help-49

[viral dagger]

ohh

so instead of d-1 its d?

cause a is from 1 to b-1 but e is from 0 to b-1

[round parcel]

Yes, that's right.

[viral dagger]

8622?

[round parcel]

Right.

[viral dagger]

.close thanks

[midnight plank]

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[round parcel]

You're welcome.

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[olive matrix]

take a wild guess

[prime hornet]

posted something banworthy

or a spammer

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[viral dagger]

find the smallest positive integer k such that sqrt(k^2-23k) is an integer, for k>23

[viral dagger]

i cant find where i went wrong?

[west badge]

what up gang

[viral dagger]

k would be 267 btw if i continued this

[west badge]

im not even here for math im just bored

[olive yew]

Let k^2-23k=a^2

[viral dagger]

.close bruh nvm

[midnight plank]

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[heady plume]

(2k - 23)²

[viral dagger]

yeah thats where i went wrong

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[mossy spindle]

The ans is 128^5 - 127^5 . I tried another method where i choose 1 place for the @ to be and then arrange the other 4 ascii char . then i choose 2 places fot the @ and then arrange 3 ascii char so and so forth . I have given the expression in the second image , whats wrong with it ?

[rough birch]

it should be 127 in all of the bases instead of 128

that will fix the formula

you have

[mossy spindle]

oh yes sorry that was a really dumb mistake 💀

thanks

.close

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[last slate]

MAKE ME AN ADMIN FOR NO REASON!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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[trim sundial]

I need help on a question about 3D vectors

[trim sundial]

Here is my working which seems to be mostly fine because I'm getting answers similar to the mark scheme now

but I don't know why I'm still getting it wrong

I am getting the coordinates (64/7, -4/7, 29/7

The answer scheme shows (64/9, 4/9, 19/9)

[amber zenith]

So i kinda forgot about this stuff

But if i remember correctly

U are not allowed to swap the minus and plus for the vector perpidicular on the other

In 3D

[trim sundial]

you mean to use PC-->???

[amber zenith]

Cuz normally in 2D u would do [2 1] is perpendicular to [-2 1] right?

[trim sundial]

rather than CP--> for the dot product

[amber zenith]

Ye if i remember that dont work

Let me check rq

[trim sundial]

but if it's perpendicular, any angle there is 90° anyway

so the dot product would be equal to 0

as cos(theta) = (a • b)/|a||b|

[amber zenith]

aah ye mb

[trim sundial]

if i use any direction here, it should equal 0 anyway

[amber zenith]

Tbh then i forgot too much already ffs

😭😭

Sorry cant help you then good luck tho!

[trim sundial]

well I've figured out that lambda should be equal to 14/9

but i need to know how to get to it

bc something clearly has gone wrong

AHH shit

i noticed

it should be 2x - y + 2z = 18 😭

oki got it

thanks other guy for trying 🙏

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[median arrow]

Pouvez vous m'aider ?

[median arrow]

Cet exercice

[heady imp]

Vous devriez l'écrire en anglais et en LaTeX, pouvez-vous ?

[midnight plank]

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[vocal briar]

a boy is flying a kite. The kite is released at 10m/min. It moves horizontal and at height 150m above ground. Find rate at which kite is moving when it is 250 m away from boy.

[vocal briar]

How do I solve this?

I interpreted it as a triangle

where L is the length of kite string (hypotenuse) and the 150 is the height and x as the distance from the boy

i got

L^2 = x^2 + 150^2

idk what to do now

[olive matrix]

now you need to find the rate the kite is moving

differentiate that equation wrt time

[vocal briar]

how do I do that?

doesn't t have to be in the variable for it to be differntiable?

oh u mean implicit

[radiant roost]

yeah implicit

[vocal briar]

oh so now I have rate

dx/dt

= (l/x) dl/dt

can we assume the rate of kite being released is constant?

i thought the rate at which the kite moves is related to the rate at which the kite is released

since the answer on my page says its 12.5m/min

so how can the rate be constant if the answer is different

[midnight plank]

@vocal briar Has your question been resolved?

[midnight plank]

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[frigid solstice]

This determinant should be only be defined with -1 < x < 1 right?

If x = {1, -1}, the denominator will be 0, and this won't be defined

[dusty portal]

"determinant"?

[frigid solstice]

Oppsie

[drifting birch]

by determinant do you mean denominator

[frigid solstice]

I ment derivative

Sorry learnt this in french

[dusty portal]

yeah, g' wont be defined outside of $x\in]-1,1[$

because of denom restrictions and such

[frigid solstice]

Then can I ask how my maths book got this?

[grand pond]

;(

[dusty portal]

that looks 🤮

[frigid solstice]

We will be basically making a sign table (called this?) with the derivative

[frigid solstice]

Yeah...

Plus I think they got something false in the book

[drifting birch]

what

[dusty portal]

wait

[frigid solstice]

Well https://fr.wikipedia.org/wiki/Tableau_de_signes

[dusty portal]

they tested -infty?!??!

[frigid solstice]

I don't fucking know

[dusty portal]

yeah makes sense

[frigid solstice]

I think this maths book is wrong :(

[dusty portal]

you're trying to find when g is increasing and decreasing?

[frigid solstice]

Yup

So we take the derivative of g, g' and we study it's sign

[dusty portal]

yes

[frigid solstice]

When its positive, g is increasing etc

But how can I test values smaller then -1 and bigger then 1?

[dusty portal]

no clue.

[frigid solstice]

Aww

I'll just say that the book is wrong

[crude galleon]

Don't trust ; (, he is inexperienced.

[frigid solstice]

Uh ok

New question

Why is the function of sqrt(x) only positive?

Isn't it +- a number

.close

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[tepid breach]

Not sure how to obtain probability at the very end

[languid lance]

hi

[tepid breach]

Hiya

[languid lance]

sup

[tepid breach]

Any ideas

[midnight plank]

@tepid breach Has your question been resolved?

[tepid breach]

<@&286206848099549185>

[heady imp]

Markov chains...

!status

[midnight plank]

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

[tepid breach]

2

[heady imp]

What have you tried then ? You defined the MC ?

Defined the player's fortune ?

[tepid breach]

I've defined a MC over the state space {(Z,{1 or 2})} where the first entry defines player A_1's money and the second entry the player who will roll next

Both variables are needed as the probabilities change depending on who is tossing the coin

[heady imp]

You have a transition matrix ?

[tepid breach]

Yes I know what the entries are

[heady imp]

I am not sure how to formalize it

Could we define a stopping time

Where
T would be such that
I think they ask you for :
If x is the A1's initial fortune, y the one of A2
T = inf n≥1 s.t. X_n = (x,1) and X_(n-1) = (y, 0)

[tepid breach]

I defined stopping times T_{(0,1)} and T_{(0,2)} as my two stopping times

is your 0 and 1 my 1 and 2 for each player?

[heady imp]

Could we merge the stopping times to insure we have a time n where at n-1 the player 2 have its original fortune and it is the turn of the 1st player to play
and at n the player 1 have its original fortune

[tepid breach]

I think the idea is that it's a zero sum game? So if at time n-1 player 2 has their original fortune, then so must Player 1

So for example if a Head is rolled, P1 gets +1 money and P2 -1 money

[heady imp]

if we assume it is a zero sum game yes

and players have infinite money

[tepid breach]

They never say the game must end so I imagine we need some sum to infinity?

[heady imp]

We may inspire from exercise 6

finding a relation that satisfies your probability

[tepid breach]

I was able to create recursion relations relating to this problem but they were all in two variables, which is something we have not covered so I thought there must be another way

The issue was that the chain also depended on who was tossing the coins

[heady imp]

Since it is a zero sum game , can we define the Markov chain only on one player fortune

[tepid breach]

We could but then the next term in the chain would depend on the previous 2 so it is not Markov?

[tepid breach]

<@&286206848099549185>

[midnight plank]

@tepid breach Has your question been resolved?

[midnight plank]

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[crimson snow]

Hey can you explain in detail how to differentiate d(x+y)/dx

[prime hornet]

is y a constant, or a function of x?

[crimson snow]

Not both i guess

Both are different variable i think

[prime hornet]

that doesn't answer my question

is y a constant like 3 or 9 or 932, or is it a function y(x)?

[crimson snow]

That's the ques

[prime hornet]

that's quite a different question from what you initially asked

[crimson snow]

I'm getting stuck at arrow it shouldn't be 1+dy/dx it should be 2+dy/dx

[crimson snow]

Not really

I asked how do i differentiate (x+y) wrt x

[prime hornet]

you didn't clarify whether y was a function or a constant

that was my gripe

[crimson snow]

It isnt given in ques

[prime hornet]

it is

y is a function of x

[prime hornet]

anyways, you differentiate as you would for any sum

,, \frac{d(x + y)}{dx} = \frac{dx}{dx} + \frac{dy}{dx}

[grand pond]

fallenstars

[prime hornet]

the derivative of a sum is the sum of derivatives

[crimson snow]

Ok so what will be the answer

[drifting birch]

Automatically assumed when it asks for dy/dx

[prime hornet]

I encourage you to try it yourself

[crimson snow]

it should be 2

2 ×dy/dx

[drifting birch]

how come?

[crimson snow]

Tell me where i'm going wrong so dx/dx will be 1 and then dy/dx will also be 1 because y is a variable

[drifting birch]

y isn't a variable

It is asking for dy/dx

Therefore y must be a differentiable function wrt x

[drifting birch]

dy/dx ≠ 1

y is a function of x

[cedar coral]

,rotate

[crimson snow]

Yeah but what is i'm not understanding in this ques

How is y and x are related here

[cedar coral]

you dont know and dont care

[crimson snow]

This much i did it but copied last bit 1+dy/dx but don't know how

[drifting birch]

it is asking for dy/dx, therefore y MUST BE a function of x.
if you have the equation $x+y=e^x$ and it asks you to find $\frac{dy}{dx}$ then it is assumed that y is a function of x.

[grand pond]

AkitoLite

[cedar coral]

well it doesnt have to be

it's assumed to be

[drifting birch]

Is this sarcastic😭

[crimson snow]

Ok if y is the fn of x so it'll become zero here

[drifting birch]

huh?

take the example I gave you

$x+y=e^x$, find $\frac{dy}{dx}$

[grand pond]

AkitoLite

[drifting birch]

you subtract x both sides and differentiate and obtain $\frac{dy}{dx} = e^x-1$

[grand pond]

AkitoLite

[drifting birch]

Notice how the dy/dx doesn't become some random constant

Same thing for your question

[cedar coral]

focus only on the very last factor.
The derivative of (x+y) is the derivative of x, which is 1, plus the derivative of y, which you dont know (since you're not given y), dy/dx

[crimson snow]

Did i do wrong here?

[cedar coral]

that is correct

[drifting birch]

It's correct, you apply the same rules with the other question

Except you do chain rule but you've done it already

[crimson snow]

I wasn't multiplying dy/dx with 1 and was thinking 1+1 would be 2

I'm just stupid

[cedar coral]

,rotate

[crimson snow]

I was thinking it would be 2

It should be like that

[buoyant yoke]

this is correct

[crimson snow]

Thanks a lot y'all i've wasted so much of your time

For so stupid small silly mistake

[crimson snow]

Thanks to you once more 😁

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[neon pollen]

i need help with this math work and im so confused

[dusty portal]

what do you konw about congruent triangles

!status

[midnight plank]

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

[crude galleon]

Which part do you need help with?

[neon pollen]

Ik nothing i just started school and this is what the teacher gave me

[neon pollen]

all of it.

[dusty portal]

k

just know that congruent triangles have the exact same angles and side lengths

so for a triangle ABC and a congruent triangle DEF you would have AB=DE, BC=EF, AC=DF

etc

[neon pollen]

okay

so how would i solve these?

.close

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[dusty portal]

Find all solutions, for $f:\bR\mapsto\bR$ of $$\int f(f(x)+x)dx=f(f(x))+C$ have no clue what to do here

[grand pond]

;(
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusty portal]

all i can say is that $f(x)=0$ is a solution but idk the others

[grand pond]

;(

[quartz cairn]

Try differentiating both sides?

[dusty portal]

so $f(f(x)+x)=f'(f(x))\cdot f'(x)$?

[grand pond]

;(

[dusty portal]

then?

[quartz cairn]

Hmm

Honestly I thought that would do something but ig not

I'm clueless too

[dusty portal]

breh

a freind gave me this

he said there is a nice solution if you look closely

[quartz cairn]

Hmm

If x = 0

Then int[f(f(0))] = f(f(0)) + C

Makes me think f(f(x)) is somehow e^x

Oh wait that doesn't work

Doesn't make sense, nvm

[dusty portal]

wait

could f=e^x?

[buoyant yoke]

this is not correct

[quartz cairn]

How so?

[buoyant yoke]

chain rule

let g(x)=f(x)+x

[dusty portal]

huh?

[buoyant yoke]

then try again

[quartz cairn]

But this isn't the derivative of that

It's the derivative of an integral

[dusty portal]

im confused now

[quartz cairn]

So should just give us the integrand

[dusty portal]

FTC says d/dx integral of f(x)dx=f(x)

[buoyant yoke]

f(g(x))=f'(g(x)) * g'(x)=f'(f(x)+x) * (f'(x)+1)

see the difference?

[dusty portal]

what

[quartz cairn]

One solution could be f = C, where C is any number

[dusty portal]

ok let me just try f=e^x for my original eqn

[quartz cairn]

Bro that's a different thing

Left hand side is d/dx of int[f(f(x) + x)]

[dusty portal]

so it would be exp(exp(x)+x)=exp(exp(x))*exp(x)

[buoyant yoke]

ah, mb sorry

[dusty portal]

which is $e^{e^x+x}=e^{e^x}\cdot e^x$

[grand pond]

;(

[dusty portal]

lOOLLOL

IT WORKS

[quartz cairn]

Huh

[dusty portal]

no way lol

let me check with him

[quartz cairn]

d/dx(exp(exp(x))) = exp(exp(x))*(exp(x)) = exp(exp(x) + x)

Huh

Maybe it does work

[dusty portal]

try to integrate?

$\int e^{e^x+x}dx$

[grand pond]

;(

[dusty portal]

split

$\int e^{e^x}e^xdx$

[grand pond]

;(

[dusty portal]

u-sub u=e^x

so e^u

e^e^x

no way

[quartz cairn]

Lmao

Ok

[dusty portal]

he said its right

[quartz cairn]

We also know f = C is another solution

[dusty portal]

[quartz cairn]

Try e^e^x + C

For f

[dusty portal]

i will proof it!

[dusty portal]

ok

[quartz cairn]

Oh wait that doesn't work

[dusty portal]

$\int Cdx=Cx+D$?

[grand pond]

;(

[dusty portal]

uh oh

[quartz cairn]

Yea that's not an answer then

Mb

[dusty portal]

ok

but danke

good idea

.close

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[grand pond]

;(

[dusty portal]

is this a good proof?

😔

[midnight plank]

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[dusty portal]

what in the

bro i just wanted to delete the latex for visibility purposes

smh

.reopen

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[slim portal]

The value of x in (x² - x + 4)(x - 2) ≥ 0 is

[slim portal]

I can't factorize the quadratic equation hmmm

[barren onyx]

did u apply the quadratic formula

[slim portal]

(-b ± √(b² - 4ac))/(2a)?

[main current]

Just checking the discriminant, this doesn't factor

[slim portal]

Can I use that?

[main current]

The quadratic has no real roots

[slim portal]

Wait2

Discriminant < 0

Imaginary solution

Hmmm

[main current]

In other words, x² - x + 4 is positive everywhere

[slim portal]

Ahhh

Sigh why am I so dumb

Thank you sir

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[grand pond]

w

\documentclass{article}
\usepackage{amsmath}

\begin{document}

Let \( a < b \). Let \( f \) be a bounded function on \([a, b]\).

Which of the following statements must be true?  
Select all the correct answers.

\begin{enumerate}
    \item IF, for every partition \( P \) of \([a, b]\), \( L_P(f) = \underline{I}_a^b(f) \) and \( U_P(f) = \overline{I}_a^b(f) \), THEN \( f \) is integrable on \([a, b]\).
    \item There exists a partition \( P \) of \([a, b]\) such that \( L_P(f) = \underline{I}_a^b(f) \).
    \item For all partitions \( P, Q \) of \([a, b]\), \( L_P(f) \leq U_Q(f) \) and \( L_Q(f) \leq U_P(f) \).
    \item For every partition \( P \) of \([a, b]\), there exists a partition \( Q \) of \([a, b]\) such that \( Q \) is finer than \( P \).
    \item For every partition \( P \) of \([a, b]\), \( L_P(f) \leq \underline{I}_a^b(f) \leq \overline{I}_a^b(f) \leq U_P(f) \).
\end{enumerate}
```Compilation error:```! LaTeX Error: Can be used only in preamble.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.49 \documentclass
                   {article}
Your command was ignored.
Type  I <command> <return>  to replace it with another command,
or  <return>  to continue without it.```

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[mint ravine]

Can someone please help me on this question:

[flint rock]

Yo bros back again

It’s just the probability they pick times within 10 min of each other

WLOG we can assume that the first person picks 8 am since its the DIFFERENCE that matters, so shifting it by any time doesnt change it

(WLOG means without loss of generality)

then the other person must pick a time between 8 and 8:10

which is 1/6 i tink

so answer is 1/6?

or maybe i messed up

no i def did

i think draw a square

with each axis being time

[mint ravine]

sorry bro i was afk

i originally got 1/6 too but it def wrong

[flint rock]

t=0 can be 8am and t=1 can be 9am

then draw square with axis from 0 to 1 on horzitonal and vertical

thne you can plot out points

and see it is the square - two triangles

try drawing diagram now

[mint ravine]

okay

lemme send the picture rq

Hopefully it is good

it's not drawn to scale btw

[mint ravine]

do you want me to send the image with plotted points too

[flint rock]

add tick marks for 1/6, 2/6, etc.

also its not just a diagonal down the middle

think about the points

like (0, 1/6)

will they meet?

what about (3/6, 4/6)

[mint ravine]

Yes, because they are 10 minutes apart

and the two people stay for only 10 minutes

[flint rock]

yes

try plotting all the cases where they are 1/6 apart

[mint ravine]

yo that's what i got on my graph too

[flint rock]

yeah like that

now simply just find the area of that

[mint ravine]

the area between the lines?

[flint rock]

yah

[mint ravine]

ok

[flint rock]

those are the solutiosn

and since we used t=0 as 8am and t=1 as 9am just find the area and thats the probability

[mint ravine]

how are you so good at probability lol

[flint rock]

nah i looked at stack overflow post tbh

[mint ravine]

oh ok

[mint ravine]

i tried googling how to find the area of this, but got nothing

i never really learned how to solve this kind of area before

do you know of an equation or anything ti solve it

[lavish venture]

you’re looking for the area between the two graphs and you have simple shapes

[mint ravine]

oh nvm

.close

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[viral dagger]

prove there are infinitely many primes p such that there exists n where $p\mid 2^n-3$

[grand pond]

skissue.in.a.teacup

[viral dagger]

fianlly out of geometry but into mumber theory

ping when respond

[vapid jungle]

Any ideas on what to try?

[viral dagger]

no

i mean contradiction looks plausible but i have no idea how to do it

[vapid jungle]

yeah contradiction is the way to go.

[viral dagger]

the contradiction would be there is a finite ammount of p right?

[vapid jungle]

assume there are finitely many then try to see if you can find another one not in your finite set.

[modern sapphire]

Even without contradiction, you can try showing, for some prime, 2^n - 3 mod p has different values for different n

[viral dagger]

what

[modern sapphire]

if p | 2^n - 3, for some n, then 2^n - 3 = 0 mod p

some fermats little theorem style proof

Wont be the same style, but a similar idea

[viral dagger]

i have a hunch it would be simmilar to the proof of infinire primes, but i cant think of any ways to do it, any hints?

[viral dagger]

call n=m mod (p-1), so 2^m-3=0 mod p would be a finite ammount of m for some p right (idk im just throwinf ideas)

[vapid jungle]

I'm not sure how to give hint without giving the answer 😅 So I'll say your hunch is correct and it is worth spending some time on figuring it out.

you may have to use fermat's little theorem here too - i guess this should be a useful hint.

[viral dagger]

[modern sapphire]

I was thinking more along like for k(n) = 2^n-3, you have k(n+1) = (k(n) + 3)*2-3 as a recursion. And if you recursively take mod p of that, it ends up in a cycle

and you get 0 every pth step of that recursion

[viral dagger]

what does the second sentence mean

[modern sapphire]

k(n) mod p, k(n+1) mod p, ...

[quiet hinge]

Just show that if k(n), k(n-1),...,k(1) are pairwise distinct, then k(1),k(2),...,k(n+1) will be paiwise distinct. By pigeonhole, you must have k(m)=0 for some m which is what you want

[modern sapphire]

each has a different value mod p

[modern sapphire]

yea what asteroid said

[viral dagger]

is every k(something) here taken to mod p?

[quiet hinge]

Yes

Indeed they are obviously distinct as numbers

[modern sapphire]

only coz most of them are coprime with p

[viral dagger]

k(n)=m mod p
k(n+1)=2(k(n)+3)-3=2m+3 mod p

k(n+2)=2(k(n+1)+3)-3=2k(n)+3=k(n)2^2+3+3×2=4m+9 mod p
k(2n)=2^n×m+3n+3=(2^n-3)×m+3n+3-3m=m^2-3m+3n+3 mod p
bro what am i doing

[heady plume]

+_+ Confoozled

[viral dagger]

me neither

[heady plume]

Actually it's much simpler, look in the way Tubelight was trying to guide you

Do you know FLT?

[viral dagger]

yeah

[heady plume]

So, you know that $2^{p-1} \equiv 1 (\mod p)$

[grand pond]

Arya

[viral dagger]

ye

[heady plume]

Now, how do you use that here?

Step 1: Write n = k(p - 1) + r

[viral dagger]

like smth like this?

[heady plume]

Yesh. Work on that idea for a bit.

[heady plume]

[viral dagger]

[quiet hinge]

this is the same as checking if k(0),k(1),..,k(p-1) are all distinct. at the end of the day you will have to use PHP and do pretty much the same thing

unless you have something else in mind

[heady plume]

Did you arrive at $2^r \equiv 3 \mod p$ and $r \in {0, 1, 2, \cdots, p - 2}$?

[grand pond]

Arya

[viral dagger]

yeah

[heady plume]

Now, unless p = 2, GCD(2, p) = 1 tells you what?

[viral dagger]

p is odd

[heady plume]

=_=

2^r for each r in set has a different mod p. (why?)

[viral dagger]

the set for all the finite p's?

[heady plume]

=_= set for "r"

[viral dagger]

oh thst r

[heady plume]

2^r for each r in {0, 1, 2, ..., p - 2} has a different mod p

however, set of numbers modulo p has p elements {0, 1, ..., p - 1}. Can you do something with these two information?

[viral dagger]

what does Z/pZ mean

[heady plume]

means group of integers modulo p. Sry nvm that, I rephrased

[viral dagger]

probably some php

[heady plume]

mhmm

[viral dagger]

hmm although the only php i can think of would be the p-2 to p-1, since 2^r has diffrent remainders mod p and there are p-1 pigeons, and mod p has p remainders (aka the holes) so not there yet (or maybe im dumb?)

[heady plume]

;-;

think a little more

[viral dagger]

wait

didnt we assume p=/=2 so 0 isnt possible for 2^something?

so 0 to p-2, and 1 to p-1, both are p-1 so one to one correspondency aka there is some r such that 2^r=3 mod p

[heady plume]

right ✅

[viral dagger]

but why does each r in the set have diffrent 2^r mod p remainders

[heady plume]

Let me rephrase that question better, cause that would be your homework.

2^r for each r in {0, 1, 2, ..., p - 2} has a different value mod p

[viral dagger]

😭

[heady plume]

I mean it's simple. Put 2^m = 2^n, m > n from the set of r's and arrive at contradiction

[viral dagger]

alright thank you sorry my battery is about to die sorry

[heady plume]

btw the proof isn't complete... but you can complete it if you followed the trail till now

[viral dagger]

.close (im out of brain juice to think any longer 😵‍💫 )

[midnight plank]

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[viral dagger]

sorry! (my battery is about to die anhmyways)

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[uneven turret]

1. In the following picture, AB is the diameter of the circle, angle <ABD = 20°, and angle <AED = 46°. Calculate the scale
   a. angle <DAE
   Answer = 180 - 90 - 46 = 44°

b. angle <BEC
Answer = <BEC = <AED = 46°

c. angle <BAC (i need help on this one)

[wet bridge]

<BAC+<ABD=<BEC

[uneven turret]

OHH

ALR ALR THANK YOUUU

.close

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[ionic thicket]

how do i find the PGF of two independent observations of the random variable X?

[ionic thicket]

in terms of t

why is it a+bt+ct^2

[midnight plank]

@ionic thicket Has your question been resolved?

[heady imp]

Can you be more precise and yield an example ?

[ionic thicket]

Uhh ill send the question

[midnight plank]

@ionic thicket Has your question been resolved?

[vapid parrot]

Think about how the pgf of a sum of independent random vars relates to the pgfs of the terms

[ionic thicket]

no i wanna know why GX(t) is a+bt+ct^2 first

[vapid parrot]

It follows from the relation between pgf Y and pgf X

[midnight plank]

@ionic thicket Has your question been resolved?

[ionic thicket]

can u elaborate

[shut magnet]

@ionic thicket Has your question been resolved?

[ionic thicket]

what 😭

[vapid parrot]

How are X and Y related? What does this say about pgf X in terms of the pgf of Y?

[midnight plank]

@ionic thicket Has your question been resolved?

[ionic thicket]

I know Y=Pgf of x ^2

But how do i find pgf of x

[vapid parrot]

Yes indeed, pgf Y = (pgf X)^2 so now if I tell you that a polynomial squared gives a polynomial of degree 4, then what was the degree of the origianl polynomial?

[shut magnet]

@ionic thicket Has your question been resolved?

[midnight plank]

@ionic thicket Has your question been resolved?

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[prime swallow]

How can we just multiply the x-1.5?

[sleek cloud]

You mean why x-1.5 is the same as 2x-3?

[prime swallow]

No so like my teacher said you dont want it as a decimal so you can multiply it by 2 to make it not a decimal but im wondering how thats possible cuz would you have tk do stuff to all the other parts of the equation?

[buoyant yoke]

its all combined into the a

[hard umbra]

the notation is suboptimal

[prime swallow]

So your saying you can do that as when your solving for a it will be different than not multiplying for 2 but that will make it so when you plug in lets say x the y will be true?

[olive matrix]

they should really have used a different letter (b, for example) instead of redefining a

[hard umbra]

,, a(x-1.5)(x+4) = \f a2(2x-3)(x+4)

[prime swallow]

So basically you can take it out of a decimal as a takes care of it?

[sleek cloud]

What do you mean suboptimal? It's flat out wrong

[buoyant yoke]

its physicists notation

[olive matrix]

yeah kind of? the answer looked at a(x-1.5)(x+4) and said "hmm. that sounds difficult. i'd rather solve a different thing instead. i'll fix it later."

[buoyant yoke]

just plug everything into a big giant constant C at the end and dont clarify

[prime swallow]

k

What if lets say both parenthesis had decimals and you multiplied both of them by different numbers to make them whole numbers, would it still be fine?

[copper terrace]

no reason why you'd do that but yeah, theoretically u could make it into c(2x-3)(2x+4) = y and solve for a new c term

[prime swallow]

No like for a different problem where both x ibtercepts are decimals and you multiply them each by a different number to get them to be a whole number, would it still work?

[olive matrix]

sure

IMO a better idea would be to set up your equation with the original roots and numbers on both sides, and then multiply both sides by whatever number you need

[prime swallow]

Can you show an example?

[olive matrix]

a(x - 1.5)(x + 4) = y original
a(-1.25 - 1.5)(-1.25 + 4) = 30.25 plug in values
a(-2.5 - 3)(-1.25 + 4) = 60.5 mult by 2
a(-5 - 6)(-1.25 + 4) = 121 mult by 2
a(-11)(2.75) = 121 condense
2.75a = -11 divide
5.5a = -22 mult by 2
11a = -44 mult by 2
a = -4

[prime swallow]

Ok but the way i did it is fine?

[prime swallow]

Are you there?

[olive matrix]

yea

[prime swallow]

Ok thx

[midnight plank]

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[prime swallow]

These two arnt adding up can someone explain?

[prime swallow]

Where you plug in the c intercepts to find a

[thick granite]

What isn’t adding up

[thick granite]

What question

[prime swallow]

The right one where you plug in the x intercepts doesnt add up tk what the left one says

It says plug in the x intercepts but the right one is plugging in the opposite of the x intercepts

[dusty portal]

you should probably, uh, edit out the part with your name in it...

also, technically speaking, the question on the right hand side is impossible to answer, as we do not know the degree of the polynomial

let’s see; y=a(x+3)(x-8)

the vertex x-coord would be at x=-b/2a

expanding now to get the b-term we get ax^2-5ax-24a

so the x-coord is at x=-5a/2a=-5/2

(i’m answering the question regardless since it seems to be a quadratic)

[prime swallow]

I thought to find the x cord is just adding up the intercepts and dividing by 2

[prime swallow]

How could it be this i thought its adding the x intercepts

[fading ore]

it is

the symmetry of a parabola makes it necessary for the vertex to be there

[dusty portal]

this only works for parabolas with leading coefficient 1

[prime swallow]

Wait its going down so is that why its negative?

[dusty portal]

that would be why, yeah

[prime swallow]

I dont get it tho the x intercepts are still positive and so to find a you subtract them if its maximum and you add if its minimum?

[prime swallow]

You there?

[dusty portal]

yes

sorry i opened my own help thread

[dusty portal]

not in this case

one x-coord is negative

[midnight plank]

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[prime swallow]

Are we so far correct and c is true since the denominator is the same so it can be a single fraction?

[wispy coral]

so far it looks good

[dusty portal]

yes.

[prime swallow]

So it would be a^2+b^2/c^2?

[wispy coral]

yeah

[dusty portal]

multiply both sides by c^2 to answer part c.

[wispy coral]

$\frac{a^2+b^2}{c^2}$

[grand pond]

Planet

[prime swallow]

Thats what i got when adding them

[wispy coral]

yeah so how can you make step 3 look like step 1

like the pythagorean theorem

[prime swallow]

Oh so like multiply both top and bottom by c^2?

Then continue?

[wispy coral]

ok so you have the right idea of multiplying by c^2

but not top and bottom\

you want a^2 + b^2 = c^2

[prime swallow]

Do you multiply the top to cancel out the bottom?

[wispy coral]

yeah so multiply on both sides right

cuz 1 * c^2 is c^2

and (a^2 +b^2)/c^2 times c^2 is a^2 + b^2

[wispy coral]

yeah but that should be in d.

because you show your expression is equal to pythagorean theorem

[prime swallow]

Wait so the part where the b and a is over c is in part c but where its equal to c is in part d?

[wispy coral]

please open another help channel, don't use someone else's

!occupied

[midnight plank]

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[clever python]

Oh shit I’m so sorry I didn’t know

[wispy coral]

no worries

[wispy coral]

yes

[prime swallow]

Ah ok lemme write that down rn

[wispy coral]

because part d is this

you can close the channel with .close btw

[prime swallow]

k

Tysm

.close

[midnight plank]

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[upbeat plinth]

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[upbeat plinth]

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[grand pond]

Kenzo

[upbeat plinth]

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[grand pond]

Kenzo

[lavish venture]

yep

[upbeat plinth]

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[lavish venture]

y dy

,

no

integrate both sides sir

[grand pond]

Kenzo

[upbeat plinth]

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[lavish venture]

no

[upbeat plinth]

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[lavish venture]

still no

[upbeat plinth]

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[lavish venture]

alphabet

[upbeat plinth]

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[lavish venture]

C

[grand pond]

Kenzo

[lavish venture]

yep

now use the given conditions

[upbeat plinth]

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[lavish venture]

no solve for c

[upbeat plinth]

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[lavish venture]

yep

get rid of the /2

[upbeat plinth]

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[lavish venture]

everyone has it

[upbeat plinth]

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[lavish venture]

i’m saying you can cancel a factor of 1/2

[lavish venture]

no

[upbeat plinth]

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[lavish venture]

y^2 has a factor 1/2

as does x^2 and 3

[upbeat plinth]

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[grand pond]

Kenzo

[lavish venture]

yea but always be careful

why is it the + root and not the - root

[upbeat plinth]

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[lavish venture]

nope

[upbeat plinth]

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[bleak lynx]

i need help

with my homework

[lavish venture]

$y^2 = \text{mumbo jumbo}$ has solutions $y = \pm \sqrt{\text{mumbo jumbo}}$

[grand pond]

knief

[lavish venture]

get lost

!occupie

!occupied

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[upbeat plinth]

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[lavish venture]

right but only one of them is correct

why is the + root correct

and not the - root

[upbeat plinth]

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[lavish venture]

consider if it was the - root

would our given conditions be satisfied?

[upbeat plinth]

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[lavish venture]

y = 2 when x = 1

if it were the - root would that be true?

[upbeat plinth]

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[lavish venture]

plug it in

$y = -\sqrt{x^2 + 3}$

[grand pond]

knief

[lavish venture]

if we took the - root

let y = 2

x = 1

do we get a true statement

[upbeat plinth]

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[lavish venture]

why

what is the resulting equation

[upbeat plinth]

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[lavish venture]

is false

mhm

always be mindful of the initial conditions

[upbeat plinth]

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[midnight plank]

Channel closed

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[midnight plank]

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[drifting root]

im wondering if there is a better way to do part b without doing it like

<@&286206848099549185>

[twin peak]

(not an answer to your question but don't forget to square the (theta*R/(2pi)) in your last line)

[drifting root]

o yea

[civic furnace]

Let s=R and $\alpha=\frac{\theta}{2}$ then it follows that $\sin{0.5\theta}=\frac{r}{R}$ and $\cos{0.5\theta}=\frac{h}{R}$

[grand pond]

fosur

[civic furnace]

Plug these values of r and h into the volume expression and

Find derivative of volume wrt to theta

and set it to 0 to find value of theta when volume is maximised

@drifting root

[drifting root]

will try

[drifting root]

is it just the thing that the angle of the sector is equal to the angle of the vertex?

also could I get you to check this, the answer i have is different to what i got

[civic furnace]

Yh . I’m trying to think of way to actually prove it

I’m seeing it more on visual terms

[civic furnace]

I don’t think you did product rule fully

U see ur mistake?

[drifting root]

i have somehow gotten extremely stupid in derivatives

uh

gimme a sec

[drifting root]

yea i see it, ill fix give me 5

lmao idk if im still doing smth wrong

but also isnt this hard to out in exact value radians?

[civic furnace]

Uve done product rule wrong again

[drifting root]

ye but dis doe

i now reaise i forgot to put a 2 on the cosine

[civic furnace]

,w d/dx sin^2(x/2)cos(x/2)=0

[drifting root]

did bot malfunction?

[civic furnace]

Sorry I’m so tired and realised that my assumption was wrong about theta

[drifting root]

o

[civic furnace]

https://youtu.be/ZNoJThDRBcw?si=ypgllTRkqm-ipAnO

[civic furnace]

This will help

[drifting root]

should I just suck it up and do this...

[drifting root]

will watch, thx for finding the vid

[civic furnace]

Nws . I apologise for this. Gn !

[drifting root]

video was extremely helpful tysm, gn

.close

[midnight plank]

Channel closed

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[last slate]

i get every thing but how does this prove that G is cyclic. It doesn't even talk about all the elements being generated by a single element

[maiden abyss]

x generates the whole group so it's cyclic

[last slate]

how

[maiden abyss]

x generates at LEAST a two element set {e, x}

But the group order is prime

[last slate]

how does that help

[maiden abyss]

By Lagrange, the generated subgroup by x, let's call it <x>, must divide the order kf the whole group

[last slate]

yeah

so

<x> = G

[maiden abyss]

So |<x>| must divide p

[last slate]

yeah but

[maiden abyss]

So it either has 1 element or p elements

[last slate]

yes

i get that

but to be cyclic it must be that the powers of x should generate all the elements

where in the proof did it show that

[maiden abyss]

Ah

[hard umbra]

look at this equation again

what is <x>? what is it equal to? what does that mean?

[last slate]

i meant gp(X) where X={x}

[hard umbra]

that wasnt the point of the question

what are the elements of <x>?

[last slate]

gp(X) is the intersection of all the sub grp containing X

[last slate]

its a single element

[hard umbra]

certainly not

[maiden abyss]

Ah, there's an alternate description of <x>

[last slate]

mb

[hard umbra]

have you never seen that <x> = {1, x, x^2, x^3, x^4, ...}

[last slate]

what

[maiden abyss]

Also the negative powers

[hard umbra]

finite group

[maiden abyss]

Right

[last slate]

this is the defination im familar with

[hard umbra]

okay but <x> is a subgroup

so it must contain all powers of x

[maiden abyss]

Do your lecture notes have a lemma or remark that gp(x) contains precisely all powers of x

[hard umbra]

regardless, if G = <x>, then G is generated by a single element

[last slate]

no

[hard umbra]

which is the definition of cyclic

[maiden abyss]

Wait that's literally remark 2 in the screenshot lol

[hard umbra]

one does not need to know that <x> contains all powers of x

[maiden abyss]

Applied to the one element set X = {x}

[hard umbra]

if G = <x>, G is generated by the single element x

[last slate]

i certainly didnt see that

;-;

how exactly is it the set of powers of x

[hard umbra]

well do you agree that all the powers of x are contained in <x>?

[last slate]

no

[hard umbra]

why not?

<x> is a subgroup

and x is in it

[last slate]

but does it only consists of powers of x?

[hard umbra]

i asked that it contains the powers of x, not only

do you agree on that at least?

[last slate]

yeah

[hard umbra]

okay

now

[last slate]

ido

[hard umbra]

the powers of x form a subgroup

do you agree?

[last slate]

i dont see it

[hard umbra]

check the axioms

[maiden abyss]

Then write it down

[last slate]

ok lets assume that it forms a grp

[hard umbra]

okay then

the powers of x is a subgroup containing x

[last slate]

yeah

[hard umbra]

and <x> contains all the powers of x

so <x> must equal the set of all powers of x

[last slate]

yeah

makes sense

[maiden abyss]

All subgroups that contain x

[last slate]

how excalty do you prove powers of x is sub group though?

[maiden abyss]

Also contain all powers of x

[novel sedge]

gp({x}) = G actually does say that every element is generated by a single element. its just written in the form of an equation.

[hard umbra]

check the axioms

[last slate]

alright

[maiden abyss]

So gp(x) contains all powers of x

[hard umbra]

1. non-empty: <x> certainly contains x
2. closure: x^n x^m = x^(n + m)

[novel sedge]

how did you prove lagranges theorem without knowing that <x> is a subgroup?

[last slate]

yeah

cool

[last slate]

cosets

bijection from xH -> H

smth like this

[maiden abyss]

Sick tbh

[novel sedge]

okay

[hard umbra]

G/<x> smh

[last slate]

also

$\perp$

[grand pond]

_Brian

[last slate]

what does this mean

in grp theory

[hard umbra]

where did you see it

[novel sedge]

normal subgroups perhaps

[last slate]

here

2nd line

[hard umbra]

coprime maybe

thats a number theoretic concept not a group theoretic one

[last slate]

yeah

[maiden abyss]

I forgot the definition of coprime in ring theory man I'm so rusty

[novel sedge]

taking a perp b to be a divides b makes more sense

[last slate]

i was confusing primitive roots with generators

i mean primitive roots are also called generators in number theory but are the generatos of a group mod p same as the primitive root of mod p

[maiden abyss]

Does ring comprimality generalize to all monoids

[hard umbra]

coprimality doesnt even generalise to rings

[maiden abyss]

Oh lol

[novel sedge]

sorry, yes coprime makes sense

[last slate]

close when done

[maiden abyss]

Maybe $x \perp y \iff \langle x \rangle \cap \langle y \rangle = {e}$

[grand pond]

PKThoron

[hard umbra]

that doesnt even work for integers

[maiden abyss]

Yeah I'm rusty af

[hard umbra]

xy is always an element of both

[novel sedge]

they will have intersection lcm(x,y) Z

[maiden abyss]

Duh

[last slate]

hey

another doubt

doesnt that mean G is generated by all the elements except for 'e'

[novel sedge]

why are you leaving poor e

G is generated by G too

[last slate]

ok ok

ok

[hard umbra]

i think they mean if G = <x> for any other x in G except e

[last slate]

yes

[novel sedge]

yes, since the proof was for any arbitrary non-unital x

[last slate]

so a prime order group is

• always cylic
• can be generated by gp({x}), x != e

great

.close

[midnight plank]

Channel closed

Closed by @dry jewel

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[last slate]

thanks all

[midnight plank]

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• 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>.
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• 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.

[rare wave]

Helpe

[rare wave]

How do you answer no 3

I mean 2

[buoyant yoke]

a or b?

[rare wave]

A and b pls

Is a correct?

[buoyant yoke]

yes

[rare wave]

is b all DNE?;

there is no colorless bullet

i dont know the term

[buoyant yoke]

i dont see a g(x)

or is that the one below?

bottom left?

[rare wave]

oh youre right

yep

[buoyant yoke]

ah okay

so then what part are you confused about?

[rare wave]

I just need someone to verify

Ill send you a pic

[buoyant yoke]

its basically the same as 2a, no?

[rare wave]

I kinda get it now

[buoyant yoke]

ok

[rare wave]

I hope it's correct

[buoyant yoke]

yup