help-10

[pallid willow]

you're not allowed to use calculus, right?

[coarse birch]

No

[pallid willow]

good

the way I have involves translating the function left to exploit symmetry but I am very sorry that I don't think I am in the position to explain this

[coarse birch]

It's ok, most of exprienced people here have failed to explain many things to me

Im special case

[pallid willow]

please don't say that. it's not you being bad or anything, I promise.

if you think I'm thinking that you're free to report me, idm

[coarse birch]

What

[old lily]

im crine

help

[coarse birch]

Whot

[pallid willow]

?

[mild sandal]

you're not allowed to use calculus?

[coarse birch]

Nope

This is function transformation worksheet, didnt learn calculus yet

[mild sandal]

nah chill, you're good

yeah no idea how to do this lmfao

[coarse birch]

No way

[pallid willow]

my idea involves shifting the function left 1 unit

[coarse birch]

Ok if people from here can't even solve this then I either need calculus or the author of the worksheets made these while drunk

[pallid willow]

so we use f(u + 1) instead

no it can be solved via algebra, but the process is of course longer

[mild sandal]

nah im not as goated as the helpers

[coarse birch]

Ill wait for a helper then

[pallid willow]

but if you're willing to try, you can do this

[coarse birch]

You need to expand

[pallid willow]

yes there is a crapton of expansion to be done

[knotty crow]

you can use the fact that the inflection point is the center of symmetry of the extrema (i.e. the smallest and largest values)

[coarse birch]

But I mean you need to expand your expalnation

[pallid willow]

oh lmao

[coarse birch]

What

[pallid willow]

I'll let Modus try first, cuz my method is bad

[coarse birch]

[mild sandal]

hey dont give up yet

[coarse birch]

Im not giving up

[knotty crow]

But still, you would have to be familiar with the formula for the inflection point of a cubic function (like x = -b/2a for quadratics)

Here it's simply x = -b/3a

[coarse birch]

I haven't been taught for cubic formula

In fact I never had cubic formula introduced yet

[mild sandal]

but we got this from calculus yes?

[pallid willow]

I don't think Modus is talking about the cubic formula

you also will not see the cubic formula in most syllabi

for a good reason

[knotty crow]

you can use Vieta's formulas, but calculus would be faster

x1 + x2 + x3 = -b/a

[coarse birch]

I haven't been taught vieta's formulas

[mild sandal]

💀

[pallid willow]

sum and product of roots?

ok maybe not for a cubic but surely for a quadratic?

[echo zenith]

do you know am gm inequality

[coarse birch]

How does it look like

[echo zenith]

oh mb it's maximum

[pallid willow]

for a general quadratic $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, then
$$\alpha + \beta = -\frac{b}{a}$$
$$\alpha \cdot \beta = \frac{c}{a}$$

[warm shale]

Towa

[pallid willow]

not useful here though cuz not a quadratic, but yes these are Vieta's formulas

[coarse birch]

No, don't kno wthis

[pallid willow]

hm, alright then. sorry for bringing it up

[coarse birch]

Buddy, don't be sorry

[pallid willow]

anyway I won't clog this channel any further. I'll let Modus handle this

[coarse birch]

Why are you sorry for

[pallid willow]

sorry for intruding and flooding

[coarse birch]

You aren't flooding buddy

[mild sandal]

hey man you're good 😭

[coarse birch]

You are showing me the things I am missing

[pallid willow]

wrong place and time

[mild sandal]

did they teach you that stuff?

[coarse birch]

No I don't know vietas something

[mild sandal]

idk how they're expecting you to solve this

[coarse birch]

I love it

[echo zenith]

I think we can find the inflection point and use symmetry from there?

[coarse birch]

TBH I don't know it shouldn't be that complicated

[echo zenith]

like the distance between the point of inflection and maxima is the same for minima

[coarse birch]

I wasn't taught that yet

[echo zenith]

damn

[coarse birch]

You know what I am just going to close this channel because this stupid question is creating destruction in my mind

I love everyone that appeared here and tried to help me

.close

[obtuse pebble]

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[obtuse pebble]

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[zenith crest]

\textbf{(c) Prove that if $a \neq b$, then $(a, b)$ has precisely two elements.}
\\
Choose any $a, b$, where $a \neq b$. Then by the Axiom of Extensionality, $a \neq b$, iff $\exists z(z \in x \land z \notin y) \lor (z \in y \land z \notin x)$

[zenith crest]

I am trying to prove this proposition

am i allowed to make logical negations like this?

Choose any $a, b$, where $a \neq b$. Then by the Axiom of Extensionality, $a \neq b$, iff $\exists z(z \in a \land z \notin b) \lor (z \in a \land z \notin b)$

[warm shale]

toast

[zenith crest]

eh

[trim portal]

i think that you got the second part wrong

[zenith crest]

yeah

i meant z \in b and z \notin a

[trim portal]

But yeah, you are allowed to say that

[zenith crest]

kk

[trim portal]

it follows just by logic

[zenith crest]

for this proof the eventual goal is to show that ${a,b} \neq {b}$

i think

[warm shale]

toast

[zenith crest]

so as long as i show there exists an element in the first set not in the second set, i am done right

[trim portal]

yeah

you must suppose that a =/= b though

[zenith crest]

\textbf{(c) Prove that if $a \neq b$, then $(a, b)$ has precisely two elements.}
\\
Choose any $a, b$, where $a \neq b$. Observe that the sets ${a,b} \neq {b}$ by the Axiom of Extensionality, since $a \notin {a,b}$, but $a \notin {b}$, as $a \neq b$.

[warm shale]

toast

[zenith crest]

is this too unrigorous

i mean its obviosu but idk if i formalyl proved anything

oops

\textbf{(c) Prove that if $a \neq b$, then $(a, b)$ has precisely two elements.}
\\
Choose any $a, b$, where $a \neq b$. Observe that the sets ${a,b} \neq {b}$ by the Axiom of Extensionality, since $a \in {a,b}$, but $a \notin {b}$, as $a \neq b$.

[warm shale]

toast

[zenith crest]

do i need to show $a \ne b \implies a \notin {b}$?

[warm shale]

toast

[trim portal]

Hmm, what do they mean by "has precisely two elements"?

how did they define number of elements?

[zenith crest]

not exactly sure ngl

but i guess there are 2 things

in the set

\textbf{(c) Prove that if $a \neq b$, then $(a, b)$ has precisely two elements.}
\\
Choose any $a, b$, where $a \neq b$. Observe by the Pairing Axiom, we can generate some set ${b, b} = \forall w ( w \in {b,b} \iff (w = b \lor w = b))$. This is equivalent to $w \in {b,b} \iff w =b$. By the Axiom of Extensionality ${b,b} = {b}$. so $w \in {b} \iff w = b$. However, because $a \neq b$, then $a \notin b$. But $a \in {a, b}$. Thus by the negation of the Axiom of Extensionality, ${a,b} \neq {b}$, so the set ${{a,b}, {b}}$ contains precisely 2 elements.

[warm shale]

toast

[zenith crest]

here is a more rigorous proof i wrote

[trim portal]

Eh, if they didnt define it rigorously, why bother proving it rigorously

[zenith crest]

fair enough 😂

[trim portal]

because a =/= b, a not in b?

[zenith crest]

oiops

typo

\textbf{(c) Prove that if $a \neq b$, then $(a, b)$ has precisely two elements.}
\\
Choose any $a, b$, where $a \neq b$. Observe by the Pairing Axiom, we can generate some set ${b, b} = \forall w ( w \in {b,b} \iff (w = b \lor w = b))$. This is equivalent to $w \in {b,b} \iff w =b$. By the Axiom of Extensionality ${b,b} = {b}$. so $w \in {b} \iff w = b$. However, because $a \neq b$, then $a \notin {b}$. But $a \in {a, b}$. Thus by the negation of the Axiom of Extensionality, ${a,b} \neq {b}$, so the set ${{a,b}, {b}}$ contains precisely 2 elements.

[warm shale]

toast

[trim portal]

have you proved that {{a, b}, {b}} exists?

[zenith crest]

i guess we cant set {b,b} = some logical statement eitehr

[zenith crest]

we assume it exiss i think

but i did it in a prior question

\textbf{a) Prove that for any two sets $a, b$, there is a unique $c = {{a,b}, {b}}$. We define this $c$ to be the \textit{ordered pair} $(a, b)$}
\\
Choose any sets $a$ and $b$. By the pairing axiom,, there exists a set ${a, b}$. Now by the pairing axiom again, we have the set that contains exactly the sets ${a,b}$ and ${b}$, namely $c = {{a,b}, {b}}$. Observe by the Axiom of Extensionality, if there exists $c'$ such that $c'$ contains the same elements as $c$, then $c = c'$ so $c$ is unique.

[trim portal]

Sure, then why bother using the pairing axiom

[warm shale]

toast

[zenith crest]

oh true

[trim portal]

(a, b) = {{a, b}, {b}}, which has elements {a,b}, {b}. All you need is to check that they are distinct

[zenith crest]

i jut wanted to make sure i cant assume $a \ne b \implies a \notin {b}$?

[warm shale]

toast

[zenith crest]

oops

[trim portal]

yeah, thats false

[zenith crest]

i jut wanted to make sure i cant assume $a \ne b \implies a \notin {b}$?

[warm shale]

toast

[trim portal]

oh you mean not in {b}

[zenith crest]

ye

like

can i just say this? then its alot easier

[trim portal]

Yeah, you can use that I'd say

[zenith crest]

thats the only reason i used pairing axiom

i think

oh ok

[trim portal]

The {} notation tells you all the elements the set has

so {b} has only b and nothing else (so it cant have a, since a != b)

[zenith crest]

ah right

yeah it should be obvious

i never really know what i can or cannot assume 🤣 especially cuz these questions are asked

so

"unrigorously" idk

[trim portal]

if they didnt even define what "has 2 elements" means, or what the {} notation means, I wouldnt focus on full rigour that much

all you need is convince your prof that you know what you're talking about

[zenith crest]

kk

[zenith crest]

yep!

alr tysm

.solved

[obtuse pebble]

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[obtuse pebble]

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[fringe blade]

Prove that the set of all of polynomials with coefficients in F is a vector space of F^n

[fringe blade]

The set of all polynomials with coefficients in F is a vector space over F

[obtuse pebble]

@fringe blade Has your question been resolved?

[fringe blade]

$F[x] = { a_0 + a_1 x + \dots + a_n x^n \mid n \in \mathbb{N},\ a_i \in F }$

[warm shale]

andre_al

[fringe blade]

<@&286206848099549185>

[solemn mural]

if F is a vector space, do you agree that if you sum polynomials with coefficients in F you obtain a new polynomial with coefficients in F ?

[fringe blade]

Oh

Damn it

So obvious

I feel so stupid now

[fringe blade]

Yes

[solemn mural]

no no it's ok , if you have other question i can help

[fringe blade]

It is closed under that

Addition

Multiplication is the same

[fringe blade]

I was learning about linear combinations, I guess

@solemn mural, the zero polynomial

[solemn mural]

?

[fringe blade]

It is said that a polynomial is of degree m when a_m is not zero

a_m * x^m

So that means m is negative infinite?

I am confused by -infinity < m

The zero polynomial, basically

I have not taken abstract algebra courses

[solemn mural]

the only confusing thing for you is " -infinity < m " ? I want to make sure

[fringe blade]

“The polynomial that is zero is said to have a degree of negative infinity”

[solemn mural]

yes that is correct

[fringe blade]

That is confusing me

[solemn mural]

it's a convention, the purpose is to make some manipulations easier, the same as x^0 =1 , 0! =1 ....

i can show an example

[fringe blade]

Mmm, OK

[solemn mural]

if P,Q are polynomials deg(P*Q)=deg(P)+deg(Q) is still true with deg(p=0) = -infinity if we agree that -infinity + m = -infinity

When dividing one polynomial by another, if the divisor is of degree 0 (i.e., a nonzero constant), this simplifies the analysis of the quotient and remainder, as it ensures the degree of the divisor is always well-defined.

[fringe blade]

I must take a break off abstract linear algebra and go study abstract algebra, I think.

Is that a good move?

[solemn mural]

it depends how necessary is abstract algebra to linear algebra , do what you enjoy doing !

[fringe blade]

Sure

Thank you

I think I will

.close

[obtuse pebble]

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[undone quest]

For complex number, alpha and beta, when
____
alpha * beta == 2
, _____
alpha + (1 / alpha) == 3i
What is the value of
____
beta / 2 + 2 / beta
?

[pallid willow]

do you have a picture? looks like very poor formatting here

[pallid willow]

Given complex numbers a and b such that $a\overline{b} = 2$ and $a + \frac{1}{\overline{a}} = 3i$. Find the value of $\frac{b}{2} + \frac{2}{\overline{b}}$.

[warm shale]

differential Towametry (Towa)

[gaunt flume]

write a= 2/ bconjugate. replace a by b in the second equation from the first.

take conjugate of the first equation and similarly replace a conjugate.

[obtuse pebble]

@undone quest Has your question been resolved?

[nocturne shard]

how does everyone know fluent korean wtf

[undone quest]

It would only be a == 2 / (a b with bar)

[nocturne shard]

can u translate pls?

[nocturne shard]

is this the right translation?

[undone quest]

When two complex numbers alpha , beta exists and alpha * (a beta with bar) == 2 and alpha + 1 / (an alpha with bar) == 3i
Find the value of beta / 2 + 2 / (a beta with bar)

((an alpha with bar) and (a beta with bar) is KyeolLaeBokSsoSu of alpha and beta each)

[whole bone]

sans do you know how to express y' = y in terms of dy/dx

[undone quest]

Is it 1` == 1 by division?

[nocturne shard]

ok in the second equation write 1/a(bar)= a/(|a|^2)

[undone quest]

Or does the symbol ` has something else I don't know

[nocturne shard]

@undone quest

did u understand this?

[whole bone]

y' means derivative

[nocturne shard]

..

[undone quest]

No idea how the abs function appeared

[nocturne shard]

its a property

u can try ur self

a.(a with bar)=|a|^2

[undone quest]

But the abs function does not work on complex numbers

[nocturne shard]

the abs of a complex number is the distance of the complex number from origin

thats the meaning of abs function

in argand plane an complex number x+iy will have the cord (x,y)

so the abs value of said complex number will be (x^2+y^2)^1/2

[undone quest]

Is it simplified to (x² + y²) / 2 ?

[nocturne shard]

no sqrt of (x^2+y^2)

[obtuse pebble]

@undone quest Has your question been resolved?

[plucky mountain]

apologies for the intrusion

uhm

@undone quest is this a competetive programin thing

or r u jus doin maths

[obtuse pebble]

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[flat moon]

solve using similarity / bpt/ converse bpt please

[flat moon]

thanks in advance

ive tried a lot but no progres

[carmine spear]

ayy this came this year in the board exams didn't it

so, I'll tell you the first step

[flat moon]

yea it did

[carmine spear]

draw a line frm D that's parallel to BY

Label the point it intersects with AC as Z or whatever

that'll give you your solution pretty fast

[winter cargo]

whatever you did yesterday followed by drawing a line from Y parallel to AD

insane for a board question though

[obtuse pebble]

@flat moon Has your question been resolved?

Channel closed

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[flat moon]

u think its tough

aswell

no one was able to solve it

in our whole class

their whole class

[obtuse pebble]

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[white pendant]

.reopen

Ok so here's my question

[white pendant]

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

Is there any other way to prove that the angles of the quadrilateral pqrs is 90°

Other than using the perpendicular diagonals property of the rhombus

<@&286206848099549185> help me guys

[inner sierra]

!15min

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[inner sierra]

anyways what do you use to prove PQRS is a rectangle?

[white pendant]

By using the perpendicular diagonals property of the rhombus

Yo

Where are u?

[inner sierra]

specifically how

how do you use the diagonals?

[white pendant]

Lemme send you a photo

[fickle sail]

If you prove the diagonals of the quadrilateral are equal on top of there being equal opposite sides, it is enough to prove it is a rectangle i think

[white pendant]

Like this

[inner sierra]

midpoint theorem is the first thing that comes to my mind ngl

[white pendant]

It is

I am just asking any alternate ways of doing this question

[white pendant]

Well i also wanted to ask if to prove a rhombus do we have to prove all the sides as equal

[inner sierra]

how about proving PQRS is a parallelogram first

and then prove one of the angles are 90

[white pendant]

First step is easy

Just use midpoint theorem

[fickle sail]

Once you have that it is a parallelogram, you can prove it's diagnals are equal instead of proving angles are 90

I think it is easier

I mean idk

[white pendant]

But how to do that

We cannot use Sss

[inner sierra]

well

we have multiple ways

one way is that 2 pairs of sides are equal

i meant parallel

[white pendant]

Yes

.close

[obtuse pebble]

Channel closed

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[fickle sail]

But does it make sense intuitively that it is true to you?

[fickle sail]

Sorry my internet is very bad

[obtuse pebble]

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[split sluice]

Heres my working i dont know how to continue the second sum i posted this before but the bot auto closed the help

[split sluice]

<@&286206848099549185>

[distant knot]

Hello

[split sluice]

I need help with dumb math 😅

[distant knot]

Generating response

[oblique glen]

do u know the addition of two exponential random variables is another exponential random variable?

[split sluice]

Yes

[oblique glen]

just calculate cumulative distribution function of X1 + X2

[split sluice]

Thats not the question tho

[warm shale]

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

[oblique glen]

you'll see a pattern

[split sluice]

T_n is the sum of exponentials

But then u want K_m

Which is the sum of T_n

Up to m

[oblique glen]

first find distribution of Tn

then Km

[distant knot]

$\sum^{n}_{i=1}$

[split sluice]

All i need to do is compute the var of k_m the distribution of t_n isnt necessary

[warm shale]

Noah

[oblique glen]

Km is the sum of Tns bruh

[split sluice]

Yea but u dont need to find the distribution of t_n

To find var of k_m

[oblique glen]

u need to

[split sluice]

No?

[distant knot]

No

[split sluice]

Var(sum) = sum i=1 to n of var + 2 * double sum i<j of cov

All i need help with is computing my second sum

Which im not sure how to do

[oblique glen]

k my brain is not functioning properly

[split sluice]

.

Someone help me 😭

[quick nova]

Only thing I can think of is using the fact that Cov(X,Y) = E(XY) - E(X)E(Y) but idk how would that help

[split sluice]

They gave a hint

Im just not sure how u use it

Cov(x,x+y)=var(x)+cov(x,y)

[quick nova]

Maybe they want you to show T_n and T_i is also independent so you get Cov(T_n,T_i) = 0?

[split sluice]

is T_n and T_i always independent?

They're both sums of independent exponentials

[quick nova]

Hmm yeah, I'm currently procrastinating my ass in tutor while looking at this so I can't give much thought 😭

But I do think it might be that

[split sluice]

Hmm ok

[split sluice]

😭

No wait it cant be independent since if T_1=X_1 then T_2 literally contains T_1 =X_1+X_2

[obtuse pebble]

@split sluice Has your question been resolved?

[obtuse pebble]

@split sluice Has your question been resolved?

[split sluice]

Yo can someone help me like hello

[quick nova]

Okay, i'm back. I think I managed to get it. Let me just write it down (might take a while though)

[split sluice]

🙏

Thx for coming back lol

[quick nova]

yeah my tutorial just finished

I decided to make K_m be rewritten in terms of X_i as it's useful since X_i variables are independent. This isn't the full solution I came up with, just a hint.

excuse the big ass shadow though 😭

though my solution didn't use the fact that cov(x, x+y) = var(x) + cov(x,y)

it did use i^2 identity though

[split sluice]

Wait so how would i find the var from this

[split sluice]

Its fine lol 😭

I didn't think abt writing k_m as X_i's hmmm

Would it be sum i=1 to m of the bracket squared* var(x_i)

[quick nova]

Yes

[split sluice]

Alr ill try it one sec

[quick nova]

But yeah as you said T_n can't be independent so I made it into x_i terms

[split sluice]

Yea thats smart ima try when i get home thx

Hopefully i get the same answer as the markscheme

[quick nova]

Alright nice, good luck with it though

[split sluice]

Yea it worked thanks

.close

[obtuse pebble]

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[outer dagger]

what does this question mean? and how do you solve the 3rd one?

[fickle sail]

You're supposed to find eigenvectors y of P and show that Py are eigenvectors of A

And also eigen vectors of PAP^-1

Sorry for you this looks tedious

This is what i understood at least

[outer dagger]

wait... IT'S THAT LONG?? first finding P inverse after that eigen vector of P? and then find eigen vectors of A???

[fickle sail]

I mean there do be a formula to find the inverse of 2×2 matrices but i don't remember it

[outer dagger]

i found the gauss inverse to be easier than finding determinant and adj of matrix

[fickle sail]

Idk what that means but if it works it works

Don't hesitate to ask quastions if you struggle

[vast cape]

chaos

[obtuse pebble]

@outer dagger Has your question been resolved?

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[fierce harbor]

Does anyone knows Maths?

[glad vapor]

Yes

You're welcome

[inner sierra]

-# this is a math server so 99% of us knows math

[mild sandal]

maths? hardly know her

[mild sandal]

why are you whispering?

[neon vector]

I don't think anyone does. Those guys like Pythagoras, Thales, Euclid just tried acting cool by making stuff up

[fierce harbor]

Any one know how to learn maths

[inner sierra]

[inner sierra]

learn by understanding

[mild sandal]

you start from numbers

[inner sierra]

you should understand why those rules work

[fierce harbor]

I want to make stuff up like them

[neon vector]

What's stopping you

[inner sierra]

they used existing rules to make them

[inner sierra]

you can even make up new rules and from that you can make more rules

if you wanted to

[fierce harbor]

1m = 0.001 b

[inner sierra]

that's why boolean algebra was born

[mild sandal]

ayo

chat is this helping?

[fierce harbor]

I am billionaire I have $ 0.00000000000001b

[mild sandal]

i dont think this is helping...

[obtuse pebble]

@fierce harbor Has your question been resolved?

[obtuse pebble]

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[slim ibex]

i need help with this question

[slim ibex]

im not use to the wording

[inner sierra]

let's try question 6 first

a fixed cost here, is just like the starting point

imagine you're on a taxi

then you would have a starting cost, for example $15

that $15 is the fixed cost

[foggy haven]

I can help u

also

[slim ibex]

yes

[inner sierra]

and the cost that is proportional to blah blah blah

[slim ibex]

that is the initial value

[inner sierra]

that's like how much more money is being added for each kilometer

or miles if you're in usa

[slim ibex]

yeah

[inner sierra]

so knowing that

we can assume the initial cost as $t$ or something

[warm shale]

1 divided by 0 equals Infinity

[slim ibex]

right

[foggy haven]

I remember talking to u 1 year ago

[inner sierra]

uhm

[foggy haven]

U helped me 1 year ago

[inner sierra]

you need a new name idk

and the cost for making each soap is $c$ for instance

[slim ibex]

so far we got t as something

[warm shale]

1 divided by 0 equals Infinity

[slim ibex]

right

[inner sierra]

so then the cost of making $x$ soap is?

[warm shale]

1 divided by 0 equals Infinity

[slim ibex]

it costs 2

[inner sierra]

if there's no initial cost, it would be $xc$

[slim ibex]

2 bucks

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

hold up

i think you're in another question

[slim ibex]

i am?

[inner sierra]

uhm

the 2 bucks is in q2

each bar was sold for 2 bucks

[slim ibex]

oops

well

there is no initial cost

[inner sierra]

well we assumed the initial cost was $t$

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

so the total cost is?

[slim ibex]

cx + t

?

[inner sierra]

yes

so $y = cx + t$

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

first question done

[slim ibex]

isnt this just

a linear relationship

y = mx + c

[inner sierra]

okay

okay

yea it's a linear relationship

[slim ibex]

i see okay

ok next question

well

c would be = 2?

so itll be

y = 2x+t

[inner sierra]

no

$c$ is the cost to MAKE each bar

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

$2$ is a different story

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

it's the cost of SELLING each bar

[slim ibex]

@inner sierra my wifi turned off im back

[slim ibex]

okay

im not sure what to do next

[inner sierra]

for the first information, you can plug in $x = 800$ and for the second information, you can plug in $x = 300$

[warm shale]

1 divided by 0 equals Infinity

[slim ibex]

i dont understand why one is positive and the other is negative

part b

[inner sierra]

wdym one is positive and other is negative?

[slim ibex]

this part

[inner sierra]

for the 1st info

it was originally derived from $y + 500 = 2(800)$

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

2nd info it was derived from $y - 80 = 2(300)$

[warm shale]

1 divided by 0 equals Infinity

[slim ibex]

ohhh

[obtuse pebble]

@slim ibex Has your question been resolved?

[obtuse pebble]

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[fallow frigate]

.reopen

[zenith crest]

hey I’m trying to figure out what the last line is saying

does anyone have an idea?

[daring ravine]

z is in b?

[zenith crest]

oops before that

Also z in y, and y. in b, ????, => z in b

[oak bolt]

so alpha ordered

maybe

ordinal?

[daring ravine]

ordinal smth

[zenith crest]

i thought it said so alpha ordered but like tht doesnt make sense since alpha is already claimed to be an ordinal

also this step kindof confuses me

oh ig its not that confusing

we need to show z, y, b are elements of alpha, and we know alpha is an ordinal so the elements of alpha are transitive, We already claimed z in y and y in b, so because of this, z in b

then b is transitive

[oak bolt]

maybe they mean "so alpha ordinal implies z in b"

as opposed to "so alpha ordinal, and therefore z in b"

[zenith crest]

ohok

make sense

.solved

[obtuse pebble]

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[obtuse pebble]

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[hidden dove]

What is the span of an arch?

[tardy epoch]

what's an arch

[molten sleet]

I don't see no span

oh wait do you just mean the distance between the two endpoints?

or is there some algebra thing Im unaware of

[obtuse pebble]

@hidden dove Has your question been resolved?

[hidden dove]

Uhh my board exam asked me as there is a bridge

[hidden dove]

Adjust like this

[molten sleet]

are you in 10th?

[hidden dove]

And it give me points

[hidden dove]

Yeah

[molten sleet]

I heard the exam was cooked

[tardy epoch]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[hidden dove]

(-288.5,0) and (288.5,0) as the end of radius of semicircle

And it asked the span of arch

[hidden dove]

It gave concepts of class 11 and even the Google can't solve

[tardy epoch]

A picture or screenshot is best

[molten sleet]

Can you send the original pr-

yeah

[hidden dove]

Yess wait

[molten sleet]

I mean google can solve it

[hidden dove]

36 (ii)

[molten sleet]

I believe they just asked the distance between the two endpoints

[hidden dove]

Span of arch means this dude

[molten sleet]

yeah

so isn't it just 238.5+228.5

[molten sleet]

oh?

I don't think they'd ask that

[hidden dove]

I did that

[molten sleet]

you did that?

[hidden dove]

Can you answer (i)

[hidden dove]

Yess i crote pie r

And solved it

[molten sleet]

isn't it a parabola

you'd need calculus to find the arch length

[hidden dove]

That's what the students are saying

It is out of syllabus

Answer (i)

[molten sleet]

yeah so they probably meant distance between endpoints no?

[hidden dove]

Yess maybe

😔

[hidden dove]

..

[molten sleet]

to solve 1 u can just use that vertex formula

they teach

it's in book

[hidden dove]

I can't see that in book

Answer it bro

[tardy epoch]

!noans

[obtuse pebble]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[molten sleet]

this thing

[hidden dove]

Parabola is not in topic😭

[hidden dove]

Shitt got mark 1 cut

[molten sleet]

otherwise you can use symmetry ig

u see how they gave endpoints?

[hidden dove]

Bro in (i) I used logic as taking o is midpoint

[molten sleet]

O as the origin?

if u did that it's wrong

[hidden dove]

I answer it (0.288.5)

[hidden dove]

Why

[molten sleet]

cause it's -238.5 and 228.5

so it's off centre

true midpoint is further to the left than 0

at -5

[hidden dove]

Literally the fucking cnse

Cnse

Cbse

I left 4 marks in their hand and

These case studies are also wrong

OMG😭😭

[molten sleet]

I mean they're not wrong..

[hidden dove]

Why

[molten sleet]

well it's not wrong logically

like you can still get the right answer

[hidden dove]

But this type of questions are not in any book

And it suddenly appears in exam we just got hyper and thought wrong

🥲

[molten sleet]

I do feel like not labelling the parts of an arch was a bit scummy on their part

[hidden dove]

If I didn't know vertex formula then how can I find the coordinates

Is cbse mad or what

[molten sleet]

symmetry then right

u did it yourself

I mean you got it wrong

but like you had the right idea

[hidden dove]

Yess

[hidden dove]

Still I can't find Y coordinates of A

[molten sleet]

why

[hidden dove]

I can just find X

[molten sleet]

they gave the formula for the parabola

so if u find x just plug it in

[hidden dove]

Where?

[molten sleet]

p(x)=-0.0025x^2+...

at the end of the para

[hidden dove]

But there is no Y

Only x brother

[molten sleet]

y=p(x)

[hidden dove]

Fuck😭

[molten sleet]

do they teach functions in 10th gng

they teach sin and cos right

[hidden dove]

No functions

[hidden dove]

Yes

[molten sleet]

usually they cover concept of a function there

[hidden dove]

No

[molten sleet]

just the basics

[hidden dove]

Not at all bro

[hidden dove]

No brother

[molten sleet]

they do, depends on the school

[hidden dove]

It's extra knowledge

[molten sleet]

how else would you graph stuff

[hidden dove]

Depends on cbse syllabus

[molten sleet]

do you have coordinate geometry in 10th?

[hidden dove]

Like there is no functions basic

[hidden dove]

Yess there is just distance and section formula

And no other things right there

Getting my point?

[molten sleet]

yeah i just checked ncert and there isn't anything about the function notation

[hidden dove]

Yess bro

That's what I'm saying

[molten sleet]

so it is unfair

[hidden dove]

🙂

Like even I studied for jee advanced from 9th i know some fucktions but forgot them

[molten sleet]

skell

[hidden dove]

As not revised since 6 months

[distant wyvern]

U in which class

[hidden dove]

10th bro

[molten sleet]

yeah if you'd revised u would have got that question

[distant wyvern]

Cool jee 2028 aspirant?

[hidden dove]

But I didn't thought cbse would make me feel like beggar of marks

[hidden dove]

Right! Got admission in allen kota

[molten sleet]

they should make the papers percentile based

[distant wyvern]

how was the maths exam lol

[hidden dove]

Fucking bad bro

[distant wyvern]

demm broo all the best yrr tske care

[distant wyvern]

lol

[molten sleet]

so if a paper is hard your overall grade doesn't look worse

[solar stump]

,redir

[hidden dove]

Today is english literature exam

I didn't even slept

[solar stump]

fok the thing doesn't work

[distant wyvern]

yea lol

[molten sleet]

why are you discussing math then 😭

[hidden dove]

Thanks

[molten sleet]

sleep gng

[solar stump]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[solar stump]

This ^

[hidden dove]

Bcz itnis striking my mins

[solar stump]

<@&268886789983436800>

[distant wyvern]

gg

[solar stump]

This isn't the right place to spam videos here

[distant wyvern]

fr

[solar stump]

btw, assuming you're done...?

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[hidden dove]

.close

[obtuse pebble]

Channel closed

Closed by @hidden dove

Use .reopen if this was a mistake.

[obtuse pebble]

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[steel night]

Im a bit confused from the statement underlined in red

[steel night]

I drew the blue and red arcs to maybe see if thats what they mean

[kind hawk]

if the answer you got was for example 3pi/2 you are instead supposed to say -pi/2

[steel night]

I see

but the question is saying find theta between 90 degrees and -90 degrees if cos(theta)=-1/sqrt(2)

[kind hawk]

between -180 degrees and 180 degrees

[steel night]

ah right yes

so practically its from 0 to 360 degrees

its actually hard to visualise it from a unit cirlce, do I need to use a graph?

[steel night]

also I assume ive drawn the unit cirlce correctly

[obtuse pebble]

@steel night Has your question been resolved?

[obtuse pebble]

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[inner sierra]

!status

[obtuse pebble]

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

@spice trellis Has your question been resolved?

[trim portal]

Can you work with complex numbers

oh nvm i suppose its not very helpful, my idea was to use roots to get a contradiction but there just too many options

probably just compare the coefficients then

oh it works

[trim portal]

If you want a hint on how to work with this:
||Suppose r is a root, using the equation, what else is a root?||

[obtuse pebble]

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[trim portal]

Have you tried anything? If so, whats the issue?

[shell oar]

one sec

[trim portal]

can u send it as an image

[shell oar]

ok

not sure what to do next

[trim portal]

im not sure about your base case

[glad vapor]

let the neighbours of the removed vertex be u and v

you want to apply the induction hypothesis

[trim portal]

do you allow loops in your graphs like that?

[shell oar]

in the definitions in my notes, for 'regular', and 'connected', it doesnt say anything about it needing to be simple

[glad vapor]

right now d(u) = d(v) = 1 and also you need some way to argue that the graph remains connected, can you see a quick way to get two birds with one stone

[shell oar]

doesnt say anything about it needing to be simple

[shell oar]

so, theres the possibility that u and v may already have an edge between them

[glad vapor]

hmm actually you may want to assume m > 2 so you don't end up with the degenerate case where u = v

[shell oar]

ok

[shell oar]

you could add an edge between the two?

which results in k edges

[shell oar]

but what about what i said here ..

[trim portal]

and also in a 2-regular graph, importantly

[glad vapor]

there is no problem with that

since you allow multigraphs, you can adjoin a distinct edge to the one that already exists

[shell oar]

ok

[glad vapor]

otherwise if you're working with simple graphs your base case would be |G| = 3 anyway

[trim portal]

In that case, the only way that could happen is if these 3 vertices (including the removed) one were in a triangle and weren't connected to anything else, so that contradicts connectedness

[shell oar]

but isnt a triangle connected?

so how does that contradict connectedness?

[glad vapor]

did you mean to reply to them

[trim portal]

connectedness to other vertices

[shell oar]

yes sry mb

[shell oar]

oh ok

[trim portal]

For simple graphs, we would have m = 3 as a base case and then m > 3 for the induction, so there is 4th vertex to which that triangle wouldnt be connected

[shell oar]

understood

Is there anything else required? Do we not still need to show that the graph with K + 1 edges is a cycle?

acc no that would be impossible right

since that doesnt make it 2 regular anymore

[trim portal]

You should probably add something, yeah. At least sth like "Reversing the process, we can see that G is also a cycle"

[trim portal]

wdym?

to reconstruct G from H, you would remove that edge and replace it with the edge-vertex-edge (the ones that were removed)

And that will obviously still be a cycle

[shell oar]

idk if im tripping but if u have a 2 regular graph with k edges, surely adding an extra edge to make it k+1 would not make it 2 regular anymore

since two of the degrees would increase

[trim portal]

its not adding an edge, you have to reverse the construction

When you went from G to H, you first removed a vertex and its edges and then added an edge between the neighbors

Now to go from H to G, we reverse the process

[shell oar]

no but im saying that how does these starting lines make sense. if u have a 2 regylar graph with k edges, then how can u have a 2-regylar graph with k+1 edges. because the degree of two vetrices will increase by 1 by adding an additional edge

[trim portal]

we remove that added edge, and add the 2 edges and the original vertex back in

nvm worded that poorly

youll add a new edge

that increases the number of edges by 1

[shell oar]

yes

[trim portal]

but youll also add a new vertex

[shell oar]

wouldnt it need to be a loop tho

so that the new vertex is of deg 2

[trim portal]

you have to add it by first removing an edge, and then putting the vertex in between that

not just randomly append it

[glad vapor]

you're subdividing one of the edges

[trim portal]

take a look at this (read it in reverse if you want)

[shell oar]

ohh ok

sure yes that makes better sense

thanks

ok so if we reverse the process

ie

remove the edge, add the removed vertex back along with its two edges, we create a bigger cycle

hence G is a cycle

and that concludes the proof

[trim portal]

indeed

!help

[obtuse pebble]

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

[shell oar]

perfect, thanks a lot for ur help and also to @glad vapor

.close

[obtuse pebble]

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[trim portal]

. @marble laurel you can ask your quesiton here now

[trim portal]

do you know anything about infinite geometric series? (mainly the formula a/(1-r))

[gusty oxide]

Binomial theorem

Wait actually nope

[trim portal]

can you send your progress?

[gusty oxide]

I thought it goes to infinity

[trim portal]

oh okay

i wouldnt do that

1+x+x^2+x^3 appears in a certain algebraic identity

are you familiar with the x^n - 1 factorizations?

well then its probably the right time to learn about that

try multiplying it by (x-1) and see what happens

here is the identity btw

indeed

try using it to simplify the fraction

simplify is kind of subjective here, but just express it in a slightly different form that you think could be useful, so that we get rid of that long 1 + x + x^2 + x^3

[gusty oxide]

Using power series after no?

[trim portal]

yeah

not quite

(x-1)(1 + x + x^2 + x^3) = (x^4-1)

dont forget that (x-1)

and dont factor x^4 - 1 for now

no, its correct like this i think)

$\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{\frac{x^{4}-1}{x-1}}=\frac{1-x}{1-x^{4}}$

[warm shale]

MathIsAlwaysRight

[trim portal]

okay now it looks slightly more like the geo series formula

[silent bloom]

what is the question

[trim portal]

the only issue is the numerator

but you can split it into 2 fractions

[silent bloom]

mb bro u solve this guy's doubt but i want to know the q.

[silent bloom]

integrals?

[trim portal]

.

[pallid willow]

check pins

[silent bloom]

thx boss

[trim portal]

do you know what i mean by this?

[silent bloom]

binomial i believe but i shall no longer talk further. mathisalwaysright is goated

[trim portal]

can it be done via binomial? i honestly dont see how, but maybe im just blind lol

[silent bloom]

i have encountered a similar problem of this sort (i forgot) but they did it via binomial.

[gusty oxide]

If it went to infinity it could be solved with binomial

I thought that at first

[trim portal]

yeah, sure, that works

and you can use geo series now

[trim portal]

wouldnt that be just 1/(1/(1-x)) = 1-x (the function)?

[silent bloom]

im not sure it has to go to inf

but ig it can be solved with binomial.

idk, gimme a sec

[ornate sable]

Split into fractions

[silent bloom]

dont give the ans.

im pretty sure math left it like that for bro to figure it out.

[ornate sable]

Even idk the ans

I’m speculating

[silent bloom]

math gv it like that for him to figure it out

[trim portal]

It should work like this as well, though i too originally intended it to be split

[silent bloom]

ik this may be stupid as heck but @trim portal if i diff it first, then 2nd time, then third then i may find a patter right

that may help

[gusty oxide]

1/(1-x)
If I remember correctly, 1+x+x^2+x^3+.... Is (1-x)^-1

[ornate sable]

||eventh derivative of an odd function is zero at 0||

[trim portal]

yeah, so 1/ that is gonna be 1/(1-x)^-1 = 1-x

[silent bloom]

is it so?

[ornate sable]

Easy to prove

[gusty oxide]

Oh yeah I I got confused lol

[silent bloom]

not as deep in maths as yall

[trim portal]

if only it was odd

[gusty oxide]

It is gonna be 0 then

[ornate sable]

One part of the fraction is, if you split

[gusty oxide]

Kinda fancy ngl

[silent bloom]

any itegration sums

[trim portal]

well, true. Or you can just get the series explicitly, it shouldnt matter. It's just
||1-x + x^4-x^5 + x^8-x^9 + ...||

[ornate sable]

GENIUS

you just need the expansion for the left part of the fraction! ||since the right fractions derivative is 0||

1+x^4+x^8+…

Now you can just read off the derivative at 0

[trim portal]

yeah, its not

sounds about right

[ornate sable]

Perfect

[trim portal]

np, i noticed the 1 + x + x^2 + x^3 and immidiately remembered the expansion (when u see sth like this, 99% of time you're gonna use it). When I tried plugging it in, i got that simple (x-1) / (x^4 - 1) and that looked like it could be solved with power series quite easily

the cruicial steps were noticing that expansion and then using power series

[obtuse pebble]

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[obtuse pebble]

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[trim portal]

how many attempts do you have

[kind hawk]

what have you tried

[glad vapor]

this is not really a maths problem is it

[trim portal]

no, its a guessing problem

one picture cant be enough to guess what they mean

i'd literally just try numbers around 10 one by one

[kind hawk]

huh?

there is an obvious first guess

[trim portal]

okay im blind then

[kind hawk]

count triangles

[trim portal]

ah, i didnt think of the big one

[green lintel]

infinite

[trim portal]

try this then

[green lintel]

random numbers above 15

[trim portal]

unfortunate choice of a first guess

[green lintel]

who said 13

oh wrong server

[obtuse pebble]

@green lintel Has your question been resolved?

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[weak tinsel]

guys is this correct or is ts question done in some other way

[mild sandal]

is this a question off a youtube thumbnail titled "oxford entrance exam" with an einstein pic on it?

[weak tinsel]

no

<@&286206848099549185>

[velvet quail]

Ig u can make a square inside the sqrt

[weak tinsel]

huh

[hidden compass]

I also thought so, but seems not to work

[weak tinsel]

so is it not answerable

[hidden compass]

Because 50 = 49 + 1 = 7² + 1 but there's that √6

[velvet quail]

Mhm ye

[hidden compass]

,w calc sqrt(50 + 14√6)

[warm shale]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Feb 21, 2026, 5:08 PM

[velvet quail]

Do u think it cud come out to be a+b+c) ^2

Hmm idts

[hidden compass]

https://www.wolframalpha.com/input?i=sqrt(50+%2B+14√6) seems not, indeed 🤷‍♂️

[weak tinsel]

i see ty

.close

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[brave halo]

Can someone tell me what to think when i see a question like this

[analog thorn]

uhhh

i mean u just use the log property for this one

[opaque mesa]

express the logarithm in terms of P, Q and R

[hidden compass]

In general, what is $\log{\left(\frac{A}{B}\right)}$ equal to?

[brave halo]

This question is multiple choice but maybe its not

For the other

[warm shale]

Alberto Z.

[glossy basalt]

the first thing I'd think is to see why the Find part has a 7 which didnt appear in the previous conditions.
which may have some trick on the 7 when using the specific based log

[brave halo]

Umm

[analog thorn]

someone here good with calculus?

[brave halo]

Log a + log b?

[analog thorn]

log a - logb

[brave halo]

Sorry yea

Its divide

[glossy basalt]

you can go to the (available) help channels

[brave halo]

I think he meant for me

[analog thorn]

and ask my doubt there?

[glossy basalt]

yes e.g. #help-33 is okay now

[analog thorn]

ok thank you

[brave bramble]

If you're not good with your log properties yet, keep a table of properties in front of you. You don't need to guess at them, you need to memorize them.

[brave halo]

I know it i just have a problem with a question like this

[brave halo]

U see this one is pretty easy

But

[glossy basalt]

yea?

[brave halo]

But the other are difficult

Isnt it

A

[glossy basalt]

yea

[brave halo]

Yea but the other one idk where did the 7 came from

[glossy basalt]

think why 7 appears with log base 7

[brave halo]

Okay why

[glossy basalt]

hmm... lets say log base 10

[brave halo]

Okay

[glossy basalt]

what is log_(10) 10

[brave halo]

1

[glossy basalt]

so similarly

what is log_7 7

[brave halo]

1

[glossy basalt]

there we go

[brave halo]

What do i do with the 9 then

I cant just cancel like that can i ?

[glossy basalt]

the 9 one is given above

like we have log(7/9) so
log7-log9
with base 7 of course

[brave halo]

Haaaa yaaaa

So log 9 will be left?

[glossy basalt]

yep

[brave halo]

Okay and which is it

[glossy basalt]

oh well, it's your turn

with log base 7,
log(7/9)=log7-log9=1-log9

which is?

[brave halo]

C

HAAAAA

WAIT I FORGOT

ABOUT THE

when the 7s cancel each other it will equal to 1

[glossy basalt]

oh um..

I'll type

[brave halo]

Is it not true?

[brave halo]

Okay

[glossy basalt]

$y=\log_a x$ is like asking what number $y$ is needed so that $a^y=x$

[warm shale]

Biscuity

[glossy basalt]

e.g.

[brave halo]

Yeaa okay ik that property

[glossy basalt]

$\log_2 16$ is asking $2^?=16$

[warm shale]

Biscuity

[brave halo]

4

[glossy basalt]

so basically $\log_a a$ is asking $a^?=a$

[warm shale]

Biscuity

[brave halo]

1

[glossy basalt]

that's the reason

[brave halo]

Wait wait i dont get it

[glossy basalt]

so $\log_7 7 =1$

[warm shale]

Biscuity

[glossy basalt]

or $\log_6 6 =1$