help-28

[atomic blade]

(It also helps knowing when ln(x) = 0)

[hot herald]

doesn't hurt to also consider the domain

[proven heron]

Okay then

Lemme try

[hot herald]

try not to overthink again

[proven heron]

lmao

okay

[hot herald]

and in case it wasn't clear,
when I mentioned do not try to find for the inverse,
you shouldn't be attempting to isolate x in the equation 2 = 2x + ln(x) either

[atomic blade]

If you're confused on what inspection is:

||The thing is this: it's really convenient that you have a term 2x, and you have to make it equal to 2. It would be very convenient if we let x = 1 and hope that ln(1) = 0.

Which it does. This is what we mean by "inspection". The question's rigged such that you can make that correlation.||

[proven heron]

@random mica ty for the explanation

[random mica]

found it
is 1
Am I right
because if we put 1
in the original function
we see
2x + lnx
2+ ln 1
2 +0
=2
and as I said

[atomic blade]

Lil late but yeah

[proven heron]

and also ty @atomic blade and @hot herald

[random mica]

what is ty

[proven heron]

I think I have found out the anwser

[atomic blade]

"Thank you"

[proven heron]

thank you

[random mica]

ok

so how to write the answer

[hot herald]

Ty is a name

[random mica]

in the answering page

[proven heron]

Hmm

I think that

I will let f^-1(2) be a random variable

such as "a"

and then

I will have f(f^-1(2)) = a

which leads to 2 = 2a + ln a

[hot herald]

no

[proven heron]

and a = 1

what

why

[hot herald]

f(f^-1(2)) is just 2

a isn't 2

[random mica]

what ????

[hot herald]

I will let f^-1(2) be a random variable
such as "a"
I will have f(f^-1(2)) = a
you're saying a is both f^-1(2) and f(f^-1(2))

[proven heron]

wait

[hot herald]

you should have
f(f^-1(2)) = f(a)

[proven heron]

f(f^-1(2)) = f(a)

[random mica]

the rule

[proven heron]

yeah

and f(f^-1(2)) = 2

[random mica]

f o g (x) = g o f (x) = x

[proven heron]

true

but the answer for f^-1(2) is 1, right?

[hot herald]

yes

[proven heron]

okay then

[random mica]

so how do we write

the answer

in the

aswering page

answering *

[hot herald]

[insert work]
f^-1(2) = 1

[proven heron]

here's my draft

[hot herald]

ideally you'd write f(a) instead of fa

seems sufficient

[proven heron]

oh

yeah

okay

[proven heron]

how'd u think? @random mica
I think it's possible to write like this in the answering page

[random mica]

let me

try

in my wok sheet

and then i will

return

[proven heron]

okay

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[proven heron]

@random mica dm me privately, I wanna see your sheet, now I gotta do the other problems hehe

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[viral cargo]

I have a feeling I have to show the van Hall theorem here. But I have no clue how to approach this question.

[fast peak]

well (a) is the easy direction of that theorem. if there is a matching that means every node in U is connected to a different note in N(U). surely then there can't be less nodes in N(U) than in U

[viral cargo]

I can see that principal yes, but how would I incorporate the Van Hall theorom in a proof like that?

[fast peak]

ok what statement of halls theorem do you have

but well I would assume that if anything you have to use your version of hall in some later task

[viral cargo]

We didn't get any statement or anything. And our lecturer works without a book, he just gives exercises for us to figure out and chooses a few the following lecture that he will explain.

[fast peak]

how are you supposed to use a theorem without knowing what it says

[viral cargo]

He gave that as a hint. He likes us finding out things via the internet, so we can create our own material to use for studying for the exams. So all exercises together will form our syllabus for the exam.

[fast peak]

well that's a bit weird

[viral cargo]

Yes, lots of people struggle with this class...

[fast peak]

a very classic statement of halls theorem is: a bipartite graph like this has a matching if and only if for all U, |U| <= |N(U)|

and like you notice, a is just the direction =>

which is easy

the hard part of the theorem is the other direction

[viral cargo]

Ah, okay. I see

[fast peak]

the setup of the question sounds to me more like they want you to use max-flow-min-cut somewhere

[viral cargo]

I have only ever used that in actual given graphs, never with proofing.

But looking at question b, you might be right.

b) proof that every matching M in G gives a motivation for an s-t flow in D under c of value |M|.

[fast peak]

max-flow-min-cut and halls theorem are equivalent

you can prove each one using the other

[viral cargo]

Ah cool!

[fast peak]

https://en.wikipedia.org/wiki/Hall's_marriage_theorem#Logical_equivalences

here is a list of a couple theorems actually like this

[viral cargo]

Awesome! Thank you so much. I am gonna give it another go with this new knowledge! If I get stuck again, I will open a new channel 🙂

[fast peak]

👍

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[silk mirage]

solve the equation graphically

[silk mirage]

task e and f need help with how to make the left side = 0

[hot herald]

apply the same operation to both sides of the equation

what could you do to 8x that would result in 0
( apart from multiplying by 0)

and do that to both sides of the equation

[silk mirage]

so -8?

and -x?

the answer is supposed to be 4

[hot herald]

so -8?
and -x
no

[silk mirage]

alright

[hot herald]

8x - 8 - x isn't 0

[silk mirage]

8x/8x?

thats 1

so

how do you make it 0

[hot herald]

don't overthink this

[silk mirage]

i am

[hot herald]

lets use a simpler example,
k = 5,
how would get 0 on the right side of the equation

[pearl ocean]

Antar at du skal løse det grafisk. Bruk geogebra til det.
Hvis det er for hånd, bruk kvadratsetninger.

[hot herald]

what could you do to 5 that would result in 0
( apart from multiplying by 0)

[silk mirage]

det er grafisk, men jeg skjønner ikke hvorfan jeg skal gjøre det med geogebra det er det som er problemet

[pearl ocean]

@hot herald is it okay for you that I take over?

[hot herald]

sure if they're more comfortable and more likely to understand in their native language. try to avoid feeding the answer though if possible

[silk mirage]

jeg trodde jeg skrev det her at det skulle gjøres grafisk

[pearl ocean]

Er du vant med funksjoner?

[silk mirage]

er ganske så ny med algebra

[pearl ocean]

Jo, det gjorde du. Det var jeg som ikke så.

[pearl ocean]

Funksjoner er resultatet av endringer.

F.eks. f(x)=2x+1

Du har y-verdier (y-akse) på venstre side, og x-verdier (x-akse) på høyre side.

[silk mirage]

ja har forstått det

men ikke det første du sa

vent du mener at på venstre side av = er y akse og høyre x-akse?

[pearl ocean]

Ta utgangspunkt utifra denne funksjonen.

[pearl ocean]

Ja.

[silk mirage]

åja okok forstått

men er det meningen at jeg skal klare å forstå det i hode med hvordan d blir det samme?

[pearl ocean]

.

Du har variabler, de er de som endrer verdien av funksjonen.

[silk mirage]

skjønner

[pearl ocean]

Så du har bare y på den ene siden, og x på den andre siden.

[pearl ocean]

Prøv å få e til det som jeg sa, y på den ene siden og x på den andre.

[silk mirage]

hva mener du nå

[pearl ocean]

Nei, jeg sa det forvirrende.

Bare ha verdier på den ene siden.

Og den andre siden ingenting.

[silk mirage]

jeg misforstår

[pearl ocean]

Se på løsningsforslaget i eksempel 26.

Samme algoritme.

[silk mirage]

f(x)=x^2+16?

[pearl ocean]

Hva er likningen i eksempel 26?

[silk mirage]

10x+2x^2=0

[pearl ocean]

Hva har du på høyre siden?

[silk mirage]

0

[pearl ocean]

Samme gjør du med e og f.

[silk mirage]

what

[pearl ocean]

$x^2 +16=8x$ , gjør det som du sa, høyre side er 0.

[glossy valve]

Potet

[pearl ocean]

Det skal være 0 på den ene siden.

[silk mirage]

men på eksempel 26 starter likningen med 0 på høyre side så hvordan gjør jeg det når e har 8x på høyre side

[pearl ocean]

Er du kjent med likninger?

[silk mirage]

kan ikke si meg godt kjent

[pearl ocean]

Det er som en skålvekt.

Begge sidene skal være like.

Du tar en mengde av noe fra den ene siden, som du samtidig gjør på den andre siden.

[silk mirage]

ja men kan jeg gjøre -8 -x?

[pearl ocean]

8x er en, ikke to.

[silk mirage]

-8x?

så da blir det x+8=0

?

[pearl ocean]

.

Du tar 8x fra høyre siden. Det betyr at du også skal ta 8x fra venstre siden.

[silk mirage]

ja var det ikke det jeg gjorde?

[pearl ocean]

Hva er det her da?

[silk mirage]

jeg er helt dø i hodet

[pearl ocean]

Skålvekt. Det er en vekt som måler forholdet mellom to plattformer.

Når du har 8 epler og 2 appelsiner på venstre plattform, og 9 epler og 5 appelsiner på høyre plattform.

Du skal ta 2 appelsiner fra venstre plattform, hva er det du må gjøre for at plattformene som fortsatt i balanse?

[silk mirage]

tar 2 fra høyre

men blir der 16-6x=0 på e da?

[pearl ocean]

$x^2+16=8x$

[glossy valve]

Potet

[pearl ocean]

Du tar 8x vekk fra høyre siden. Hva er du skal ha på venstre siden?

Det skal være balanse.

Det som skjer på side a skjer det samme på side b

[silk mirage]

ja

har skjønt det

bare vet ikke helt hva svaret skal bli

[pearl ocean]

Skjønner du hva jeg prøver å få deg til å gjøre?

[silk mirage]

vell ja jeg bare skjønner ikke hvordan du skal regne det ut

[pearl ocean]

.

[pearl ocean]

.

[silk mirage]

går det ann å skrive x^2+16-8x=0

[pearl ocean]

Der gjorde du det.

[silk mirage]

åja det er det jeg har prøvd å gjøre hele tiden bare at jeg har prøvd å ta fra tallene som allerede er der 💀

[pearl ocean]

Når det er gange mellom noe, så er de sammen som en.

[silk mirage]

skjønner

[pearl ocean]

8x er en, 8+x er to.

Nå gjør du bare som løsningsforslaget, og du får svaret.

[silk mirage]

tusen takk

unnskyld for at jeg var helt idiot

[pearl ocean]

Null stress.

[silk mirage]

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[stuck fiber]

Is there any way to work out cos^2(7pi/3) without a calculator

[stuck fiber]

I know that cos(7pi/3) is 1/2

but how do I get cos^2

[hot herald]

cos^2(7pi/3) is convenient notation to represent
(cos(7pi/3))^2

knowing the value of cos(7pi/3) = 1/2
square that and you'll have what you need

[hot wadi]

7pi/3 = 2pi + pi/3

The 2pi can be discarded as it brings you back to first quadrant

[hot herald]

they've already done that

[hot wadi]

So it's ultimately cos(pi/3) which u know already

[stuck fiber]

Cool thank you so much guys!!

❤️

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[stuck fiber]

.reopen

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✅

[stuck fiber]

I have a follow up question

What about cos^2(4pi/6)

because (4pi/6) is -1/2

but does the minus cancel after being squared?

[short siren]

Yep

[hot herald]

*cos(4pi/6) is -1/2

[stuck fiber]

so its always going to be a + no matter what quadrant?

[hot herald]

missing the cos

4pi/6 itself is not -1/2

[short siren]

If you square it, yes. It's gonna be positive, assuming you're in the real numbers

[stuck fiber]

okok cool

[hot herald]

same idea as earlier

[stuck fiber]

thanks again guys!

[hot herald]

square (-1/2) and you'll have what you need

[stuck fiber]

thanks so much guys

❤️

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[short siren]

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[short siren]

[novel sparrow]

Small question: What does the "a arrow to a" mean?

My notation would be h=>0 and then a+h and b-h as integral borders

[short siren]

That's a peculiar way to write limits

[novel sparrow]

I got it from wikipedia

[nocturne creek]

Alpha arrow to a

[novel sparrow]

but what does that mean? is it identical to what I wrote or something different?

[short siren]

Oooo

[short siren]

And the upper one is b from the left

Yeah

[novel sparrow]

ahh thank you

Am I deducing correctly that this would be the same as h=>0 and then a+h and b-h as integral borders?

[topaz valley]

obviously you dont know if f is defined outside of the interval (a, b)

so you cant even take the limits from the other sides

no

[novel sparrow]

fk

hmmm

[topaz valley]

unless you specify h to be positive

[novel sparrow]

oh yeah thats what I meant

wait a second

[topaz valley]

yes that works

[novel sparrow]

okay okay

[topaz valley]

although

terminals look sus

[novel sparrow]

terminals?

[novel sparrow]

ahh yeah I know what you mean

these examples are unrelated to each other and were used in a situation where only one border is critical

[topaz valley]

no as in

[topaz valley]

the upper limit should still be b here

[topaz valley]

and lower limit should still be a here

[novel sparrow]

but wait a second

[novel sparrow]

lets say we are integrating over the intervall (a,b) which means both borders are critical

= we have to divide the integral in 2 seperate intervals

the first one is from a+h to c

the second one is from c to b-h

h being lim h=>0+

is that correct?

[topaz valley]

yeah sure

[novel sparrow]

(and c being somewhere between a and b)

okay okay

but the 2 examples you posted would be in used in a case where only one border is critical right?

[novel sparrow]

a here

[topaz valley]

yeah

[novel sparrow]

and b here

okay

I am just asking to make sure I understood correctly, its a bit confusing tbh

but thanks for clearing it up

[topaz valley]

its just weird behaviour => take limit

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[west spire]

Does anyone know why the step function has t-1 and what to do when the a value which is 2 does not match with 3

[west spire]

And why is there not a sin

[gritty rose]

Yea that all just looks wrong unless U(t) relates to sin

[west spire]

Yea idek, I understand every other example shown but this example just seems weird

[west spire]

<@&286206848099549185>

[gritty rose]

Find the definition of U

[west spire]

What do you mean like u(t-2)?

[gritty rose]

Any definition of U

[gritty rose]

The one used in the example

[west spire]

There is one given saying, Ua(t) = U(t-a)

[gritty rose]

That looks like a property and not the definition. Screenshot the definition.

[west spire]

I can’t find a definition definition, my notes just has plenty of examples

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[west spire]

Does anyone know why cos does not become (s/(s^2+9pi^2))

[west spire]

<@&286206848099549185>

[worthy ginkgo]

Because you evaluate it on one point

$\cos(t) \delta(t-3\pi) = \cos(3\pi) \delta(t-3\pi)$

[glossy valve]

black_couscous

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[west spire]

I don’t understand that

What does it mean to evaluate on a single point. I am trying to connect this to step functions and this does not make sense

[gritty rose]

Do you know the definition of delta function?

Write out the integral formula for the Laplace transform of the product cos and delta

[west spire]

According to like the notes, it’s a function which is zero everywhere except a point where it takes the value inf

[gritty rose]

https://miro.medium.com/max/1100/1*42ch09EgkdizjzfRaDF8jA.webp

Use equation 3 with f=cos

[west spire]

Ohhh i didn’t see that theorem thank you

I am actually going insane with this course but truly appreciate your help

Thank you @gritty rose and @worthy ginkgo

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[rugged raven]

sin sin x θ . find the value of θ

[torn jolt]

[short crest]

?

what?

[rugged raven]

ye thats the question in my exams and I cant figure it out

[light sonnet]

You're taking an exam?

[rugged raven]

its already done

[forest widget]

you don't even have an equation

missing symbols?

[torn jolt]

Theta isn't even in the expression that we are meant to magically find for you

Or u somehow have $\sin(\sin(x\theta))$

[glossy valve]

♡LexQa♡

[torn jolt]

Which is wild

[rugged raven]

f

sin (θ sin (x))

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[nocturne trench]

#help-31 message
I'm trying to handle problem B
https://i.imgur.com/Uxl5BpU.png Here are the answers path to them?

[onyx glen]

https://media.discordapp.net/attachments/486841085210132490/1053641414866632785/IMG_0139.png?width=648&height=515

to repost the original question for everybody else's convenience

[nocturne trench]

Oh I didn't link it properly

sorry

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[mild veldt]

How do you work out the probability of A and P

B

i got P(A) = 31/216

Because u do 25/216 + 5/216 + 1/216

e.g for 25/216 you have 1/6 * 5/6 * 5/6

[forest widget]

pretty sure you're severely undercounting

why is the odds of rolling at least one 6 going below 1/6 when you have more dice

[mild veldt]

What

U roll 3 dice

and whats the porbability i get at least one six?

So the probability i get one 6 from rolling 3 dice

is 1/6 * 5/6 * 5/6

the probability i get 2 6 from rolling 3 dice is

1/6 * 1/6 * 5/6

and 3 6's from 3 dice is

1/6 * 1/6 * 1/6

and u add the overall probabilities up

@forest widget

[forest widget]

what if the first dice isn't the 6

[mild veldt]

its 5/6 * 5/6 * 1/6?

[forest widget]

the 6 can be any of the 3 dice

[mild veldt]

How do u know

if its ordered sampling tho

[forest widget]

does your question mention anything about order

[mild veldt]

I sent the question

It doesnt mention anything abt order

[forest widget]

ye, so if you only get one 6, it can be any of the 3 rolls that got the 6

so you have to consider cases where {the first roll is the 6}, {the second roll is the 6}, {the 3rd roll is the 6}

[mild veldt]

so what i did

1/6 * 1/6 * 1/6

[forest widget]

that's the case where all 3 dice come up 6s

[mild veldt]

so what i said was right

at the beginning

[forest widget]

yes, and the other 2 cases you're undercounting

[mild veldt]

How am I under counting tho

I dont get it

[forest widget]

because for example, {6,1,2} is a valid arrangement where you get one 6

but so is {6,2,1}

and so is {1,6,2}, {2,6,1}, {1,2,6}, {2,1,6}

[mild veldt]

bruh

so what do i do

😂

[forest widget]

come back to this

when you multiply probabilities together you're implicitly defining some form of ordering

[mild veldt]

ah ok

im doing it rn

ill lyk the answer

ok

is it 91/216

[forest widget]

ye

[mild veldt]

nice nice

so how do i know when to use order

or no order

[forest widget]

multiplying probabilities together always implicitly define some form of order

sometimes it doesn't matter due to the problem's restrictions, but in this case it does, since (6,1,2) and (1,6,2) are considered distinct

and you judge whether to account for permutations based on whether different orders are considered distinct or identical

[mild veldt]

ah ok i get u

so for P(B)

im rly confused on that

because i wasnt sure how the ordering works

e.g can i do 112,121,211 ?

[forest widget]

they're considered distinct, so that's 3 permutations

[mild veldt]

yep

so if theres 3 permutations, and theres 6 numbers

is that 18 then?

[forest widget]

no

[mild veldt]

yh this is my problem

idk how to solve them

[forest widget]

usually you try to calculate the probability of getting a valid combination then permute each combination

but that's not exactly intuitive to grasp until it clicks

in this case, try to think through:
i want a pair, what's the probability of getting a pair in my first 2 rolls?

i then want a single, what's the probability the 3rd roll isn't the same as the pair?

i now have a valid combination, how many ways can i permute this combination? (remember that you have combinations of the form {a,a,b}, but for the probability you want to calculate, the pair can be any 2 dice)

[mild veldt]

the thing is

e.g when trying to work out the prob of getting a pair in the first 2 rolls

my logic is ok, for the first roll i can choose any number

so 1/6

then the 2nd roll has to be the same

so 1/6

and the third can be anything so the probability is 1

is that the correct way of thinking

[forest widget]

no

first we're not worried about the 3rd dice yet

second are you sure the odds of rolling 2 dice and getting a pair is 1/36

for 2 dice the sample space is small enough that you can realistically list out all outcomes or look up a chart

then you might want to think about where your line of reasoning is flawed

[mild veldt]

so u can do 1,1 2,2 3,3 4,4 5,5 6,6

out of 36

so 6/36

which is 1/6

no?

[forest widget]

oh that's what you're trying to express, ok

it just looks weird, because the first roll can be any number, so the probability the first roll is acceptable is 1

the second roll must agree with the first, so you accept the second roll with probability 1/6 (since the first roll is already determined to be some number between 1 to 6)

now continue

[mild veldt]

continue to what bit now

with the 3rd dice?

[forest widget]

what's the probability you accept the 3rd dice, such that the 3 dice form a pair + a single

[mild veldt]

the 3rd roll can be anything

[forest widget]

can it be equal to the first 2 dice?

[mild veldt]

apart from the double

so 5/6

so its 1 * 1/6 * 5/6

so its 5/36

[forest widget]

yes, so the probability of getting a valid combination, where in the exact order, you get a pair and a single, is 1 * 1/6 * 5/6 = 5/36

but this is only the case where you roll the single on the 3rd roll, how many ways can you permute each of these dice arrangements (a,a,b)?

[mild veldt]

u can have (b,a,a),(a,a,b),(a,b,a)?

[forest widget]

yes, so the final probability is?

[mild veldt]

15/36

[forest widget]

and that's the correct answer

[mild veldt]

is that it

for B

[forest widget]

ye

[mild veldt]

tf

thats more simple than i thought

Thanks a lot

So to work out P( A and B)

So how do I work out P(B|A) now

[forest widget]

probably get P(A and B) first

using the same line of thinking as above

[mild veldt]

I got it

Idk how to work out probability of A and B

I know the cardinality

is 30

but i dont know what its over

[forest widget]

conditional probability formula

[mild veldt]

so A and B = (116) (226) (336) (446) (556) (661) (662) etc until (665)

then u have 3 permutations for each

so thats 30

[forest widget]

correct, so what's P(A and B)

[mild veldt]

idk what u divide by

is it 30/36?

[forest widget]

how many possible results are there when you roll 3 dice

this shouldn't be the hard part i think

[mild veldt]

6 * 6 * 6?

[forest widget]

ye that's it

[mild veldt]

so 30/216

ah ok cool

so its 30/216 divided by P(A)

Alright I did the question

Thanks a lot

for ur helpp

Acc 1 more question

What is P(A and B and C) the same as writing

[forest widget]

what

define the event C

[mild veldt]

nvm

i did it

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[forest widget]

factorise the expression

[solid nova]

ok

so i got n(n^2+1)

[forest widget]

let n be an odd number, then investigate the behavior

[solid nova]

so say

let n= odd integer 3 or 5

then like expand?

[forest widget]

that might help

[solid nova]

how do u do it so its true for all odd values tho?

what if u make n= 2x+1

[forest widget]

then you can start the proof

[solid nova]

ok cool

so i made n=(2x+1)

then expanded n(n^2+1)

so i got

8x^3+12x^2+8x+2

@forest widget

[forest widget]

good, what does this expression mean

[solid nova]

so first i also divided by 2

but then how do i show its not a multiple of 4

could i replace x with like an odd number and see result?

[forest widget]

this is the expression for n²+n
can you notice anything about its divisibility by 4

[solid nova]

u cant divide by 4

only by 2

[forest widget]

that completes the proof

[solid nova]

oh

ok

so i just say

but at the start it says

n^3+n is an even number which is not a multiple of 4

[forest widget]

and you just showed both

[solid nova]

ok so my fianl sattement will be

this expressions cannot be divided by 4 so therefore its not a multiple of 4

[topaz valley]

can't you do something like

n^3 + n = n(n-1)(n+1) + 2n

[solid nova]

why plus 2n

[topaz valley]

expand the rhs and check it's equal

,w expand n(n+1)(n-1) + 2n

[glossy valve]

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[solid nova]

why would u add 2n to that factorsied bracket

[topaz valley]

then because n is odd, n+1 and n-1 are both even so the first term is divisible by 4

but 2n is not

[solid nova]

but why would u put the 2n in the equation

[topaz valley]

because it's what makes it equal to n^3 + n?

otherwise you get n^3 - n

i mean you can't leave it out

it's just there because that's how the algebra plays out

[solid nova]

right i see ok thanks

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[kind kraken]

I do not know how to write a proof for this

It seems like very trivial but how to write a proof can someone help me?

[carmine minnow]

It's showing that two sets (of solutions) are the same

[full forum]

@kind kraken Has your question been resolved?

[kind kraken]

so should i take some sytem of two equation

and such that the second one is the linear combination of the first

and then prove that they have the same solutios

[carmine minnow]

you dropped a condition

[kind kraken]

each equation is a linear combination of the equations in the other?

Can you tell me how I should prove it?

[carmine minnow]

each equation in one system is a linear combination of the equations in the other system

[kind kraken]

I have no Idea how and where to start?

[carmine minnow]

It's showing that two sets (of solutions) are the same
To show two sets are the same, show that one is a subset of the other, then a superset of the other

[kind kraken]

Could you give me an example?

should I do it with an example?

[carmine minnow]

an example would be say I want so show that the span of (1, 2, 0) is {(t, 2t, 0) : t real}

both of these are sets

so if we take an element in the first set, say a(1, 2, 0), we see it is (a, 2a, 0), and substituting t=a shows it is in the second set - so the first set is a subset of the second

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[torn jolt]

how to do this problem???

[torn jolt]

i don't understand what the lines mean and how to find the radius/height

[carmine minnow]

equal lengths?

[torn jolt]

it's a weird way to show equal lengths but prob

so basically a condition for triangle similarity is AA

so using this information

it's possible to find the volume i think

[verbal mantle]

It is indeed

You get radius

And then height with some trig

[clear lily]

no trig

[verbal mantle]

Pithagoras

[clear lily]

just pyramid height formula

[verbal mantle]

Still

Triangles

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[torn jolt]

ty

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[torn jolt]

Please help me how do I do 43

I tried doing it here but apparently I'm wrong

I'm so confused bros

[full forum]

@torn jolt Has your question been resolved?

[shrewd hamlet]

Start at the step: x^2 - 2x - y^2 - 4y - 4 = 0

Then complete the square for x and y correctly

to complete the square for y, try factoring out a -1

-(y^2 + 4y + 4)

Then see if u still need to complete the square

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

I dont even know what completing the square means🗿

Also I factored out -1

This is what I got

By the way this is the answer

[shrewd hamlet]

https://youtube.com/shorts/wHauKOchRgk?feature=share

[torn jolt]

Thank you

[torn jolt]

I did this

And this is what I got

I dont know what I'm doing wrong

[shrewd hamlet]

Did u watch the video?

It shows what ur doing wrong

[full forum]

@torn jolt Has your question been resolved?

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

Yes

Idk if your there

But I got this

But that isnt right since the y+2 shouldnt have -1 under it

I seriously dont know what I did wrong

1/-1 is -1

I need help please I'm begging

🙏

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[brave star]

why does it have to be zero vector , not just any element of K ?

[clear lily]

that wouldnt be a subspace

{0} is closed under addition and scalar multiplication

so {0} is a subspace

[brave star]

.close

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[brave star]

.reopen

[full forum]

✅

[fast peak]

any vector space by definition always needs to include the zero vector

[clear lily]

but thats a consequence of closedness under scalar multiplication right

[fast peak]

so at the very least you need {0} as a subset of your subspace

[clear lily]

and/or scalar -1 and closedness under addition

[fast peak]

well by definition of (V,+) being an abelian group I would say

[clear lily]

hahah alright

[brave star]

I just thought there could be another element

that behaves similar to 0

[fast peak]

the zero element is unique

if a and b are both zero elements, then a = a+b (cause b is a zero element) = b (cause a is a zero element)

[brave star]

.close

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[slow gulch]

hi!

[slow gulch]

I need help with part ii

oh it’s okay, I can wait for someone else

I’ve tried am-gm but I’m still stuck

[alpine girder]

is this uni?

[slow gulch]

oh I’m not in uni hahah

[alpine girder]

damn

[carmine minnow]

well i guess you'll need to use part 1 to do part 2

[slow gulch]

yes

wait let me send a picture of my work

[carmine minnow]

(a) would be useful in simplifying the denominators, wouldn't it?

[slow gulch]

I tried this but I don’t know if it’s going anywhere

[carmine minnow]

hmm, maybe don't apply AM-GM too quickly

[slow gulch]

ohh

[alpine girder]

if this isnt uni what is it

[carmine minnow]

idk math competition?

[slow gulch]

oh it’s my math assignment

oh it’s not a competition haha

[alpine girder]

looks like one bro

[slow gulch]

ohh hahaha

it’s further math

[alpine girder]

so you take further math?

[slow gulch]

yes

[alpine girder]

are you a girl or a guy

[slow gulch]

umm girl why

[carmine minnow]

what happens when you use part a to simplify the denominator

[alpine girder]

I dont know many girls who take further math lmao

didnt mean for it to sound weird mbmb

[slow gulch]

let me try

[slow gulch]

it’s ok hahah

okay I’ve tried using a and b

let me send a picture

[alpine girder]

nice handwriting

is that you in pfp

[gaunt lynx]

bro fk off

why are you obsessed with her

[alpine girder]

just asking

[slow gulch]

um

[gaunt lynx]

take it to dms

[rare dock]

don't take it to dms either

[gaunt lynx]

nah take it there

she'll ghost tf outta you

[alpine girder]

bro was just asking

she’s pretty

[topaz valley]

[short siren]

Senses tingling

[gaunt lynx]

stop filling this channel with your nonsense pls

[short siren]

[rare dock]

same with you 💀

[gaunt lynx]

i am born a nonsense

alr alr soz bros

[alpine girder]

get outta here dude i wasnt doing anything

[slow gulch]

can we get back to this haha

[short siren]

Sure

[slow gulch]

tyy

I tried using a and b

[alpine girder]

@slow gulch reply dm

[rare dock]

stop

[alpine girder]

im not talking to you

[slow gulch]

hahah pls stop

[alpine girder]

just want to talk to you

[slow gulch]

I just want my question answered 🥲

[alpine girder]

ill answer all your questions in dms

[gaunt lynx]

<@&268886789983436800>

[alpine girder]

im sorry

i was playing

[short siren]

This is a math server

[normal tree]

hmm well working backward, maybe you can show that the second term <= 1/4?

actually no, that doesn't work

[slow gulch]

ohh hmm

[normal tree]

also this was fun to watch LOL

[slow gulch]

wait that seems ok

I can try

[slow gulch]

haha 💀

[gaunt lynx]

never seen a guy sliding into a girl's channel before lmao

[short siren]

Same lmao

[carmine minnow]

maybe changing the denominator back might be a good idea since it's a nice product

[normal tree]

well, you can write (b+c) = 1-a or something

[carmine minnow]

but yeah, all we need to do now is show $\frac{abc}{ab+bc+ca-abc}\leq\frac{1}{8}$

[glossy valve]

Element118

[carmine minnow]

sorry, wrong direction

[slow gulch]

oh ok! I think I can do that

[short siren]

Hmm

[carmine minnow]

i think it would be easier to do
$\frac{abc}{(a+b)(b+c)(c+a)}\leq\frac{1}{8}$

[glossy valve]

Element118

[carmine minnow]

since it's homogeneous

[slow gulch]

wait isn’t my work enough to show it is < or equal to 1/4

and I can cross multiply and use am gm I think

[carmine minnow]

yeah I think that works

[slow gulch]

yaey! ok let me try

[carmine minnow]

yeah AM-GM should work because muirhead says yes

[normal tree]

does calculus work for this one too

[slow gulch]

I think I got it!

[topaz valley]

after rewriting the denoms as (a+b)(a+c) it's already homogenous

[slow gulch]

is this ok?

[topaz valley]

then you can lift the restriction on a,b,c

if you divide through by b,c and impose another restriction jensens should finish it

unless im tripping

[normal tree]

oh yeah I forgot that AM-GM existed

does it actually work

[carmine minnow]

yeah it does

homogenising is a useful step

because it tells you "okay I don't need to continue using the condition given"

[slow gulch]

ohhhhh!

[carmine minnow]

so it's one less thing to work with

[cloud shore]

You probably can write it as a sum of squares

• 3/4

[carmine minnow]

in terms of sqrt(a), sqrt(b) and sqrt(c), eww

[slow gulch]

hahahahha

tysm for the help!!

[cloud shore]

Nvm not necessarily

[normal tree]

proof looks reasonably solid though

[slow gulch]

ok that’s a relief

ty again!

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[torn jolt]

Can someone tell me how I got this wrong? I did definite integration with an upper limit of -4 and a lower limit of 2

<@&286206848099549185>

[onyx glen]

maybe you messed up the arithmetic somewhere?

[torn jolt]

I checked online and it said I was correct

[onyx glen]

where did you check and what did you enter

[stable plover]

,w integral x=-4 to 2 of |5x^2 + 5x - 30|

[glossy valve]

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[stable plover]

@torn jolt

you're calculating the signed area

it seems they wanted the absolute area

[torn jolt]

Oh

They wanted absolute

[torn jolt]

Thank you so much

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[torn jolt]

i dont understand what this is asking exactly

[torn jolt]

nevermind i figured it out sorry

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[fallow locust]

How does the graph of the integral equation of a circle's equation look like? What are it’s properties?

[clear lily]

you would like to integrate y^2 + x^2 = r^2 with respect to x?

[fallow locust]

I mean how the graph will look like

[clear lily]

well yes but its important to be sure of what graph you mean

[fallow locust]

If you normally think of x coordinate going to the positive x then you will have two rate of changes at the same time. Like if the circle has a radius of 10 and it’s centre is at the y axis then when you start graphing the integrated function from 0 you will have two rates of changes at the same time

[stable plover]

$f(x) = 2 \int_{-r}^x \sqrt{r^2 - t^2} dt$

[fallow locust]

,w f(x) = 2( int from -r to x {r^2 - x^2})

[glossy valve]

tushar

[stable plover]

,w plot f(x) = \int_{-5}^x \sqrt{5^2 - t^2} dt

[glossy valve]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by tushar#0942

[stable plover]

lol let's see here

[glossy valve]

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[fallow locust]

,w plot f(x)=2(-1/3+x-(x^3/3))

[glossy valve]

Results provided by WolframAlpha

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Requested by Alan turing#0137

[fallow locust]

Can anyone explain me how this came?

[stable plover]

https://www.desmos.com/calculator/07awsxpiiw

[fallow locust]

I meant graph of f(x)=x

[stable plover]

at x = r, x should be pi*r^2

[fallow locust]

How the graph is continuous?

The derivative of the graph is a graph of circle so there shouldn't be any value after r for which int f(x) is differentiable

[stable plover]

oh wait

i thought you meant a function that represents the area of a circle from -r to x where x is between -r and r

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[torn jolt]

hello! one question, how would i typically calculate for example "-12,3 : 66,79" as a long division? thanks for any help :)

[onyx glen]

$-\frac{12.3}{66.79} = -\frac{12.3 \times 100}{66.79 \times 100} = -\frac{1230}{6679}$

[glossy valve]

Ann

[full forum]

@torn jolt Has your question been resolved?

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

thanks so much, but, do you maybe know how to do that as a long division? thanks :,)

anybody knows how to do that as a long division? thank you for anything :)

[onyx glen]

do you know how to do long division on integers?

if you do, then what i said should be enough for you to continue.

if you don't, then say so.

[torn jolt]

yes

[onyx glen]

ok, then you should be able to divide 1230 by 6679

[torn jolt]

ohhh, now i think i understood

yeah, understood, thank you :3

i guess i will just calculate it normally and then add the negative after

thank you again :)

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[tulip beacon]

https://gyazo.com/274308773107ab077319598a49fbf23a

[tulip beacon]

still no clue how to do it

[gritty rose]

Did you try using the condensation method

[tulip beacon]

Yea i watched a couple of youtube videos

it did something like

how

1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 ..... .

than it grouped it

like

(1) + 1/2 + (1/3+ 1/4) + (1/5 + 1/6 + 1/7 + 1/8)

[gritty rose]

Looks right. Bound each grouping by the number of terms times the smallest term

1/3 + 1/4 > 1/4 + 1/4 = 2 (1/4) = 1/2

[tulip beacon]

Yea

[gritty rose]

Do the same for the next (1/5+...+1/8)

[tulip beacon]

it changes them all to 1/8 + 1/8 + 1/8 + 1/8 = 1/2

[gritty rose]

Almost done. Count how many groupings you have

[tulip beacon]

2^n

[gritty rose]

No

That's how many terms you have in the sum

[gritty rose]

This is 4 groupings

[tulip beacon]

OH

Ok

so we wanna know how many grouping we got all togeather

idk how many will we have?

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[dapper scaffold]

How to do?

[clear lily]

did they teach you the intermediate value theorem?

if not please disregard what i said

[dapper scaffold]

i've not heard of it but maybe they have

i mean if that theorem uses the integral info i think it'll work

[clear lily]

well if they didnt teach you yet then its not needed

and probably not what they intended for you to use

[dapper scaffold]

do you know how to do it another way?

like using upper and lower bound integrals?

[clear lily]

yeah that sounds like the way

[dapper scaffold]

i just don't understand how you get that simple of an LHS and RHS

my working out doesn't show that all

[clear lily]

well if we draw 1/x

we get as lower sum 1-1/t

[dapper scaffold]

wait you do?

how did you get the minus?

[clear lily]

ehm ill draw it one second

[dapper scaffold]

like i thought you get 1 + 1/2 + 1/3 + ..... + 1/t

cool

[clear lily]

so this is 1 lower sum

or like this

https://www.desmos.com/calculator/rotmlck6dn

[dapper scaffold]

i get the diagram but i can only see how it equates to 1/1 + 1/2 + 1/3 + .... 1/t

not 1-1/t

like where does the minus even come from

[clear lily]

well you only take 1 lower sum

for the lower bound

[dapper scaffold]

wait why only 1?

[clear lily]

wait perhaps im mistaken

[topaz valley]

,w integral from 1 to t of 1/t dx

[glossy valve]

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[clear lily]

,w (t-1)(1/t)

[glossy valve]

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Requested by Gijs#0201

[clear lily]

how is that right

oh dx

nvm

[clear lily]

this area of the lower sum is (t-1)(1/t) = 1 - 1/t

[dapper scaffold]

yeah makes sense

wiat

dx

don't you have to make a substituion then? Like for dt?

[topaz valley]

,w integral from 1 to t of 1+1/t (1 - x) dx

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[dapper scaffold]

uhh aren't you taking the integral of a completley different function then?

[topaz valley]

it's literally this

how is it not true

[clear lily]

i looked at 1/x

instead of ln x

thats why lmaso

[dapper scaffold]

but isn't that an entirely different function?

[topaz valley]

integrating gives the required result

[clear lily]

$\leq *$

[glossy valve]

Gijs

[topaz valley]

[clear lily]

im just complaining about notation

[topaz valley]

[clear lily]

everybody know $\leq$ is better than $\le$

[glossy valve]

Gijs

[clear lily]

wait

[topaz valley]

[clear lily]

how did you know to add 1/t(1-x)

[topaz valley]

it's the linear approximation

if you plug in x=t you get back 1/t

[faint sigil]

Hello I need help

[topaz valley]

,w 1 + 1/t (1 - x) where x=t

[glossy valve]

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[faint sigil]

paimon is 2 feet tall. her feet are 4 feet above the ground. celestia is 29029 feet from the ground. the angle of elevation is 80°. how far does paimon need to fly if she wants to be directly below celestia?

What is the answer?

[topaz valley]

#❓how-to-get-help

[clear lily]

so you just found the line tangent to 1/x?

[topaz valley]

no

[clear lily]

is that the linear approximation

[topaz valley]

i found the line sitting above 1/x

it needs to be an upper bound

[clear lily]

ohhhh

yeah that makes more sense

so it is like this

https://www.desmos.com/calculator/cklt53ajt2

[dapper scaffold]

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[young bramble]

is this better for calculating the sum of an arithmetic series than just doing:

averge(min and max value) * number of elements

n = length of sequence
a = first value
d = difference between any two numbers which are next to each other in the sequence

[narrow pawn]

I would just use the formula

The second formula

[young bramble]

ok

[narrow pawn]

Its the same result but I just use the second one based on standard convention and because of memory

[young bramble]

ok

ty

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[young bramble]

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✅

[young bramble]

one question

what was the formula to calculate the length of a sequence

nvm i remember

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[calm field]

im not sure how to answer the lemma part

[calm field]

this is induction

[deft zodiac]

hais

fill in blank

hm i think just fill in from what u got right lol

a is fk+2