help-38

[stoic burrow]

Maybe

[wraith hinge]

alright

if its that define ${j=\frac{1}{x}}$

[stoic burrow]

Idk ahah

[solid kiln]

icannotdoanymorecauchy

[wraith hinge]

then refine the limit as $x \to 0^{-}$, so ${j \to -\infty}$

[stoic burrow]

1/-infinity

What's answer

How

[solid kiln]

icannotdoanymorecauchy

[wraith hinge]

sorry for confusion

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[quick bronze]

Why is this wrong

[quick bronze]

I’m gonna crash out I swear this is the right answer

[limpid dawn]

that's actually weird

[quick bronze]

And the teacher says it’s 100% wrong

[limpid dawn]

well g(x) is not continuous at x=-1

[trail ingot]

:hmmge:

the red X is what you're referring to?

[quick bronze]

Yes

[trail ingot]

for all x > 2, g(f(x)) = 2

so i agree with the red X

it should be 2

[wraith hinge]

its just 2

[wraith hinge]

ye

[quick bronze]

Wdym

I don’t quite get it

[trail ingot]

what is g(f(2.1))?

what is g(f(2.2)? g(f(2.0001)?

nope

[quick bronze]

g(-1) right

[trail ingot]

yep

it's always g(-1)

[quick bronze]

Ok yeah but how does it get to 2 as it’s a limit and it describes the motion as it’s approaching

[solid kiln]

bacc

[trail ingot]

that's not what's going on here

[trail ingot]

g(f(x)) evaluates to 2 for all x > 2

it's like taking a limit of a constant function

[quick bronze]

Ohh

OHHHHH

SHIIIIITTT

[trail ingot]

$g\circ f$ is constant (on $(2,\infty)$)

[solid kiln]

layla is not harper

[quick bronze]

It’s because it’s not approaching

It just hits the same value of 2 over and over again

When it approaches from right

Ahhhhh okkk thanks

[trail ingot]

[quick bronze]

how to close ticket

.close

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[quick bronze]

Okkkkk

Ty all

[limpid dawn]

I thought about it, you were right

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[exotic pine]

if (x-5)^2+(y-2)^2=7^2, find the maximum of x^2+y^2

[exotic pine]

how fo you do this?

[violet gust]

try plotting this

[frozen plover]

parametrise in terms of sin and cos

differentiate then

[exotic pine]

i dont know differentiation

[frozen plover]

complete the square then

[violet gust]

In 2D, what does this (x-5)^2+(y-2)^2=7^2 represent?

[exotic pine]

circlr

[violet gust]

Great

[violet gust]

and whats the eulidean distance?

[exotic pine]

huh

[violet gust]

d^2 = (x1-x2)^2 + (y1-y2)^2 right?

where d is the distance between (x1,y1) and (x2,y2)

so, you can think of x^2 + y^2 as distance from origin

[exotic pine]

hm

[violet gust]

So can you rephrase the problem in context of distance from origin?

[exotic pine]

wait

is it 7+sqrt(29)

[violet gust]

yep

|| (radius of this circle + distance between centre and origin) ||

[exotic pine]

wait is it 7+sqrt(29) or (7+sqrt(29))^2

[violet gust]

squared is the thing you want

[exotic pine]

so its 49+29+14sqrt29=78+14sqrt29?

[violet gust]

should be the answer for max x^2+y^2

[exotic pine]

ok thank you

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[exotic pine]

for positive integers a,b find (a,b) such that
$$\frac{1}{a}-\frac{1}{b}=\frac{192}{4009}$$

[solid kiln]

Skill_Issue

[exotic pine]

i tried to combine the fractions and make a diophantine equation but got $$xy+y\frac{4009}{192}-x\frac{4009}{192}=0$$

[solid kiln]

Skill_Issue

[violet gust]

you gotta factorize 4009, that would be huge progress

so you can directly correlate b-a to 192 and ab to 4009

I cant find the factors mentally, so id leave it to you

[exotic pine]

19×221

[violet gust]

I guess you are done ¯_(ツ)_/¯

221-19=202 so none exist

wait no, 221 is divisible by 13

4009 isnt

its 211

[exotic pine]

oh yea

[violet gust]

so 19,211 is the pair

[exotic pine]

wait what

[violet gust]

211-19 = 192

[prime crystal]

and 211*19 = 4009

[exotic pine]

oo

thanks

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[sly basalt]

not related to maths, but I need help from someone who studies.

when solving a question, I do mistakes sometimes. but my matter is that, suppose I apply a wrong concept in middle of solution. and I before that I had already applied another wrong concept in beginning of solution. and somehow those two wrong concepts cancel out each other, and I end up with correct answer. and when I recheck, I realise that I've done a huge blunder. but I'm seeing this trend in various subjects, and it happens all the time. I do more than one mistake and I get right answer.

and why I get right answer, is due to those complement things. like 1/sin60=cosec60, and sin60=cos30, etc and more things like this in subjects other than maths.

it happens in my physics related subjects. but concept is still wrong, and solution should be treated as wrong

[cosmic meadow]

The only logical way for me to solve this problem is by knowing the definition of what you are using at that moment.

[sly basalt]

I'm seeing this since many years

different subjects

[wary basin]

usually I'd think a teacher would mark you wrong

or you would notice the mistake within other questions

in my experience most of the time I wouldn't get the right answer making mistakes

I would also suggest checking for solutions if they have them

[sly basalt]

I solve a question with full enthusiasm and get correct answer. when I check it after some time, (maybe few hrs or days), I realise that I've understood the question incorrectly. but my concepts are still wrong! I don't want this kind of luck thing 😭

[wary basin]

as long as you have the solutions

I would probably check more regularly

but if you do enough practise examples this shouldn't be so much of a problem

because when you do get a question wrong you should be able to identify the error at that time

all in all the best way to avoid it is probably to be more proficient with both practise and checking your solutions imo

[sly basalt]

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[hoary glacier]

can someone help me in this please

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[next vapor]

I have the sequence an+1=root(1+2a1) where a1 is 1

[next vapor]

I need to study its mono

Monotony *

[whole coral]

Do you mean $a_{n + 1} = \sqrt{1 + 2a_n}$, rather?

[solid kiln]

@whole coral

[next vapor]

Yes

And a1 =1

@whole coral any updates?

<@&286206848099549185>

[tacit sleet]

yes

[next vapor]

Can you help me solve this problem

[next vapor]

.

[whole coral]

Maybe induction might help?

[next vapor]

Good idea but how do i use in this example

[whole coral]

I would play around with the first few terms, see if you notice any order in those

Then if you suspect that it's increasing/decreasing, try and make an induction argument with that and use one of the examples you found as your base case

[next vapor]

Yes i tried that

And i am sure that it is increasing

But i just don't know how to continue 😢

[whole coral]

@pulsar dust hiii

[pulsar dust]

hello chartbit

[whole coral]

And, well, if you think it's increasing, then for the induction argument, assume, say, that $a_{n - 1} < a_n$, and from there, try and show that $a_n < a_{n + 1} (= \sqrt{1 + 2a_n})$

[solid kiln]

@whole coral

[next vapor]

Maybe we can do it together step by step?

I found a2 to be root3

And a3 root(1+2root3)

So a3 is bigger than a2

So how do i continue?

[whole coral]

Have you done proof by induction before?

[next vapor]

Yes

But not like this

[night patio]

is it bounded above (the sequence)

nvm ignore i was responding to wrong channel oops

[whole coral]

Well, let's take this as the base case, it's pretty alright that a2 = sqrt{3} > 1 = a1

The idea is that we now wanna show that if for a given $n$, we have $a_{n - 1} < a_n$, that we also then must get that $a_n < a_{n + 1}$

[solid kiln]

@whole coral

[whole coral]

For which, it's actually quite easy, remember how you recursively define the sequence here, so the argument would be something like-

We assume that $a_{n - 1} < a_n$. Now,
[
a_{n + 1} = \sqrt{1 + 2a_n} = \ldots
]

[solid kiln]

@whole coral

[next vapor]

Ok

[whole coral]

Finishing off the proof should hopefully be alright, it's practically almost immediate and there are few hints I can think of that don't immediately give it all away

[next vapor]

Ok give me an other hint?

And i will try to do it myself

[whole coral]

There's only really one place you can make use of the fact that a_{n - 1} < a_n...

[next vapor]

Do i need to extend the RHS utill i get a1

Is that it?

[whole coral]

No, you don't, you literally want to have a sequence of steps here that lead to $a_{n + 1} = \ldots > a_n$

[solid kiln]

@whole coral

[whole coral]

That's induction, using the truth of the statement a_{n - 1} < a_n to then get to a_n < a_{n + 1}

[next vapor]

Maybe an other hint?

[night patio]

chartbit cant relaly hint anymore without giving the answer

it is just performing the induction now

[next vapor]

Ok

[night patio]

are u confused how to do induction ?

[whole coral]

I will say, to think about the square root and how it behaves...

[next vapor]

Yrs

Do i need to square both sides?

Ok

[whole coral]

(you can, but you really don't need to-)

[next vapor]

Do i need to replace an+1 with root(1+an)

Like so?

[night patio]

that is your RTP essentially

[whole coral]

(but be very careful, that requires justification, and ideally should depend on the fact that a_n > a_{n - 1}, something you haven't yet used)

[next vapor]

Ok i think i got it

Yes i got it

I think i did it correctly?

Pls confirm if i did it right

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[delicate belfry]

Factorize $$z^3+z^2-4z+6$$ which has a root of $1+i$

[solid kiln]

Good

[delicate belfry]

I got $(z-1-i)(z^2+2z+iz+3i-3)$, is it able to factorize it further?

[solid kiln]

Good

[night patio]

do you notice anything about the coefficients

of the polynomial

||if all coefficeints are real, then complex roots occur in conjugate pairs||

[delicate belfry]

Not really. I tried quadratic formula, and it gave some negative imaginary inside squareroot.

[night patio]

this is just a theorem so probably should be familiar with it

[delicate belfry]

Alright, I'll look at into that. Thanks:)

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[tardy hemlock]

This looks fine, right?

[somber ginkgo]

yeah seems fine

[somber ginkgo]

i think you meant |a_n - 0| rather than |a_n - a| for the first line

[tardy hemlock]

Yeah, sorry

Thanks

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[somber ginkgo]

nw!

[nova spire]

wrong way around

it's N_epsilon'

since you're trying to show that for every epsilon', sqrt(a_n) < epsilon' after some N_epsilon'

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[simple haven]

Did you have a question?

[trim jolt]

@lavish nebula Has your question been resolved?

[lavish nebula]

not really. im just trying to figure out how to solve these two

[simple haven]

!status

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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

[lavish nebula]

i don't know where to begin

[simple haven]

Ok, you can begin the first by completing the square and getting a circle formula in the form of (x-k)^2 + (y-h)^2 = r^2

From there you know the radius and center of the circle.

[lavish nebula]

yes i did finish upto that point

[simple haven]

If you have two tangent circles what can you say about their centers and the tangent point?

[lavish nebula]

equal radius

[simple haven]

No

[lavish nebula]

the distance from the center to both tangents are equal

[simple haven]

Both tangents? There's only one

[lavish nebula]

ah yes my bad

point is the distance from center would be equal to the radius

[simple haven]

Specifically what can you say about the line from the center of circle C to the tangent point, and the line from the other circle to the tangent point?

[lavish nebula]

the distance between the centers is equal to the sum of radius

[simple haven]

True which implies that this is the same line

The three points are colinear

[lavish nebula]

wait so

[simple haven]

So if you make a line equation using the center and the tangent point, what other piece of information can you use to determine the circle C?

[lavish nebula]

create an equation then find perpendicular distance?

[simple haven]

Perpendicular distance to what?

[lavish nebula]

the equation of that line would be the radius subtracted from the total distance between the centers

im sorry im a bit confused

[simple haven]

The equation of the line would be y = mx + b

[lavish nebula]

yes

[simple haven]

We don't yet know the distance between the centers

Because we only know one of them

[lavish nebula]

wait it also passes through the other point

(4,0)

we substitute that

[simple haven]

The circle does yes

Into what?

[lavish nebula]

never mind

[simple haven]

Ok so

We have a line

We know the center of the circle is on this line

[lavish nebula]

yeap

yes

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[simple haven]

And we know that two different points are equal distance from the center

[lavish nebula]

.reopen

[simple haven]

.reopen

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✅

✅

[lavish nebula]

equal?

[simple haven]

So we have three unknowns, h, k, r for circle C

[simple haven]

Both the tangent point and the point (4,0) are on the circle

[lavish nebula]

yes

[simple haven]

So they are equal distance from the center of the circle

[lavish nebula]

yeap

[simple haven]

So we have 3 unknowns, r,k,h for circle C

And we have three equations, the line from the center of the first circle through the tangent point, the distance formula to (0,4), and the distance formula to the tangent point

You think you got it from here?

There's still a lot of math math math and number crunching, but conceptually at least is the path to the solution clear?

[lavish nebula]

yes i think i got it

thank you. any help with the last one?

[simple haven]

Hmmm

I have a notion give me a moment

So you can construct a circle that passes through any three points you choose

[lavish nebula]

yes

[simple haven]

However, I think touching means tangency

Which means the center is on the line x = 3

[lavish nebula]

i was wondering if we can make an equation out of the given information so we can substitute the options to see which one satisfies the equantion

[simple haven]

You can find the entire circle equation with just these two points.

I think

Yes

So notice that k = 3, h = r.

So a distance formula from (1, -2) to the center at (3, h) will equal h

That's one equation in one variable

Then you just plug in each of the points and see if it satisfies the equation

[lavish nebula]

hmmmm alright i think i got it from here

thank you!!

[simple haven]

Though

One second

[lavish nebula]

yes

[simple haven]

You can use a symmetry argument to avoid all math

Circles are symmetric about any line passing through its center

So in particular it is symmetric about x = 3

Which of the points in the multiple choice satisfy (1, -2) when reflected about x = 3?

[lavish nebula]

unfortunately i gotta show the math;-;

[simple haven]

This is math

It's just smart math

Work smarter not harder math

[lavish nebula]

lol ok you're right i'll try that

actually i got two more problems can you help with that as well?

[simple haven]

I need to cook myself some lunch, so I will be a little more terse, but I'll be happy to stick around for a bit

[lavish nebula]

sure sure i won't take much of your time

number 8 and 10

[simple haven]

For 8, the key is the chord will be perpendicular to the radius

For 10 you can get the circle equation in the same way as problem 15 (5?) from the last set

[lavish nebula]

i know the length of the chord and the center of the circle but i dont know how that helps me determine the radius

[simple haven]

The radius passing through the center of the chord (and therefore the center of the diameter) is perpendicular to it

[lavish nebula]

but i don't know the coordinates of the center of the chord how does that help me with the distance?

[simple haven]

The center of the chord is the center of the other circle

[lavish nebula]

yes

[simple haven]

Big circle C, little circle D.

[lavish nebula]

but there is still a portion left after the point of intersection between the center of chord and the center of the circle

[simple haven]

A radius of C passes through the center of D

The diameter of D that is a chord of C is perpendicular to this

Giving 2 points on C

Third point on C is at the end of the radius from before.

Well, actually you don't need the third my bad

You just need the radius, and once you have either endpoint of the diameter you have everything

[lavish nebula]

ah alright. thanks a lot, sorry for stalling you, you can go cook some lunch up for yourself :)

closing the ticket

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[solid kayak]

Idk what to do next in part ii

[night patio]

can u find a value of x such that the derivative yields -51

[solid kayak]

the answer i got in the calc was in decimals

[night patio]

ok so we know the derivative is a quadratic right?

[solid kayak]

yea

[night patio]

3x^2-18x+30 from what u wrote

we can simplify this by just factorising

3(x^2-6x+10)

[solid kayak]

but the values of x would still be in decimals

[night patio]

oh wait i got tripped up

ur original derivative is 3x^2-18x-21 = f'(x)

[solid kayak]

yuh

[night patio]

or if we factorise it 3(x^2-6x-7)

what is the vertex of this quadratic

recalling that vertex of any quadratic is found when x = -b/2a if the quadratic takes on the form of ax^2 + bx + c

you will realise the vertex y value of this quadratic is above y=-51

therefore it cannot reach it

which suggests ...

[solid kayak]

so theres no points

[night patio]

yep

at least from a glance, i havent done this on paper so maybe i made a calculation error in my head

[solid kayak]

wait how does this show that there are no points on the graph that show the normal that is parallel to the line

[night patio]

the line gradient is 1/51 right?

[solid kayak]

yea

[night patio]

so therefore the normal of the tangent at some point of the curve f(x) has to be 1/51

the normal of the normal = gradient of tangent at some point of the curve

so the gradient or derivative has to produce a function value that is -51

so if there exists this point, we should be able to find f'(x) = -51

however, analysing f'(x) as a quadratic will show that the lowest value it will take greater than -51

therefore it will never reach -51

hence this function value doesn't exist

and hence this point doesnt exist

[solid kayak]

how is this supposed to be written down as working

especially the part where f'(x) lowest value is greater than 51

i mean -51

[night patio]

ie:
"The gradient of a line parralel to the given line will take have value of 1/51. If this is the normal to a tangent at a point on the curve f(x), then the gradient value at that point must take on -51 to satisfy the fact that gradient tangent * gradient normal = -1

Equivelantly, we must find a value of x such that f'(x) = -51

however, f'(x) = 3x^2 - 18x - 21, analysing the vertex which is (fill in the blank here), the y value of the vertex is > -51 and the quadratic is concave up as its leading coeffcient is positive.

Therefore, the derivative will always be greater than -51 which means it cannot be the normal to a line parralel to the line 51y-x=19

Hence this point does not exist"

[solid kayak]

ohh i get it now

tyty

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[real birch]

Hi I'm currently practicing finding eigenvectors, I'd like someone to review my solution so I can see if I'm doing it right

[dapper swift]

careful: you should have - (-1)(4 - lambda) = 4 - lambda in the 4th line of your working

[real birch]

It's + because -1 is the third element

wait what

[dapper swift]

the cofactor still has a positive sign there

it alternates, so + - +

[real birch]

I did put a positive

Um wait can you point out where my error is

[dapper swift]

the determinant of (-1, 4 - lambda) (-1, 2)

[real birch]

Ah yea

[dapper swift]

$= -1 \cdot 2 - (-1)(4 - \lambda)$

[solid kiln]

higher's secret brother

[real birch]

That determinant is the third term tho

Which is why I put positive

[dapper swift]

I know so that's not the issue I'm talking about now

[real birch]

Ok so is it correct?

[dapper swift]

no

you wrote down -4 + lambda

when it should be 4 - lambda

[real birch]

oh I got that cuz -(4-lambda) = -4+lambda

The term is multiplied with -1

Oh wait lemme try again

[dapper swift]

and you multiply by -1 again

yeah try it again on a new piece of paper

[real birch]

YOOOO

yea I got it lol

silly mistake 💀

[dapper swift]

yeah miraculously your eigenvalues are still correct

,w {{2, 0, 1}, {-1, 4, -1}, {-1, 2, 0}} eigenvalues and eigenvectors

[solid kiln]

Results provided by WolframAlpha

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

Requested by sooouth.

[real birch]

I assumed I already had 4-lambda in my mind so my steps seemed correct after lol

Ok thanks tho

[dapper swift]

so the lambda = 1 part is fine

[dapper swift]

lambda = 2 is not correct cause of a really dumb mistake, don't beat yourself up haha

$2x_2 = x_1$ means that $x_1 = 2t, x_2 = t$ and so on

[solid kiln]

higher's secret brother

[dapper swift]

so (2, 1, 0) not (1, 2, 0)

[real birch]

why did you introduce t here

[dapper swift]

that's the general form, cause any $t \ne 0$ will also be an eigenvector

[solid kiln]

higher's secret brother

[dapper swift]

any scalar multiple except 0 of the eigenvector works

[real birch]

so if I need x1 I substitute x2 with t?

[dapper swift]

well you can actually just let x2 = 1, so x1 = 2

then (1, 2, 0) * t

[real birch]

doesn't that mean 1,2,0 is correct

wait no

uh yea I don't get it 💀

[dapper swift]

that's only one eigenvector

there are infinitely many

say, (2, 4, 0) also works

[real birch]

I did get one out of infinite eigenvectors for the eigenvalue

I don't understand why it's not correct tho

[dapper swift]

I'm saying it's correct

[real birch]

oh ok

But wolframalphas giving a different answer 😭

[dapper swift]

also another mistake in lambda = 3

you get to -2(x1) + (x2) = 0

again, if you plug in x1 = 1, you get x2 = 2 and x3 = 1

so you should have (1, 2, 1)

or (1/2, 1, 1/2) also works

oh yeah you did the same kind of mistake in all 3

[real birch]

💀💀💀

WHAT

[dapper swift]

yep

[real birch]

so all my evecs are wrong?

[dapper swift]

no

your eigenvalues are all correct

[real birch]

ok so I don't need to change anything right? 😭

[dapper swift]

no, you need to change how you approach things like -2(x1) + (x2) = 0

[real birch]

or did I get the right answer but in the wrong approach

[real birch]

Ok

[dapper swift]

that's your only mistake

everything apart from that is correct

just sub in 1 value for x1

get x2 and x3

then multiply what you got by $t$, where $t \ne 0$ ofc

[solid kiln]

higher's secret brother

[real birch]

So I multiply x2 and x3 with t?

And x1 would be any value, say 1?

[dapper swift]

and x1 also, but yes

[dapper swift]

yeah you can choose any value that isn't zero

[real birch]

Ok

I'll try that approach

[dapper swift]

npnp

[real birch]

Wait holon lemme try again

@dapper swift 😭

How does t help me make my eigenvector

[dapper swift]

great so your eigenvector is (t, t/2, 0), when t is not 0

[real birch]

ok but it's not 2,1,0 tho

[dapper swift]

so when t = 2

you get 2, 1, 0

[real birch]

AHHHHH

DAMN so I was correct anyway

Wait this is like echelon form

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[jolly lion]

need help with boolean expressions
Using the Identities and Laws of Boolean Algebra, simplify the following expressions. List
the specific law used for every simplification.
A∗B∗~C+A∗B+A∗C+A

[violet star]

Using the distributive law we know that your thing is equal to A * (B*~C + B + C + 1). It is basically like taking A out and since A*1=A, we leave a 1 in the last place. The null law states that 1+A=1, which means that the entire thing in the brackets simplifies to just 1. we are therefore left with A*1 which is equal to A by the identity law

[jolly lion]

oh

[jolly lion]

i started out with t3

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[winged epoch]

Hello, for solving a question, I used this website to build my explanation for what I was trying to do. I'm not sure if I worded it correctly, though, since my explanation of it isn't really in matrix form. Anyone know a better way for me to word this or is this just fine?

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[hushed trail]

I just wanted somebody to quickly tell me if my answer is correct or not. Only question on a quiz that I am unsure of.

[dusty sleet]

If you sub x = 1 what do you get for the angle marked as 15(x+1)?

[prime lynx]

Vertically opposite angles

[prime lynx]

@hushed trail

[hushed trail]

Sorry, I figured it out I no longer need help. My bad for opening a help channel.

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[prime lynx]

No problem

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[river willow]

is p^r the size of the normalizer of P?

[amber python]

no

[river willow]

:(

why do they write it like this then???

why not just skip to the last part

because we know that the normaliser would form a subgroup and subgroup always divides the order of the group so we could just skip to the |N(P)| * k no?

[amber python]

[river willow]

???

[amber python]

if all they want to use is that p doesnt divide k then its only necessary to say that P is contained in N(P)

idk what the proof is gonna do though

[river willow]

no idea i havent read so far

spoilers

[river willow]

yea so true

[amber python]

huh

i guess thats one way to do it

[river willow]

ive no idea what the proof is doing

isnt Q... by definition in S

[amber python]

no

[river willow]

as Q is a p slow subgroup

so it is of a p group of maximal order

thus in S

[amber python]

Q is sylow but not necessarily a conjugate of P

[river willow]

and i dont get whats clear about the following line :(

yea but S contains all sylow p groups

[amber python]

thats not known a priori

thats the conclusion of the theorem

[river willow]

no isnt S defined as containing all sylow p groups

[amber python]

no

[river willow]

oh oops sorry

its defined as the set of all conjugates of P

[amber python]

any group action partitions the set being acted on into orbits

same with Q acting on S by conjugation

[river willow]

sorry

Q conjugacy class of each Pi means

QPiQ^-1 or PiQPi^-1?

im guessing the first?

[amber python]

{q P_i q^-1 | q \in Q}

[river willow]

ty

one of these equivalence classes must contain only a single Sylow
p-subgroup, say Pj.

how did they get this

[amber python]

because the conjugacy classes are powers of p

[river willow]

yea but wouldnt they all be the same order

[amber python]

no

[river willow]

why :(

[river willow]

what does this mean anyway

[amber python]

because the orbits don't have the same cardinality

[river willow]

do the other conjugacy equivalence classes contain multiple sylow subgroups? why?

[river willow]

what is the cardinality of an orbit

[amber python]

|orbit|

[river willow]

number of elements in the orbit

[river willow]

how do we know that

[amber python]

conjugacy classes don't all have the same size

[river willow]

yea

we have orbits of different sizes but how do we know one has a conjugacy class of exactly one

[amber python]

they all must be powers of p

[river willow]

yes

[amber python]

S has cardinality k

the conjugacy classes partition S

p doesnt divide k

so p does not divide the cardinality of at least one conjugacy class

so one conjugacy class is p^0 = 1

[river willow]

wait first

the cosets of the normalizer of P is the number of elements in S?

which is exactly the idex of N(P) or in this case k

[river willow]

but what if say we have 8 groups of order 9 where we are looking at sylow groups of order 3 (3^2 is the sylow 3 group). then wouldnt we have like 2 groups in the last partition

or 2 groups of 1 im not sure whats happening

how do we know we would have exactly 1 partition of 1 at the end

[amber python]

nani

are you saying k = 8?

[river willow]

for instance yea

[amber python]

all the orbits must have size powers of 3

so an orbit of size 2 is impossible

[river willow]

cant we have 2 orbits of size 1?

[amber python]

sure

[river willow]

oh

but any orbit of size 1 would satisfy the condition they talk about which is to keep itself invariant under conjugation by Q (normalizer of Q)?

[amber python]

it means its a Q fixed point

[river willow]

elements of Q are fixed points for the Pi under conjugation

[amber python]

no

[river willow]

:(

[amber python]

theres some P_j which is a fixed point under the Q-action

thats what it means for the orbit to be a singleton

[river willow]

theres some i such that elements of Q are fixed points for the Pi under conjugation

[amber python]

no

[river willow]

does that work

:((((((((

[amber python]

Q is acting on P_j

P_j gets fixed

q P_j q^-1 = P_j for all q in Q

[river willow]

sorry how do u defined a being a fixed point of b

ab=a?

or ab=b

[amber python]

when a group G acts on a set X, the G-fixed points of X are X^G = {x in X | gx = x for all g in G}

here, G = Q and X = S

the action is conjugation

[river willow]

the Q fixed points (under conjugation) of some Pi is Pi

[amber python]

no

you just say that P_i is a fixed point

[river willow]

Pi is a fixed point of Q?

[amber python]

of the Q-action

[river willow]

Pi is a fixed point of the Q action

ok thank you

!!!!

❄️

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[glad grotto]

yo

[glad grotto]

I need help with these 3 questions

Anyone?

@sweet grotto ?

Yk how?

[sweet grotto]

I can but I'm preparing for a test myself you can ping helper role and someone will surely help you

[glad grotto]

Alr thanks

<@&286206848099549185>

[limber wharf]

I can help a bit

[glad grotto]

Alr

[limber wharf]

So x intercepts is just where the line crosses the x axis

[glad grotto]

yea

[limber wharf]

Sign of the leading coefficient depends on whether the function points up or down

Least possible degree is how many times it “curves”

Sorry I mean based on

Don’t count the curves

[glad grotto]

Oh

What do u mean by based on?

[limber wharf]

Go on Desmond

Desmos

[sweet grotto]

like sometimes it doesn't curve

[glad grotto]

Yea idk how I'd graph that on desmos

[limber wharf]

Plot a random x^3

[sweet grotto]

it stops then moves

[limber wharf]

Wait I’ll send a video

I suck at explaining

[glad grotto]

Alr

It has 2 turning points

Plus 1 shouldn't that be the leading coefficient?

Or degree

The minimum degree is 3

And it's positive

X intercepts at -4, -0.5, and 1

[sweet grotto]

m sry i rly gtg :(

[glad grotto]

Ur good dw

[glad grotto]

Nvm I got it

This is easy

[limber wharf]

👍👍

[glad grotto]

Ty tho

[limber wharf]

No problem

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[north plaza]

Investigate the effect of the term on simple interest amortized auto loans by finding the monthly payment and the total interest for a loan of $15,000 at
9 7/8%
interest if the term is the following. (Round your answers to the nearest cent.)
(a) three years
(b) four years
(c) five years

[north plaza]

Ik the equation is (P)(r(1+r)^h/(1+r)^n-1))

P=15,000
Anual interest rate=0.09875

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[north plaza]

.close

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[trim cradle]

can we add a 1x4 matrix to a 4x1 one?

[lusty delta]

no

[wary basin]

you can perform matrix multiplication iirc

[trim cradle]

Yeah I got confused by MATLAB sorry

We are required to solve some question but it didn't make sense

[glacial pebble]

I have 1 question, in transformation functions would a horizontal/vertical strecth or shift come first

[trim cradle]

I'll close then you can ask again

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[glacial pebble]

in transformation functions would a horizontal/vertical strecth or shift come first

[wary basin]

I don't believe it matters

but I assume it would be notation of the order assigned

[glacial pebble]

so like h(x)=f(1/3(x-3))-1

it doesent matter if i strecth by 3 or move right 3 first

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[plush breach]

Hello, someone could help me to understand this Ordinary Differential Equation (ODE)?

Basically I want to know why the dt disappear when I do the integral and the ds becomes s(t).

[wary basin]

the dt disappears because it's been integrated

[wraith hinge]

you integrate 1ds

[wary basin]

^

and then the integral of 1 ds is just s or s(t) in this case for notational purposes

[wraith hinge]

seperation of variables is pretty confusing some times because u kind of multiply by dx and stuff.. but it works so u gotta just accept it lmao

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[echo jasper]

pls

[echo jasper]

an example

john dawton

like the bottom of the slide says this rule is not so true

so idk why i need to know it

pls i cant understand it

like is it trying to say there is diffferent elements because of atoms

[night patio]

this feels like it isn't that well fitted for math discord

[river willow]

math

yea it isnt true

[echo jasper]

oh

[ionic pendant]

every element has a different average mass per atom. but individual atoms of the same element can have different masses due to being different isotopes

[river willow]

so elements are defined by the number of protons

[echo jasper]

Oh

[river willow]

but u have isotopes and so on

[echo jasper]

oh the isotopes make this rule not true

[river willow]

isotopes have a different number of neutrons thus a different mass

[echo jasper]

Ok

[river willow]

theres also if u have a different number of electrons, electrons have mass as well though its a thousandth or so of a neutron

with different number of neutrons and electrons they have different properties as well

[ionic pendant]

dalton wouldn't know about this because it all averages out over an uncountably large amount of atoms

[river willow]

what

[echo jasper]

So whats the point of knowing dawton ahah

dalton

[river willow]

are we going into chemistry history D:

[echo jasper]

lets not 😄

[high heron]

surely it's countable... /j

[ionic pendant]

carbon is 99% carbon 12 and 1% carbon 13 making an average mass per atom of 12.01 g per mol. there are 6 x 10^23 atoms in a mol so an individual mol randomly having a few more or a few less atoms of 13C would not affect the average mass per atom very much (it would have to be a substantial difference in isotope composition)

so it's a distinction which is very important for individual atoms and less so for the amount of substance you can see

[river willow]

i see that is cool

well i would say that they have similar mass but the key point here is the quantifier all at the start

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[mystic goblet]

Why is x^3 - 2x + 1 not an odd function? what does the C value change?

[bright quarry]

consider f(-x)

if it were odd then it would be equal to -f(x)

is it?

check

[mystic goblet]

no

[bright quarry]

then that answers your question

[mystic goblet]

in these questions it says use property f(-x) = f(x) to determine if its even and in another question to determine if its odd use f(-x) = -f(x), will it always be these two? f(-x) = -f(x) for odd and f(-x) = f(x) for even?

[bright quarry]

yea

that’s precisely the definition of even and odd functions

[mystic goblet]

ok thanks

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[inner seal]

I dont want to just substitite it. I want to manipulate the question itself can anyone teach me how?

[wraith hinge]

equare the two equations using a^2 + b^2

i suppsoe?

[lusty delta]

multiply the second equation by 2/3

[inner seal]

I did but

[inner seal]

Thats fucking clever

[lusty delta]

👍 problem should solve itself from there

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[gloomy garnet]

yo

[river willow]

yo

[wet apex]

yo

[gloomy garnet]

yo

someone

anyone

[wet apex]

I got u

[gloomy garnet]

tanks

[wet apex]

Do u know pythagorean theorem

Ofc u do

[gloomy garnet]

ya

a sqaure b sqaure equal c 2

c sqaure

but the area thing

and perimeter

is confusing

[wet apex]

Ok i get it

What is a square?

[gloomy garnet]

wym

a shape

[wet apex]

A square is a quadrilateral with 4 equal sides

[gloomy garnet]

ya ya

yep

[wet apex]

Do you know what perimeter means

[gloomy garnet]

ya

add all of the sides

[wet apex]

So if each side is equal and the total is 100

[gloomy garnet]

10

wait

25

oops

25?

[wet apex]

👍👍👍👍🔥

[gloomy garnet]

ok

so

that’s c sqaure right

um

[wet apex]

Wait

[gloomy garnet]

i got 18

[wet apex]

Wait no

[gloomy garnet]

18. something

[wet apex]

Don’t write 100 inside the square

Write 25 on each side of the square

[gloomy garnet]

ok

what next

[wet apex]

replace 10 with 25 btw

[gloomy garnet]

ok

[gloomy garnet]

15 stays?

[wet apex]

Yes

[gloomy garnet]

ok

[wet apex]

so you alr have 2 sides of yhe triangle

right

[gloomy garnet]

yes

i have c sqaure and a sqaure

i’m looking for b sqaure

or the area of 3

[wet apex]

Red is 25

Green is 15

[gloomy garnet]

ya

so i erase

100

and put 25

and do the same for 225

and put 15

[wet apex]

No

[gloomy garnet]

o

[wet apex]

What’s 25 squared

625

[gloomy garnet]

ye

[wet apex]

That’s the area of square1

100 is just the perimeter

[gloomy garnet]

625-225

[wet apex]

👍👍👍

[gloomy garnet]

400 the answer

thanks

[wet apex]

Square root

[gloomy garnet]

20

but it’s asking for the area so it is 400

nice

[wet apex]

Never mind

Yeap

Yep

Nice job

[gloomy garnet]

yep yep thanks

thanks for helping

can u friend me incase i need help again

[wet apex]

Np

Sure

[gloomy garnet]

ok

bye

.close

[trim jolt]

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Closed by @gloomy garnet

Use .reopen if this was a mistake.

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[rich stag]

2. , 4. and 6.

idk how to find horizontal and slant asymptotes

[hybrid girder]

I was dead asleep when I was doing this and got this entire section wrong.

I just need someone to double check my answers on the orange sticky note if it’s correct, and which one is wrong. Thank you!

[rich stag]

Grr

[trim jolt]

@rich stag Has your question been resolved?

[rich stag]

<@&286206848099549185>

[west estuary]

Im here now...daddys here...

Ok so

SO WE GPTTA DO

PARENTHESIS FIRST OK

WE GONNA DO PEMDAS WITH THIS

Alr bro

[west estuary]

I got you

OH

Uh

So we gotta divide though because its a fraction i think

[west estuary]

Sp for the

So we gonna

[rich stag]

i think we r supposed to divide everything by |x| and -|x|

but idk y

[west estuary]

Well we about to find those

So y is

Y is kind of yknow rude and standoffish

So when we take

Stay with me now

I hate math with letters when am i gonna use this

LETS USE PHTOPHTOMAYH

PHTO

fart

[west estuary]

Vrother use phtomath

[rich stag]

<@&268886789983436800>

[west estuary]

Will an image of a cat help u rocus

WHAT

Bro pinged the mods on me aint no way

[lusty delta]

[west estuary]

Gusy..

She ask for help

Fine u wont get none

Yeah

Ok guys

[knotty oriole]

This is unnecessary. Keep things civil here.

You're muted for the day because of this

[knotty oriole]

If you don't know how to help somebody please try to avoid clogging up their channel.

[west estuary]

I did know i was about to tell her bro

[knotty oriole]

Seems a lot like you're just trolling in the help channels based on your chat history

If you are that's pretty shitty.

If you're not you should really evaluate whether you're explaining things as well as you think you are.

[west estuary]

Why are you the one getting mad

Please do not swear at me

[knotty oriole]

You'll survive.

[west estuary]

Guys im feeling targeted...

[knotty oriole]

You do come across disingenuous here.

[trim jolt]

@rich stag Has your question been resolved?

[subtle fulcrum]

mitch…..

[subtle fulcrum]

i can help u

horizontal asymptotes are just based on the power

for question 4

[rich stag]

Yayaya

plz question 2

[subtle fulcrum]

sorry for going out of order it’s what i saw first 😭

alr alr

[rich stag]

Yayaya

[subtle fulcrum]

alr so for 2

ur looking at long run behavior right

[rich stag]

wat

[subtle fulcrum]

sorry i’m reposting

so i don’t have to scroll

[subtle fulcrum]

since the x is going towards infinity you’re just trying to determine what the graph is doing as it reaches infinity

aka long run

the horizontal asymptote of 2 is just 0

[rich stag]

bos

how

[subtle fulcrum]

bc when you take the square root of the denominator and make it (4x^2 + 3)^1/2

ur essentially simplifying to 4x + 3^1/2

and since 4x is leading term in denominator

and 2x is leading term in numerator

you can compare degrees

and if the degrees are equal (they’re both 1 here) then the horizontal asymptote is 0

wait

no it’s not 0

[rich stag]

Fr

[subtle fulcrum]

it’s the ratio of the coefficients

okay i had the right idea i was just side tracked

so the horizontal asymptote should be 2/4

[rich stag]

i heard ur suppose to divide every term by |x| and -|x|

[subtle fulcrum]

absolute value?!😨

[rich stag]

Umm no

umm I don't think I got it right Omg

[subtle fulcrum]

freak

that was my next guess!

but yes think abt the degrees

[rich stag]

Oh ya it is 2x/|2x| and negative

[subtle fulcrum]

i’m not locked in enough

[rich stag]

so 1 for the infinity things

and neg

idk why we divide everything by the degree of the denominator

[rich stag]

<@&286206848099549185>

wat

[rich stag]

<@&286206848099549185>

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@rich stag Has your question been resolved?

[trim jolt]

@rich stag Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hot ore]

number 55

i got for the tangent line

1.354x + 0.646

which is correct

i verified in desmos

but i now need the normal line

how can i find the normal line from this

im not used to decimals

[sacred sapphire]

tangent line slope * normal line slope = -1

[hot ore]

multiply?

so then

1.354 + 1 = 1/normal line slope

did i do that right

$1.354 + 1 = \frac 1 {normal line slope}$

[solid kiln]

Devil Wears Prada

[sacred sapphire]

+1?

[hot ore]

oh wait

cant it be

normal line slope $= -\frac 1 {tangent line slope}$

[sacred sapphire]

-1/tangent yes

thats the slope for normal

[solid kiln]

Devil Wears Prada

[sacred sapphire]

then just use point slope formula to get normal line