help-23

[drowsy karma]

but i also count on your intelligence, i explian it

the formula you show before

fo rth evolume is:

do you agree on it ?

that is also very known formula

[misty ember]

I agree! I'm going to take a short break for now, I'm about to return home from college

[drowsy karma]

ok no problem) but it is night at me so I may be off )

[safe radish]

@misty ember Has your question been resolved?

[drowsy karma]

I leave the calculations to you 🙂

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[dark marlin]

I’m just confused on how to go about it

.close

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[tiny heart]

hi

[tiny heart]

how do i do this question with long division

[drowsy karma]

note that the number 3 is the zero of the numerator so this can be used to use the product form in the numerator

[tiny heart]

??

long division..?

[terse lichen]

set up the long division first

[drowsy karma]

but if you cant see it, then use long division

[tiny heart]

yeah ik how to do that

but my teacher told us to do long division

[drowsy karma]

ok)

[terse lichen]

you still have a -3x² btw

[tiny heart]

o ye

so what do i do after

i’m stuck on the after step

[terse lichen]

continue

[tiny heart]

how tho there’s no x2

[terse lichen]

if theres no x²

means its 0

so you have a 0x²

0-(-3)

[tiny heart]

oh

ty i got it now

,close

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[lyric bloom]

im lost and dont know how to use the difference of cubes formula

[calm bridge]

a^3 - b^3 = (a-b)(a^2 + ab + b^2)
so you can use (a-b) = (a^3 - b^3)/(a^2 + ab + b^2)

where a and b are those cubed roots

[lyric bloom]

ok ill try that

[lyric bloom]

sorry am i rearranging the formua for (A-B)?

[calm bridge]

yes

[lyric bloom]

ok thank you

[vagrant ice]

Multiply top and bottom by a^2 + ab + b^2

It will work out if the denominator has the surds instead of the numerator

[lyric bloom]

but where does the (a^3 - b^3) go

[vagrant ice]

You'll get a^3 - b^3 in the numerator

[safe radish]

@lyric bloom Has your question been resolved?

[vagrant ice]

Hey do you still need help?

[lyric bloom]

yeah sorry im still working on it ill show you what i have in 2 seconds

[lyric bloom]

i got this but idk what to do with it

[vagrant ice]

Right, now you can sub h = 0 in

Also the h in the numerator cancels with the h in the denominator

[lyric bloom]

wait so the a-b = (a^3-b^3)/(a^2+ab+b^2) just becomes my equation now? assuming i put h in the denominator at that last step and cancel them out

[vagrant ice]

Yes you're left with 1 over your denominator

[lyric bloom]

oh

i didnt know i could replace the equation like that thank you

[vagrant ice]

Basically you couldn't sub 0 in at first cause it would be a 0/0 form

But after multiplying top and bottom by a conjugate

You can sub 0

Npnp

[lyric bloom]

uh idk what to do now

did i skip a step

oh nevermind i just calculated wrobg

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[past birch]

how do i do q8b?

[past birch]

in the form of asin(b(x-c))+d, i know a = 7, 2pi/b = 12.4 -> b= 5pi/31, but i dont know how to find c, i know d = 9.2

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[modern pier]

To find all complex solutions to the equation
z^3=−8i.........how to solve this

[dry mural]

,, x

[flat frigate]

ilyhedges

[dry mural]

,, z^3 = -8i
\ z = \sqrt[3] {-8i}

[modern pier]

its z^3

[flat frigate]

ilyhedges

[thin bridge]

convert -8i to polar, mod arg, cis form

[dry mural]

or that

[thin bridge]

either of those will work

[modern pier]

okk...thankss

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[modern pier]

.reopen

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✅

[modern pier]

So i got this nd then?

@thin bridge

[thin bridge]

consider the periodicity of CiS

CiS(t) = CiS(t + 2pi) = CiS(t + 4pi)

then cube root / DeMoivres each case

[modern pier]

Okk...thankss🔥

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[coarse kelp]

shouldn't x[-n+2] be more advanced than x[-n]?

[coarse kelp]

something like this?

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@coarse kelp Has your question been resolved?

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@coarse kelp Has your question been resolved?

[plucky elk]

No it's right

-n+2 = -(n-2)

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[rigid acorn]

can someone help me find convegence or divegence using limit comparison test or integral test

[lean otter]

sure!

can you think of a sequence that is closely related to what you have?

[rigid acorn]

the only series our teachers gave us to use for known divergence or convergence were the harmonic series and 1/n^2

the basel i think

[lean otter]

do you not know the convergence/divergence of $\ds\sum_{n=1}^\infty \41{n^p}$

[lean otter]

for any number p bigger than 0

[rigid acorn]

no we didn't learn that yet

hes also not allowing us to use the normal comparison test

[plucky elk]

What comparison test can you use then

[rigid acorn]

only limit comparison

intergral test

and nth term test

[plucky elk]

Slow the exact statement of your limit comparison test

[rigid acorn]

like my notes?

[plucky elk]

Yea you can use n^3 + 2 > n^3 and then integral test

[rigid acorn]

im not following sorry

[plucky elk]

This tells you the sequence to compare to

[rigid acorn]

oh so i compare it to 1/n^3?

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[wheat temple]

heloooo how do I factorise this?

[wheat temple]

<@&286206848099549185>

[dense harness]

yea?

[wheat temple]

how do I factorise this?

[dense harness]

factorize what?

[dense harness]

this?

[wheat temple]

yupp

[dense harness]

hmm

[dense harness]

i did this before with my friend

I cannot remember it

how to do it

sorry

[wheat temple]

it ok ill tag helpers again

<@&286206848099549185>

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[last wren]

I'm not sure there's a very satisfying answer to this

I don't think there is

oh wait you can do a little

start by factoring out an x^2

[wheat temple]

(x�1)(x+1)(y�1)(y+1) is the answer

yup

[last wren]

so you know how to do it?

[wheat temple]

nope

i just wanna know how to get the answer

[wheat temple]

and i thought if you saw the answer it might help u remeber how to get there

[last wren]

factor those separately

to start

[wheat temple]

im not gonna do ^ jsut pretend i do

x2(y2 -1) - 1(y2 -1)

[last wren]

sure

okay that's the first, now what about the second part? how do you factor -y^2+1

oh nice

see where to go now?

[wheat temple]

(y^2 - 1)(x^2 - 1)

?

[last wren]

yep

[wheat temple]

hm that does not look like the answer though

[last wren]

well you're not done yet

use this: $a^2-b^2 = (a+b)(a-b)$

[flat frigate]

doaby

[wheat temple]

so you think square the 1 s

[last wren]

yes, think of x^2-1 as x^2-1^2

[wheat temple]

ah okay

(y -1)(y + 1)(x - 1)(x + 1)

[last wren]

:D

[wheat temple]

ahhh

thank you

[last wren]

any time

[wheat temple]

:DDD

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[pearl pollen]

Can someone guide me through the next step?

[icy lance]

factorise

[pearl pollen]

Ohh

-9 and -3

[icy lance]

seems so
(m-9)(m-3)=0, m=9 or m=3

[pearl pollen]

mathway says 1,2

[icy lance]

then just undo your subsitution and solve

[pearl pollen]

I'm confused l

[icy lance]

well, yeah

[patent flame]

That’s because 3^x = 9 or 3

[pearl pollen]

it's been simplified right

[icy lance]

m=9 or m=3 means 3^x=9 or 3^x=3

[pearl pollen]

oh yeah I get it now

About here, -3 multiplied by 9 is -27, not positive 27, what to do?

Oh nevermind, -3 × -9 will

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[median flare]

I dont underatand the method used in this worked example

[median flare]

I mean i do

They solved for x in a normal graph for 0-360 and then the other uses 2 as the period and its domain is 0-180 so i get that

But i feel this shouldn’t always work

Because if u had an equation like sin(2A-pi/4)=0.5 for A is 0-180 and u solve with the same method for A you’ll get -7.5 and 52.5

I checked on demos the answer is 52.5 and 172 something

I just wanted to know as long as its asinbx with this method its ok but if its a sin(bx+c) where c is an integer then this may not always work am i right?

[stoic dune]

,w solve sin(2x - π/4) = 0.5

,calc (0-11/24)180

[flat frigate]

Result:

-82.5

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kaynex

[median flare]

These answers r in degrees

[stoic dune]

Recheck your work on desmos

[median flare]

Hey im talking about the method

[stoic dune]

I'm not sure why you think the method wouldn't work?

You just said "desmos doesn't agree". But I did check using that method and got Wolfram's answer

[median flare]

So i used sin graph 0-360

And my equation is

sin(2A+45)=0.5

For 0-180

So at 30 and 150 there’re intersections

[stoic dune]

Alright, so this time it's +, we can change our work a little

[median flare]

So ill take those vals and equate to sin(2A+45)=30 and 150

2 answers

-7.5

For 30

And 52.5

[stoic dune]

You mean
2A + 45 = 30

[median flare]

For 150

[median flare]

Ye

[stoic dune]

,w solve sin(2x + π/4) = 0.5

[flat frigate]

Results provided by WolframAlpha

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Requested by kaynex

[median flare]

The answers im supposed to get are

53.5 and 172.5

52.5*

This is using the method from the example

[stoic dune]

Ah I see. You're worried about the domain

Note that
172.5 = 180 - 7.5

[median flare]

Ig u could like check minimum and maximum val ull get inside the sin of the equation sin(2A+45)

And then use like that domain for sin x graph

[stoic dune]

More generally though, when does sin(x) = 0.5?

You're correct it's at 30 and 150, but it's also at 390, 510...

[median flare]

Ye u add the period and get answer within domain

[median flare]

Ye

[stoic dune]

Using 390 here should give that answer directly

[median flare]

Does this way work?

[stoic dune]

I'm not sure I understand. Minimum and maximum of 2A + 45?

[median flare]

Oh

Wait lemme explain

So u got sin(2A+45)=0.5

For 0-180

So input 0 and 180 oncr in 2A+45

So ull know what values are coming inside the sin function of the equation

So ig use the sin x graph for the values that are coming inside the sine function yes?

So fot eg

2(0)+45

=45

And 2(180)+45

=305

So yk the values that are coming inside sin function of the equation

Use that as fomain for sin x graph

Check intersections fot x like in the example and then solve for A

That correct?

[stoic dune]

I'm sorry, I'm really not following. I suggest trying it on something a bit more exotic, like sin(20x + 3π/2) = 0.5

[median flare]

Ok wait

Try what i just said for the wuestion

Feel like u should get right andwrr

Take

2A+45

[stoic dune]

Note the more "common" method is to solve for all solutions, then fix for the ones that fit in the domain.

[median flare]

Replace A with 0 and 180(as thats the domain for the equation i said before)

[stoic dune]

You might be having trouble with this, because it's difficult to do in degrees.

[median flare]

The andsers u get should be the domain for sinx graph

Not should i mean use it as the fomain for sinx

Chrck intersections with 0.5

And then solve for A

Using intersection values

@stoic dune do u get it?

[safe radish]

@median flare Has your question been resolved?

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@median flare Has your question been resolved?

[median flare]

<@&286206848099549185>

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@median flare Has your question been resolved?

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[flat frigate]

random

[hot thistle]

how does pugh define the partial on a metric space

how is division defined in the metric space

my point is that it seems odd to bring out a partial on an arbitrary metric space

what makes you say that you should use thm 14?

im inclined to say it isn't, and i would suggest trying something with sequences

take a sequence yn -> y in Y
show that F(yn) -> F(y)

[safe radish]

@plain owl Has your question been resolved?

[hot thistle]

you'll have that
\begin{align*}
|F(y_n) - F(y)| &= \left |\int_a^bf(x,y_n) - f(x,y), \dd x\right |\
& \le \int_a^b|f(x,y_n) - f(x,y)|,\dd x
\end{align*}

[flat frigate]

maximo

[hot thistle]

@plain owl do you see how i got to this

now for any eps > 0

by continuity of f

there exists a delta > 0

so that |f(x,y_n) - f(x,y)| < eps, provided |y_n - y| < delta

so for the proof of continuity, let (\varepsilon > 0) and define (\delta) to be such that
[|y_n - y| < \delta \implies |f(x,y_n) - f(x,y)| < \frac{\varepsilon}{b-a}]
which exists by the continuity of (f)

[flat frigate]

maximo

[hot thistle]

well that's basically the whole proof

so which part do you not get

there aren't really multiple variables here

since F : Y -> R

also for some reason i am mixing sequences and epsilon delta and it is very poor notation

sorry about that

a better way to write all this is by fixing y_0 and instead of y_n using y

e.g. [|y - y_0| < \delta \implies |f(x,y) - f(x,y_0)| < \frac{\varepsilon}{b-a}]

[flat frigate]

maximo

[hot thistle]

but that's beside the point

[hot thistle]

once you have picked the delta, you can continue the inequality here

which gives you the integral of eps/(b-a) from a to b, which is just eps

yeah once you note that F goes from Y to R it is a bit more straightforward

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[safe radish]

@white oak Has your question been resolved?

[white oak]

<@&286206848099549185>

?

[thin bridge]

!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

[white oak]

<@&286206848099549185>

[white oak]

2

[thin bridge]

if you have work show it,
don't wait for someone to ask

and don't spam helper pings

[white oak]

ur mom

also sending working

[thin bridge]

ty for giving me a reason to bail

[white oak]

aw man

aight then ur choice

imma still send the work

<@&286206848099549185> anyone??

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@white oak Has your question been resolved?

[white oak]

.close

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[naive lily]

ok

so see

i understand ur concern

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[stiff tinsel]

could someone help me with example 42 and example 40 ii?

[hallow shuttle]

help in the last exercice 3

[edgy ore]

completely not visible

[hallow shuttle]

<@&286206848099549185>

[edgy ore]

Its occupied by someone else, you should check out #❓how-to-get-help

[frozen marlin]

!occupied

[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).

[frozen marlin]

and also

!15m

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[edgy ore]

i mean they don't own this channel so why not

[frozen marlin]

shrugs

[stiff tinsel]

could you help me with example 42\

example 40 ii

and this?

[stiff tinsel]

<@&286206848099549185> Hey guys, could you help me with this question?

I'll have to go to sleep as it is really late for me and I have the maths test in the morning.

If you can, would you be able to type up answers to all the questions so I could skim through them before my maths test?

[unique salmon]

@stiff tinsel

Close it

Before u go

[stiff tinsel]

will I still be able to get maths help if I close it?

[unique salmon]

i mean yeah

u open a new channel

i guess

[stiff tinsel]

wdym?

How do I do that?

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@stiff tinsel Has your question been resolved?

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[weary osprey]

I need find an isomorphic between $\mathbb{Z}_{p-1}$ to $U(p)$

[flat frigate]

Dubs

I need find an isomorphic between $\mathbb{Z}_{p-1}$ to $U(p)$

[weary osprey]

My idea is to find generator ‘a’ and map phi(1)=a

the do the following computations phi(1+1)= phi(1)* phi(1)

phi(1+1+1)=(phi(1))^3

will this work everytime?

[peak estuary]

yes. all cyclic groups with the same number of elements are isomorphic

[weary osprey]

so by just getting one generator of the other cyclic group, i can establish an isomorphism right?

[peak estuary]

yes

[weary osprey]

can I count how many of such possible?

I feel like when I establish one choices of other generators are reduced

i mean the number of isomorphism between Z_{p-1} and U(p)

[safe radish]

@weary osprey Has your question been resolved?

[peak estuary]

well as soon as you pick where you send 1 to, your isomorphism is fixed

so you just need to count generators

[weary osprey]

Makes sense

[safe radish]

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[marsh ibex]

Hi

[marsh ibex]

In ABCD square , we draw a line from A that faces DC in M .
And we draw the bisector of BAM and it faces BC at K .
Prove that
AM= BK+DM

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[marsh ibex]

<@&286206848099549185>

[meager dagger]

Can you give picture?

@marsh ibex

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[marsh ibex]

I can't send pictures in discord . The problem doesn't have a picture though

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[fierce cairn]

hi

[fierce cairn]

16

.close

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[pliant gorge]

How do u solve this

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[mellow gulch]

@pliant gorge It looks like your answer is right

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[brazen pecan]

Hi!

[brazen pecan]

I am having trouble with this assignment right here:

I think I have to start by putting the given data of each university that is provided out it into excel or other resources and graph it. From there I think I will get a general trend if which uni is going up or down. From there I’m not really sure what I’m supposed to do. Help would be really appreciated!

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<@&286206848099549185>

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[safe radish]

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[uncut blaze]

Part (b)

Is what I am struggling with

I was able to solve it using the disk method but I am not sure what the setup would be for (b), I know this is the reference formula

But the x = ln(4) is tripping me up

[safe radish]

@uncut blaze Has your question been resolved?

[uncut blaze]

<@&286206848099549185>

[solemn hearth]

change your bounds

also having e^-x makes it easier

[uncut blaze]

ok so like

from 0-1?

[solemn hearth]

@chilly pasture wrong channel

[chilly pasture]

sorry

[uncut blaze]

I have this graph on desmos to represent it

so if bounds are flipped we go 0 to 1 right?

[solemn hearth]

yeah you can

[uncut blaze]

how do we deal with the x = ln(4)

Because this is it without the x = ln(4) I think, but I don't know how it interacts with this

[solemn hearth]

you can integrate over y

rather than x

y goes from 0 to e^-x

x goes from 0 to ln(4)

[uncut blaze]

the problem is the question wants an x-axis rotation

So 0 to ln(4) would work you're right but for y-axis rotation right?

[solemn hearth]

oh yeah i misread that

also you dont need to evaluate the integral

[uncut blaze]

yeah thank god

[solemn hearth]

your horizontal component is just ln(4)-e^-x

the x = ln(4) is just a bound

[uncut blaze]

so this?

if it's written in terms of x

because of horizontal rotation

could you explain why it's ln(4) - e^-x and not e^-x - ln(4)

[solemn hearth]

what is $e^{-ln(4)}$

[uncut blaze]

oh my bad, I mean

[flat frigate]

pacmanboss256

[uncut blaze]

then - ln(4)

so like this

[solemn hearth]

oh yeah youre coming from the right side

[solemn hearth]

thats why

[uncut blaze]

coming from the right side?

what do you mean exactly

[solemn hearth]

ok so your y value is determined by where your x value is

[uncut blaze]

ok so is that supposed to mean that by doing ln(4) - e^-x we chop off anything after x=ln(4) and only get the integral for the region between 0 to ln(4)

[solemn hearth]

yes

its like flipping the integral upside down

and then adding the extra rectangle

[uncut blaze]

so x=ln(4) is our top function

in that case?

I should imagine it like this?

[solemn hearth]

let me make a graph real quick

[uncut blaze]

If I understood right this is it

for the final thing

[solemn hearth]

youre rotating around the x axis arent you

[uncut blaze]

yeah

[solemn hearth]

your graph was sideways

[uncut blaze]

ok so I see your graph here

and since its a horizontal rotation our bounds are 0 to 1

and since our x bound is ln(4) to the right, we do that ln(4) - whatever our function is written in terms of x?

[solemn hearth]

yes

[solemn hearth]

this isnt correct i dont think

[uncut blaze]

https://cdn.discordapp.com/attachments/903481106819612714/1206602799761195048/image.png?ex=65dc9b6d&is=65ca266d&hm=1731cfdd4dc4eec882fc5277742421df37ec492ff5477d80b7b243ba7897d887&

It seems to match this though right?

in terms of x*

[solemn hearth]

let me try something real quick

[uncut blaze]

ok

[solemn hearth]

i might be misunderstanding it

btw your first answer is correct

ok im lost

i keep getting the bounds confused

sorry

<@&286206848099549185> can someone else help I have to go

[uncut blaze]

First answer as in the very first one I posted? Or the one with ln(4) - (-ln(y))

[solemn hearth]

the first problem

it is 15pi/32

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[uncut blaze]

.close

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[timber violet]

how can i count the terms with multinomial theorem

[timber violet]

.close

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[gilded scroll]

why is that half positive

jesus christ i need help

.close

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[elfin zealot]

How to set this up?

$V=2\pi \int_c^d y g(y) dy$

[flat frigate]

🌸 Katsune

[lean otter]

if its about the y axis using the shell method it's going to be in terms of x

so [
V = 2\pi\int_a^b x\m fx \dd x
]

[lean otter]

you already know f(x)

so your job now is figuring out the bounds a and b

[elfin zealot]

0 to 2

[lean otter]

yea

that's it

just integrate

[elfin zealot]

$V=2\pi \int_0^2 x(x^3)dx$

[flat frigate]

🌸 Katsune

[lean otter]

yes

[elfin zealot]

no, this is wrong

I get 2pi * x^5 / 5 |_0^2

Thus 2pi * 25/5 = 10pi

Answer is somehow 40.21

[lean otter]

2^5 isnt 25 tho

[elfin zealot]

kill me

[lean otter]

lmao

[lyric wraith]

do horizontal slicing instead

try that

[lean otter]

the answer is correct

she just miscalculated

[elfin zealot]

Prof wants me to do shell method

[lyric wraith]

yeah horizontal slicingw ith shell method

[elfin zealot]

Tf is x-axis

x minus axis

[lean otter]

just y = 0

[elfin zealot]

Oh yeh

[lyric wraith]

x-axis

[lean otter]

anyways thats like exactly the same as ur original question

just copy paste the same procedure

[elfin zealot]

I did it again and it failed jfc

$V=2\pi \int_0^4 x(\sqrt{x})^2 dx$

[flat frigate]

🌸 Katsune

[lean otter]

wait

why are u squaring it lmao

[elfin zealot]

Oh uh

Idk :u ok

[lean otter]

yea its just sqrt(x)*x

[elfin zealot]

okay I got it all the way

[lean otter]

like, correctly?

[elfin zealot]

Yes :D

[lean otter]

ayee

[lean otter]

oh boy lmao

[elfin zealot]

I'm thinking I should shift this to the left 6 and use shell method

[lean otter]

yeah pick your poison I guess. Give it a try and show me your work after

[elfin zealot]

No

This is better to use washer method ?

No it's shells method since it's a vertical rotation with a hole

[lean otter]

I'd personally use disk

[elfin zealot]

Christ why

Disk is for horizontal rotations

That don't have holes

And this is the exact opposite

[lean otter]

i mean you can make it work it's not like there is a hard rule for it

[elfin zealot]

☠️ Why would you do disk when disk is recommended for horizontal rotations without holes, and this question is a vertical with a hole, where shell is recommended

oml i can't do what we did earlier with shifting it to the left

[lean otter]

you can always do [
\pi\int_a^b f^2(x) - g^2(x) \dd x
]
to account for the hole

[elfin zealot]

this is washer method

[lean otter]

yeah its just a more general disk method. They are the same with washer just having a hole

[lean otter]

show mayb

[elfin zealot]

No, I didn't work it, i realized it

[lean otter]

oh oki

[elfin zealot]

This is for rotations about the x axis ye? But for a rotation on x=6, I can do

$\pi \int_c^d f^2(y) - g^2(y) dy$

Ye?

[lean otter]

with a minus in the middle yeah

[flat frigate]

🌸 Katsune

[elfin zealot]

Oh no, I'm getting something weird :P

$V=\pi \int_0^3 x^2 - (x-3)^2 dx$

[flat frigate]

🌸 Katsune

[elfin zealot]

How to account for the rotation at x=6

also are the limits right >w<

[lean otter]

hmmm

i think you need to like make the (x-3)^2 be first minus x^2

oh wait

x = 6 is like a shifted y axis rotation yk? @elfin zealot

so your integration variable needs to be dy

[elfin zealot]

Okay

[lean otter]

i mean it doesn't matter because you have y = x but just so u dont like confuse yourself and all

[elfin zealot]

$V=\pi \int_0^3 x^2 - (x-3)^2 dy$

[flat frigate]

🌸 Katsune

[elfin zealot]

:u

How would you set this up using washer/disk

[lean otter]

i think its probably a good idea for you to graph the region. If you want to learn for your intuition. Or we can do it the faster route if u want

like

graph it first i suggest and we can go from there

[elfin zealot]

https://cdn.discordapp.com/attachments/1121986300774465628/1206673686967615528/IMG_3885.jpg?ex=65dcdd71&is=65ca6871&hm=94a436f7b505c19e87d4c96f1f1b827150233fbde50abf88d8f8c0268ec11192&

FINALLY I managed to do this (problem 7.3.3 is what I was just doing)

[lean otter]

oh AYEEE good job you did it on your own

[elfin zealot]

I think I want to change majors to housewife

I want to cry

[lean otter]

[elfin zealot]

LOL

[lean otter]

head empty return to monke

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[lone arch]

#1195800467880558672 message

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[urban vale]

so ik there are 4 quadrants. What quadrant is line c in?

[hardy lion]

its on the -y axis

no quadrant

[urban vale]

ok

so I would just say its not in any quadrant?

[lean otter]

yes

[hardy lion]

you would say its on the -y axis

[urban vale]

alright, thanks

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[dense tiger]

can someone help with this stats proof

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[inner tulip]

You can't take the inverse of A right?

[last wren]

not really no

[inner tulip]

so saying theyre undefined is correct right?

[last wren]

I'd say so, as long as you haven't learned about pseudo inverses lol

[inner tulip]

we have not 0_0

[last wren]

then you're right :D

[inner tulip]

thank you! I just wanted to make sure that was right

[last wren]

🫡

[inner tulip]

.close

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[full yarrow]

on pc you cant tell but there are 3 images attached here :3

[wheat locust]

in part a

you have for all x in R

is that what you meant?

also the delta looks like epsilon/3 and i think you meant epsilon/m

[full yarrow]

oops yes

[full yarrow]

idk how necessary it is i was kinda just mimicking the format from stuff online

does my work look fine besides the epsilon/3?

[wheat locust]

it’s not right

[full yarrow]

will remove 🫡

[wheat locust]

look back at the definition

also what is happening with the letters in the definition you wrote on the top right

it’s all mixed up

[full yarrow]

i just had that there in case i needed it ig i wrote it wrong lol

[wheat locust]

i wasn't saying just remove the for all x in R

[full yarrow]

im aware but i was unsure what to replace it with (like for all espilon >0 or such)

so i just opted to go without

[wheat locust]

find a correct definition of limit

[full yarrow]

will do but is the actual math/logic correct?

[wheat locust]

there is some logic missing

[full yarrow]

from which one? what is missing

[wheat locust]

i was going to explain after you looked at the definition of a limit

[full yarrow]

is it not for all epsilon>0, there exists a delta>0 such that if 0<|x-a|<delta then |f(x)-L| < epsilon?

[safe radish]

@full yarrow Has your question been resolved?

[wheat locust]

if 0<|x-a|<delta is the important part there

[full yarrow]

okay nice

so how does that tie in to the logic im missing

[wheat locust]

lol sorry i keep forgetting about this, ping me if you don't reply immediately

this should hold for all x with 0 < |x-a| < delta

which should be written

[full yarrow]

do you mean i should write that it holds for all x with 0 < |x-a| < delta? or just incorporate it into my work

[wheat locust]

you should write for all x in R with 0 < |x-a| < delta or something

[full yarrow]

ah okay that makes sense

no alterations besides that though?

[wheat locust]

yes it is ok otherwise

[full yarrow]

many thanks 🙏

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[elfin zealot]

How do I square this?

[elfin zealot]

$(\frac{x^4-1}{2x^2})^2$

[flat frigate]

🌸 Katsune

[elfin zealot]

Like how do I solve this

[carmine glade]

the equation becomes

(x^4-1)^2 / (4x^4)

[rustic goblet]

you square the numerator and denominator seperately

[carmine glade]

(x^4-1)^2 = (x^8-2x^4+1)

[elfin zealot]

$\frac{(x^4-1)(x^4-1)}{(2x^2)(2x^2)}$

[flat frigate]

🌸 Katsune

[elfin zealot]

$\frac{x^{8}-2x^4+1}{4x^4}$

[carmine glade]

Well

[rustic goblet]

this looks a bit wrong

[elfin zealot]

In what way

[carmine glade]

(x^4)^2 is x^8

[rustic goblet]

the middle term of the numerator

[elfin zealot]

Oh ye

[rustic goblet]

also this

[flat frigate]

🌸 Katsune

[elfin zealot]

Do you guys do calculus arclengths in the calculator or do you do them by hand

I'm looking at this

$\int_1^3 \sqrt{1+{\frac{x^8-2x^4+1}{4x^4}}}dx$

omfg

what is the katex for sqrt

[flat frigate]

🌸 Katsune

[elfin zealot]

This is the arclength integral for a curve y where (y')^2 was just calculated

How the HELL do you do this without the calculator, or should I just do this in calculator

I see what you're saying but I am allowed to use a graphing calc during the test and this integral seems to be extreme unless I'm forgetting some nifty shortcut

[rustic goblet]

you can do this with a sum i think?

[elfin zealot]

Is this just integral sqrt{u} deals? I can do like

2u^{3/2}/3

and put all that shid under the root as the u

Would that work or nah

[rustic goblet]

er, what is u here?

[elfin zealot]

Turn this: https://cdn.discordapp.com/attachments/903481106819612714/1206801594881019904/195837132370214912.png?ex=65dd5491&is=65cadf91&hm=769a29d10dd7c8d5843b6795a425771ebb0b330e394de44c266567e8d0c3f6af&

Into this: \int_1^3 sqrt{u}dx

where u is just all of that in the root

then simply integrate u^{1/2}

Then replace it when integrated

[rustic goblet]

well, you'd have to change the differential then

to du

and that's gonna be really hard

if not impossible here

[elfin zealot]

??? Hard how

Forget the small stuff

du/dx w/e

Will what I'm askign for work? or is that not a shortcut

[rustic goblet]

^^

u sub looks a bit futile here

perhaps theres a clever one?

[elfin zealot]

Ok ye so forget integrating then just use the graphing calc

New question... how do you square this

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

[flat frigate]

🌸 Katsune

[rustic goblet]

square each factor

[elfin zealot]

x^2 * (x^2 + 2)

[rustic goblet]

yeah

[elfin zealot]

Curious but how do you integrate a multifactor function again? so like

$\int_0^1 x\sqrt{x^2+2}dx$

[flat frigate]

🌸 Katsune

[rustic goblet]

probably by parts?

[elfin zealot]

Can you remind me of that rq

tfw forgot integration by parts

I've been too busy doing washer/shell bs

[rustic goblet]

because i cant latex

[elfin zealot]

Holy mother I've completely forgot about that

I'm gonna have to relearn that completely

[rustic goblet]

yeah u sub probably works too

why did i say probably

it does work

[elfin zealot]

UMMMMMMMMMMMM WTF?

Why did nobody tell me you could do that???

??? Uhhh is this right??

How did that x left of the interior square root just vanish

This is from https://www.emathhelp.net/en/calculators/calculus-2/arc-length-calculus-calculator/?ft=x&f=(1%2F3)(x^2%2B2)^(3%2F2)&a=0&b=3

And https://www.emathhelp.net/en/calculators/calculus-2/definite-integral-calculator/?f=(x^4+%2B+1)%2F(2*x^2)&var=x&a=1&b=3

[rustic goblet]

i feel like some trig sub just happened here

but i cant be sure

[elfin zealot]

Is sqrt{4 + 9} = 2 + 3

No wait

LOL

How did they do this then

sqrt{A + B^2} = B where A is 1

https://cdn.discordapp.com/attachments/903481106819612714/1206804892719775785/image.png?ex=65dd57a3&is=65cae2a3&hm=496a0b0131687f85cb1a488204cb9121a48804af73e82a90143d61ba7c05f3c3&

[rustic goblet]

it has to be some trig sub

[elfin zealot]

How can they just remove the positive one like it never existed

[rustic goblet]

my guess is tan(theta) = everything under the square

[elfin zealot]

Oh wait a sec

[rustic goblet]

because then we have sec^2(theta) under the root which is nice

[elfin zealot]

OOF I just noticed that in the left side it's x^4 - 1, on the right side it's x^4 + 1

I will assume it's advanced substitution method I do not know

I do not know any trig sub

How do I do this?

Surface area formula:

$2\pi \int_a^b r(x) \sqrt{1+f'(x)^2} dx$

[flat frigate]

🌸 Katsune

[elfin zealot]

Do I solve the function for X and use that for the r(y) and f'(y) ?

It kind of looks like from what I'm saying on emathways that

doing a y-axis version is just replacing r(x) with x

and that's literally it

Why is this the surface area formula for rotating a solid along the x-axis, but when it comes to the y-axis rotations then r(x) is replaced with x?

[elfin zealot]

Meaaow, hello Joanna. Soo it is true that y-axis rotations use x, while x-axis rotations use the original function, for surface area of a revolved curve?

[drowsy karma]

precisely, use this formula, yoru integral is easy,, no need to spend too much time on it 🙂 smiles

[elfin zealot]

Okay. Thank you also to ING, Scarecrow, and bethest for helping earlier too :3

.close

[safe radish]

Channel closed

Closed by @elfin zealot

Use .reopen if this was a mistake.

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[little glacier]

what is the equation from the points (0,0) (1,1) (3,2) (6,3) and so on

[little glacier]

i cannot see the relationship

in human terms i see it as the incrament of y increases every "step"

but i dont see how it could be related to X

(10,4) (15,5) (21,6) (28,7)

cause im just trying to graph it on desmos

[marsh walrus]

whhat kind of formula are you expecting

you can resolve 4 unknowns

or less if you get lucky

[little glacier]

what is an unknown

[marsh walrus]

i mean what kind of fit do you want here

exponential?

[little glacier]

somewhat

i could maybe have it as a function?

[marsh walrus]

its up to you what kind of function you want, yea you can get a function

[little glacier]

the most accurate one

[marsh walrus]

i think youre over determined for exponential equation

so youll need to do regression

[little glacier]

oh and theres another one i need to graph

like recursion?

im trying to explain roblox simulators to some people

the problem this stems from is

[marsh walrus]

regression means like ... find the equation that gets the closest to going through each point

for when you cant get an equation that doesnt go through every point, or don't want to

[little glacier]

a machine produces 1$ per second, m2 produces 1 machine per second, m3 produces 1 m2 per second

at what time does m2 surpass a different m2 where it gives a flat 10$ per second?

and m3?

[marsh walrus]

you just came up with this question?

its not from a textbook or something

[little glacier]

no

[marsh walrus]

im not certain this is exponential growth

id have to think

[little glacier]

it would be like

[marsh walrus]

i think its actually more like $n^n$

[flat frigate]

jan Niku

[little glacier]

cause im just thinking of what i learned from mathisfun

which is acceleration

[marsh walrus]

lemme see

[marsh walrus]

im not sure i understand this question

[little glacier]

ill rephrase it them i guess

[marsh walrus]

do you start with 1 of each of those machines?

ive never played roblox so

[little glacier]

you start with 0 of each

[marsh walrus]

hi Namington

[little glacier]

when it refers to having a machinez you have 1

[warped roost]

so i'm assuming they're asking

[marsh walrus]

but then what creates the m2

[warped roost]

given two polynomials tx₁(t + x₂(t + x₃)), 10tx₄, for what domains of integration and what values of x₁, x₂, x₃, x₄ does the right hand polynomial exceed the left hand polynomial?

[marsh walrus]

[warped roost]

i.e. a polynomial idle game problem

[marsh walrus]

oh, but this is the composition i was meaning

[little glacier]

Machine A produces 1$ per second.
Machine B produces 1 Machine A per second.
Machine C produces 1 Machine B per second.

Machine B¹ produces 10$ per second.
Machine C¹ produces 100$ per second.

Starting out with one Machine B, compared to having Machine B¹ replace Machine B, after how much time needs to elapse before Machine B outproduces Machine B?

and then that scenario but add Machine C and C¹

[marsh walrus]

i guess n^n doesnt make sense

[warped roost]

easiest way to compute this is via just programming it fwiw

rather than trying to express everything in formulas

[marsh walrus]

you start with nothing but one machine B. why does it matter what machine C produces?

[warped roost]

for the second part of their question

it seems clear enough to me

[marsh walrus]

oh

[warped roost]

anyway ill write a quick python script

[marsh walrus]

sorry they edited

[warped roost]

you could do this formulaically but that takes effort

[marsh walrus]

i was going to draw

[little glacier]

i do not understand 70% of these words

[marsh walrus]

i mean the first problem is obvious

its 10 seconds

well, 11, i guess

[little glacier]

oh

[marsh walrus]

the second one is more fun

[little glacier]

thats not the problem im asking

uhh

how much time until the total money produced from machine B exceeds machine B¹

not the current money

[marsh walrus]

but machine B doesnt produce money

[marsh walrus]

right, i think i get what you mean

[little glacier]

but it produces machines that make money

[marsh walrus]

I mean B' makes 10 dollars a second

so you really are looking at total production

how long until you have more than 10 A's

in the first scenario its 11 seconds, in the second its 6

assuming im understanding you which IDK i can be pretty dense

[little glacier]

let me just first bruteforce

[warped roost]

machine_a_count = 0
machine_b_count = 1
machine_c_count = 0
static_production = 10

totalpolynomial = 0
totalstatic = 0

for time in range(0, 10000):
    totalpolynomial = machine_a_count
    totalstatic = static_production
    if (totalpolynomial >= totalstatic):
        print(time)
        break

    machine_a_count += machine_b_count
    machine_b_count += machine_c_count

prints 10

[marsh walrus]

10 to equal

[warped roost]

machine_a_count = 0
machine_b_count = 0
machine_c_count = 1
static_production = 100

totalpolynomial = 0
totalstatic = 0

for time in range(0, 10000):
    totalpolynomial = machine_a_count
    totalstatic = static_production
    if (totalpolynomial >= totalstatic):
        print(time)
        break

    machine_a_count += machine_b_count
    machine_b_count += machine_c_count

prints 15

[marsh walrus]

👀

[warped roost]

so buying machine B breaks even relative to B' after 10 time steps

and machine C breaks even to C' after 15 time steps (seconds)

[marsh walrus]

Oh, i didnt see the new value

shown up by a pinkname

i guess it was fate

i did ask for help

now show the formulaic way

[warped roost]

this assumes that the amount of machines increases in the order i assumed

[marsh walrus]

i guess i can google

[warped roost]

for example, i assumed that, if you have 2 machine Cs at time step 0, then you'll have 2 machine Cs and 2 machine Bs at time step 1 (i.e. the machine Bs haven't "had a chance" to produce any machine As yet)

and have made $0

if the order is different then you can reorder some lines

the formulaic way is setting up a polynomial inequality and solving it

which isnt much harder but more annoying

[marsh walrus]

hmm

okay

[little glacier]

i have no clue what im doing when i bruteforce

[warped roost]

oh whoops

just realized i made a mistake

[little glacier]

im just doing 0+1+3+6+10+15+21+28 and so on

until the sum is more than the total amount of terms in the sequence, minus one, times 10

[warped roost]

correct values are 20 and 25

not 10 and 15

my bad

i did = where i shouldve +='d

[little glacier]

let me uhhh

[warped roost]

this is the polynomial-based approach yes

you are computing the minimum value of $k$ such that [\sum_{n=1}^{k} \frac{n(n+1)}{2} \geq 10k]

[flat frigate]

Namington

[warped roost]

this rearranges to $\sum_{n=1}^{k} n^2 + \sum_{n=1}^{k} n \geq 20k$

[flat frigate]

Namington

[warped roost]

the LHS obeys known formulae

namely $\frac{k(k+1)(2k+1)}{6} + \frac{k(k+1)}{2} \geq 20k$

[flat frigate]

Namington

[warped roost]

this can then be solved algebraically but its a pain in the ass

[little glacier]

B B¹
(0,0) (0,0)
(1,1) (10,1)
(3,2) (20,2)
(6,3) (30,3)
(10,4) (40,4)
(15,5) (50,5)
(21,6) (60,6)
(28,7) (70,7)
(36,8) (80,8)
(45,9) (90,9)
(55,10) (100,10)
(66,11) (110,11)
(78,12) (120,12)
(91,13) (130,13)
(105,14) (140,14)
(120,15) (150,15)
(136,16) (160,16)
(153,17) (170,17)
(171,18) (180,18)
(190,19) (190,19)
(210,20) (200,20)

[warped roost]

yes

[warped roost]

,

[little glacier]

yes

[warped roost]

so buying B breaks even relative to B' after 20 seconds

and buying C breaks even relative to C' after 25

[little glacier]

also uh heres a response from another server i asked this in

[warped roost]

im not sure what theyre saying "works", but this is a polynomial relationship yes

[little glacier]

im going to be honest

i have no clue how the sum function works

uhhh

im going to read this entire conversation again

[warped roost]

ah i screwed up the indexes in my math

its still the correct idea but there should be a lot of -1s