help-17

[tidal dock]

because your original statement is t^3 = 3t. notice that t=0 works here (0^3 = 0, and 3*0=0, so 0=0)

$t^3 = 3t$ and $t^2 = 3$ are not equivalent because going from the first one to the second one means dividing both sides by $t$ which is not possible when $t=0$

[twin meteor]

artemetra

[tidal dock]

because we can't divide by zero

[frank pier]

yes

[tidal dock]

so substituting back we get that log(x) = 0, log(x) = sqrt(3), log(x) = -sqrt(3)

for which value of x is log(x) zero?

[frank pier]

0?

[tidal dock]

no

x=1

log(1) = 0

[frank pier]

yes

[tidal dock]

so x=1 is one of your solutions

[frank pier]

ohh

[tidal dock]

in your textbook, is $\log(x) = \log_e(x) = \ln(x)$ or $\log(x)=\log_{10}(x)$?

[twin meteor]

artemetra

[frank pier]

$\log(x)=\log{10}(x)$

[twin meteor]

𓉸ྀི

[tidal dock]

great

so

[tidal dock]

so, if we have $\log_{10}(x) = \sqrt{3}$\
how do we solve for x?

[twin meteor]

artemetra

[frank pier]

divide by log10?

[tidal dock]

no, we can't do that

log10 isn't a number, it's a function

[frank pier]

oh

wait

uh

convert to exponent?

[tidal dock]

yes

[frank pier]

10^sqrt3=x

?

[tidal dock]

yep

$10^{\sqrt{3}}$ is another solution

[twin meteor]

artemetra

[tidal dock]

by same logic $10^{-\sqrt{3}}$ is also a solution

[twin meteor]

artemetra

[tidal dock]

so your final answers are $x=1,10^{\sqrt{3}},10^{-\sqrt{3}}$

[twin meteor]

artemetra

[frank pier]

Ohh

tysmm , no other questions for now..

[tidal dock]

no problem

[vocal sleet]

@frank pier Has your question been resolved?

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[upper haven]

hey guys i have a math question that isnt really a question,im buying a supplement and doing the general bulk calculation and as of rn i got a whole graph plotted and have been looking at the costs and im stuck between buying 60 or 120,anyone down to help?
ill give you the values

x=y
x=number of capsules
y=price per capsule

30=0.32
60=0.27
90=0.27
120=0.23
150=0.23
180=0.21
210=0.21
240=0.21
270=0.21
300=0.21

[tranquil trellis]

I suppose it depends on how much you'll need

[upper haven]

hmm i mean i wanna do 120 or 60

because based on da math they da best

[vocal sleet]

@upper haven Has your question been resolved?

[tranquil trellis]

Why?

60 has one of the highest price per unit

[vocal sleet]

@upper haven Has your question been resolved?

[vocal sleet]

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[severe steppe]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[severe steppe]

hi

i dont understand

where did they get the extra 1 from

inside the [ ] brackets

[twin meteor]

MethIsAlwaysRight

[peak matrix]

now its just factoring (1/8)^x out

[severe steppe]

oh

(\left(\frac{1}{8}\right)^{-2} = 8^2 = 64)

why is this true

[twin meteor]

chillies

[severe steppe]

how?

[autumn trail]

a^(-1)=1/a

So (1/8)^(-2)=((1/8)^(-1))^2=8^2

@severe steppe

[autumn trail]

Not sure how clear this is

[autumn trail]

But it's basically just this and some mucking about with rules of exponents

[severe steppe]

its hard to read

i see

[autumn trail]

$(\frac{1}{8})^{-2}=((\frac{1}{8})^{-1}))^{2}=(\frac{1}{8})^2$

[twin meteor]

DootDooter

[severe steppe]

i should learn the rules of exponents then

ok thx

i see

so whenever theres a negative exponent thats even

it swaps to a positive

or does it matter if its odd or even

as long as its negative its always swaps to a positive?

but how is 1/8^2 of 64

wouldnt it just be some super small decimal

[autumn trail]

So the key idea is that an exponent of -1 flips a number, so a^(-1) is 1/a

It doesn't matter if the exponent is even or odd

[severe steppe]

so

if its always negative

it always swaps to a positive

so just think as if negative exponents dont exist only positive ones

[autumn trail]

Mmm I wouldn't think like that exactly

[severe steppe]

but

is my statement true

if its always negative it always truns positive

[autumn trail]

What I think when I see a number to a negative power like a^(-n) is that the negative sign can be replaced if I swap out a for its reciprocal.

[autumn trail]

Well you don't have to do it.

You can just leave the negative exponent sometimes. So the wording "always" here is a little too strong imo.

[severe steppe]

so

u can if u want to

so if you wanted to you always could

[autumn trail]

If you remember this rule though it always gives you a way to swap out negative exponents for positive ones tho

[severe steppe]

ok

so how is 1/8^2 64

i think it should be a super small decimal

[autumn trail]

It's not 64 you left off the - sign on the 2

[severe steppe]

man

this is hard

[autumn trail]

Just takes a bit of practice. Thankfully these rules won't change much as you do more math.

[vocal sleet]

@severe steppe Has your question been resolved?

[severe steppe]

doot dooter

i figured it ut

out

[vocal sleet]

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[wild meteor]

when i multiply the bottom by 3 both x and y will =0 when i combine them

how do i solve it if both =0

[leaden walrus]

my math is a bit rusty when it comes to systems but

couldn't you just multiply the bottom equation by -3

and cancel out the -15y and 15y

[wild meteor]

yeah thats what i meant

but then -12 will cancle 12

so i will have two things that =0

[leaden walrus]

ah yeah

hm

[wild meteor]

so idk what you do in this case

[leaden walrus]

i've tinkered around a bit

the answer is Ø

no solutions exist

whatever you try they all just cancel out

does the answer have to be word for word or is it checked manually by the teacher

[wild meteor]

wait is there infinate solutions or not a solution

[leaden walrus]

there isnt any solutions

[wild meteor]

wait is there infinite solutions or not a solution

[leaden walrus]

if i am correct no solutions

[wild meteor]

if the lines are on top of each other there would be infinite though?

since they are constantly touching each other

im not sure though

we might want someone else to verify

[leaden walrus]

these are not the same value afaik

if you multiply the bottom one by 3

it becomes

12x+15y=-6

they're not the same

yeah im correct

they're parallel

checked on desmos

[wild meteor]

i checked on desmos too

and i only see one line

doesnt that mean one is under the other

[wild meteor]

wtf

i only see one

[leaden walrus]

zoom in

[wild meteor]

nvm

i put it in wrong

😭

[wild meteor]

how do i show my work for this one

[leaden walrus]

can you submit images?

or does it have to be just a written thing

and is it a type of test where you have to enter an exact answer or you can enter something and it is checked later by the teacher

[wild meteor]

just written im not allowed a graphing calc for this assignment

well i am

its just

you cant use it as evidence

[leaden walrus]

ah

[wild meteor]

well its checked by the computer first

then the teacher verifies

that the computer marked it right

so normaly i get like a 50%

then my teacher marks it and its a 90

or above

[leaden walrus]

well i mean you could write that the unknowns cancel each other out if the second one is multiplied by -3 or if both are multiplied by a common factor ( one of the signs flipped of course due to being forced to use the method of elimination)

i dont know if the computer will like that though

if you could send an image into that prompt this proof would be much easier

its only like a few lines of actual math

[leaden walrus]

i guess i would just submit this

since it is checked by a teacher too

[wild meteor]

ok

[wild meteor]

can i just write 0x+0y=-9 on my paper then write that the lines will be parrel without the same y intercept

or how do i explain i know they will be parrelel

without using a graping calc

[leaden walrus]

hm

when the 2nd unknowns are multiplied by -3 the yielded answer is smaller than the 1st one

so i guess because the 2nd is not equivalent to the 1st

the yielded graphs are parallel and as such they have no meeting point

their slope ( i think thats what its called i am from croatia my english is not top notch when it comes to math jargon lol)

is the same

[wild meteor]

can i write "since both 0 and y =0, and they do not = 0 it means that they are on the same slope with a different y-intercept, this means they are parallel while not overlapping"

[leaden walrus]

since both x and y equal to 0, and the 2 equations arent equivalent to each other, it means that they are on the same slope with a different y-intercept, which means that they are parallel yet never overlap nor meet

[wild meteor]

that should be good?

[leaden walrus]

think so

[wild meteor]

did everything i do so far here look corrct?

just want to make sure i didnt make mistakes before i do the big work

[leaden walrus]

yeah it is correct

[wild meteor]

how do i do elimination

so i need to make s

33

by multiplying the first one by 11

and the 2nd one by 3

[leaden walrus]

-3

or -11 the first one its optional

one of them has to be negative

basically when you do that and cancel the unknown all you're doing is adding the equations together

[wild meteor]

ok, but is that the correct thing to do?

[leaden walrus]

yeah when you're asked to do elimination

there is multiple methods of solving but they want explicitly elimination

[wild meteor]

ok

[wild meteor]

sorry my power went out so i might be a bit slow now

im using my hotspopto

hotspot*

hopefully my laptop doesn't die

[vocal sleet]

@wild meteor Has your question been resolved?

[wild meteor]

Is this correct?

M=10 s=5

[wild meteor]

<@&286206848099549185>

i did something wrong

but what

i need help

wait no i didnt

.close

[vocal sleet]

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[vocal sleet]

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[river kettle]

What are the next 5 terms in the sequence 1,1,2,3,5,8,13?

I can’t c the pattern

[lavish river]

Fibonacci sequence

[river kettle]

oh 💀

tru

.close

[vocal sleet]

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[vocal sleet]

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[rare oasis]

Is this correct so far and how would I do the top two questions?

[pale perch]

,rccw

[pale perch]

that doesnt look like proper sets

and they would be wrong even if they were

how did you come to your domain and range exactly?

we'll start there

[rare oasis]

i created a set for the domain based off of each x value for the points, then i did the same for the Y

[pale perch]

that would work if the function were just those points

but it isnt

i can see that f has point at -6, -2.1, 4,5, etc

[rare oasis]

isnt that what each point is?

[pale perch]

no

youve tried to say the domain is only those points

but f is defined outside of them also

[rare oasis]

i dont understand

[pale perch]

like you can say the domain of 1/x is (-inf,0) U (0, inf)

that encloses all values 1/x can take as inputs

what youve done is like saying the domain of it is (-3, -2, -1, 1, 2 ,3) just because its defined for those points

which isnt the true domain

[rare oasis]

okay, how do i do this correctly then?

[pale perch]

do you believe f is undefined for any x value?

[rare oasis]

im not sure i understand

[pale perch]

what do you think the domain of a function is?

[rare oasis]

a set of x values for each point on the graph

[pale perch]

not really no

the domain of a function is the set of values a function can take as an argument

more practically its the set of all values that if you input into the function youll get a defined value as the output

[rare oasis]

alright

[pale perch]

from the graph of f, do you think it has any points where it doesnt have a defined value?

[rare oasis]

i mean anything outside of the line

[pale perch]

?

what does that mean

[rare oasis]

im not sure what your question means

you're making this seem incredibly complicated

from what i've been told and taught i take the set of x and put that down to get the domain

[pale perch]

but what you wrote isnt the complete set of x is what im saying to you

what about 4.1? 4.1111111111111111? 4.24224434?

those are all in the domain

[rare oasis]

(negative infinity, infinity)

[pale perch]

yeah

or just R

[rare oasis]

so then when would it be numbers instead of infinity

[pale perch]

you mean if the domain was literally just something like {1,2,3}?

[rare oasis]

yes

[pale perch]

youd just have 3 points on a graph

no line or anything

[rare oasis]

ah so a line means infinity

what about the range

[pale perch]

no

if the domain was (-2,5) that could still be a line

its just not discrete values

[rare oasis]

so i can have a line or wave and have it still not be infinity?

how do the graph the points then

[pale perch]

it could be any shape, as long as an x doesnt map to two values (for a function)

[pale perch]

thats function specific

[rare oasis]

okaay

what about the range here

[pale perch]

what do you think it is

[rare oasis]

(negative infinity, 5)

[pale perch]

nearly

do you know the difference between ( ) and [ ]

[rare oasis]

(negative infinity, 5]

[pale perch]

perfect

you need to revise the rest of them too

[rare oasis]

what about the increasing and decreasing intervals?

[pale perch]

you seem to have the idea of what they mean, wanna have another go?

[rare oasis]

negative infinity to 5, then 1 to four

[pale perch]

is that increasing then decreasing?

[rare oasis]

no just the increasing points

[pale perch]

not quite

first off, -inf to 5 contains 1 to 4

and there is decreasing between -inf and 5

eg at -2

[rare oasis]

but there are two rises

[pale perch]

there are, but your bounds are wrong

[rare oasis]

how so?

[pale perch]

starting from the far left

where does it stop increasing?

[rare oasis]

5

[pale perch]

5 is on the far right

[rare oasis]

no Y

5 Y

[pale perch]

you dont write y values as your intervals

its in terms of x

rewrite them

so i can see what you really meant

[rare oasis]

ah so negative infinity to -4 then 1 to 5

[pale perch]

much better

[rare oasis]

for decreasing -4 to 1 then 5 to negative infinity

[pale perch]

good

can you write each case as sets?

[rare oasis]

(-4, 1) (negative infinity, 5)

[pale perch]

good

put U between them

or just a comma

[rare oasis]

what about these? how would these be done?

noted!

[pale perch]

those should be the easiest honestly

you just find the y value for each input

youve already done that

[rare oasis]

oh 0 and 3?

[pale perch]

yup, that simple

[rare oasis]

thanks! thats not bad at all

[pale perch]

no worries

[rare oasis]

If I may continue, how would I do this, graph the function?

[pale perch]

are you aware thats a parabola?

[rare oasis]

nope lol

[pale perch]

,w graph x^2

[twin meteor]

Results provided by WolframAlpha

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

Requested by aldrnari_

[pale perch]

that shape is a parabola

[rare oasis]

how did you do that so fast

[pale perch]

magic

anywho, all quadratics have that shape, or a flipped one

you can identify the transformations and then just move the x^2 graph about to match it

do you know where the vertex is?

if youre opting to sub in specific values in a table, youll want to center around that

[rare oasis]

no i dont, i was asking how to graph it

im not sure about the other stuff but i feel graphing it is a good place to start

[pale perch]

y=(x-1)^2+2 is a parabola in vertex form
y=a(x-h)^2+k which means the vertex is (h,k)

[pale perch]

this is the shape

[pale perch]

you can get some good values if you want to by doing this

[rare oasis]

im not sure i understand

[pale perch]

which bit?

[rare oasis]

all of it.

im just asking how i would actually graph it

[pale perch]

thats what ive been telling you though

theres literally no more i can give you on how to graph it

its all there

apart from finding roots and y intercept

thats it

[vocal sleet]

@rare oasis Has your question been resolved?

[vocal sleet]

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[vast shale]

: Bingo is a game in which each player has a square card with k
rows (horizontal) and k columns (vertical), where k^2 distinct numbers taken from
set {1, 2, 3, . . . , N}. Numbers are drawn one by one and players must mark on the
card the drawn numbers. We have two game types, A and B.

Mode B: Here, whoever completes a line on the card wins the game. From this
At this point, let's consider this type of game.
(a) Two cards are said to be similar if the two cards always win the game together,
that is: for any line on the first card there is a line on the second card that
have exactly the same numbers. Once a card is fixed, calculate the number of cards
similar to her.
(b) You want to make bingo so that each person plays with a single card and not
there are 2 similar cards. What is the largest number of people who can participate?

[hybrid flicker]

!status

[vocal sleet]

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

[vast shale]

4

a) (k!)^2
b(n!)/((n-k^2)!*(k!)"2)

[hybrid flicker]

how did you get (k!)^2 for a)? just interested in the reasoning

also are both columns and rows considered lines?

or only rows

[vast shale]

a) k! * k!

and yes

both

[hybrid flicker]

seems to be correct

[vast shale]

the a and b are correct?

lol

the a is just that?

wow

[hybrid flicker]

well yeah

I think

mmmh

didn't say anything

[vast shale]

ok ok, thank you so much

[hybrid flicker]

just wanna be sure

[vast shale]

mm

[hybrid flicker]

you reasoned as k! ways to permute rows and k! ways to permute columns?

[vast shale]

yes

[hybrid flicker]

because

i'll take an easy example and see if there's something wrong or now

take grid
1 2 3
4 5 6
7 8 9

permute rows 1,2

4 5 6
1 2 3
7 8 9

permute columns 1 and 2
5 4 6
2 1 3
8 7 9

ok no maybe there's nothing

[vast shale]

yes, its correct, not?

[hybrid flicker]

yes, but we need to argue a bit about why

so two cards are similar if there's a finite amount of row permutations and column permutations that link them

[vast shale]

rows: you can permute k rows of the card in k! ways
permuting columns: for each permutation of the rows, you can permute the k columns in k! ways

[hybrid flicker]

And that simplifies down to 1 row permutation and 1 column permutation

because

a row permutation followed by column permutation

is the same as column permutation followed by row permutation

it gives the same result

[vast shale]

yes

[hybrid flicker]

so if you had

(row permutation) then (column permutation) then (row permutation) then (column permutation)

you are allowed to regroup the row permutations together, same for column

because it's commutative

(only between a row permutation and a column permutation, since two row permutations won't commute)

[vast shale]

oh

yes

[hybrid flicker]

so we regroup them into

(1 big row permutation) then (1 big column permutation)

which leads to what we did here

k! possible row permutations

k! possible column permutations

wait I realized

[vast shale]

so it gives the samething, no? but with a complex reasoning

[hybrid flicker]

there's another operation we can do

transposing

ie
1 2
3 4
can become
1 3
2 4

and I'm not sure if you can get that with row/column permutations

since the diagonal needs to stay in place

so 2 (k!)^2

but at the same time it doesn't make sense, because when you only have 1 number it should be only one possibility

but here we're saying it's 2

[vast shale]

wait

let me think

[hybrid flicker]

ok in the meantime I'll show what happens for small values of k

k=1, one way

k=2
1 2
3 4

1 3
2 4

yeah 8 ways for k=2

[vast shale]

so

is 2(k!)^2?

to be honest

I understood your reasoning, but I don't understand why mine is wrong

[hybrid flicker]

two cards are similar if they share the same lines

however we interpreted that as "they have the same rows and they have the same columns"

[vast shale]

oh

i understood

thank yo so much

[hybrid flicker]

so I think it's 2(k!)^2

only works for k >=2 tho

[vast shale]

yes

but

it cant be anything with k=(0,1) correct?

[hybrid flicker]

k=0 makes very little sense so avoid it

k=1 would just give 1

I'm making a python script to compute the number of ways with other values of k so if you have some time to wait

[vast shale]

yes, no problem

and thank you so much

[vocal sleet]

@vast shale Has your question been resolved?

[hybrid flicker]

sorry for the wait @vast shale the code is working

for k=2, k=3 the value 2(k!)^2 is correct

The code has been running for 2 minutes computing for the case k = 4

[vast shale]

so, its not correct for k=4?

[hybrid flicker]

no it's just that my code is so inefficient that it takes too long 😢

[vast shale]

oh, np

thank you so much

[vocal sleet]

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[zinc heron]

This one seems really intuitive, but I cannot prove it.

Let A in R^n be an open set, with n >= 2. Given a in R^n - A, the set A U {a} is open if, and only if, a is an isolated point in the boundary of A. Equivalently: there exists a ball B = B(a,r) such B-{a} is contained in A.

[zinc heron]

Well, what I really want to use is the fact that A is in R^n with n>= 2. We can find a sequence of points in the interior of R^n - A if such ball does not exists, but that doesn't seem to help me getting more points in the boundary.

[lavish river]

wait if a is on the boundary of A then this should be false no?

[zinc heron]

a is on the boundary of A

[half imp]

No, consider an ball with a single point in the middle missing

Let a be that point in the middle

[zinc heron]

yeah this is was my drawing as well

the thing is this is not true for n = 1.
A = (-inf,a)

[lavish river]

yea but as its on the boundary if A, part of the ball would be outside A no?

[half imp]

No, read my example again

Anyways

Equivalently: there exists a ball B = B(a,r) such that B-{a} is contained in A

This is pretty much the definition of open right

Like, you already know that property holds for every point in A

if it holds for a as well then A U {a} is open

[zinc heron]

no, but a is not in A

[half imp]

Yes I know

[zinc heron]

well yes, one direction is clear

the other one is harder

[half imp]

Which one is harder?

[lavish river]

ohh yea i got it sorry

[half imp]

np

[zinc heron]

if a is an isolated point from the boundary of A, then AU{a} is open

[half imp]

well we already know if we choose a point in A then it'll have a neighborhood that's in A and thus in A U {a}

all we need to do is check that that property holds for a as well

[zinc heron]

yes

[half imp]

That's the property I quoted here

From your initial message you seem to know that that is equivalent to a being isolated

[zinc heron]

hmmm not really, for me isolated means that there exists a ball such that a is the only point from its set in that ball. But in this case we could "hipothetically" cointain points from A and int(Rn - A)

as in the case of (-inf,a)

[lavish river]

yea so all points in the ball excluding a would be in A right

[half imp]

Oh, it sounded like you knew it was equivalent

[zinc heron]

no no, the questions says it's equivalent. if you prove this proposition the if and only if one follows and vice versa

sorry bad english i'm translating from my portuguese book

[half imp]

oh ok

no worries

What you want to do is imagine a neighborhood of an isolated point on the boundary @zinc heron

Or rather an open ball of radius r

For the sake of contradiction assume every such ball contains points not in A.

Because it's on the boundary, it also clearly contains points in A.

Remove a from the ball, the resulting space is still path connected. Find a path from the inside of A to the outside of A, this path must contain a boundary point

[zinc heron]

oh

ok

that's smart

i haven't read the chapter on connectivity yet

[half imp]

Ah well you can construct the specific path

[zinc heron]

i'll take a closer look into it. thanks

[half imp]

Np

[zinc heron]

yeah there's a theorem here that may help

well thank you very much

sometimes books do that lol

.close

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[half imp]

No problem

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[slow atlas]

can someone please explain to me why the eigen values are -3 and 1

shouldnt it be 3 and negative 1?

[pale perch]

-1+-2

[slow atlas]

wait what

im confused

[pale perch]

how far back in the work is your confusion

or is it just the last line

[slow atlas]

just the last line

[pale perch]

sqrt(1+3) is 2

[slow atlas]

shouldnt the eigan thing factor to -3 +1

sorry

the line before that then

how did he get to that line

[pale perch]

(x+3)(x-1) thats fine sure

[slow atlas]

because when I factor that i get eig -3 eig +1

ok

[pale perch]

they used the quadratic formula

i think anyway

[slow atlas]

oh im dumb

I did it backwards

ok srry mb

[pale perch]

nw

[vocal sleet]

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[jagged cargo]

$\frac{(|z|-1)(1+iz)}{z-\frac1{\bar{z}}} = i$

[jagged cargo]

$(|z|-1)(1+iz)=\frac{i\bar{z}}{|z|^2}$

[jagged cargo]

not sure what to do next

[astral pilot]

what do you want to do?

find z?

[jagged cargo]

solve for z

[astral pilot]

oh ok

|z|^2 = z * z-bar

do you know this?

[jagged cargo]

yeah, thats how i got the |z|^2

[astral pilot]

reduce both sides to be in form of a + bi

and compare real and imaginary parts

[jagged cargo]

looks messy, but sure let me try

,,
\begin{cases}
|z|-b|z|-1+b=\frac{b}{|z|^2}\
a|z|-a=\frac{a}{|z|^2}
\end{cases}

[jagged cargo]

well?

oh wait

i can reduce the a can i

but it doesnt specify a \neq 0 though

[hybrid flicker]

you mean $(|z|-1)(1+iz)=iz(1-\frac{1}{|z|^2})$

[twin meteor]

rafilou2003

[jagged cargo]

ahhhh

the damn -1

let me revise my work

alright, whats next

[astral pilot]

what is |z|

[jagged cargo]

absolute value of z...?

[astral pilot]

yes, but it's value?

[jagged cargo]

what

z value is what im solving for

[astral pilot]

|z| = \sqrt{a^2 + b^2}

z = a +bi

if you get a and b, you'll get z

[jagged cargo]

i dont think i want to solve a system of equations with square roots

[astral pilot]

wait a minute till i try to solve it

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[jagged cargo]

.close

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[astral pilot]

i think this has no solutions

[hybrid flicker]

if you want to find a solution, first note that |z| is different from 1, then write z = f(|z|)

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[bleak grove]

hi

[mild flower]

hi

[bleak grove]

HI How to get domain and range of (x+4)/(x-3) and all asympotes

[plucky trout]

for domain, what values can x not be

[bleak grove]

0

[plucky trout]

if x=0

we have -4/3

so x can be 0

but x cant be 3

since we will have 7/0

which is undefned

[bleak grove]

oh thanks

[silk hollow]

and as x approaches infinity, the +4 and -3 becomes insignificant so the only thing that would matter would be x/x = 1

so the horizontal asymptote would be y = 1

[bleak grove]

oh thanks I understand now

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[dull sphinx]

hello

[dull sphinx]

I tried to calculate the area under a x² curve

y = x²

x² is rotating

so that means

[hybrid breach]

yeah what did you do

first of all from where to where

[outer warren]

what interval and axis

[dull sphinx]

V = pi * Integral ( y² ) dx

[outer warren]

post the full question

[hybrid breach]

are you trying to calculate the volume when rotating around the y axis

[dull sphinx]

"Show the formula for calculating the volume under a x² rotating curve from point a to b"

I said area my bad

[hybrid breach]

the hell does that mean

[dull sphinx]

exactly what it says

[hybrid breach]

define a rotating curve

[dull sphinx]

a formula for calculating the volume under a x² curve rotating from a to b

[hybrid breach]

also the volume can't be "under" something

a curve can't rotate from a to b

[dull sphinx]

assume x² is like a cup

its a rotating

function

and x² is like the radius of the cup

[hybrid breach]

you're using the wrong terms but i get it

ok then what's the formula

are you rotating around the x or the y axis?

[dull sphinx]

V = pi * Integral (y²)dx = pi * integral(x^4)dx

[dull sphinx]

x axis

man I think thats my problem

should be around y axis no

[hybrid breach]

then it's correct

oh

[dull sphinx]

anyways

[dull sphinx]

the solution would be

[hybrid breach]

if it is the formula is different

[dull sphinx]

pi * x^5/5 no?

so from a to b the formula is

[hybrid breach]

you're messing up the formula

[dull sphinx]

pi*b^5/5-pi x a^5/5

[hybrid breach]

you use that formula when rotating you don't get "holes"

you do get holes here

wait let's clarify

[dull sphinx]

thats the formula

[outer warren]

assuming rotation around x-axis, sure that works

[dull sphinx]

for this volume

under the x² curve

[hybrid breach]

ahh ok

i misunderstood then

[dull sphinx]

my problem is that

[hybrid breach]

then it's correct

[dull sphinx]

This would be the formula from a to b

[hybrid breach]

the rotation is like this?

[dull sphinx]

yes

around x axis

[hybrid breach]

ok then the formula is correct

[vast shale]

About the x axis ?

I see

[dull sphinx]

yes

I get negative values

is my problem

[hybrid breach]

well

[outer warren]

negative how?

[dull sphinx]

for the volume

[hybrid breach]

from a to b means you have to do b-a

[dull sphinx]

wait

[outer warren]

setting the integral up like this
a <= b

[hybrid breach]

is a or b bigger?

if a is bigger, it's a-b

[dull sphinx]

I stored in the wrong order in my calculator (b smaller than a)

kill me

[hybrid breach]

sad

[dull sphinx]

im sorry guys

im really sorry

didnt intend to waste time

it should always be positive lol

.close

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[hybrid breach]

no problem

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[earnest swan]

Helo
I play a minecraft World and im trying to transform a rectangular area to a ellipse.
Length of the rectangle is 168 and width is 129

I don't know how to calculate an ellipse in which the rectangle fits perfectly.
If possible give me an easy explanation how I can calculate that.
Thank you guys for your efforts

[bitter pilot]

,,a \cdot b \cdot \pi = 168 \cdot129

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[bitter pilot]

there are many possiblities to construct an ellipse that has the same area as that rectangle

[earnest swan]

Bruh is it that easy

Thanks alot 🙏

.close

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[keen echo]

How can I show the monotony intervals of a function?

[mild flower]

find its critical points

[keen echo]

this is what i remember from class

[keen echo]

how ?

[mild flower]

,rcw

[mild flower]

critical points are:

• points where the derivative is equal to 0
• points where the derivative is not defined
• points where the function is not defined or discontinuous

[keen echo]

so i need to find f(0) and f(2)

?

[mild flower]

what are you trying to find, exactly?

[keen echo]

i need to determin the intervals of monotony

[mild flower]

okay. in this one it looks like you've found those?

you can see that on (-oo, 0) it's decreasing, for example

[keen echo]

so I just need to make the graf, noting more ?

[mild flower]

i suppose that depends on the format you're expected to put it in

[keen echo]

is good, a friend texted me and said i need to '' matematicaly write'' the graf too

thx for the help

.close

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[polar vine]

.

Hello

[polar vine]

i Know this is not a true question. But i want any good source that covers all lines integrals topic and explain like I am five. I did not find any good ones

I am more into videos

[vast shale]

khan academy ?

[polar vine]

does khan academy covers everything in calclus 2 level ?

[polar vine]

like green's theorem and all stuff?

[peak matrix]

it has multivar calc i think

[polar vine]

okay thanks

[peak matrix]

https://www.khanacademy.org/math/multivariable-calculus

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[bitter pilot]

I am kinda lost right now haha, but how can I convert this integral in terms of polar coordinates?

$$\iiint_V 1 : \dd V = \int_0^9 \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \dd y \dd x \dd z $$

I tried to parameterize the volume $$\vec{v}(x,y,z) = \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x \ y \ 9-x^2-y^2 \end{pmatrix} $$ into
$$\vec{x}(\theta,r) = \begin{pmatrix} r\cos\theta \ r\sin\theta \ 9-r^2\cos^2\theta-r^2\sin^2\theta \end{pmatrix} = \begin{pmatrix} r\cos\theta \ r\sin\theta \ 9-r^2 \end{pmatrix} \quad r\in [0,3], : \theta \in [0,2\pi]$$

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[full arch]

how u gonna access the other dimension with only one varying angle? :D

[bitter pilot]

haha yeah

i thought of that too

so idk what to do about it

[full arch]

introduce another free angle but consider the bounds on it, or rather the angle you currently have written

[bitter pilot]

ok let it be phi

uhm

do i need spherical coordinates?

[full arch]

yeah... but I was just gonna ask im slightly confused by the original integral written

is it a volume of sphere question?

[bitter pilot]

$$\vec{x}(\theta,\varphi,r) = \begin{pmatrix} r\cos\theta\cos\varphi \ r\sin\theta\cos\varphi \ (9-r^2)\cos\varphi \end{pmatrix} \quad r\in [0,3], : \theta \in [0,2\pi]$$

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[bitter pilot]

haha it was a task of calculating the flux

Flux of $S={\vec{x}\in \mathbb{R}^3 : : : z = 9-x^2-y^2, : z \geq 0 }$ with $\vec{\textbf{F}} = \begin{pmatrix} -x \ 2xz \ 2z+1 \end{pmatrix}\$

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[bitter pilot]

and I used gaussian theorem

But I was rather interested on how to do it with other coordinates

[full arch]

oh

what you were doing was fine then with the cylinderical shebang

[bitter pilot]

really?

I just cant set up the integral bounds haha

It's basically a circle

and then I am integrating over height

so maybe I need a second radius, or like 2nd angle

,,\int_0^9 \int_0^{2\pi} \int_0^3 r : \dd r \dd \theta \dd z

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[heady ibex]

you did it using divergence theorem?

[bitter pilot]

yea

[heady ibex]

wait how

don't you need a closed surface?

[bitter pilot]

that's why

hold on katharine

[bitter pilot]

\begin{align*}
\iint_S \vec{\textbf{F}} \circ \vec{\textbf{N}} + \iint_{S'} \vec{\textbf{F}} \circ \vec{\textbf{N}}
&= \iiint_V \vec{\grad} \circ \vec{\textbf{F}} : \dd V \
\end{align*}
where $S'={\vec{x}\in \mathbb{R}^3 : : : x^2+y^2 \leq 9, : z = 0 }$

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[bitter pilot]

They added the circle "cap"

which would make it a closed surface

right?

[heady ibex]

yes

that would make it work

[bitter pilot]

ok puh that was another question i had haha

good that you came

[heady ibex]

you can split it up into arbitrary sections

with some limitations on things

you can't have jumpy around bits

you can split something like a sphere into two halves

but you can't take some crazy subset of the sphere

anyway

[bitter pilot]

IOH

it's like a half ellipsoidal object

[heady ibex]

ok

and you want to calculate it directly?

[bitter pilot]

SO HE was right

yea

[heady ibex]

you want to find a normal vector to the object

to the surface

[bitter pilot]

cross prduct

[heady ibex]

and you wanna find a good way to write down dA

[bitter pilot]

of the tangent vectors

[heady ibex]

cuz you have $\iint \vec{F} \cdot \hat{n} dA$

[twin meteor]

Katharine

[bitter pilot]

yea

[heady ibex]

so you find n in the same coordinate system you have F in

and get the inner product

[bitter pilot]

oh i forgor the differentials

[heady ibex]

and then you find dA

[bitter pilot]

hmm okay

but how about the triple integral

[heady ibex]

there's no triple integral

as you're doing it directly

you integrate over the surface

triple is when you use divergence theorem

and calculate over the volume contained in the closed surface

i say that but you can use triple integrals but i wouldn't as that requires the integration limits to be funky

[bitter pilot]

oh sorry

english not my first language

[bitter pilot]

yes i wanted this 🙏🏻 sorry

[heady ibex]

you want to use divergence theorem?

[bitter pilot]

yes

i was interested if can somehow switch the coordinates

\begin{align*}
\iint_S \vec{\textbf{F}} \circ \vec{\textbf{N}} : \dd S + \iint_{S'} \vec{\textbf{F}} \circ \vec{\textbf{N}} : \dd S'
&= \iiint_V \vec{\grad} \circ \vec{\textbf{F}} : \dd V \
&= \iiint_V (-1+0+2) \dd V \
&= \iiint_V \dd V \
&= \int_0^9 \int_{-3}^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \dd y \dd x \dd z \
&= \int_0^9 \int_{-3}^3 2\int_{0}^{\sqrt{9-x^2}} \dd y \dd x \dd z \
&= \int_0^9 2\int_{0}^3 2\sqrt{9-x^2} : \dd x \dd z \
&= \cancel{4}\int_0^9 \underbrace{\int_{0}^3 \sqrt{9-x^2} : \dd x}_{= \frac{1}{\cancel{4}}\cdot \pi \cdot 3^2} : \dd z \
&= \int_0^9 9\pi : \dd z \
&= 81\pi
\end{align*}

[twin meteor]

coolseph (𝔸dωn𝓲²s)

[bitter pilot]

I did it this way

but was wondering

if I could somehow switch from dydxdz to polar coordinates or something like that

[heady ibex]

you could use cylindrical

polar but also z

dz

dr dtheta dz

if you want to find the volume version

[bitter pilot]

yea i did that above

.

but is that all?

like couldnt I also somehow switch dz

[heady ibex]

i mean you could

but it would make it worse

because the shape you want is not symmetric in any way along z

it is constantly different as you go along the z axis

[bitter pilot]

hmm okay!

thanks for your time Katharine

[heady ibex]

if you wanna calculate the integral directly i think something like dtheta dz

or dtheta dr

and then have z be a function of r

or vice versa

that will give you dA

[bitter pilot]

yea

.solved

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[vast shale]

The first question

[pallid zenith]

parts maybe hmm

,w integrate t^m/(2+t^m) dt

[twin meteor]

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Requested by janniku

[pallid zenith]

ah geez

[vocal sleet]

@vast shale Has your question been resolved?

[twin horizon]

Will changing it from $\int$ t^m/(2+t^m) dt to $\int$ 1/(2t^{-m}+ 1) dt be useful in this case

[pallid zenith]

you have to ❌ or it will lock

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[twin horizon]

What

[vocal sleet]

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[somber holly]

How do I find the axis of symmetry using either the table or graph?

[pale perch]

(its the vertex)

or the line going vertically through it anyway

[covert python]

to find it using the table you find two x values with the same output (y value) and average them

put x= to it and that's the axis

[somber holly]

Hmm okay

Is it viable if I use the vertex formula?

Okay I solved it thx for info!!

.close

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[orchid pier]

is the horizontal/vertical axis/dy/dt and dx/dt respectively?

[pallid zenith]

no

[orchid pier]

then what is it

how can i read it

[pallid zenith]

usually x(t) and y(t)

it makes more sense if you look at one with arrows

maybe something like this

[orchid pier]

so if arrow goes to left dx/dt is negative

?

[pallid zenith]

https://homepages.bluffton.edu/~nesterd/apps/slopefields.html

this used to be and still kind of is a decent site for these

you should try playing around

click on the upper left, and select system

[orchid pier]

but this?

[pallid zenith]

well, it goes in that direction laterally

but also dy/dt at that point is going to impact the a;rrow

[orchid pier]

yeah iknow

[pallid zenith]

so if dy/dt is 0 at that point

and dx/dt is negative

then it will point directly left

[orchid pier]

but if it points to the left even slightly

then dx/dt is negative no?

[pallid zenith]

yes

because only the sign of x' is going to contribute to its leftness or rightness

[pallid zenith]

theres a nullcline here that is very helpful

a nullcline is a curve along which at least one of x' or y' is equivalently 0

[orchid pier]

y=2?

[orchid pier]

why is it helpful?

[pallid zenith]

the curve is like ....

so these purple lines, do you get what they mean?

[orchid pier]

there vectors for dy/dx

[pallid zenith]

well, i dont know if vectors is the most helpful word

[orchid pier]

but why would a nuciline be usefull?

[pallid zenith]

so a purple line is maybe better thought of as a trajectory

if you pick an initial point, so ( x(0), y(0) ) along a purple line

it will move along that purple line, it will stay on it

[pallid zenith]

like the blue and red lines here

[pallid zenith]

so try to interpret what is going on at y=2

lets say we pick an initial point ( x(0), y(0) ) = (0,2)

what happens?

[orchid pier]

y will not change for all x @pallid zenith

[pallid zenith]

yea, i think that at least within the range of the plot, it looks like y will stay about the same

can you express this mathematically?

maybe an equation

maybe some kind of equation that expresses the change of y

[orchid pier]

dy/dt=0

dx/dt = +_ inf?

+-

[pallid zenith]

yea, maybe more specifically, if y=2, then y'=0

[pallid zenith]

we dont know anything about it, really

[orchid pier]

but why is it useful for this plot

[pallid zenith]

well, at least not from this observation

[orchid pier]

yh for the ranges of the plot atleast

[pallid zenith]

they give you some candidates

right

so the usual strategy is one of two

either you try to plot each, and see if they match

trial-and-error

or you try to identify features you know must exist

and rule some out

process-of-elimination

were doing the second

we know that y'=0 where y=2

(at least within the range of the plot, yadda)

so, can you identify any systems where this isnt the case?

then, we can reject them as candidates

[orchid pier]

ah okay

thx a lot

.close

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[pallid zenith]

you got it?

[orchid pier]

sort of

can i pm you for more later on?

[pallid zenith]

id prefer you ping me in a new channel

but feel free to ping

[orchid pier]

ah ok

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Remember:
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[uncut tendon]

<@&286206848099549185>

[heavy yoke]

<@&268886789983436800> advertising

[tranquil trellis]

!status

[vocal sleet]

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

[uncut tendon]

2

[uncut tendon]

What does this mean? Is it a command of some sort?

[tranquil trellis]

ok so it looks like it's oriented clockwise

[uncut tendon]

May you please help me?

[tranquil trellis]

I'm trying the problem rn

[uncut tendon]

Okay

[tranquil trellis]

what part did you get stuck on

[uncut tendon]

Okay just a sec I'll show you my work...

Sorry I have so many papers idk where I put my work. It may take me some time.

Okay I think I'll just change it to 1 and start all over again.

[tranquil trellis]

oh

[uncut tendon]

I used exactly this it did not work out.

[uncut tendon]

I don

[tranquil trellis]

So basically, the point is that instead of evaluating the line integral directly, we can instead calculate the double integral, which is often easier

[uncut tendon]

oh okay

thanks!

How would you set it up?

[tranquil trellis]

So here $L=e^{-x}+y^2$ and $M=e^{-y}+x^2$

[twin meteor]

Ari

[tranquil trellis]

and these are just taken from the vector field given

So $M_y - L_x = 2x - 2y$. So that's what goes inside the double integral.

[twin meteor]

Ari

[tranquil trellis]

@uncut tendon were you able to get this far

[uncut tendon]

I am in the process of doing it okay?

I'm just tired and exhausted