help-42

[calm coral]

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[mortal fern]

I am confused about what counts as a distinct zero of a polynomial.
Let's take f(x) = (x+1)^2 as an example. This has a single zero as far as I know. i.e. -1 but apparently that's not the case? From what I'm getting it has two zeros, -1 and -1.

Sum of zeros of f(x) is α + β = -b/a will not hold true if -1 is seen as a single zero. But if you but f(x) on a graph then it clearly has a single zero of -1. So which one is it? One zero or two?

[tall moon]

one

-1 and -1 is a single root

[mortal fern]

then α + β = -b/a will not hold true since -1 != 2

[swift laurel]

there is a single root of multiplicity 2

some formulas involving roots count multiplicity

[mortal fern]

multiplicity 2? my book never mentented such thing. is that an actaul comcept?

[swift laurel]

the multiplicity of a root is the power that the corresponding linear factor has in factored form

so for example in the case of (x-1)(x+2)³(x-4)² you have x = 1 with multiplicity 1, x = -2 with multiplicity 3, and x = 4 with multiplicity 4

[mortal fern]

https://en.wikipedia.org/wiki/Multiplicity_(mathematics)

it's this thing right?

[swift laurel]

under "multiplicity of a root of a polynomial", yes

[mortal fern]

looks bit more complicated on wikipedia

[mortal fern]

ok thanks

.close

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[calm coral]

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[last loom]

Hi

[calm coral]

Please don't occupy multiple help channels.

[last loom]

Am not understanding where to ask my aur

Question

[mental tendon]

#help-45

[last loom]

Tq

[swift dragon]

.close

[calm coral]

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[tall moon]

if $a,w,x,y,z$ are complex numbers then find $(w,x,y,z)$ such that $w(x+y+z)=x(w+y+z)=y(w+x+z)=z(w+x+y)=a$

[potent lotus]

skissue.in.a.teacup

[tall moon]

bro what is this

[ancient sorrel]

who needs help

[tall moon]

everyone in the math help (occupied) section..?

[ancient sorrel]

yeah but sometimes people are already through something and don't need an extra person to get involved

[weak rapids]

the equations are symmetric in w,x,y,z

when you see symmetry like this, it often suggests that the solution might also be symmetric ie w,x,y,z might all be equal

[tall moon]

errr

hold on cant any complex number w=x=y=z hold for this what

does it want all w,x,y,z or smth or is w,x,y,z supposed to be distinct

(i cant really post it cause its in another language and has a watermark on it)

23. Diberikan bilangan kompleks 𝑎. Tentukan solusi komplek (𝑤, 𝑥, 𝑦, 𝑧) dari
    (𝑤 + 𝑥 + 𝑦)𝑧 = 𝑎
    (𝑤 + 𝑥 + 𝑧)𝑦 = 𝑎
    (𝑤 + 𝑦 + 𝑧)𝑥 = 𝑎
    (𝑥 + 𝑦 + 𝑧)𝑤 = 𝑎

heres the original, basically roughly translates to what i wrote, the wording implies theres only a single solution to (w,x,y,z) but its not true?

[alpine lynx]

I understand malay.

[tall moon]

thats indonesian

[weak rapids]

the wording implies theres only a single solution to (w,x,y,z) but its not true?
this is only true when a is not equal to 0

[alpine lynx]

They all the same.

[tall moon]

its diffrent(ish)

[weak rapids]

the problem doesn't explicitly say whether w,x,y,z must be distinct or if they can be equal. if they can be equal, the symmetrc solution w=x=y=z= whole root a/3 works. however, if they must be distinct, then:

if a \neq 0, there are no solutions where w,x,y,z are distinct

if a = 0, there are solutions where three variables are zero, and the fourth is arbitrary, but these don't satisfy the distinctness condition

[tall moon]

also how do you prove this?

[calm coral]

@tall moon Has your question been resolved?

[tall moon]

.close im gonna go to sleep

[calm coral]

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[calm coral]

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[calm coral]

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[calm coral]

@next geyser Has your question been resolved?

[calm coral]

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[tawdry sphinx]

Hey I was wondering which function should I use to find out my side s

I did it and got 48.32 with the tangent function so I don’t think that’s right

[dull wagon]

,rcw

[dull wagon]

sounds about right

refer to
soh cah toa
if you aren't sure

[tawdry sphinx]

Ok and I’m getting 151.4 for my r

The way I’m doing it is doing functions for both of my sides and then using Pythagorean for my check but it doesn’t match my check

[dull wagon]

there may be rounding errors

[tawdry sphinx]

Ok I’ll check on that thanks

[dull wagon]

(also draw your right angles properly)

[tawdry sphinx]

Letters in the wrong place?

Or the triangle itself

[dull wagon]

the triangle itself

[tawdry sphinx]

It’s supposed to be flipped?

My teacher drew it like that

[dull wagon]

i mean the right angle itself

the core component of the right triangle should at least look around 90°

[tawdry sphinx]

Gotcha

Ty

.close

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[calm coral]

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[wary aspen]

Hello! I am stuck on question 2 and i have not done this for a long long time and its not in my workbook anywhere. Could you please help?

[wary aspen]

Sorry Question 2.2*

[river helm]

u can use the prime factorization methode

[wary aspen]

Ok bu5 would you mind telling me how it is done?

But*

[river helm]

this is an example

[wary aspen]

This really helps thanks i have got a test coming up on monday and need to revise but this does really help Thanks

That is all of my questions.

I'll close the chat

Cya

.close

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[calm coral]

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[lilac spoke]

I dont understand how to do this question and I want someone to explain me how

[glossy pulsar]

Use pythagoras

Also what is your name bro?

[lilac spoke]

simon

[glossy pulsar]

what do you have against ducks?

[lilac spoke]

nothing

i just put it a while ago and stuck with it

i dont remember now

[glossy pulsar]

oh ok lol

damn must be a long time ago tne

[lilac spoke]

idk like 4 years maybe

or 3

[glossy pulsar]

well do you understand how to do this problem now?

[lilac spoke]

not that long

[glossy pulsar]

oh ok

[lilac spoke]

honestly

im not really sure why there are right angles

isnt incenter the intersection of the bisectors

of angles

[glossy pulsar]

There are right angles because it the lines connected to the incenter

[glossy pulsar]

Yes but the incenter touches all sides right?

[lilac spoke]

i thought it was just the angle bisectors

[glossy pulsar]

yea its more than that lol

its the biggest circle that can fit in the triangle

[lilac spoke]

what are the properties of those lines with right angles

[glossy pulsar]

So since the incenter is biggest circle incribed in triangle the lines from the incenter that touches to the sides of the triangle are tangent

you know circle tangent rules right?

[lilac spoke]

no

[glossy pulsar]

well if not it basically states that a line tangent to a circle will always give right angle

[lilac spoke]

alright

[glossy pulsar]

tangent is touching at one point

ye

[lilac spoke]

so the circle is going to be surrounding those tangeant lines

[glossy pulsar]

no its incribes so its inside the lines

its the biggest circle that can fit in the triangle lines

[lilac spoke]

ok

are they equidistant from the point?

[glossy pulsar]

Yes

[lilac spoke]

all the time, even with right triangles and obtuse ones?

[glossy pulsar]

yes

whatever triangle its still the same

[lilac spoke]

oh i get it now

thanks for help

[glossy pulsar]

np

[lilac spoke]

.close

[calm coral]

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[calm coral]

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[atomic python]

does my venn diagram look correct?

wait it doesnt even add to 100%

im cooked

[plain bone]

Some of the people dont like any sports

[void umbra]

Add a universal set

[atomic python]

ohh so uhh

dang i got it

thanks guys

.close

[calm coral]

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[calm coral]

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[simple musk]

\textbf{Exercise 2} Let $A = \begin{pmatrix} 2 & 1 \ 0 & -6 \ 4 & 5 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \ k & -k \ 3 & k+1 \end{pmatrix}$. Determine all values of $k \in \mathbb{R}$ for which there exist infinitely many matrices $X \in \mathbb{R}^{2 \times 2}$ such that $AX = BX$, and for each of these values of $k$, find all matrices $X \in \mathbb{R}^{2 \times 2}$ that are solutions to the equation.

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[sharp narwhal]

if X != 0 is a solution then so is lambda*X for any lambda

so if there's 1 non-zero solutions there's infinitely many

[simple musk]

wdym..?

[swift dragon]

bro we were almost there, why didnt you just solve the matrix we had?

[simple musk]

Bong please forgive me

I fell asleep

[sharp narwhal]

wdym wdym

[swift dragon]

classic southern european genes

[swift dragon]

Bong 😂

#linear-algebra message

solve this

[quiet swan]

if there exists a non-zero matrix $X$, where $AX=BX$, then you know for any $\lambda\neq 0$, $A(\lambda X)=B(\lambda X)$

[potent lotus]

qwertytrewq

[quiet swan]

so one non-zero solution produces infinitely many solutions

So it suffices to determine all $k\in \mathbb R$ for which there exists ONE non-zero matrix $X\in \mathbb R^{2\times 2}$ such that $AX=BX$

[potent lotus]

qwertytrewq

[swift dragon]

X={{0,0},{0,0}}, done

[quiet swan]

nonzero.

[swift dragon]

where does it say nonzero?

[quiet swan]

[swift dragon]

it says X in R^(2x2)

{{0,0},{0,0}} is in that

[quiet swan]

infinitely many matrix X <-> one non-zero matrix X

just reducing the problem into finding one matrix

[swift dragon]

ah wait, i see it lol

anyway, we already almost had the solution but renato fell asleep and for some reason he wants to start back at the beginning

[quiet swan]

wtf

[swift dragon]

.

[simple musk]

k = 2 is for a nontrivial kernel

[simple musk]

wdym?

[quiet swan]

read the explanation before lmk where you got stuck.

[quiet swan]

here

[simple musk]

yeah

[quiet swan]

so you agree that if there is one non-zero X, you can find infinitely many?

and that if there are infinitely many X, then one of them has to be non-zero?

[simple musk]

yeah

non trivial kernel

[quiet swan]

so just find one solution: bash it out

let $X=\begin{bmatrix} a & c\ b & d\end{bmatrix}$

[potent lotus]

qwertytrewq

[quiet swan]

you get something like $2a+b=a-b, 0a-6b=ka-kb, 4a-5b=3a+(k+1)b$

[potent lotus]

qwertytrewq

[swift dragon]

#linear-algebra message

[quiet swan]

yeah and you get something like k=2, and everything works out.

[simple musk]

,, \operatorname{Ker(A-B)} = \left\langle\begin{pmatrix} -2 & 0 \ 1 & 0 \end{pmatrix} , \begin{pmatrix} 0 & -2 \ 0 & 1 \end{pmatrix}\right\rangle

[quiet swan]

you probably meant Ker(A-B).

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[quiet swan]

but yeah you are just done. Ker(A-B) non-zero if and only if there exists infinitely many X

and you deduced Ker(A-B) is non-zero if and only if k=2

lastly the set of solutions X is precisely Ker(A-B)

so all is good

[swift dragon]

my point is, we did all this before

[quiet swan]

ik lol

[simple musk]

no need to fight bonk

[quiet swan]

not fighting, but do you get the solution?

[swift dragon]

im just disappointed, thats all

[simple musk]

I fell asleep bro

[swift dragon]

does that mean you forgot the whole derivation?

[simple musk]

lowjey

[swift dragon]

we were like 1-2 steps away

[quiet swan]

it just makes Bonk more disappointed

[swift dragon]

and that

[calm coral]

@simple musk Has your question been resolved?

[swift dragon]

did oyu fall asleep again?

[simple musk]

I think its solved

[calm coral]

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[calm coral]

Available help channel!

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[cosmic jetty]

$\int\frac{1}{25x^2-20x+53}dx$

[potent lotus]

BlackWizard

[cosmic jetty]

im not sure how to start this

[drifting seal]

square

[bitter maple]

oh

[drifting seal]

complete

[radiant sluice]

complete the square

it'll be in some form of 1/(x^2+a^2)

[drifting seal]

\dd

[cosmic jetty]

this isnt an integration by parts?

[drifting seal]

whet

[radiant sluice]

you can't make factors of denominator

[cosmic jetty]

yeah

[radiant sluice]

thats why you gotta do what knief said complete the square

[drifting seal]

shoutout yoobie

[cosmic jetty]

im not sure how to do that so i guess i should look up a video or something

[drifting seal]

why are you learning integration before completing the square

this is an algebra technique

[cosmic jetty]

ask my school bro

[radiant sluice]

which country are you from?

[drifting seal]

ax^2 + bx + c
a(x^2 + b/a x + c/a)
a(x^2 + b/a x + b^2/4a^2 - b^2/4a^2 + c/a)
a(x+b/2a)^2 - b^2/4a + c

[cosmic jetty]

literally never seen this before in my life

[cosmic jetty]

canada

[drifting seal]

🤔

vertex form??

equation of a circle??

never did that

🤔

[cosmic jetty]

is the equation of a circle like root of two squares?

[radiant sluice]

oh

[cosmic jetty]

not rlly

[drifting seal]

what does this mean?

[cosmic jetty]

idfk

we really didnt

[drifting seal]

(x-h)^2 + (y-k)^2 = r^2

[cosmic jetty]

yeah

we did that

[drifting seal]

so you did complete the square

stop lying

[cosmic jetty]

i didnt

on king colton i didnt

[drifting seal]

brother

fuck king colton

you just forgot how to do it

[cosmic jetty]

id have forgotten how to do it if it was taught in 3rd grade

[drifting seal]

you probably forgot long division too

[cosmic jetty]

yeah man

[drifting seal]

do you remember your times tables

[cosmic jetty]

yes

dude

[drifting seal]

dude

[cosmic jetty]

cmon

what country are you from

[drifting seal]

guess

[cosmic jetty]

usa

[drifting seal]

good guess

[radiant sluice]

they didn't teach completing the square in schools in my country explicitly either

they taught it in prep school

also didn't teach root gain and root loss

[cosmic jetty]

we skipped some stuff too cause my high school was unorganised and because of covid

[radiant sluice]

lot of things they didn't teach in schools but taught in prep school

[drifting seal]

@cosmic jetty you’ll be an american soon enough

first you have to learn how to complete the square

[cosmic jetty]

great

i guess im gonna go look at flippin algebra

so i can do this

[drifting seal]

shouldn’t take long

the concept is quite simple

take for example x^2 + 10x + 69

to make things simpler

let’s make the coefficient of x even

x^2 + 10x reminds me a hell of a lot like (x+5)^2 doesn’t it

except the constant term is different

expanding (x+5)^2 yields a constant term of 25

which leaves us with a deficit of 69-25 = 44

so to make the expressions equivalent we need to add on the 44

x^2 + 10x + 69 = (x+5)^2 + 44

in general if the coefficient of x^2 is 1 just add do (x+half the coefficient of x^1)^2 + the shit you need to make the constant terms equal

[cosmic jetty]

so do i have to find out what to add to (25x^2 - 20x + 53)^2

[drifting seal]

well what does 25x^2 - 20x look like

i know it should be (5x + some shit)^2 + some other shit

because (5x)^2 is 25x^2

then set -20x = 2(5x)(some shit)

so some shit = -2

[cosmic jetty]

so all i need to add is 3

[drifting seal]

expanding (5x-2)^2 gives 25x^2 - 20x + 4

[drifting seal]

what

[cosmic jetty]

dewd

[drifting seal]

you lied bro

you forgot your times tables

[cosmic jetty]

I KNOW MY TIMES TABLES

[drifting seal]

sure

[cosmic jetty]

its just the first TWO terms on the bottom?

or all of them

[drifting seal]

🤔

say what

we want

25x^2 - 20x + 53 = (5x - 2)^2 + some shit

because the first two terms look like they’d be from (5x-2)^2

but the constant term is off

so we compensate

by buying a big pick up truck

and then find out what constant needs to be added to the square of -2 to get 53

[cosmic jetty]

so (5x+2)^2+49

[drifting seal]

nope

(5x+2)^2 + 49 = 25x^2 + 20x + 53

not quite mate

[cosmic jetty]

i wrote the wrong shit

sooooo (5x-2)^2+49?

$\int\frac{1}{(5x-2)^2+49}dx$

[potent lotus]

BlackWizard

[drifting seal]

brilliant

look familiar?

[cosmic jetty]

absolutely not

[drifting seal]

🤔

$\int \frac{1}{u^2 + a^2} \dd{u}$

[cosmic jetty]

is this a u sub

[potent lotus]

knief

[drifting seal]

not at all familiar

[cosmic jetty]

alright bro

[drifting seal]

it’s a question

is it familiar

ring any bells

[cosmic jetty]

well yeah actually that picture kinda looks a little similar!

[drifting seal]

,av blackwizard11

[potent lotus]

blackwizard11's Avatar

Click here to view the image.

[drifting seal]

hard

[drifting seal]

yes but do you know the anti derivative for the integral i showed

[cosmic jetty]

nah i dont

[drifting seal]

do you know the derivatives of inverse trig functions

[cosmic jetty]

not by memory

[drifting seal]

oh boy

are you by chance in calc 2?

in college

or is this high school

[cosmic jetty]

calc 2

[drifting seal]

makes sense

you forgot the derivatives from calc 1

did you take it in the fall

[cosmic jetty]

it was a while ago

[drifting seal]

yep

you’re going to want to revise

[cosmic jetty]

apparently yeah

[drifting seal]

knowing your derivatives by memory makes this class much much easier

but

this should remind you of arctan

[cosmic jetty]

im pretty sure i purged them from my memory when i left the exam room

[drifting seal]

$\dv{x} \arctan x = \frac{1}{1+x^2}$

[potent lotus]

knief

[drifting seal]

yep

so if my input is u/a i get

$\dv{u} \arctan(u/a) = \frac{1}{1+(u/a)^2} \cdot \frac{1}{a} = \frac{a}{u^2+a^2}$

[potent lotus]

knief

[drifting seal]

but to make that last a go away just multiply arctan(u/a) by 1/a

and we get

$\dv{u} \frac{1}{a} \arctan(u/a) = \frac{1}{u^2+a^2}$

[potent lotus]

knief

[drifting seal]

bam

so

integrating 1/(u^2 + a^2) gives 1/a arctan(u/a)

@cosmic jetty

[cosmic jetty]

so u is 5x-2

[drifting seal]

yep

[cosmic jetty]

and a is sqrt 49

[drifting seal]

but then we need to do du shenanigans

[drifting seal]

you forgot your times tables again bro

but yes

sqrt(49) =7

[cosmic jetty]

49 is weird actually not my fault

[drifting seal]

🇨🇦

[cosmic jetty]

so its $\int\dv{u}\frac{1}{a}\arctan(u/a)\dd{x}$

[drifting seal]

\arctan

no italics mate

and \dd{x}

and \dv{u}

still i have no idea why there’s a d/du

or a dx

here

what’s going on

[cosmic jetty]

thats what u wrote

[drifting seal]

nope

not what i wrote

[cosmic jetty]

so why is there a d/du

[drifting seal]

differential operator

to signify i’m taking the derivative

use $\dv{u}$

[potent lotus]

knief

[drifting seal]

and $\dd{x}$

[potent lotus]

knief

[cosmic jetty]

seems readable to me

but ok

[drifting seal]

nah

but also what is u and what is a?

[potent lotus]

BlackWizard

[drifting seal]

cooked

[drifting seal]

i mean i guess you’re getting at the right idea but this again is just meaningless

[cosmic jetty]

is this wrong or not

without subbing the values of u and a

[drifting seal]

it shouldn’t be dx

there is no x in the arctan

[drifting seal]

this only has u

you’ll have to convert u = 5x - 2

then du = 5dx

[cosmic jetty]

ok but its not wrong?

[drifting seal]

write that on your test and see what you get

[cosmic jetty]

dude

[drifting seal]

dude

[cosmic jetty]

im going to sleep

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[keen flower]

for 52 i got p^r - p^(r-1)

and for 54 i got gcd(n,m)

does that seem right?

[calm coral]

@keen flower Has your question been resolved?

[west fractal]

hm

[west fractal]

yeah 52 is good

[keen flower]

thanks

[west fractal]

54 appears fine too

[west fractal]

i think you're good to go

anything else?

[keen flower]

thank you! 🙏

that's all

.solved

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[foggy garnet]

hello I have a relatively quick question about particular solutions for non-homogenous linear ODE

[foggy garnet]

I am confused what I have donw wrong here

for 1 I chose B because thats the form for exponentials, for 2 I chose A because its an n=2 degree polynomial, for 3 i chose C because its the form for a trig function, and for 4 I chose D because it is the trig form multiplied by n=1 polynomial (because of the x)

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[visual bluff]

the graph on on left is f’(x) and question is asking to graph f(x) with reasonable accuracy. Given point on f(x) is (-1,-2). does my graph look correct?

[visual bluff]

i used area under the curve

roughly

to find max value

on f(x)

and i know infection point must be at 2.1=x ish

[untold compass]

looks good with the slopes and areas, except I'd make (-1,-2) completely flat

[untold compass]

👍

[visual bluff]

ik it doesnt rlly go thru -1,-2 gonna fix that lol

but thank u very much

+close

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[sour parrot]

Ello, am I adding capacitants correctly? It feels like I'm doing something wrong cause I went back to the same value after adding capacitor 3

[clear delta]

just work in uF the whole time

personally i'd also recommend staying in fractions

rather than decimals

[jolly pilot]

what is that addition in C6,7,8 step?

1e-6 + 1.43e-6 = 2.43e-6

not 1.43e-12

[clear delta]

@sour parrot ^

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[opal thunder]

Find x please I don't get it

[dull wagon]

draw a line at the 130° angle that's parallel to the existing parallel lines

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[blissful carbon]

.close

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[atomic grove]

Is my book wrong? It says that the values whereby a misses real solutions is when a>4. But I got it to a>0.
Equation: x²+a=4

[atomic grove]

Also ignore that it says $a_2$, it's just because I have two variables named a.

[potent lotus]

Elliot Pixel

[swift dragon]

x^2+a-4=0

so $\Delta^2=0^2-4(1)(a-4)>0$

[potent lotus]

Bonk

[atomic grove]

What does delta mean in this?

[atomic grove]

Wait you're totally right

I forgot to move it to the left side 🙏

@swift dragon

Whoops

I forgot to multiply the -4

Okay thanks for your help

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[swift dragon]

this is wrong

actually, nvm

[atomic grove]

Yeah I figured it out

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[wary haven]

How to solve this I terms of e^nlogx pls

[fringe reef]

$n \log(x) = \log(x^n)$

[potent lotus]

artemetra

[fringe reef]

use this

on all of them

[wary haven]

Ok

Then ?

Should I write e^logx^n as x^n

[shy swallow]

i'd presume you'd mostly want to eliminate most of the terms

[fringe reef]

yes

[shy swallow]

you can factor out e^3logx from the numerator and denominator to maybe simplify the function

[wary haven]

Ok guys .. thanks for the help

Got the answer

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[amber plinth]

Can I get some explanation on how #8 here is a washer problem

[amber plinth]

a sketch would be nice

[eternal shard]

you can use desmos, besides that you have two radii, one inner and outer

[amber plinth]

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[olive lion]

I think I made a mistake cuz wolfram has a different answer

[eternal shard]

show

[olive lion]

<@&268886789983436800>

oh mb mods ;-; he only posted it here i think my accident ;-;

dont hurt me please

[eternal shard]

16/15

,w -64/12+32/5

[potent lotus]

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[olive lion]

fucking chatgpt

[eternal shard]

what

[olive lion]

he said it was something else

-64/12+32/5

[eternal shard]

bro u better than chatgpt

[slate field]

i know that this is posted with good intention, but this isn't really the place for it.

[pearl basin]

IMPORTANT

[drifting seal]

lmao

[olive lion]

LMAO

[hushed tartan]

[olive lion]

he straight up didnt care

[slate field]

so i just said that this was not the place for it. come back in a day.

[drifting seal]

#help-9 message

[olive lion]

damn

[slate field]

here is the last relevant message.

[olive lion]

smay went savage mode

[drifting seal]

odds on bert being his friend

[olive lion]

thanks man

so uhm my professor

she had one exercise with a triple integral

wait ill show

[olive lion]

so she started with the outside integral

i suppose that is only allowed here cuz you dont have any variable theta

[eternal shard]

fubini

[olive lion]

aaaah

and then u have constant so u put it in the front

got it

[eternal shard]

i suppose

[olive lion]

smart

[eternal shard]

the two inner integrals converge you can basically pull them out like a constant

[olive lion]

wait

ur not allowed to use fubini when you have a variable in the bounds right?

[eternal shard]

yea to change the order

but basically if you integrated the dzdr you will end up with a constant

[olive lion]

aaah ye true

[eternal shard]

if it converges foc

[olive lion]

ye exactly

thank you again

I need help again

I can definitely not draw this

so uhm

what then ;-;

is it just

0 to 1 dy

y to 1 dx

1 to xy dz?

[eternal shard]

did you try

[olive lion]

it is 3D!

i have never drawn anything in math that is 3D

[eternal shard]

you can draw the region in the xy plane

and the z = xy is above it

[olive lion]

alright imma try

i am not sure how to draw y=<x<=1

[eternal shard]

the z = xy is the hardest

[eternal shard]

ok

so it is bounded from above and below

[olive lion]

yes

[eternal shard]

x = 1 and x = y

[olive lion]

yes

maybe

a triangle?

that is filled?

[eternal shard]

yes!

[olive lion]

lets go

[eternal shard]

and now kinda imagine above it is z = xy

[olive lion]

maybe like a hill that is going upwards to the top right?

anywhere on the x-asis will z be 0

and top right it will be 1

[eternal shard]

like that

[olive lion]

what

[eternal shard]

i also had trouble to draw z = xy but the triangle is possible

[olive lion]

but how do i know my bounds then?

[eternal shard]

so peep this

[eternal shard]

they are already given

[olive lion]

so what i said?

[eternal shard]

yes

[olive lion]

alright

[eternal shard]

your region is basically a solid

[olive lion]

aaaah

alright

now imma try to do this ;-;

[eternal shard]

something like this

if you project the triangle onto the surface

[olive lion]

that looks better

fuck

my answer is negative

[eternal shard]

https://www.desmos.com/3d/y8fn285rfi

[eternal shard]

show

[olive lion]

this by far the best

[olive lion]

aaaah

i think i see it now

[eternal shard]

in the end somwhere

[olive lion]

ye

it should be + integral

e/2-7/6?

[eternal shard]

ye

[olive lion]

lets goo

[eternal shard]

are you speedrunning calculus

[olive lion]

no why?

my professor is

is this calc 2 or 3?

[eternal shard]

idk

[olive lion]

or 1?

[eternal shard]

prob calc 2.5

[olive lion]

well

we started with calc 1

like eh

6 weeks ago

we are in week 6 now

so i guess we are speedrunning it after all :(

[olive lion]

but this one seems a bit sketchy

first i thought dx last

cuz of the root x

[eternal shard]

looking at your bounds how'd u think the region looks like

[olive lion]

its a cube

[eternal shard]

wow pro okay

soo

[olive lion]

right?

[eternal shard]

yes

[olive lion]

great

[eternal shard]

so your best friend is fubini

[olive lion]

yes!

[eternal shard]

so like you could integrate with respect to x in the end

[olive lion]

ye cuz we dont like that one

so we just put it away

and simplify it right

[eternal shard]

well tbh

i dont think you can by pass ibp

[olive lion]

wait

does IBP work with a root?

[eternal shard]

actually

oh

[olive lion]

maybe just try smth

hope it simplifies later on?

[eternal shard]

the sqrt(x)e^x part can be turned into 2 * u^2 * e^(u^2) which is also shit

[olive lion]

true

oooh

hmmm

u will get 1/(root(x) * z) tho

after the first one

[eternal shard]

I just considered sqrt(x) = u

[olive lion]

with respect to y

[olive lion]

i know i know

but that doesnt seem to simplify it really right?

[eternal shard]

how did u get it

[olive lion]

i didnt solve it yet

i am just trying

[eternal shard]

oh

[olive lion]

okay

now i have

[eternal shard]

i am dumb

[olive lion]

int 0 to 4

e^2x / 2xroot(x) - 2root(x) dx

then i can split them

and then uhm

i might figure it out

u figured it out?

[eternal shard]

no

[olive lion]

oh

[eternal shard]

what'd u do

[eternal shard]

yea that one is hard

[olive lion]

ye

[eternal shard]

i realized sqrt(x)e^x is related to the gamma function

but i am not sure if you can use that

[olive lion]

the what what

i dont think i am allowed to use that

[eternal shard]

,w Gamma function

[potent lotus]

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[olive lion]

oehhh

[eternal shard]

thanks wolfram

[olive lion]

maybe we can use

taylor series?

idk how tho

but i know sometimes u use them if it is not possible

[olive lion]

nope never seen this before

why is this so hard

[eternal shard]

because it isnt

[olive lion]

huh

[eternal shard]

i think i got it

[olive lion]

💀

alright teach me please

[eternal shard]

Let $\sqrt{x} = u$ then you get
[ 2 \cdot \int_0^2 \int_0^1 \int_0^2 u^2e^{u^2yz} : du dy dz ]
now apply Fubini
[ 2 \cdot \int_0^2 \int_0^2 \int_0^1 u^2e^{u^2yz} : dy du dz ]

[olive lion]

u mean u e^u² right?

[eternal shard]

no

dx = 2u du

1/2sqrt(x) dx = 1/2u dx = du

then if you integrate with respect to y u^2 cancels as a factor

[olive lion]

Hmmm

can you maybe work it out?

just the integral part with the substitution

i dont really understand the substitution yet

[eternal shard]

[ \int_0^2 \int_0^1 \int_0^4 \sqrt{x}e^{xyz} : dx dy dz = \int_0^2 \int_0^1 \int_0^2 ue^{u^2yz} \cdot 2u : du dy dz ]

[olive lion]

aaaaaaaah

i get it

damn smart man

[eternal shard]

not quite havent solved it yet

yea of course

e^z/z^2

it was a nice attempt

it's like finally dodging the punch

just to get kicked

[eternal shard]

although maybe let w = yz

is it really that stupidly simple

[olive lion]

idk

is it?

why do they give this question as homework :(

[eternal shard]

no i went circles with that one

damn

[olive lion]

the annoying part is that i cant even check the answers

[eternal shard]

i can

,w Integrate[sqrt(x)e^(xyz),{z,0,2},{y,0,1},{x,0,4}]

nice function

[olive lion]

what the hell is that

ye no idea what that s

[potent lotus]

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[olive lion]

hmmm but how do we solve it

[eternal shard]

funny

,w Integrate[ 2 * e^(u^2 * z)/z - 2/z,{z,0,2},{u,0,2}]

[olive lion]

W

[potent lotus]

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[eternal shard]

how do you continue

e^u^2 is the very same shit like e^z/z

😂

[olive lion]

no idea ;-;

[eternal shard]

,w Integrate[ 2 * e^w/w - 2/w,{u,0,2},{w,0,2u^2}]

[potent lotus]

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[eternal shard]

from where is this problem anyway

ok like fubini useless? all 3! possibilities of changing order seem to lead to nothing

unless there is some sort of substitution i have no clue anymore and i give up

[olive lion]

same

I wonder how you do this one tho

maybe someone else can help us?

[calm coral]

@olive lion Has your question been resolved?

[calm coral]

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[olive lion]

.reopen

[calm coral]

✅

[ancient grotto]

@olive lion I'm too lazy to scroll up, mind posting the original question?

[ancient grotto]

I pray for you to come out alive.

It might be over

[olive lion]

what

[calm coral]

@olive lion Has your question been resolved?

[calm coral]

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• Show your work, and if possible, explain where you are stuck.
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[unborn sentinel]

idk what to do

im so lost

[whole steeple]

Do u know the formula for circumfirence?

[unborn sentinel]

yes

pi x diameter

[whole steeple]

Ye do C=pi • d

[unborn sentinel]

i have no diameter

[whole steeple]

Also its given C = 10

[unborn sentinel]

oioh

[whole steeple]

So d = 10/pi

[violet comet]

,help

[potent lotus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[unborn sentinel]

oh ok

wait but this is a non calc paper

[whole steeple]

Ye?

[unborn sentinel]

so would i leave as 10/pi

[whole steeple]

Yes

[unborn sentinel]

oho k

so now we need the radius

yea?

@whole steeple

[whole steeple]

Sorry

[potent lotus]

hampter

[whole steeple]

Ye so radius is just d/2

[violet comet]

hety this is my sons 5th grade math

[whole steeple]

?

[violet comet]

can ya'll help me with this?

[violet comet]

right here

[whole steeple]

2pi/14

Trust

[unborn sentinel]

im at 10 / 2 pi

[whole steeple]

Yes now find the area using the area formula

[calm coral]

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).

[unborn sentinel]

so 10 over 2pi squared?

x pi

[whole steeple]

Yes

Work it out n tell me

[unborn sentinel]

is 20 over 4 pi?

[whole steeple]

100

[unborn sentinel]

i got 5 pi

in the end

wtf bro im so confused

[whole steeple]

Uhh

[unborn sentinel]

where'd u get 100 from

[whole steeple]

You should get 25/pi?

Cuz 10 squared is 100

[unborn sentinel]

im starting again

[whole steeple]

No 😭

[whole steeple]

.

Its just 5/pi squared

So 25/pi^2

[unborn sentinel]

1 sec

i got 10 over 2 pi x 10 over 2 pi

which gave me 100 over 4 pi

is that correct?

???

[whole steeple]

Ye

Thats what i said

[unborn sentinel]

ok and then after that

i did 100 over 4 pi x pi

[whole steeple]

Over 4 pi squared

Ye

[unborn sentinel]

which gave 100pi over 4 pi

[whole steeple]

So thats r^2

[unborn sentinel]

so now what

[whole steeple]

To find area

Just multiply that by pi

[unborn sentinel]

i thought i already squared it before

Ac = pi x 10 over 2pi all squared

[whole steeple]

No that was to find the radius squared

[whole steeple]

Yes

[unborn sentinel]

yes wht?

[whole steeple]

So pi x 100/(4pi x pi)

100/4 is 25

You can cancle out 1 pi

[unborn sentinel]

ok

[whole steeple]

And ur left with 25/pi

[unborn sentinel]

i got 25 pi over pi

[whole steeple]

No 😭

[whole steeple]

I think youre making a mistake here

When its all squared you also get pi squared in the denominator

[unborn sentinel]

oh

ye i forgot the squared

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[ivory parcel]

i keep getting 12 cis (-17pi/12) but its not an option

[wet urchin]

is cis cosine

or is that a term i dont know

nvm cosine +isine

[tacit moat]

it's 2π periodic

so that means you can add/subtract any multiple of 2π to the angle and still get the same point

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[whole solar]

How do I find the period here if x values for the peaks and valleys arent given?

[whole solar]

only given b>0

[plain bone]

How many periods are between x = pi and x = 2pi

[whole solar]

2

[plain bone]

So over the course of $\Delta x = 2\pi - \pi = \pi$, 2 periods pass

[potent lotus]

King Leo [Ping For Help]

[plain bone]

How much is one period

[whole solar]

is it 1 pi?

[plain bone]

But two periods pass in a 1pi interval

[whole solar]

ah so 1/2pi

[whole solar]

would the equation be y=3sin(4x)?

[plain bone]

Btw you can write \sin in latex form

[whole solar]

what do u mean?

[plain bone]

$3 \sin(4x)$

[potent lotus]

King Leo [Ping For Help]

[whole solar]

gotcha, thanks

.close

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[wicked summit]

how do i solve this?

[wicked summit]

how do i solve this?

.close

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[flat maple]

That’s insane math

[soft gazelle]

the translation is really bad

[plain bone]

TIL prathi is not a standalone word

[next geyser]

What?

Should I skip this question?

[plain bone]

No, its just that idk the answer. But i feel like the english translation is really bad, and i couldnt fully understand the hindi

[next geyser]

They are saying in hindi that: in a school everyday they have 6 hours 5 subjects and how can we adjust these hours such that all subject at least get one hour

So i guess 5 should be answer

[weak rapids]

"A school operates for 6 hours daily. There are 5 subjects that need to be scheduled within these 6 hours. In how many ways can the schedule be arranged so that each subject gets at least one hour every day?"

[dark fox]

Was really bad translation then…

There should be a smallest unit of time here right?

[weak rapids]

what do you mean by that

[plain bone]

Assume subject 1 and subject 2 take up three hours total

Lets say a + b = 3

[weak rapids]

it isn't my doubt 😭

[plain bone]

There are infinite combinations of $a$ and $b$ such that $a, b \geq 1$

[potent lotus]

King Leo

[weak rapids]

@next geyser

[next geyser]

Yess sir

Let me skip the question

Thanks all of you

.close

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[keen flower]

$\lim_{x\to a}f(x)=\lim_{x\to b}f(g(x))$ if $\lim_{x\to b} g(x)=a$

[potent lotus]

Axe

[keen flower]

does this work?

[hybrid drift]

yes

[keen flower]

are there any conditions?

just that the limit of f exists, i guess?

[plain bone]

$$\lim_{x \to \lim_{x \to b} g(x)} f(x) = \lim_{x \to b} f(g(x))$$

[potent lotus]

King Leo

[keen flower]

[plain bone]

Texit is mean

[ancient thistle]

no there are certainly conditions

[keen flower]

hmm

[ancient thistle]

some heavy handed conditions are that f is continuous at a and g is continuous at b

actually i think you can get away with just that f is continuous at a

but there are weaker assumptions under which the result still holds

one of which is the requirement that g(x) is never equal to a in a neighbourhood of b

[hybrid drift]

it does not need to be continuous at a

[ancient thistle]

did you just completely ignore what i said?

[hybrid drift]

yes

[ancient thistle]

well then don't reply ping me

[hybrid drift]

it was an accident lmfaooo