help-23

[rustic goblet]

yo adonis

[fathom jewel]

yo

[rustic goblet]

can you help

this isn't even graph theory, it's just me being unable to read a statement

[drifting knot]

Yo

[fathom jewel]

from what i read i dont think you are asked to do a proof

[desert pasture]

What about pinging Austin

[rustic goblet]

I think I should do a proof

[drifting knot]

For which n

[rustic goblet]

this question isn't good enough for Austin

[fathom jewel]

in what sense

[drifting knot]

I think they want you do something bout the n

[fathom jewel]

is he too good

[rustic goblet]

yes

[fathom jewel]

Austin looks like a graph theorist

[drifting knot]

Cz the question is based on it

[rustic goblet]

to be clear, I've already found a condition and proven =>

For which $n$ is there a graph $G$ with $|G| = n$ such that $G$ is isomorphic to $\overline{G}$?

\begin{proof}
We claim that there exists a graph $G$ such that $G$ is isomorphic to $\overline{G}$ if and only if $n = 4k$ or $n = 4k + 1$. for some $k \in \N$.
\medskip\newline
$\implies$: Suppose that we have a graph $G$ with $|G| = n$ such that $G$ is isomorphic to $\overline{G}$. Then we must have that $e(G) = e(\overline{G})$. It follows that $G$ contains half the number of possible edges in the graph $K_n$, and $\overline{G}$ contains the other half. Thus, $e(G) = {n \choose 2}{2} = \frac{n(n - 1)}{4}$. This implies that either $n$ or $n - 1$ is divisible by 4, since we cannot have a fractional number of edges. But this means that either $n = 4k$ or $n = 4k + 1$ for some $k \in \N$, which is what we wanted to show.
\end{proof}

ah

[drifting knot]

For which n is there a graph with mod G =n for which G is isomorphic to G bar

Yes

[flat frigate]

higher!

[rustic goblet]

I'm wondering if the question is worded as an if, and not an if and only if

cause if it is then I'm done, but if not, then I have way more to do

[drifting knot]

They just want the n

[rustic goblet]

and it's not clear to me whatsoever if the other implication is even true

[rustic goblet]

which is an if or iff?

[drifting knot]

U got the condition for n

And it's done

n=4k or n=4k+1

[rustic goblet]

I'm not sure

which is why I'm asking

[drifting knot]

I think they only want the n condition

Cz they have asked with a question mark

Eg.

[fathom jewel]

I have a question

Is this easy for you to prove stuff

[rustic goblet]

@raven wind ping

I can't read this statement for the life of me

[drifting knot]

@devout shale

[devout shale]

Hello sir

[rustic goblet]

why the Austin ping

[drifting knot]

Hello sir

[devout shale]

We have a doubt?

[drifting knot]

Yes higher has it

[rustic goblet]

just me being unable to read a question

[devout shale]

Is |G| the vertices or amount of automorphisms of G

[rustic goblet]

former

[devout shale]

Yes TTepa has the right interpretation

[rustic goblet]

so the question is phrased as an iff?

I have a condition which I think works in that case

but then I don't know how to show <=

We claim that there exists a graph $G$ such that $G$ is isomorphic to $\overline{G}$ if and only if $n = 4k$ or $n = 4k + 1$. for some $k \in \N$.

[flat frigate]

higher!

[rustic goblet]

=> is proven, but <= eludes me

thank you for clarifying

[drifting knot]

Fr Austin the 🐐

[rustic goblet]

I appreciate it

[drifting knot]

I told him this too

[rustic goblet]

@devout shale do you have an idea on how I should prove <=

[devout shale]

Currently driving

But

I’d construct the isomorphism somehow

No real ideas

[rustic goblet]

ic

[devout shale]

Off the top of my head

[rustic goblet]

okay

let me attempt it then

stay safe!

[safe radish]

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[blazing dew]

i dont get why it would simply be 18? I swear this has to involve some kind of combinatorics, can someone help me understand the language used here so I know what math to apply and when

[marsh walrus]

its honestly less about language, and more about seeing enough of these that you understand the types of problems

in this case you can draw a picture

it might help to simplify the problem or lower the numbers a bit

[blazing dew]

how do I draw a picture of this

[marsh walrus]

use a tree diagram

[blazing dew]

how

[marsh walrus]

simplify the problem

[blazing dew]

like now I got permutations & combinations

[blazing dew]

in what way

[marsh walrus]

say 3 days and 2 choices each day

[marsh walrus]

discord sucks and i cant open this image, sorry

[blazing dew]

crazy

[marsh walrus]

there we go jesus

[blazing dew]

okay but there are so many different ways of doing 2 choices

[marsh walrus]

this problem is talking about arranging

every possible option is used

do you see how thats different?

[blazing dew]

nope, dont know how you deduced that from the question itself

[marsh walrus]

we could swap the language over to be like the one in the original problem you posted

[marsh walrus]

There are 3 sightseeing tours, 2 restaurants, and 2 tourist spots. In how many distinct orders could a person see these attractions if they didn't distinguish between different attractions of the same type?

everything gets used up

[marsh walrus]

in this problem, we have a limited number of slots, we select a tour each day and exclude all of the others

you could think about this using choose, if you wanted, but usually people dont

since $\binom{6}{1}$ is not something people usually say on its own

[flat frigate]

jan Niku

[blazing dew]

so if they did distinguish that means many of the combinations wouldnt be distinct anymore because they've already been used right

i think the word 'distinguish' is the key here to understanding the difference in what to use here

[marsh walrus]

i guess id need a more clear problem statement here

if they did distinguish the number would go up

but thats only because now we can discern differences between outcomes that we couldnt before

[safe radish]

@blazing dew Has your question been resolved?

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[blazing trench]

Is it allowed to ask for checking work?

[safe radish]

Please don't occupy multiple help channels.

[pale river]

send it

[pale river]

i'm too stupid to help sorry

[blazing trench]

Dang

I’m too stupid to know if im right or wrong

[rigid inlet]

seems fine

[blazing trench]

This one is much longer but does it look right?

Actually nvm

.close

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[proud zealot]

I need help with understanding how to solve for b

[proud zealot]

or at least, what i was doing wrong for b

[vale oriole]

how did u get ur answer for b

[proud zealot]

I think I added two of the years together, then I subtracted the 3rd year from that

im not too sure tbh

[proud zealot]

yeah i added 272+267=539, then I had 539-252=287

im starting to think it should have been as 272-267=5

then 15-5=10

or is it 272-252= 20?

[vale oriole]

doing numerous subtractions gets tedious if you want to find the average over a much longer peroid

have you tried using the formula provided on the right?

[proud zealot]

Ah no.

[proud zealot]

i think im still trying to figure out which numbers to plug into the formula

[vale oriole]

well essentially if you wanted to use the formula on the right, you would find the total amount of change, which you worked out as 272-252, then divide it by the peroid of time it spanned across, which was 2 years, thus you end up with 20/2 = 10

[proud zealot]

oh.

[proud zealot]

dam.

Cant believe it was that simple lol

[proud zealot]

thanks.

[vale oriole]

[proud zealot]

.close

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[broken wyvern]

If

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

What is the value of

$$\frac{2}{x} + x$$

[flat frigate]

raymondclie

[wispy monolith]

Try to solve the first equation and find value of x, it will probably give you a quadratic equation

[broken wyvern]

Alright

[flat frigate]

raymondclie

[broken wyvern]

$$2x - 2 + x = x^{2} - x$$

[flat frigate]

raymondclie

[broken wyvern]

$x^2 - 4x + 2$

[flat frigate]

raymondclie

[broken wyvern]

Is the answer 4?

[wispy monolith]

Check again, what is value of x ?

[wispy monolith]

From this equation?

[broken wyvern]

Either $2 + \sqrt{2}$ or $2 - \sqrt{2}$

[flat frigate]

raymondclie

[broken wyvern]

I subsituted it to $\frac{2}{x} + x$, and rationalized the fraction, both cancel out to 4.

[flat frigate]

raymondclie

[broken wyvern]

Is there another way to solve this without the quadratic formula?

[plucky elk]

No

[wispy monolith]

Then it's fine

[broken wyvern]

Alright

Thank you so much!

.close

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[simple gazelle]

help

[vagrant ice]

just ask

[simple gazelle]

do u guys answer physics too?

im hvinf a rough time answerinf this

finding the displcement of the box after falling

[safe radish]

@simple gazelle Has your question been resolved?

[magic junco]

What have you tried so far?

Do you know the ultimate speed of the box right before the box leave the roof?

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@modern bloom Has your question been resolved?

[safe radish]

@modern bloom Has your question been resolved?

[obsidian oracle]

here's a hint : since f(x,y) doesn't change too much from f(0,0) when close to (0,0) (continuity), then the integral over a small area is just approximately area*f(0,0)... Up to you to find a margin of error that suits you

[safe radish]

@modern bloom Has your question been resolved?

[random hazel]

What are you studying bruh? Why are your questions like this

[fathom jewel]

real

[random hazel]

Is it??

[fathom jewel]

what is this

you are having a limit where your region gets smaller and smaller to 0 integrating basically over nothing

[short topaz]

yeah so think intuitively

let's go back to a normal single integral

if i integrate say

$\int_{-\delta}^{\delta} f(x) dx$

[flat frigate]

LY

[short topaz]

(where $\delta$ is really really small)

[flat frigate]

LY

[fathom jewel]

,, \delta \approx \dd x

[flat frigate]

bacc

[short topaz]

then if f(x) is continuous, this integral 'effectively equals' $2\delta f(0)$

[flat frigate]

LY

[short topaz]

i.e. the area of the rectangle times f(0)

diagram for reference

now this still holds for double integrals too

[short topaz]

hence this hint

you do have to formalise it with analysis

[short topaz]

bcus this isn't true if f is not continuous

anyway that's how you deal with the numerator

[modern bloom]

but what about the denominator

[short topaz]

u can probably use taylor's theorem

the error term -> 0 so we're good

[modern bloom]

use taylor to g(r^2)??

[short topaz]

actually wait we don't even need taylor's theorem

we know that g(x) = g(0) + xg'(0) + E(x)

where E(x)/x -> 0 as x -> 0

[short topaz]

so use this for the denominator

[short topaz]

lmk if u need any more hints

for which part? @modern bloom

this is more analysis than calculus

[modern bloom]

ngl this is really a calculus assigment

[short topaz]

^

[modern bloom]

maybe just teach me how to do this one, this q is completely beyond my ability I would say, I need to learn how to do it

[fathom jewel]

me2

i wish i could understand tensors

[short topaz]

right so our general idea of the proof is as follows:

[short topaz]

read my discord status

[fathom jewel]

yes exactly

differential geometry was

it was

[short topaz]

we know that the numerator is 'approximately' pi*r^2 f(0,0)

and that the denominator is 'approximately' ar^2

do u understand the intuition for these? @modern bloom

[fathom jewel]

this seems some calculus ~~technique ~~ intuition to approximate it like that

[modern bloom]

btw does this technique have a name @short topaz

[short topaz]

it's not really a technique, it's just intuition

[short topaz]

the first comes from what integrating a continuous function around a very small area looks like

[short topaz]

and this just comes from taylor expanding g(r)

[short topaz]

this means that the limit should be pi f(0,0)/a

now that we have an understanding of what the limit should be, we should prove it rigorously

[fathom jewel]

yea

[short topaz]

firstly is to formalise this idea that integrating a function about a very small area gives area*f

here, think about what it means for f to be cts at 0,0

and use this to get bounds on what our integral equals for small r

then again use the fact that the derivative gives us the best linear approximation

to get another bound on what g(r^2) equals

then you just have to put it together and algebra manipulate

i've attached a full solution if ur still unsure

[fathom jewel]

sigma algebra

[modern bloom]

@short topaz I saw it somewhere else too, what's this? Why did you take the min?

[short topaz]

because we want both our inequalities to be true at the same time

intuitively, we know that the integral is very close to pi*r^2 f(0,0)

and g(r^2) = ar^2

but our definitions don't guarantee the same delta for both of them

[fathom jewel]

i am playing mc

i understand the intuition

[short topaz]

that's our error term

[fathom jewel]

LY is prob doing some rigorous approx

[short topaz]

that just comes from the definition of the derivative

do you know how to differentiate a double integral?

as in like can you differentiate the numerator?

we're differentiating w.r.t r tho

well you either need to be able to differentiate the numerator w.r.t to r to apply l'hopital's

or we need to take a rigorous approach using epsilon delta

i think if u haven't really done analysis then there isn't much point doing a question like this

i'm ngl i've kinda forgotten how to differentiate the numerator lol

[fathom jewel]

you first understand what that double integral actually means

what you are actually calculating

what

volume of the region under your surface function f(x,y)

idk what you said, something like "volume in a circle area" or what you meant

yea

np

dont be mad at me

🥹

no

stop doing math

also whats the point of helping if you are not being helpful

[blazing dew]

🤣

[modern bloom]

I looked over your proof carefully again. I'm sure I can't do such a rigorous analysis, but I came up with a simplified approach. can you take a look and see if this makeshift method is okay ? @short topaz

[safe radish]

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[short topaz]

just send it in

[safe radish]

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[short topaz]

yeah so that's what i was trying to say with intuition

[short topaz]

depending on how rigorous you need to be, this might not be allowed basically

[modern bloom]

thank you, I really canot do that eps del stuff so I rewrite it like this lol

[short topaz]

but that is the idea behind why the limit is what it is

.close

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[regal cypress]

Hello ppl, so for this sum I foud the mean which is 31 (217/7) we use the measure of variation to find the answer but I'm stuck after that.

[shadow sparrow]

kristina walking weird distances :(

[regal cypress]

lol true

[shadow sparrow]

can u find k?

[regal cypress]

nope

[shadow sparrow]

try to find it first

how much does she walk on day 1?

[regal cypress]

so I tried finding k using this quadratic discriminate method (b^2 - 4ac) and the answer is -1

[shadow sparrow]

.

[regal cypress]

I'm not quite sure abt that but will assume its 31?

[shadow sparrow]

how did u get 31?

[regal cypress]

217/7

[shadow sparrow]

i see

what does the question say about how much she walks on day n?

[regal cypress]

k + 2n - n^2 km

[shadow sparrow]

and on day 1 that would be?

[dusk steppe]

You can solve for k to find out the individual distances she walks on each day

[shadow sparrow]

yes we are going to that

but to solve for k we first need to know how much she walks on day 1

[dusk steppe]

Hmmm

You could just sum the 7 days and equate to 217 no?

[shadow sparrow]

yes u can

i would say its more complicated but both works

[shadow sparrow]

no

just substitute, n=1

[regal cypress]

are we finding k or n?

[shadow sparrow]

we are just writing out day 1

[regal cypress]

???

[shadow sparrow]

good job u found k

can u find standard deviation from here?

[regal cypress]

nope, that's exactly where I'm stuck, I found the mean which is 31 (217/7) but how do I substitute for the SD formula and IQR?

[shadow sparrow]

u use the k value u have found and sub it into the previous stuff u have written

eg for day 7 we have 43+14-39

[regal cypress]

ohh okayy

[shadow sparrow]

as u have the value of each day and the mean u can use that to find the variance, and std

[regal cypress]

could I assume the answers of these is sumation of x?

[shadow sparrow]

summation of x?

[regal cypress]

nvm, got the answer, thank you sm!

.close

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[sour fog]

Need help with physics problem

[sour fog]

.rotate

[frozen marlin]

,rccw

[sour fog]

I tried it again and got -22.414 m/s

[frozen marlin]

use the equations of motion

specifically

[flat frigate]

ren

[sour fog]

okay ill try them, my professor hasn't even gone over those

[frozen marlin]

really lol

i'll write them down for you

[sour fog]

yeah just kinematics and different free fall variations

[flat frigate]

ren

[sour fog]

s being final distance?

[frozen marlin]

yes

displacement actually

not distance

[sour fog]

oh okay

and a still equals (-g) / (-9.8 m/s^2)

[frozen marlin]

?

yes -g

[sour fog]

sweet

[frozen marlin]

i thought for a second you meant -g/-g = 1

[sour fog]

o the / wasn't division yeah lol

idk why i did that

[frozen marlin]

is fine

[sour fog]

okay I got it wrong again but i did it sloppily so I'm retrying rq

[frozen marlin]

ok

[sour fog]

I got 26.88 m/s both times

-26.88

[frozen marlin]

oh mb

i miswrote smth

in this it's not s

tf

[sour fog]

change in s?

[flat frigate]

ren

[frozen marlin]

yeah

[sour fog]

so 84-41?

im assuming

okay so I think for some reason s has to be negative

I did it and got 27.6 but if I change the sign to negative and add everything I get the right answer

LETS GOOO

okay thank you im gonna watch a vid on motion equations

[safe radish]

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[clever rock]

Help

[clever rock]

$/sqrt 3+2$/sqrt2$

$\sqrt 3+2$\sqrt 2$

$\sqrt 3+2 \sqrt 2$

[flat frigate]

Winshing

[clever rock]

Umm yeah

That's the question

Help

[fickle monolith]

so?

what do we need to do?

[clever rock]

How would I simplidy this

Ist a question on polynomials

Pls help

[fickle monolith]

is this the original question?

[clever rock]

Yeah

Just the 3 root goes over 2 root 2

Wait I got it

[frozen marlin]

what?

[clever rock]

Sry sry for disturbing

[safe radish]

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.

[clever rock]

I got it

[frozen marlin]

aight

[safe radish]

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[clever rock]

.close

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[bitter sinew]

hi just quick question

[bitter sinew]

is f(x) = 5000x
x + 5

can have uh graph

.close

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[rain dagger]

this a mistake?

[rain dagger]

obv this is not a correct premutation

[hardy lion]

why not?

[short topaz]

remember that you multiply right to left

so a_1 goes to a_2

a_2 goes to a_1 which then goes to a_3 etc.

[rain dagger]

well not necessarily, since it's associative

i dont see how sigma(a1)=a2=a3, while a2 != a3 is a possibility

[short topaz]

wut

that's by convention

you multiply out from right to left

i think you mean commutative

if it's commutative then it doesn't matter whether you go left to right vs right to left

but the permutation group is definitely not abelian

[short topaz]

i'm not sure what you mean by sigma(a1) = a2 = a3?

the LHS says sigma(a1) = a2

[rain dagger]

no i mean i could calculate the cycles from left to right first, since it's associative (we proved cycles are a group for premutations)

[rain dagger]

yes

[rain dagger]

but this means sigma(a1)=a3

[short topaz]

wait nvm i'm being dumb lol ignore me

[rain dagger]

yes so i started to calculate the cycles where i want

(a*b)*c, instead of a*(b*c), like you want to

[short topaz]

but i mean you could do it like that but that'll just be extremely unwieldy

like how on earth are you multiplying from left to right?

[rain dagger]

well i still dont see how the (a1, a2)*(a1, a3)=(a1, a2, a3)

[short topaz]

(as in going from left to right)

[short topaz]

a3 goes to a1 first, then it goes to a2

[rain dagger]

and a1 goes to

• a2
• a3
  imposible

[short topaz]

a2 goes to a1

[short topaz]

?

[rain dagger]

definitions of the cycles

[short topaz]

a1 goes to a3

[short topaz]

we're multiplying the cycles

we're doing 1 after the other

so let's say we have A B C

doing (3 1) first means we swap the first and 3rd letters

so then we get C B A

then we do (1 2) which means swapping the first and 2nd

so then we get B C A

if you look at what our final thing is

1 -> 3 -> 2 -> 1

so it's (1 3 2)

hence, (1 2)(1 3) = (3 2 1)

[short topaz]

what i meant is you apply the cycles right to left

so if we have a1

(a1 a2) is applied first

then (a1 a3) etc.

which is why you should be thinking about it 'right to left'

[rain dagger]

okay i got it now

ty

.close

[safe radish]

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[merry swift]

omg car seat headrest!

whats the question

to simplify this expression?

[wheat badge]

yes

[merry swift]

pull the largest square factors you can out of each square root

[wheat badge]

so 16 squared and three squared

for the first one

and 25 squared and three squared

[merry swift]

sounds right

[wheat badge]

ik the answer i just cant figure out how one of the xs dissapeared

answer is 23x√3

[merry swift]

and you're unsure why it isn't -12xsqrt(3)+35xsqrt(3)?

[wheat badge]

yes

[merry swift]

unrelated-seeming but humour me, what's -12y+35y?

[wheat badge]

23y

i get it

[merry swift]

awesome

all good now?

[wheat badge]

yeah

.close

[safe radish]

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[modest wraith]

sinx times 2x is 2 sin x?

i thouhht it would be sin 2x^2

[obtuse plover]

sin(x) * 2x = 2xsin(x)

The reason u only see a 2sinx is cuz they factored out an x from the top and bottom and canceled

[modest wraith]

ohhh

[modest wraith]

i dont get this

how is equal

[obtuse plover]

$\frac {a-b}c = \frac ac - \frac bc$

[modest wraith]

bro

why am i so slow

thanks bro

[obtuse plover]

np

[modest wraith]

wait

for the first term

3x/x^1/2

they canclled x^1/2

to make it

ohh

ok

thanks

@obtuse plover

how is this euqal

[obtuse plover]

first factor out a 2

2(-x^-2 + x^-3)

then we can do it as so

$2(-\frac 1{x^2} + \frac 1{x^3})$

[obtuse plover]

now make common denominators, then add the fractions

lmk what u get

did u get it?

@modest wraith

[modest wraith]

hol up

imr eading

[modest wraith]

Ohhh

that makes sene

lol the workbook dosent show work

thanks man

[obtuse plover]

np

[safe radish]

@modest wraith Has your question been resolved?

[safe radish]

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[wheat badge]

help with 6 please

[jolly shard]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wheat badge]

1

i have no idea how to do it

[dense berry]

are you familiar with the rules of exponents?

[wheat badge]

some of them

x to the 1/2 is just x square rooted

[dense berry]

any other ones?

[wheat badge]

not really

[dense berry]

i think you should get familiar with those rules before starting the problem

[wheat badge]

have a quiz on this tommorow

[dense berry]

i wish you the best luck

[wheat badge]

i think this is right

[dense berry]

what is the question asking?

[wheat badge]

rewrite the expression with positive exponents and simplify

[dense berry]

yeah so every exponent (the little number) must be positive

are there any negative exponents in that expression?

[wheat badge]

bruh i know what an exponent is 😭

[wheat badge]

no but it’s not simplified

[dense berry]

y^(-4)?

[wheat badge]

y to the 1/4

[dense berry]

the exponent there is not a fraction

any idea on how we can make that exponent positive?

[wheat badge]

1/y^4

i might be tripping

[dense berry]

lmao ur good

[ornate slate]

could be like this

[wheat badge]

what?

i think ur doing the wrong problem

u did seven i was doing 6

[ornate slate]

have u figured out 6?

[wheat badge]

yeah

xy^7

[ornate slate]

correct

[wheat badge]

.close

[safe radish]

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[safe radish]

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[rugged valley]

I had a question regarding propositional satisfiability in discrete math.

Lets say we have something like this (P∨Q∨R)∧(¬P∨¬Q∨¬R) and we want to see whether it is satisfiable or not.

When checking for satisfiability we assign truth values to the variables. In the case that we assign it like this P = Q = R = TRUE, this compound proposition is unsatisfiable, because in the second part, all of the variables are negated which means that entire side is false, and we have an "AND" operator in the middle.

My question is when checking for satisfiability can we assign different truth values to the variables, or do the variables HAVE to have the same truth values.

For example: P = FALSE, Q = TRUE, R = FALSE. In this case, the (PvQvR) part would be true, and so would the (¬P∨¬Q∨¬R), which would also mean the entire proposition is true because we have an "AND" sign in the middle.
Im a bit confused because I was going through a youtube video online, and they only tried one combination of truth values so I wasn't sure if all of the variables need the same truth value, or if we can have different truth values.

[safe radish]

@rugged valley Has your question been resolved?

[shy lynx]

Satisfiable means that there's some configuration of truth value assignments that makes the statement true. You don't need to assign the same truth value to all the variables (P = FALSE, Q = TRUE, R = FALSE is a valid assignment). So your example is indeed satisfiable.

To prove that a proposition is satisfiable it suffices to find a single assignment under which it's true.
To prove that a proposition is not satisfiable would require either checking all that all assignments make it false, or coming up with a more clever argument for why no assignment can make it true.

[safe radish]

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[limpid estuary]

help

[limpid estuary]

(x-4)(x+5)(x^2-8x+32)

would it b e that

[pine horizon]

yes

[limpid estuary]

or this

(x-4+4i)(x-4-4i)(x+5)(x-4)

i dont think the x^2 thing is right

[pine horizon]

in which one

[pine horizon]

in this it is

[limpid estuary]

ok so just put in that answer

[severe pond]

why do you ask

it’s a test right

[limpid estuary]

what

[severe pond]

awfully suspicious thing to ask

[pine horizon]

both are right

[severe pond]

why not just put it in and find out if it’s right

[tight void]

knief is popping off rn

[severe pond]

🤷🏼‍♂️

[tight void]

🤷‍♂️

[severe pond]

no cap

[limpid estuary]

bc i get a strike

if i get it wrong

[tight void]

is this baseball

wym strike

[limpid estuary]

imma do the second

[pine horizon]

maybe put the iota one

[severe pond]

https://tenor.com/view/on-the-case-search-detective-on-it-daffy-duck-gif-26966322

[limpid estuary]

yeah sure

[pine horizon]

since it says (4-4i) factor

so maybe they want in iota

[limpid estuary]

i only have 2 strikes

ok

[severe pond]

homework

[limpid estuary]

mine

[severe pond]

i thought you would’ve sent someone else’s homework problem in a help channel so that you could learn how to help them actually

[tight void]

knief is analyzing this fr

[limpid estuary]

im so confused

yay right

[severe pond]

@tight void and i are just like that

[limpid estuary]

sure]

[tight void]

just those guys 🤷‍♂️

[severe pond]

she thought we wouldn’t notice

😹😹😹

cheating on homework

[limpid estuary]

idk how to do it bru

[severe pond]

have you tried reading the textbook or lecture notes

[limpid estuary]

ywa

yes

[severe pond]

you lying?

[limpid estuary]

no

fr

yes

can u keep this between us

[severe pond]

was just about to ping moderators actually

lol i don’t think they’re strict with homework

that’s why you cut off the top of the question though

so we wouldn’t see the title

so who knows

might actually be a quiz

🤷🏼‍♂️

[limpid estuary]

and what if it was

[limpid estuary]

pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeee dotn

[severe pond]

then we’d help you to get a great grade

homework isn’t worth my time

[tight void]

AYO???

[limpid estuary]

i have important homewrok thrusadya

[tight void]

@severe pond you should make some time on thursday

to help

[severe pond]

why is that

[limpid estuary]

bc i asked

[tight void]

the quiz

[limpid estuary]

no

[tight void]

like the one today

yk

[severe pond]

who are you?

anyways do you need help with your quiz question?

[severe pond]

how many questions are on the quiz

[limpid estuary]

5 more

[severe pond]

so why did you lie and say it wasn’t a quiz

ok let’s finish it up quick

[limpid estuary]

ok wiat

im done

wasnt a quiz

im just a girl

[severe pond]

🤔

[severe pond]

how is this relevant

[limpid estuary]

idk im just a girl

shut up

ur no help

[severe pond]

idk i’m just a boy

[limpid estuary]

that dosent work

[severe pond]

i don’t help cheaters

😂😂

[limpid estuary]

im not

ur mad

[versed wolf]

$3(x-3)^3(x+4)^2$

[flat frigate]

harshul

[severe pond]

mf

[limpid estuary]

a stupid one in that

[severe pond]

<@&268886789983436800>

[limpid estuary]

shut up hes tryna help unlike u

[versed wolf]

What happened

[limpid estuary]

hes mad

[severe pond]

it’s a graded quiz/homework

[limpid estuary]

no its not

[versed wolf]

Oh

[limpid estuary]

omg

no

[versed wolf]

Mb

[limpid estuary]

.close

[safe radish]

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[modest wraith]

haha

[safe radish]

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[fluid rover]

Helppp, I need to know whast the actual messurment of the glass is of these doors

[fluid rover]

Just the glass pls

I have tried i know one side is 42'' but I cannot find anything else

<@&286206848099549185>

Guys pleaseee

[plucky elk]

!15m

[safe radish]

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

[fluid rover]

not inches?

Wait which doors?

[soft prism]

This isn't maths

I'm unsure you'll get any help in a math help server

Someone may help you out, but the chance is low

[safe radish]

@fluid rover Has your question been resolved?

[safe radish]

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[timid crystal]

i have no idea how to go about solving this

[timid crystal]

<@&286206848099549185>

[craggy sedge]

V(x) is not 8x(5-x)^2

[timid crystal]

ik but how do i set up the equation

i js clicked it but idk what to do

[craggy sedge]

notice the diagram

[timid crystal]

V(x)=8x(10-x)^2?

[craggy sedge]

no

[timid crystal]

can u help me bru

[burnt notch]

What's the formula for the volume of a box?

[timid crystal]

lwh

OH IS IT MMx

[burnt notch]

Nice

[timid crystal]

ohhh tysm

.close

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[molten hawk]

how is this linearly dependent

[desert swallow]

Do you know trig identities involving tan^2 and sec^2 ?

[molten hawk]

no

[desert swallow]

$$ \tan^2 x + 1 = \sec^2 x $$

[flat frigate]

StrangeQuarkAL

[molten hawk]

ohhh

wow

ok thank youuu 🤣

[desert swallow]

No problem!

[molten hawk]

.close

[safe radish]

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[modern pawn]

hi guys I am from India I want to get excellent in math how do I do SO

[rugged marsh]

helo

[rugged marsh]

hlp

[wispy monolith]

By practicing

[pine horizon]

practice

[modern pawn]

THANK YOU GUYS

[wispy monolith]

[modern pawn]

i am in 11 th grade

[wispy monolith]

Any doubts you can ask in this server @modern pawn

[safe radish]

@modern pawn Has your question been resolved?

[safe radish]

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[tiny heart]

can someone pls help idont get it

[sleek sentinel]

you can rent 40 apartments at $570 each
you can rent 39 apartments at $590 each
you can rent 38 apartments at $610 each
you can rent n apartments for a total revenue of $(2280+170n-n²)

[safe radish]

@tiny heart Has your question been resolved?

[tiny heart]

??

howd u get that equation

[sleek sentinel]

for the first bit, renting 1 less apartment gives you $20 extra each

[tiny heart]

mhm

[sleek sentinel]

for the second bit, it's given as $P

[tiny heart]

where

oh

howd u get that equation tho

[sleek sentinel]

end of question a?

[tiny heart]

yea

[sleek sentinel]

yeah that's a given

idk if this is gonna go anywhere but try factoring P

[tiny heart]

arent u meant to show it

[sleek sentinel]

yeah it's up to you to explain the connection

[tiny heart]

ok thanks

.close

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[sturdy grove]

Can someone help me

[sturdy grove]

Here is questions

I had a class on this topic but questions are very different

<@&286206848099549185>

[safe radish]

@sturdy grove Has your question been resolved?

[safe radish]

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[whole forge]

Where does the 3rd inequality come from

[junior smelt]

Is that supposed to represent some form of inner product?

Oh no, I see, it's (the norm of) coordinates

[whole forge]

Ye

[junior smelt]

Does that second term then say $\norm{(x, y)}^4$ then?

[flat frigate]

@junior smelt

[junior smelt]

(hidden by the page thing, 6/10)

[whole forge]

Oh yes oops

Didn't realise

[junior smelt]

Hmmm, personally, I don't know, there is the fact at least that, say, $\abs{xy} \leq \frac{\abs{x}^2 + \abs{y}^2}2 < \abs{x}^2 + \abs{y}^2$

[flat frigate]

@junior smelt

[junior smelt]

Well, at the very least, I can get the final inequality, for what it's worth

[safe radish]

@whole forge Has your question been resolved?

[junior smelt]

Actually I see what they've (probably) done!

That numerator
\begin{align*}
\abs{x}^2\abs{y} + \abs{x}\abs{y}^3 &= \abs{x}\abs{y} \qty( \abs{x} + \abs{y}^2 ) \
&= \abs{x}\abs{y} \qty( \sqrt{\abs{x}^2} + \abs{y}^2 ) \
&\leq \abs{x}\abs{y} \qty( \sqrt{\abs{x}^2 + \abs{y}^2} + \abs{x}^2 + \abs{y}^2 ) \
&\leq \qty(\frac{\abs{x}^2 + \abs{y}^2}2)\qty( \sqrt{\abs{x}^2 + \abs{y}^2} + \abs{x}^2 + \abs{y}^2 ) \
&< \norm{(x, y)}^2 \qty( \norm{(x, y)} + \norm{(x, y)}^2 )
\end{align*}

[flat frigate]

@junior smelt

[safe radish]

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[covert talon]

is this 5x10^9?

[sturdy grove]

Hey

@covert talon

[covert talon]

hi

[sturdy grove]

could u help me?

im confused

[covert talon]

girl

[safe radish]

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).

[sturdy grove]

i had it beforre

[frozen marlin]

so

you don't have it now

[sturdy grove]

what

[frozen marlin]

#help-44｜stanley-🌲-v2-dans

[sturdy grove]

can u help me there?

[frozen marlin]

i'll see

[frozen marlin]

no, not quite

first off

[wispy monolith]

How are you getting to that answer

[frozen marlin]

10^4 times 10^5 would be 10^9 which isn't happening here

you'd just use distributivity

2.5(10^4 + 10^5)

if you wanted you could simplify that

10^4 + 10^5 -> 10^4 + 10(10^4) -> 11*10^4

[wispy monolith]

Better get (a * 10⁴ + b * 10⁴)

[frozen marlin]

so your expression would become 27.5 * 10^4

[frozen marlin]

yeah

[sturdy grove]

hey

[covert talon]

Wait huh

Is there like a visual rule for it

[covert talon]

Girl be patient

[sturdy grove]

im a guy lol

[covert talon]

Guy be patient

[valid olive]

Fr

[frozen marlin]

okay first off

[sturdy grove]

10^5 is 10(10^4)

[frozen marlin]

you are not entitled to my help

please stop asking me to come to your channel

secondly

[safe radish]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[sturdy grove]

mb

[covert talon]

🦗

[frozen marlin]

all g

lemme finish this up and i'll try to help

[sturdy grove]

k take ur time

[frozen marlin]

anyway there isn't really a visual

10^5 = 10 times 10^4

thus

[flat frigate]

ren

[covert talon]

Ohh

Yeah that makes more snes

[frozen marlin]

that's it

[covert talon]

So if its 10^6 + 10^4 would it be 12(10^4)

[frozen marlin]

no

[lime dust]

10^4(1+10^2)

[frozen marlin]

100(10^4) + 1(10^4) -> 101(10^4)

[sturdy grove]

Do you need to give answer in scientific notation?

[covert talon]

No

just the standard orm

[sturdy grove]

oh then shouldn't it be 275,000

[frozen marlin]

standard form is scientific notation iirc

[covert talon]

Oop

[frozen marlin]

yeah they're the same

[sturdy grove]

because....

2.5 x 10^4 = 25000

2.5 x 10^5 = 250000

So add them together and its 275,000

Which is 2.75 x 10^5

[frozen marlin]

you don't need to expand

[frozen marlin]

using this

[covert talon]

I get it now

[frozen marlin]

good

[covert talon]

Thanks

.close

[safe radish]

Channel closed

Closed by @covert talon

Use .reopen if this was a mistake.

[frozen marlin]

nw

[safe radish]

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[serene vector]

hi

[safe radish]

Please don't occupy multiple help channels.

[serene vector]

can someone please help me

[mossy fractal]

ask your question in the first help channel that you opened

and close this one please

[serene vector]

okay

how do i close this

!close

/close

[short topaz]

do .close

[serene vector]

.close

[safe radish]

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[safe radish]

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[safe radish]

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[crisp mason]

i am not sure where to go from here

[fathom jewel]

You can do
[ \frac{1}{2} \abs{\frac{x-1}{x+1}} < \abs{\frac{x-1}{x+1}} ]

[flat frigate]

bacc

[fathom jewel]

Now use your assumption that |x-1| < 1/2

[crisp mason]

uhhh

i'm still confused

[fathom jewel]

You can write
[ \abs{\frac{x-1}{x+1} } = \frac{\abs{x-1} }{\abs{x+1}} ]
Now you can use your assumption
[ \abs{x-1} < \frac{1}{2} \Leftrightarrow \frac{\abs{x-1} }{\abs{x+1}} < \frac{1}{2\abs{x+1}} ]

[flat frigate]

bacc

[crisp mason]

where did you get the last term from?

i have class now, i'll try and understand it but thank you!

[fathom jewel]

Multiply both sides by 1/|x+1|

[safe radish]

@crisp mason Has your question been resolved?

[fathom jewel]

Also I wanted to point out that for x = 1 the inequality is not true

[fathom jewel]

Also I realized this doesn't help much but what helps is showing
[ \frac{1}{2} \abs{\frac{x-1}{x+1}} \stackrel{!}{\leq} \frac{\abs{x-1}}{3} ]
by using the assumption that
[ \abs{x-1} < \frac{1}{2} \Leftrightarrow \frac{1}{2} < x < \frac{3}{2} ]

[flat frigate]

bacc

[safe radish]

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[modern bloom]

Let ( m, n ) be positive integers. If the improper integral
[ I = \int_{0}^{+\infty} \frac{\ln^{2n} |x-1|}{x^m} , dx ]
converges, then the possible values for ( m, n ) are?

[flat frigate]

ive

[fathom jewel]

Maybe we can bound it

basically

actually nvm

I think we should split it

[flat frigate]

bacc

[fathom jewel]

Because x = 1 is critical

Now I would try to bound it

[lean otter]

is it ln to power 2n?

[safe radish]

@modern bloom Has your question been resolved?

[modern bloom]

This method won't work. Helpers passing by, please think of another way for us maybe

[lean otter]

do you know the answer

[modern bloom]

I dont

I just tried and it leads me nowhere

[lean otter]

i am getting m=0 and n is any real number

[modern bloom]

How did you solve? I doubt this result tho

@short topaz sorry for ping..If you're free can you take a look and give some thoughts? If you're busy just ignore it's okay

[fathom jewel]

i am terribly sorry that you drew this conclusion

[short topaz]

sec

[short topaz]

why doesn't this work?

anyway that's what i'd suggest too

clearly this integral has 2 parts: one part between 0 and 1

and another between 1 and infinity

that let's us 'deal with' the || sign

now in order for our integral to converge, both parts need to converge

let's analyse the part between 1 and infinity first

note that log < polynomial < exponential

so take a guess at what the answer for when the latter integral converges

and then prove it by bounding it

(luckily our integral is nice and always >= 0)

[modern bloom]

Wait lemme show my work

[fathom jewel]

oh i am not

i was just surprised that you said it didnt work

but i am also confused how to do it correctly

[fathom jewel]

exactly my thinking too.

[modern bloom]

[ \int_{1}^{+\infty} \frac{\ln^{2n} |x-1|}{x^m} , dx < \int_{1}^{+\infty} \frac{|x-1|}{x^m} , dx]

[short topaz]

that is a bound but it's not a very tight one

we know that ln(x-1) will be 'eventually smaller' than x^2, x, sqrt(x), cuberoot(x) etc.

[fathom jewel]

ye

[modern bloom]

My internet is lagging sp hard

[flat frigate]

ive

[short topaz]

i think it's good to have an intuitive idea of what this bound should tell us

notice that the numerator is >= 1

so if m = 1, then we're integrating 1/x between 1 and infinity

which obviously diverges

that tells us m > 1

then let's consider say ln(x-1)^2/x^2

if we wanted to show that converges

we wanna bound it by an integral which converges

now $\int_1^{\infty} \frac{1}{x^k} dx$ converges for all $k > 1$

[flat frigate]

LY

[short topaz]

so ideally we wanna show that this is eventually smaller than 1/x^k for some choice of k

[fathom jewel]

and 0 to 1 1/x^k for k <= 1

[modern bloom]

Finally get my internet

[flat frigate]

ive

[modern bloom]

@short topaz Do you mean this? I'm not sure

[short topaz]

yeah basically for some suitably chosen k

[short topaz]

well actually here notice that if we just take k = 1.5

then it works cus 'log is beaten by x'

[modern bloom]

Wait

@short topaz What should I do next?

Darn my internet

[modern bloom]

So this only holds for 0 <k<2

[short topaz]

wait i'm doing smth irl now