is that a no

............................

please stop sending such mixed signals.

[celest bay]

.

f(x)-x=k(x-a)(x-b)(x-c)(x-d)
will have roots -1 -2 -3 -4

there u go

[stark wedge]

you're still reusing variables.

[celest bay]

what did i reuse

[stark wedge]

also k=1.

a, b, c, d are given in the problem. you can't just assign them new meanings out of the blue.

[celest bay]

but how

[stark wedge]

(x-a)(x-b)(x-c)(x-d)
variables a-d are part of the original defn of f(x)...

there's a possibility you're being dense here idk

[celest bay]

im sorry im bad at math

[celest bay]

..

ill rethink for a moment

[lyric sundial]

It's not math, it's logic and comprehension I'd say

[celest bay]

im bad at all of them then ig

is it 24

the answer

@lyric sundial

[lyric sundial]

No, 29

[celest bay]

uhhhhhhhhh

DANG ITT

i didnt add 5

[lyric sundial]

If you sub x = 5 here

[celest bay]

got itt

thanks guys

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[lilac mountain]

Hey is there a name for the notation 5 6/2 meaning 5 + 6/2?

[hearty belfry]

the + sign?

that just means addition

[ebon glade]

mixed fraction

[hearty belfry]

oh you're talking about that

[timid wagon]

mixed numbers

[lilac mountain]

Oh great thanks guys

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[slender plover]

this are 2 functions which are basically the same things, recently i started to learn derivatives and came a question to me what if i tried to solve the same functions with 2 different methods, i also added the result of the derivatives for comparison,
my problem is trying to figure out what went wrong

[hearty belfry]

,rotate

[slender plover]

rotate?

nvm haha

would it help if i showed you my way of solving it?

[river sequoia]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[river sequoia]

I dont think u should

[hearty belfry]

sure, but it would be easier just to split the fraction and apply index laws

[hearty belfry]

?

[river sequoia]

Ohhh

Helpp i didnt realize hes the one asking the qs

Ok sry

[hearty belfry]

then you can just apply power rule when doing the derivative

[river sequoia]

I also didnt know u were here ahhhh

[slender plover]

so the left function can be written as (x+3) * x**-1/2

sorry the right function

[hearty belfry]

I meant doing something like this for the left function $\frac{x}{\sqrt{x}}+\frac{3}{\sqrt{x}}$

[rocky lotus]

឵឵MxRgD

[hearty belfry]

then convert them into index form so you can apply power rule

[river sequoia]

Why not just apply the differentiation of rational functions

[slender plover]

but doesn't it complicate things?

[hearty belfry]

quotient rule?

that would take longer, I think

[hearty belfry]

Not really

[slender plover]

i am trying to solve the right function without multiplying both of the objects

[river sequoia]

Yea

[slender plover]

so keeping them separately wait let me show you my way

[hearty belfry]

,rotate

[slender plover]

i used the multiplying derivative rule

[hearty belfry]

product rule?

[slender plover]

uhmm english is not my first language so i am not exactly sure

basicaly a * b

[river sequoia]

Why do yall have weird names for them lollll

[hearty belfry]

Okay but this seems to be just overcomplicating the problem

[slender plover]

i know

but why do i get a different solution

that's what i am trying to figure out

[hearty belfry]

that's what they're called

[river sequoia]

We dont give them names

[river sequoia]

Ohhh u get different solutions when u derive in different ways for the same function and wondering what u did wrong ?

[slender plover]

yea

[river sequoia]

Ok lets see what u did

[slender plover]

i was thinking maybe it's just a different way of expressing the same derivative of the function

[river sequoia]

We’ll try to pinpoint where things went south

[slender plover]

do you want to see the way of the other function?

[slender plover]

.

[river sequoia]

Yup

[slender plover]

could the derivative functions be equal to each other?

no i just checked they aren't 😭

[slender plover]

the minus?

[river sequoia]

Yea where did u get it from

[slender plover]

ohhhh

[river sequoia]

The derivation of sqrt(x) is 1/2sqrt(x)

[slender plover]

wait let me redo it

so the result is (x -3) / (2x * sqrt(x)) right?

but they still aren't the same if i place x = 4 at both equations i get different results...

[river sequoia]

Also this

U did this using the chain rule

[slender plover]

let me google the chain rule

[river sequoia]

Wait its acc called the product rule

This

[slender plover]

i don't think i did i didn't multiple by the inside function

oh this yea i did use it

but in the other function i didn't use it

wait so was my derivative functions correct?

[river sequoia]

Here x^(3/2) is sqrt(x^3) not the other way around

[slender plover]

ohhh so the divider was wrong

wait isn't it the same

if we place x = 4 in both sqrt(x^3) and (sqrt(x)) ^3 we get the same answer of 8

wait it's not the same because what if x < 0

wait actually?

[river sequoia]

Hold on lets get something clear

X^(2/3) != x^(3/2)

[slender plover]

because if x < 0 right?

[river sequoia]

And these represent the same fraction which is x^(3/2)

[river sequoia]

It kinda gets tricky with x<0 and even powers
But this mainly comes from exponent rules
(X^a)^b = x^(a*b)

Which makes it
=(x^b)^a)(X^a)^b = x^(a*b)

In this case u used a=3 and b =1/2

[slender plover]

so it's ( x ^ 1/2)^ 3

[river sequoia]

Switch them wouldn’t be a problem

[river sequoia]

Yes

[slender plover]

so they are both the same?

[river sequoia]

Yes but i wanted to point out that this can be a bit tricky to work with

Since it does not always work with negative numbers ..

Lets say x=-1 snd i put in (x^2)^3

Its fine id get 1

However if u try to switch it to (x^3)^2 u will run into problems

[slender plover]

at x = -1

they are both 1 no?

[river sequoia]

Oh wait mb

[slender plover]

wait my older brother came he helped me see the problem do you want to see it?

[river sequoia]

Sure

[slender plover]

i haven't checked it though

[river sequoia]

Yes this is correct

[hearty belfry]

just use this method as I've said before, it's a lot simpler

[hearty belfry]

yeah that's the quotient rule

[slender plover]

do you mean multiplying them?

[river sequoia]

He basically showed you in general sense how both of the functions end up as the same expression

[slender plover]

my problem was i didn't use the chain rule

i didn't multiple 1 / sqrt(x) by the inner derivative of square root of x

i think

so i think i am correct but it doesn't make sense why would i use the chain rule

[hearty belfry]

,rotate

[hearty belfry]

It looks like your brother is showing you how to derive the quotient rule using the product rule

but as i've said before, it's just overcomplicating things

[slender plover]

i think i understand

anyway thank you both @river sequoia and @hearty belfry for your time and help i really appreciate it

[hearty belfry]

np

[slender plover]

have a nice day!

.close

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[river sequoia]

Yt!

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[wary panther]

We have two different random samples, X and Y. Both are on normal distributions with expected values mu_x and mu_y and standard deviations sigma_x and sigma_y. We want to find a 95% C.I. given that sigma_x = 2, sigma_y = 1.5. We also know that X has 15 points of data, the sum of X_k = 21.8, the sum of X_k^2 = 67.3. Y has 12 points of data, the sum of Y_k = 29.5, the sum of Y_k^2 = 84.4.

[wary panther]

They're both ofc iid

How do we get a distribution that is nice to deal with when making the confidence interval

Do we first say that the sum of X_k is N(mu_x, sigma_x) distributed, do the same for Y_k and then make a new r.v called delta or something, defined as Y-X and find the distribution for it?

If we say delta = Y_avg - X_avg, is it true that delta ~ N(mu_y-mu_x, sigma_y + sigma_x) distributed

How do we know what std to divide with when we don't have a 'simple' R.V? How does it become sqrt(sigma_x^2/n_x+sigma_y^2/n_y)?

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[void cedar]

can someone help?

[stark wedge]

ok you should probably post these one at a time but w/e

which one would you like to begin with

@void cedar

[void cedar]

idk from 1st?

[river shale]

Any idea in advance?

@void cedar

[stark wedge]

this uhhhhhh hold on

[void cedar]

I have no any ideas can u help?

[stark wedge]

try to find the dimensions of this rectangle in terms of these square cells it's made of.

if you were given something like

The area of a rectangle is 3179 square inches and its length is 2.75 times its width. Find both dimensions.
would you know what to do?

@void cedar

[void cedar]

oh get it ty

[stark wedge]

as you say...

okay, on to problem #2? same question to you -- do you have any progress?

[vale dock]

@void cedar Has your question been resolved?

[river shale]

@void cedar still need help on question 1? It's been a long time since you started trying

[void cedar]

no I have solve everything thanks

[vale dock]

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[sharp whale]

.close

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[thorn flame]

Does anyone know how to derive this formula (product moment correlation coefficient

[thorn flame]

I can use it but i would like to know how this formula appeared/the steps to get to it

[jovial scarab]

do you know a covariance/correlation is in the context of random variables

The RHS is just the definition of sample correlation in that sense

[thorn flame]

I need to remind myself on covariance

mb

[jovial scarab]

if we have 2 random variables X and Y, their covariance is described as E[XY] - E[X]E[Y]

what this means is “multiply each x coordinate by the y coordinate and find the average product, then subtract the average x value * average y value”

(E just means expectation)

This is like describing how strongly the two variables behave together

[thorn flame]

this is seems similar to the formula for variance; am i correct on that?

[jovial scarab]

and we approximate these averages with our samples, we just average the samples based on whatever function we’re trying to get

yes! Variance is a special case of covariance where the two variables are the same

they both are centered statistics, and after you do the centering (by subtracting out means) what’s left over describes the variations

Approximating E[XY] over a sample is basically just taking the individual xi * yi products and averaging them

same idea for E[X] and E[Y] etc

correlation is basically a normalized version of covariance

After taking the covariance, you divide by the square root of the product of the variances of X and Y

if you’ve seen dot products before it’s the exact same principle there, the covariance is like the dot product, and dividing by the standard deviation (sqrt of variance) of each variable is the same idea as dividing by the norm of each vector in a dot product, to isolate the similarity

in the same fashion, dividing by the standard deviations (the norms of the variables) gives us just the similarity between the two variables, which will lie between -1 and 1 as a normalized version of covariance

[thorn flame]

I am going to save this paragraph onto mu computer as a word document if you dont mind 🙏

[jovial scarab]

I don’t mind!

[thorn flame]

ALso

I will try right it down on paper

to what i understand of it

just checking on this one

because idk if i understood this correctly

can you say that covariance is;

( sum(xi x yi) / n ) - ((sum(xi) x sum(yi) ) / n^2 )

(sum being uppercase sigma)

Hmmm

Why does N become the coefficient of sum of the summations?

[jovial scarab]

finish the denominator first, and then you’ll see that we can simplify the divisions

We still need to divide by the standard deviations

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@thorn flame Has your question been resolved?

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[sour reef]

If d has a right angle to l1 and l2 then shouldnt l1 and l2 be parallel?

[distant pulsar]

not in 3d

more clearly

[sour reef]

ahh

I see thank you

Also, how did you have that image up so quickly

[distant pulsar]

showed up on google ("skew lines")

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@sour reef Has your question been resolved?

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[regal forge]

h(x)=2^x
find h^-1(x)=10
need help it is a non-calculator question

[open abyss]

Apply $h(x)$ to both sides

[rocky lotus]

StrangeQuarkAL

[regal forge]

okay so the answer is 1024?

[open abyss]

yh

[regal forge]

thank you

[open abyss]

np

[regal forge]

.close

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[safe fulcrum]

everyone asks this IG question at some point

lmao

[regal forge]

at least im not alone😂

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[junior ore]

wow

x=3

-_-

how not so?

tf

2+1 is 3

how do u get pi involved in that

whattttt

[tidal terrace]

!xy

[vale dock]

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.

[junior ore]

2.13?

[tidal terrace]

......

......

[junior ore]

bruh

englighten us then

[noble anvil]

<@&268886789983436800>

[stiff lily]

.close

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[chrome beacon]

was about to mute them for 2.13 days

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[north scarab]

Hiii, what is lnx?

[glass kelp]

[ \ln x = \log_e x]

[north scarab]

oh

[rocky lotus]

k

[north scarab]

i heard of log

but we never learnt about it

[merry crystal]

you havent learnt log?

[glass kelp]

It’s jost notation

[glass kelp]

what

[north scarab]

nah but next semester probably

[glass kelp]

In the big calc 🥀

[merry crystal]

and you are doing integrals?

[north scarab]

yea

[glass kelp]

Interesting choice of curriculum

Anyhow

[north scarab]

fr

[glass kelp]

[ a^b = c \iff b = \log_a c]
So,
[ e^b = c \iff b = \log_e c = \ln c]

[rocky lotus]

k

[north scarab]

whats the arrow?

who gave me the actve role 😮

[merry crystal]

implies in both direction

[north scarab]

thanks 🙏

.close

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[lime vapor]

Is this valid?

[stark wedge]

what's your goal? is it to show that N-P (ie the set of composite naturals) has the same cardinality as N?

@lime vapor

[lime vapor]

Kind of

I want to show that they are statistically identical

For example if in N the even numbers are 1/2 of the set, the same applies on N-P

[stark wedge]

what is 'statistically identical'

[lime vapor]

.

[stark wedge]

gonna need a formal definition there...

[lime vapor]

I see

How do I do that?

[stark wedge]

where did you get this concept of "statistically identical" from?

do you have a formal definition of "sets A and B are statistically identical" on hand, yes or no?

[lime vapor]

I need it for smth I'm working on. I don't remember the original name but I've seen it before

[stark wedge]

that sounds like a no

[lime vapor]

Mhm

[stark wedge]

no definition no proof

you cannot possibly prove that two sets satisfy foobar property if you don't even know any definition of foobar property

[lime vapor]

Ah ok

Thx I'll try to find the formal definition

.close

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[north scarab]

I need help with a

[tawny moon]

what have you tried and what do you not get

[north scarab]

I just tried random stuff

[junior ore]

u have to integrate

[tawny moon]

?

[hearty belfry]

first integrate 3-2x

[tawny moon]

why did you try random stuff? integrate it first

[junior ore]

u will get 2 values of k

[north scarab]

I had no idea how to approach this

[glass kelp]

Then evaluate normally

[junior ore]

integrate it and tell me what u get

[glass kelp]

And set the final expression to 2

[tawny moon]

i mean just because k is a bound doesn't mean you can't integrate with it yk

[junior ore]

yeah

u will get 2 values

[tawny moon]

yes it's an unknown, but you can treat the unknown like a number and just sub it in

[junior ore]

yup

[north scarab]

actually this what I have tried before

[mild cipher]

,w x^2= -1/9

[tawny moon]

don't use this in somebody else's help channel?

[junior ore]

wrong

thats not the values u would get

[mild cipher]

srry

[junior ore]

k=1 and k=2

[nimble crypt]

Wrong because you have 2 on the other side not 0

[tidal terrace]

you have #bots use the command there

[tawny moon]

did you forget to bring the constant term over to the left

[mild cipher]

alr thxs