help-23

[shadow cloud]

i want to know the error

[flint vessel]

many errors actualy

reattempt the second line

and send me what you get

[shadow cloud]

wait

[jagged belfry]

its helpful to look at this, as this is your goal

x^2 + b/a*x + c/a     = 0
(x+d)^2 + h           = 0
x^2 + 2d*x + d^2 + h  = 0

then all you want is to solve b/a = 2d and c/a = d^2 + h

[flint vessel]

try to do the completing the square part again

[flint vessel]

yup

[shadow cloud]

this really hurt my mind

[flint vessel]

the best way would be to turn the coefficient of the second degree term to 1

then try to complete the square

that should lower your error rate i think

[shadow cloud]

bro

how do i do it

tbh i can just ask my home teacher on saturday :(

i can only solve average quadratic equations but this no. I gotta ask my teacher cuz hes a master in math

[jagged belfry]

hold up i messed up

[shadow cloud]

do u mind if u solve it on a piece of paper and send me?

i asked google and chatgpt, both are the sames and idk how they do it.

[jagged belfry]

We get:
d = b/(2a)
c/a = d^2 + h => h = c/a - b^2/(4a^2)

Hence:

x^2 + b/a*x + c/a = (x+b/(2a))^2 + c/a + b^2/(4a^2)
                  = 0

so

(x+b/(2a))^2 = b^2/(4a^2) - c/a
             = (b^2 - 4ac)/(4a^2)

[dusk gyro]

ur just confusing him atp

he doesn't even know the fundamentals of how to solve quadratics

based on what i've observed

[shadow cloud]

I dont know how to read those 😭

[jagged belfry]

oh my bad

[dusk gyro]

@shadow cloud i think you should just use the quadratic formula

[shadow cloud]

i know hot to solve but this no

[jagged belfry]

that doesnt help him

to use the quadratic formula you should derive it first

[shadow cloud]

that doesnt work

[burnt notch]

Huh?

Why not?

Quadratic formula works always

[shadow cloud]

I only use (a + b)^2 formula and (a-b)^2 formula and a^2-b^2 formula

[flint vessel]

https://www.youtube.com/live/MHXO86wKeDY?si=_lbgmayHVfE6uosY

here

[jagged belfry]

it works but you cant blindly apply it

[shadow cloud]

to solve quadratic equations

[flint vessel]

i dont think you have a good grasp on the basics

[flint vessel]

ye

[shadow cloud]

U guys can giv eme a quadratic equation work and i try my way to solve it

[burnt notch]

This is not the quadratic formula for solving second degree (aka quadratic) equations

[flint vessel]

the video has some problems in it to

just pause and solve em

[burnt notch]

Sure, x² - 3x = 0

[shadow cloud]

Hmm give me a work and i use the 3 formula

[jagged belfry]

i think if you solve this youre very close to derive the quadratic formula

[flint vessel]

trust , just watch the lecture , you gain nothing from mugging up formulas you dont understand

[jagged belfry]

in its generality

[shadow cloud]

thats easy bruh

[dusk gyro]

he's kinda close, he already tried to complete the square just that his fundamentals are really weak

[shadow cloud]

x= 0. x=3

[jagged belfry]

what is the square of 5/2

[burnt notch]

Well, what can I do lol? Better for you that I gave you an easier one 😅

[jagged belfry]

remember (x+y)^2 = x^2 + 2xy + y^2

[shadow cloud]

My home teacher havent teach me this but he teach me alot of quadratic equations that doesnt have 3x^2 or 4x^2

[shadow cloud]

(radical 5/2)^2

[jagged belfry]

what

square

not square root

[dusk gyro]

i can write out the proof of quadratic formula for you if you want

[dusk gyro]

ok how about i guide you through proving the quadratic formula

[shadow cloud]

Ok

[shadow cloud]

this also useful solving quadratic equation

[jagged belfry]

its wrong tho, it should be 5^2/2^2

[shadow cloud]

???

just cross it?

is the same way

[jagged belfry]

you have x^2 + 5/4x

you want to complete the square

the square is (x+a)^2

[shadow cloud]

i can solve that

[jagged belfry]

(x+a)^2 = x^2 + 2ax+a^2

[dusk gyro]

@jagged belfry what are you doing rn

[shadow cloud]

equal 0?

[jagged belfry]

your 2a = 5/4

[shadow cloud]

bro what

[jagged belfry]

so a = 5/8

[shadow cloud]

write it on a piece of paper

[jagged belfry]

a^2 = 25/64

[shadow cloud]

where does the 2 go?

[jagged belfry]

divide by 2

both sides

[dusk gyro]

@jagged belfry gang what are you doing rn i dont wanna confuse him

[shadow cloud]

why not 2a/2 = 5/2?

a=5/2

[jagged belfry]

ok you are messing up fundamentals indeed

[shadow cloud]

let me check

@jagged belfry is it realte to this?

Im cambodian.

[dusk gyro]

are you just gonna continue ignoring me or what

[shadow cloud]

@dusk gyro

[jagged belfry]

yes

[shadow cloud]

can u do it on a piece of paper?

[shadow cloud]

yeah?

[jagged belfry]

instead of 3

you have 5/4 right?

[shadow cloud]

5

Yeah

is just an example

[jagged belfry]

yes but you can use that

if you follow as you already did with your teacher

[shadow cloud]

so like (x-5/4)^2 - 25/16- 36/16???

[shadow cloud]

yeah

I have 2 math teacher

[jagged belfry]

no

[shadow cloud]

one is school teacher one is home teacher

[jagged belfry]

in your example you have 3

in your current problem

you have 5/4

[shadow cloud]

yeah

[jagged belfry]

what did you do with the 3

and why

[shadow cloud]

@jagged belfry can u do the work on a piece of paper???

[jagged belfry]

no its your job to do the work

[shadow cloud]

aight

is 10 pm

[jagged belfry]

otherwise you will not learn \

this is why chatgpt never works

[shadow cloud]

I never use it

[dusk gyro]

clearly ur not working time for me to take over

@shadow cloud where are you at right now

[shadow cloud]

living room

aight i will show u my teacher work on saturday

uh add me

[dusk gyro]

bro no

this will only take 30 minutes

[shadow cloud]

is 10 pm

😭

can u add me

[jagged belfry]

take a rest, sleep or walk a bit. you wont solve anything by force

[shadow cloud]

u guys are useful

only if i understand

[jagged belfry]

you will

it takes some time to mature

digest

[shadow cloud]

vyasa add me 😭

[safe radish]

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[safe radish]

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[versed wave]

for (a), is it possible to break down M into smaller statements?

[mossy lotus]

well, it is exactly as small as it needs to be

any smaller and you would need many more logical steps, and any larger in scope, and it would be difficult to even write the statement

[versed wave]

i see, thanks

.close

[safe radish]

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[versed wave]

.reopen

[safe radish]

✅ Original question: #help-23 message

[versed wave]

for (b), what should be my universe of discourse?

should it be the set of all people, or the set of all countries?

[wild copper]

Well in this case, different quantifier quantifies different set

[versed wave]

ah i forgot i can have multiple domains of discourse

thanks

.close

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[potent oar]

"find out if these sequences are definitively monotone"

[potent oar]

how the hell do i even approach it?

[flint vessel]

I'm pretty sure a montone sequence is one that either constantly increases , decreases or remains constant

[potent oar]

yeah the definition is clear, im not sure how to actually prove it if it is or not

[flint vessel]

so just differentiate it and check if the derivative is always positive , negative or zero

if not then no

[potent oar]

right
what is "differentiating"?
i should mention this is the first excercise of the course, so idk if we need to imply such black wizardry

this is also n in N

[flint vessel]

hmm then just try the values and check if the expression is constantly increasing or not

[flint vessel]

ohh that makes sense

[potent oar]

also what the hell is this? why is it missing the fraction line??

[spark marlin]

It's combinatorics

A combination, C(2n,n)

2n!/n!*n!

[flint vessel]

you can also see it as 2nCn

ye

[spark marlin]

I hate that notation.

[potent oar]

right, ill take it as is for now
to get back on track, what do i need to do to say if a sequence like those are monotones?

definitively

[spark marlin]

"Easiest" way is if f(n+1)-f(n) is higher, lower or equal to 0

Actual easiest way is first degree differentiation, and compare the result with 0.

[potent oar]

isnt differentiation impossible in N

like natural numbers N

[spark marlin]

No

[potent oar]

oh

[spark marlin]

You have elementary functions

[flint vessel]

just put in values and check in manualy

you dont need to diffrentiate here i think

[potent oar]

manually... damn

[spark marlin]

I don't think that's accepted

[potent oar]

i hope it doesnt come up on the exam...

[flint vessel]

hmm

[potent oar]

it's a homework excercise not an exam drill so idk

[spark marlin]

I can attach photos, can I?

Let me get a pen and paper

[flint vessel]

another way to do it could be to check the sign of Tn+1 - Tn

that could tell you if it's a monotonous function

and it would be a valid method as well

i think

[spark marlin]

It is, but not always

It works if the function does not suddenly change how it works

[flint vessel]

i mean it is just a larger scale diffrentiation

but yes thats true

[potent oar]

how do you check the sign of a function then? like let's say for 1/n
then it's 1/(n+1) - 1/n
so that's
... no i got stuck

[flint vessel]

1/(n+1) - 1/n

now simplify it

-1/(n+1)(n)

right

now the denominator is always positive

but the numerator is -ve

so it will always decrease

so it's monotonous

[potent oar]

so the function is negative... okay i think i get this one, lets' see for another one

[spark marlin]

I solved the first one, it decreases, then increases

[flint vessel]

ye for values greater than root 10 its increasing

[potent oar]

it needs to be definitvely monotonous, not monotonous

[flint vessel]

im pretty sure they are the same words

[spark marlin]

It's a very small change

For definitely monotonous there is a strict comparison, like greater or smaller

For monotonous, it's greater or equal or smaller or equal

It's the difference between > and >=

[flint vessel]

no thats increasing and decreasing

[potent oar]

i think the professor explained it that "definitively monotonous is a function that is monotonous after a certain point and remains so"

[flint vessel]

ohh hmm

[potent oar]

and monotonous is a function that is always monotonous

like, anything that isnt periodic is definitvely monotonous

[spark marlin]

More like, definitively monotonous, has no points in which the first derivative is equal with 0

[flint vessel]

by the way the method i gave you should work then in most of em

[potent oar]

ok imma try it with the first one and see where i get stuck

[flint vessel]

just check n+10/n

leave the ^7

[potent oar]

how so?

how did you come to that conclusion?

[spark marlin]

Because it is an exponential function

It has some properties

It increases if the base is higher than 1

[potent oar]

no wait i didnt see the -

so the function is always decreasing in this case right?

so it's monotone

[spark marlin]

It's not monotone in N

Because you have the x in the denominator spot, for n=1, 2 or 3, it's smaller than 0, that would mean it is decreasing, but for n>4, it is increasing, that means it is not monotone in N

[potent oar]

no sorry i meant definitively monotone

so after a while it's monotone

[spark marlin]

But it does not have the same behaviour in the entire N

You need the same behaviour over the entire N

[potent oar]

no i there to be an n0 so that for each n>n0, the sequence a(n) is monotone

that's how the professor defined definively

[spark marlin]

Yes, but you have numbers that do not have this property

As long as there is a single counter, the definition would not stand

Keyword: each

[potent oar]

again, it's not for the whole of N, only after a certain point is fine, like sin(n) would not be definitively monotone but |n-2| is, altho it's only after n = 2, but after that it's monotone for the remainder of N

like pretend this is a sequence, this would be not monotonous, but definitively monotonous

so it's monotonous after a certain point, that's how the professor explained it

this is translated
"we say a succession of variable sign is:
definitively positive if there exist an n0 so that a(n) > 0 for each n > n0"

so "eventually" it's postivie

you know what i mean?

[spark marlin]

I see your point

But sign and monotony are 2 different things

[potent oar]

yeah i just made is as example to demonstrate what i meant by defintively

i think it's better if i just call it "eventually"

it's eventually monotonous

that's what the excercise wants

sorry for the language barrier

the course is not in english so it's a bit difficult for me

[spark marlin]

It's only in italian maths I've seen the use of "eventually monotonous"

[potent oar]

you've seen other examples of italian maths? or you mean just this?

[spark marlin]

Someone asked me a similar thing a couple of months ago

[potent oar]

right that's... small world i guess

[safe radish]

@potent oar Has your question been resolved?

[safe radish]

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[lapis token]

It says prove the blue statement by repeatedly applying the purple statement to the green one

[safe radish]

Please don't occupy multiple help channels.

[lapis token]

The colors are messed up on my screen
Blue is above green and purple is the lonely one

I think I understood how to determine something like A^3 but that’s kinda it

Especially the part where it’s asking me to repeatedly apply it is confusing
Am I supposed to make up random values and bruteforce the rule? How is that a proof?

[near atlas]

this is a recursive proof

no need to assume any value for n

[lapis token]

You mean the green one?

[near atlas]

n will stay as n

[near atlas]

yeah. take the determinant on both sides for the green expression

[lapis token]

Determinant?

[near atlas]

modulus

[lapis token]

Sorry I’m not familiar with English terms
My prof isn’t doing a great job either to be honest

[lapis token]

|A|?

[near atlas]

yes

these capital letters denote matrices. are u familiar?

[lapis token]

They are referring to sets

[near atlas]

oh then its different. then the bars around the A mean the size of the set

[lapis token]

Sorry should’ve mentioned that

They do refer to the amount of elements yes

[near atlas]

anyways if you do that. take the "modulus" on both sides of the green expression, then use the purple expression to simplify your right hand side

then you use the green expression again for A^(n-1)

and so on!

till n eventually gets reduced to 1!

[lapis token]

Isnt the purple expression the left side of the green one tho?

[near atlas]

|A^n|=|A×A^(n-1)| = |A| . |A^(n-1)|

look carefully. i used the purple to simplify the RHS

[lapis token]

Rhs?

[near atlas]

right hand side

[lapis token]

Also did you mean = |A| * |A^(n-1)|

[near atlas]

yes

now u can use the green expression for |A^(n-1)|

[lapis token]

I still don’t really get how
|AxB|=|A|•|B|
==>
|A|•|A^(n-1)|

[near atlas]

no the 1st doesn't imply the 2nd. we used the 1st to deduce the 2nd

its like say x + a + b =10 and a+b = 5
so we get x² = 25 but did we get x²=25 directly from a+b?

think abt it

[lapis token]

No but it implies x=5 does it not

[near atlas]

only if you're considering both expressions at the same time. not when you're considering one and not the other

[lapis token]

How does this apply here?
Sorry I’m very slow

[near atlas]

we had something of this sort on our right hand, right? :
|A × A^(n-1)|

then we used |A × B| = |A| * |B| as an identity

[lapis token]

Identity?

[near atlas]

say we assumed A = A and A^(n-1) = B

[lapis token]

A^n= A•B
Is everything I can come up with

[near atlas]

an equation thats always true

[lapis token]

Like x=y without anything else?

Or just 0x=0?

[near atlas]

this

[lapis token]

This feels kinda pointless but I wrote this at least

I was hoping to understand something when doing that but no not really

[near atlas]

see you're to prove the blue equation. so just forget about it for the moment. we can get to the blue just by using the green and the purple

[lapis token]

Alright

[near atlas]

1st two steps are okay

[near atlas]

now for the 3rd step, use the purple for the right hand side only. forget about the left side. focus on the right

on your right do you see something like |A × B|

?

[lapis token]

This?

[near atlas]

can you see that |A × B| is in a similar form as |A × A^(n-1)|

[lapis token]

I meant the second part with |AxB|
I added the purple part on the right because I wasn’t sure if it’d be useful or not

Yeah because A^n-1 is now called B

[near atlas]

yup!

so now you can write that part as |A| * |B|

but replace your B with A^(n-1)

can you show your progress?

[lapis token]

Sorry I was writing

[near atlas]

perfect!

[lapis token]

So how do I prove it by repeatedly applying it?
Do I try smth like A * A * A^n-2?

[near atlas]

now do you remember the original green expression? this time we're gonna use that as an "identity" but only for the |A^(n-1)| part!

can you see that A^(n-1) is similar to the from A^n

and from the green expression we know that A^n = A * A^(n-1)

[lapis token]

It appears on the right side yes

[near atlas]

so its pretty simple that A^(n-1) = A × A^(n-1-1)

and n-1-1 is just n-2

however do note that your expression is still between those bars. so dont forget those bars

[near atlas]

no not like that. they are similar in the sense that they're both A raised to a power. whatever that power might be

[near atlas]

yess but it was a " × "

ykim?

between A and A^(n-2)

[lapis token]

So it’s |A|•|AxA^(n-2)|

?

[near atlas]

exactly!

now what do you think should we do next?

[lapis token]

Maybe fix my definition of B

[near atlas]

note that the question stressed us to use the "original" green and purple expressions repeatedly

[near atlas]

yup!

[lapis token]

I didn’t realize I made the same mistake there too

It’s AxA^n-2 now

[near atlas]

yes. so your B wil be the 2nd term

[lapis token]

Id define a set called C

With A^n-2

[near atlas]

sure

can you see the pattern again? that this will be similar to the original purple expression

[near atlas]

yupp now sub back in your definition of C

[lapis token]

I thought I had it for a moment but this isn’t even what I’m supposed to prove

[near atlas]

yess! exactly

[lapis token]

I’ve changed it a bit

[near atlas]

now the next leap in our logic is that n here is a natural number like 1,2, 3... its an arbitrary natural number so if we keep subtracting 1 from n for a long time, we'll run out. we'll end up with 0

[near atlas]

nice. exactly what i meant to say

you can subtract 1 from a number n for n times untill you end up with 0

[near atlas]

thats it. your proof

QED

[lapis token]

Oh my god thank you

Im losing my mind

Do you know what my problem is tho

[near atlas]

wdym?

[lapis token]

This was only 20% or something of the second page

And the deadline is tomorrow

But I feel bad asking you for more help

You’ve been really patient already

[lapis token]

?

[near atlas]

its written at the end of Mathematical proofs to say "which was to be demonstrated" kinda like saying "hence, proved "

[lapis token]

Ah I see

I’ve honestly never seen that

Maybe it’s not used in Germany

To be fair I started my lectures like 5 minutes ago

Today was my sixth day

[near atlas]

i see

[near atlas]

damn you've a lot left 😭

[lapis token]

I mean it
I’m going insane

And I need to submit this HANDWRITTEN and per MAIL

My professor is an idiot

[near atlas]

yeah. im sort of heading off now. but if you have any doubts in the future man, just dm

[lapis token]

Id highly appreciate it
Is it fine if I add you?

[near atlas]

sure!

[lapis token]

Thanks!

I’ll close the channel and hopefully find another savior

[near atlas]

damn thats crazyy 😭

[lapis token]

You did a lot tho
Thank you

[near atlas]

no worries 👌

[lapis token]

Have a nice evening / day

Bye bye

.close

[safe radish]

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[trail rampart]

In the Simplex Tableau Method (Linear optimization), we generally move along the borders of the domain to get to the next vertex.
From what I understood, the basic direction is the one corresponding to the variable entering the basis. So in the image attached,
it would be (1,1) as x1 will enter the basis. However 1,1 in this case, if you represent graphically, goes INSIDE the domain and not on the border to the next vertex.
Why is that so ?

[safe radish]

@trail rampart Has your question been resolved?

[safe radish]

@trail rampart Has your question been resolved?

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[indigo scaffold]

f(g(x))

3sqrtx^2 +2x + 3 -2

3sqrt x^2 + 2x +1

now i replace x by 3? to determine the value?

[split kayak]

You can simplify it even more

[indigo scaffold]

how

[split kayak]

Solve the quadratic

x^2 + 2x + 1

[indigo scaffold]

= 0?

[split kayak]

Yeah, you can do that

[indigo scaffold]

x^2 +2x +1 = 0

-1 both sides

x^2 + 2x = -1

[split kayak]

Id advice you use the quadratic formula

[indigo scaffold]

what is it i forgot sorry been a year

[split kayak]

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[indigo scaffold]

alr

x = -1?

[split kayak]

Twice, right?

[indigo scaffold]

yeah

[split kayak]

I hope you remember that you can use this formula to also factorize a quadratic

[indigo scaffold]

ill remember

[split kayak]

So, if you have the same root twice

Youll have two equal factors, right?

(x+1)(x+1) here, right?

[indigo scaffold]

yeah

[split kayak]

You agree that thats the square of (x+1)?

as in, (x+1)^2

[indigo scaffold]

it is

[split kayak]

So:

$x^2 + 2x + 1 = (x+1)^2$

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[split kayak]

Knowing that, go look back into the problem we had

[indigo scaffold]

yes

what about it

we can replace with -1?

[split kayak]

$3\sqrt{x^2 + 2x +1}$

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[indigo scaffold]

theres a squareroot

that we can remove

[split kayak]

So how would that end up looking like?

[indigo scaffold]

3(x+1)

[split kayak]

Yeah

Thats the simplified version

[indigo scaffold]

wait but why did we have to do the quadratic thing

just for practice?

wait how did u know it was (x+1) and not like (x+3) or something @split kayak

like why 1

[split kayak]

No, because they are asking for the simplified version

Gimme a second

[indigo scaffold]

alright

[split kayak]

for any given quadratic polynomial of the form $ax^2 + bx + c\$
you can also write it as $(x-x_1)(x-x_2)\\$

where $x_1$ and $x_2$ come from the solutions of $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[indigo scaffold]

oh so (x -(-1))(x-(-1))

[split kayak]

You can always rewrite a quadratic as the product of (x- (the roots))

Since this polynomial is a perfect square

[indigo scaffold]

gotcha

[split kayak]

You can therefore write it as (x- root)^2

which cancels with the sqrt

[indigo scaffold]

gotcha

[indigo scaffold]

btw this gives (x+1)^2

[split kayak]

Yes, thats why i knew that the polynomial became (x+1)(x+1)

[indigo scaffold]

i see

wait so now we replace x by 3

[split kayak]

Yes

[indigo scaffold]

gives 12

[split kayak]

Yep

[indigo scaffold]

so to answer the question here

i just write 12?

[split kayak]

Yeah, you could write down the calculation of it

[indigo scaffold]

alright

thanks dude

[split kayak]

Well

Theres a little "but"

[indigo scaffold]

what is it

[split kayak]

when you do $\sqrt{(x+1)^2}$

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[split kayak]

The result actually isnt x+1

Its $|x+1|$, the absolute value of it

[flat frigate]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[indigo scaffold]

so its |x+1|

[split kayak]

For this particular case it doesnt really matter since x is > -1

[indigo scaffold]

got u

well thanks anyway

[split kayak]

np

good luck

[indigo scaffold]

thanks

.close

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[tranquil geyser]

https://www.reddit.com/r/UMD/s/3NLcrbLYcb

[tranquil geyser]

not sure if this counts but what even is this?? I’m actually really curious and have found nothing online

this has to be gibberish

[safe radish]

@tranquil geyser Has your question been resolved?

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@tranquil geyser Has your question been resolved?

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[still charm]

How do you rigorously prove that for a continuous 1 to 1 function f over [a,b]:
f^(-1)'(y) * f'(x) = 1

[still charm]

I mean it's intuitively obvious

But like?

I'm sure I did some illegal operations somewhere

[astral glacier]

Well show what you did

[pine hornet]

yall know about quadratic equations like finding roots and vertex??

[still charm]

@astral glacier

[empty forge]

yea

[still charm]

This is my channel lol

[empty forge]

yea I was wondering why that person was there

[pine hornet]

i need help with understanding the steps it confuses me with inequality stuff

[empty forge]

uhh ok

looks like you already opened another channel

lets move it there?

[pine hornet]

alr!

[empty forge]

damn

yea idk 😔

[safe radish]

@still charm Has your question been resolved?

[safe radish]

@still charm Has your question been resolved?

[still charm]

<@&286206848099549185>

[safe radish]

@still charm Has your question been resolved?

[lilac totem]

I think its legal if you show or mention that from continuity of f^-1 it follows when f(x)->f(c) then x->c

[safe radish]

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[west hedge]

Does anyone

[west hedge]

Want to brainstorm

I have a set of points

Data points

[magic junco]

Alright, gotten

[west hedge]

.close

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[silk cove]

x^86=6 mod 29

[silk cove]

should I use fermats?

x^2=6 mod 29

(x-6)(x+6)=0 mod 29

[timid escarp]

did you perhaps mean (x - sqrt(6)) (x + sqrt(6))?

[silk cove]

Ohh yeah

How do I solve it?

@timid escarp

. close

[austere goblet]

from here you can probably do a quick check of single-digit squares to see if they leave a residue of 6 mod 29

[silk cove]

ahha

8 wiorks

thanks

.close

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[austere goblet]

don't forget +/-!

[safe radish]

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[thin oxide]

uhh well my question is how to find an equation of a plane using a line equation and a point(x,y,z), and how do you find a equation when two planes are intersecting along a common line

[lost jewel]

equation of a plane using a line equation and a point(x,y,z)
this first, you can assume the equation is ax+by+cz+d=0 and then find the relation for x,y,z from the line equation

[thin oxide]

assume for the plane yes and what if the line's given in vector form

[honest perch]

a line and a point do not determine a plane

[lost jewel]

ehh doesn't matter really

[lost jewel]

It does if the point does not lie on that line

[honest perch]

oh right

i am off my game today

[thin oxide]

but in cartesian form a point's already given so wouldnt we use that

im kinda lost in forms of lines

[honest perch]

how many planes pass through a given line?

[thin oxide]

uhh infinite

[honest perch]

yes

[thin oxide]

but since a point's given, we assume a general point and find direction ratios

is that basically it

[honest perch]

idk what ur saying

sub in points that lie on the line into the plane equation

and find relations between a b and c

[thin oxide]

uhh but we're not sure if the point is in the plane

[thin oxide]

can we do that

[honest perch]

you have to find such a plane

i.e. u have to find a,b,c

[honest perch]

yes because the line lies on the plane

[thin oxide]

ohh

what about if 3 points are given or 2 lines are given

[lost jewel]

same thing bruh

[honest perch]

yeah u just sub in what ur given

[thin oxide]

ohh sure mb

[thin oxide]

uhh how about my other question

equation of a third plane i mean*

[lost jewel]

2 planes intersect don't determine a plane

[thin oxide]

yes but how do you find one passing through the line of intersection

[lost jewel]

hm

[thin oxide]

is there a formula or?

[honest perch]

aren't you already given two

[lost jewel]

If it is it would be lengthy

[thin oxide]

yeah but we need a new one

[thin oxide]

oh

[honest perch]

maybe like the find the general equation of a plane that passes through that line

[thin oxide]

so basically it depends on the question

[honest perch]

yeah you shouldn't memorize formulas for this

there are too many scenarios

[lost jewel]

Yeah, find the line of intersection and then do what we did for the first part

[thin oxide]

ohh alright

uhh for my final question, its a test question which had me confused for like a couple of days

[lost jewel]

you can find 3 edges AB,AC,AD using areas given

[thin oxide]

im lost as to how

we take the dot product of the pv ?

[lost jewel]

nah

ABC, ACD and ADB are right triangles at A

Do you know what mutually perpendicular mean

[thin oxide]

oh right 90 degrees

uhh let me try it quick

it works thanks

.close

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[magic junco]

,rccw

[cobalt saddle]

I just want to understand something

I got 56 as a answer for 3

[austere goblet]

yes, the helper earlier already confirmed your answer

unless your doubt is something else, in which case please raise it now

[magic junco]

@cobalt saddle

[cobalt saddle]

Just kinda bullshitted my way to a answer

So i do not really know how i got it...if that makes sense

[austere goblet]

first, show your method, bullshit or otherwise.

then we'll see if there's merit to your method that you just didn't get, or it's a lucky shot.

[austere goblet]

ok, and what was your reasoning for this calculation?

[cobalt saddle]

See that is the problem...

[austere goblet]

I mean, surely you have at least some idea of what you were doing, right?

it's ok if it sounds weird or wrong

we can work with that

(who knows. maybe your reason is actually correct)

[cobalt saddle]

Ok...well i divided the possible ways to allocate three people from a group of eight to the three positions by three people who must serve in three different positions to be left with left with possiblities where position is irrelevant?yeah i do not know what i am doing

[austere goblet]

and you got those answers from (1) and (2), yes?

[cobalt saddle]

🙌

[cobalt saddle]

Yeah

[austere goblet]

and that is the right reason

[alpine aurora]

Do you know permutations and combinatorics

[cobalt saddle]

But again i do not know what i am doing

[austere goblet]

when doing (1), you care about their positions, but in (2) you found out how many different arrangements a given group of 3 can make

so if (1) picks out a group of 3 from 8 where we care about the different arrangements a group of 3 can make, and (2) gives us the number of arrangements any given group of 3 can make between themselves, then it must follow that if we don't care about the order, you can just divide by the number of arrangements any given group of 3 can make to cut down on the overcounting

and that's really it

[cobalt saddle]

Ok...thank you

[austere goblet]

nps. anything else?

[cobalt saddle]

Other question or...

[austere goblet]

whatever. it's your channel

well, of course they must be questions

this is a help channel after all

[cobalt saddle]

https://tenor.com/view/shock-shocked-what-astounded-gif-20621468

[austere goblet]

I would recommend not sending random gifs in a help channel

[cobalt saddle]

This is a question i wanted to with yesterday

I want to understand it

Kinda used bullshit again

[cobalt saddle]

Are you sure i can ask again?

[austere goblet]

why would I tell you to if you're not allowed to?

[cobalt saddle]

Ok...

[austere goblet]

though I will be doing other stuff for a bit, so you can ping Helpers once now if necessary

[cobalt saddle]

Ok,thanks

[safe radish]

@cobalt saddle Has your question been resolved?

[cobalt saddle]

<@&286206848099549185>

[cobalt saddle]

<@&286206848099549185>

[balmy relic]

What

[cobalt saddle]

This

[balmy relic]

Let me do it rough

[cobalt saddle]

I only require a explanation

[balmy relic]

Nah I'll give u the soln

I'm done

Just wait a second

% 📘 Hotel Revenue Optimization

\textbf{Given:}\[4pt]
Total rooms = $72$, \quad Base rent = $R500$ per room.\
If rent increases by $R100$, ; 2 rooms become vacant.\[6pt]

Let new rent be $x$ ; (where $x > 500$).\[8pt]

\textbf{Rooms occupied:}
[
N = 72 - \frac{x - 500}{50}
]

\textbf{Total income:}
[
I(x) = x\left(72 - \frac{x - 500}{50}\right)
]
[
\Rightarrow I(x) = 82x - \frac{x^2}{50}
]

\noindent\rule{8cm}{0.4pt}

\textbf{Differentiate:}
[
I'(x) = 82 - \frac{x}{25}
]

Set $I'(x) = 0$:
[
82 = \frac{x}{25} \quad \Rightarrow \quad x = 2050
]

\noindent\rule{8cm}{0.4pt}

\textbf{Second derivative:}
[
I''(x) = -\frac{1}{25} < 0 \quad \Rightarrow \text{Maximum at } x = 2050
]

\textbf{Maximum income:}
[
I(2050) = 82(2050) - \frac{2050^2}{50} = \boxed{R84{,}050}
]

\noindent\rule{8cm}{0.4pt}

\textbf{✅ Therefore:}\[4pt]
The hotel achieves a \emph{maximum daily income of R84,050} when\
the rent per room is set at \boxed{R2050}.

[flat frigate]

ÆΣρx

[cobalt saddle]

Thank
.close

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@cobalt saddle Has your question been resolved?

[safe radish]

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[cobalt wedge]

I'm stuck here

[quasi bison]

,rccw

[cobalt wedge]

It's the 595 one (prove that the equation ex-1 = cosx/x has at least one root)

Idk if root is the right word for this but I mean that it touches the x'x line

Also it's e×

[carmine tulip]

I doubt that anyone here speaks Greek.

[cobalt wedge]

x is a power

[cobalt wedge]

I translated good sir.

I have to solve this using Bolzano's theory right?

[quasi bison]

wait, hang on

does syn x really mean sine?

[cobalt wedge]

Oh wait it's cosx

[safe radish]

@cobalt wedge Has your question been resolved?

[cobalt wedge]

Can someone help 🥺🥺

[silk herald]

https://tenor.com/view/huzz-seal-hi-sealutation-gif-14368511881377374735

this is the result of being a third worlder with no access to education

[cobalt wedge]

Bawwww

[safe radish]

@cobalt wedge Has your question been resolved?

[waxen cobalt]

Define f(x) by manipulating the given equation to equal 0, then determine f(0) and f(1) and apply IVT/Bolzanos Thm

Also show the f(x) you obtain is continuous but this is straightforward

[cobalt wedge]

Is that not what I've done?

[waxen cobalt]

My bad, didn't realize you already posted an attempt ; though your f(x) seems to be incorrect

[cobalt wedge]

It's just I can't find a solution for f(0) cause cos0/0 equals 1/0

[waxen cobalt]

You can multiply x to the other side first

Which gives you xe^x -x = cos x

And then subtract cos x from both sides and continue as before

[cobalt wedge]

Thank you so much 🥹

You can close this now good sir

[waxen cobalt]

.close

I think u need to type it

[magic junco]

.solved

[safe radish]

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[native flume]

hi im trying to answer this, but im not sure if i am doing it right. for the first part which involves the operations (a) and (b), i deduced that it is always possible, since if we have m<n, we subtract m-1 from both sides, such that we get 1 and n-m+1, and then we can keep doubling 1 until if we double one more if becomes greater than n-m+1 (2^a < n-m+1), in which case we keep repeating the process of slicing both down to 1 and another number, eventually when we slice both numbers, one will be 1, and the other will be a power of 2, hence, we can double 1 up to that power of 2, and then slice both to 0. although, i don't think its possible with 3 balls, since u may be left with the same method, as you may be left with 1 in one bag, and 2 in the other bag, in which case its impossible afterwards? i was wondering whether this is right, and if this can be written more mathematically? thanks🙂

[safe radish]

@native flume Has your question been resolved?

[verbal cloud]

for the second one you can find a simple invariant

[native flume]

not sure what that means

[verbal cloud]

a quantity that doesn't change when you apply the operations

then you know when a state is impossible to reach if that quantity changes

in this case it has to do with the parity of a certain value

for the first question, your algorithm probably works but that would be tricky to prove I think. There's an easier method if instead of trying to reach 1 ball in one bag, you try to make it so the second bag has exactly double the amount of the first bag

[native flume]

ok il have a think thanks for ur help🙂

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[slender cloak]

is this valid?

[safe radish]

@slender cloak Has your question been resolved?

[slender cloak]

<@&286206848099549185>

[steep lily]

You should really write e.g. the second line as [\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)] and similarly everywhere else you're using $+$ or $-$

[flat frigate]

Edward II

[steep lily]

Saying $A\cup B = A+B-X$ is meaningless; you can add/subtract the probabilities of the events but not the events themselves

[flat frigate]

Edward II

[steep lily]

this isn't as simple as just putting Ps in everywhere by the way, since you're using these plus minus inside P in the sixth line, which you'd need to resolve

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[barren geode]

Hi, is this correct?

[twilit spindle]

you duplicated negative signs

(a-b)^2=a^2-2(a)(b)+b^2

[barren geode]

The term b is negative

[twilit spindle]

but we've already account for the negative sign on the middle term, in fact: ((a\pm b)^2=a^2\pm2ab+b^2)

[flat frigate]

ΠαϳαμαΜαμαΛλαμα

[twilit spindle]

where (\pm) can be plus or minus

[flat frigate]

ΠαϳαμαΜαμαΛλαμα

[barren geode]

Ok ty, and does this applies to more formulas?

[open wedge]

Depends

[barren geode]

How can I know which formula follos this rule and which one not?

[open wedge]

One example is $a^2 - b^2 = (a - b)(a + b)$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

The plus version of it does not have a formula

[barren geode]

Why not?

Ok nvm

But is there any rule or any formula can have any rules?

[open wedge]

To figure out which formula corresponds to which

Just expand

For example: $(a \pm b)^2 = (a \pm b)(a \pm b) = a^2 \pm ab \pm ab + b^2 = a^2 \pm 2ab + b^2$

[flat frigate]

1 divided by 0 equals Infinity

[barren geode]

Ok, ty guys!

.close

[safe radish]

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[ocean hamlet]

i understand the proof except some part

this part

i dont understand how all other matrices from the cofactor contain exactly k-2 times the polynomial variabele x

[carmine tulip]

@ocean hamlet türk müsün?

[ocean hamlet]

evt

[safe radish]

@ocean hamlet Has your question been resolved?

[safe radish]

@ocean hamlet Has your question been resolved?

[night magnet]

bc the x variable is only on the diagonal

so no matter which row or column you cofactor across, there's only one x in the chosen row/column. and that x will only be multiplied to one of the determinants when you cofactor

[safe radish]

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[spring glacier]

I think it wants you to rewrite sec and csc as their 1/…

Then use cos(x) = sin(90-x)

you should have something like sin(...)=sin(...) after this

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[steep magnet]

How do we know lim(x -> 0) -|x| = 0 and lim(x -> 0) |x| = 0?

[glacial cairn]

|x| is continuous, and |0| = 0

[steep magnet]

Ahhh I see

Ok thanks!

❤️

.close

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[plain gull]

helloo

[plain gull]

ive tried factoring it out

[atomic gorge]

$$\frac{a+b+c}{d} = \frac{a}{d} + \frac{b}{d}+ \frac{c}{d} $$

[flat frigate]

Ryne___np

[spring glacier]

actually factoring works here too, what did you do exactly, so we can check your work

or, more concretely, what issues are you even having?

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[formal hazel]

genuinely how do i do these? i have a general idea but i’m just lost on the one with the square root and the fractions

[twilit moat]

do you know the chain rule?

[formal hazel]

No wait what is that

[twilit moat]

when you have a function inside another function, there's a particular rule for finding the derivative of the expression (you need it for the square root)

[still charm]

the general form of the tangent line:
y = f'(x0)(x-x0) + y0

[twilit moat]

the derivative of $f(g(x) \text { is } f'(g(x))g'(x))$

[flat frigate]

Gizmic

[twilit moat]

or alternatively, $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

[flat frigate]

Gizmic

[twilit moat]

depending on your notation

[formal hazel]

Oh yea

We usually use this one formula

Let me finf it

[twilit moat]

for the fraction you then also have the quotient rule!

however, you can also write it say as $\frac{-10}{1-x} = -10(1-x)^{-1}$ as well

[flat frigate]

Gizmic

[formal hazel]

Ohh

Thst makes sense

To gwt the general answer we usually rewrite it similar to that

Its something along those lines

But for normal ones we use a formula

Its like f(x-h) - f(x) / h or something

[twilit moat]

isn't this the first principle definition of the limit?

[formal hazel]

it sayd Its called difference quotient

Do u want To see

[twilit moat]

yeah sure

[twilit moat]

yeah this is the derivative from first principles

for normal functions you shouldn't have to use it

this is more something you use at the start of calculus to justify the derivative for some simple functions

[formal hazel]

Ohh ok

Yea that makes a lot of sense

[twilit moat]

but the rules like power rule and stuff can just be used instead if you're given them

[formal hazel]

We went over it lightly in the beginning but he barely soent five minutes on it

Differwnt rules and stuff

Thank u so much though U helped lots

.close

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[empty matrix]

I need help with geometry and finding x for supplementary and complementary angles

[empty matrix]

It’s number 11 but idk how to crop on here 😭

[swift fjord]

you know (6x + 13) + (14x-33) = 180

from there you should know how to solve

[empty matrix]

Oh I forgot to add the 180

[split kayak]

Same idea for the second part, but = 90 instead

[empty matrix]

okay

For the first one I got 10

And for the second I got 7

I do have one last one question

how would I solve this one

[swift owl]

let's say we have this angle theta

do you know how to find its angle?

[empty matrix]

Do I subtract 180-75? Since it’s a straight line

[swift owl]

that's right

so then we can use that, right?

[empty matrix]

yeah

[swift owl]

do you know the relationship between angle theta and the angle 7x

[vagrant ice]

there's something special about all of these angles

[empty matrix]

Vertical pair?

[vagrant ice]

yes, vertically opposite angles are... ?

[empty matrix]

ohh congruent

[vast comet]

since you already know the angle that's supplmentary to the angle that's 75 degrees, which is 7x...

[vagrant ice]

there's two other important facts:

• alternate angles are congruent
• corresponding angles are congruent

[empty matrix]

thank you 🙏🏾

[empty matrix]

is that the 105 I got?

[vagrant ice]

and corresponding angles are just vertically opposite angles, then alternate angles

[vagrant ice]

yes, 75 is supplementary to 105

[vast comet]

yeah

a good thing to remember is that you can find one of two angles that are supplementary by subtracting the other one from 180

[empty matrix]

okk thank you

[vagrant ice]

no worries! if you're done for now, type .close

[empty matrix]

.close

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[upper beacon]

hello people

[molten acorn]

you can usse isosceles triangle properties

[upper beacon]

huh

[molten acorn]

suppose there is a triangle ABC s.t. AB = BC (their lengths are equal), this implies the angles BAC and BCA are equal

try to find a way to show that EFG = EDH, also show that EHD = EGF

the rest should be easy

[upper beacon]

hm

[upper beacon]

cant

[molten acorn]

hint : EDF is an isosceles triangle

[upper beacon]

ohhhh

[silent nova]

What did you conclude?

[upper beacon]

i dont think i needed 10

[silent nova]

Didn't you prove it by SSS test?

[upper beacon]

how

wait

[silent nova]

DE=FE
DG=HF
EH=EG

@upper beacon

[upper beacon]

how did you get dg=hf

[silent nova]

Addition property

You have written it

[upper beacon]

alright bro

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[spring crater]

So for this problem I know that the area formula is A = 1/2absin(C). But I'm confused about angle C. Why is the supplementary angle of 28 degrees used for C and not 152 degrees?

Idk if this question makes sense

[frigid locust]

tbh no clue but sin(180 - theta) = sin theta , so its all good

[spring crater]

So you don't know why they are using 28 degrees for C?

As opposed to using 152 degrees

[quasi bison]

... a better question would be: why is the 4 m measurement not used here

[quasi bison]

smaller number looks friendlier. otherwise if you're using a calculator it won't make a difference.

[spring crater]

Yeah, maybe my professor messed up the answer

[spring crater]

So the two angles are interchangeable?

[quasi bison]

not by themselves no

but as jtp said, sin(180°-x) = sin(x) always

[spring crater]

Okay thank you

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[uneven veldt]

What am I able to do, I recognise that when I fix a partition, the Upper sum of f2 >= Upper sum of f1,

how can I show that the inf Upper sum of f2 >= inf Upper sum of f1

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@uneven veldt Has your question been resolved?

[velvet nebula]

show that "inf upper sum of f1" is a lower bound of the set of upper sums of f2

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[reef timber]

confused, unsure where to start and how to continue. Finding the value of X

[trim carbon]

google log rules

[reef timber]

very helpful

am i doing this step right

[trim carbon]

yes

there is a perfect square

so there will be 2 ans

[reef timber]

uhhh yes

but im unsure of how i can continue from where i left off

[trim carbon]

not sure what u need help with

[reef timber]

unsure of what to do for the next step

[trim carbon]

just put it to the power 3

bro

[reef timber]

do i sqr the x^2 and 48x

[trim carbon]

no

put it to the power 3

first

then subtract from both sides

[bronze gust]

The cube root can be also expressed as ^1/3

[trim carbon]

then solve for ur x

bro thats what im literally dssaying

[reef timber]

ohhhh

[bronze gust]

Well you didn't say it outright

[trim carbon]

[bronze gust]

So I'm just saying it outright

[trim carbon]

thank u

[reef timber]

am i on the right track

[bronze gust]

Mhm

Then is just quadratic

[white ibex]

yep

[reef timber]

i think i got it!! thank you guys

[magic junco]

是的

[reef timber]

.close

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[trim carbon]

はい

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[sand oyster]

I am not getting it help @everyone

[zealous ingot]

you can't ping everyone man

[sand oyster]

@everyone help me plz tm I have exam

[alpine aurora]

Please don’t do that, after 15 mins of the help channel open you can ping the helper role ONCE

[wind laurel]

we cant see the entire question

or can we idk i dont see a question mark 😭

[sand oyster]

I got u

@wind laurel here is the question

Plz solve it