help-28

[pseudo roost]

Aww ok

The book says Im right though

[sour shore]

-# Sounds right. If thts ur values for x.

[pseudo roost]

But the rule

I wanna knoe the rule for knowing which term must be made into the other

[delicate torrent]

wdym by that

[sour shore]

As in how to substitute?

[pseudo roost]

You said the x^1/4 is

[pseudo roost]

Yes

Maybe?

I’m wondering how you choose the term

[sour shore]

How u choose k=?

[pseudo roost]

Yes

That

[delicate torrent]

practice

-# practice

practice pattern recognition

[white badge]

$ax^{2p} + bx^p + c$.

[sour shore]

-# These type of ques. You're supposed to aim for a quadratic in k.

[glossy valve]

Nicole

[pseudo roost]

Ok

[pseudo roost]

K!

[pseudo roost]

I’m I suppose to try this pls?

I don’t meant to be rude if it comes out rude

[sour shore]

Perhaps

[white badge]

I don't get what you mean, sorry.

[pseudo roost]

The thing you did

With the jot

Should I try it?

[white badge]

you can save it and reuse it for other problems, if that's what you mean.

there's nothing to be tried there, it's not a question. it's a pattern you should identify for questions of this sort.

[delicate torrent]

or as i said before: pattern recognition

[pseudo roost]

I think that one’s easier

My problem is the fractions ones that have to be turned into quadratic

Also thanks for the formula

[delicate torrent]

probably turn into roots

[white badge]

you can do fractions the same way.

[delicate torrent]

so now you got pretty much all info for you to do the problem, can you do it now?

[pseudo roost]

No no I mean fractions like the one I just got helped solving

Idk which has to be l

K

[white badge]

if you want, we can take a look at an example, but I'd rather not get in the way right now so as not to break your focus.

[pseudo roost]

Oh no I’d

Idm

I need all the help I can get

Mya the difficult for me even after doing well bc I forget

[delicate torrent]

-# did you forget all golden rules i gave you?

[pseudo roost]

Not rlly

Just forgot how to use them

[delicate torrent]

Golden rule: pattern recognition

[delicate torrent]

Like this one

[full forum]

@pseudo roost Has your question been resolved?

[pseudo roost]

So should I like multiply both fractions by 2

And see which ones like is like the other?

[delicate torrent]

Wdym?

[pseudo roost]

like we have to terms with x

And those term have separate powers on the x that are reaction

Fraction

Wait nvm

Uh what I’m getting from the question you helped me with is

I should look at the bigger one

And turn it small.

[white badge]

maybe you should use another question as an example to show what you mean, so that everyone can be on the same page.

[pseudo roost]

yes please

This

Can I change the x^1/2 to 1 by (x^1/2)^2

Or do I change the x ^1

[pseudo roost]

Which will give you one right

[remote charm]

just substitute $y = /root{x}$

[glossy valve]

Ary366

[pseudo roost]

It I think I change the x^1 to x^1\2

[pseudo roost]

wha

[muted orchid]

Yes.

[remote charm]

Then it yould become y^2-5y+4 = 0

then solve it

[warped frost]

$k = \sqrt{x}$

[glossy valve]

Hanako(x, y); ∂(fox)/∂x

[remote charm]

btw the solution is 1, and 16

[warped frost]

but, let's not confuse OP with substituting for a root first

[remote charm]

OP?

[warped frost]

you do this. doing the former will change the effective exponent on the sqrt(x) term, which is not allowed

[warped frost]

the current helpee

[remote charm]

oh

[muted orchid]

!nosols

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[warped frost]

we're working based on this atm.

[remote charm]

atm?

[warped frost]

at the moment.

[remote charm]

ok, i am a nube at chatting

[pseudo roost]

oh ok ok

Tysm

[remote charm]

isn't the problem solved?

[pseudo roost]

wait how

Uh

The x^1

[warped frost]

no. the problem is not solved until the OP says it is.

[pseudo roost]

Do I do x^1/2)2

[muted orchid]

if u meant $(x^{1/2})^2 = x$, then yes.

[pseudo roost]

Ok!

[warped frost]

if you mean (x^(1/2))^2, correct

[glossy valve]

Kaladin.

[warped frost]

alright I think there are too many ppl here, so I'll excuse myself. gl with the problem OP!

[torn jolt]

-# lot of ppl here what's the matter

[pseudo roost]

tysm

[warped frost]

nw

[pseudo roost]

Like this?

Ty’sm everyone

🥹

.close

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[torn jolt]

Open more channels in the future. help me grind helpful.

[torpid stag]

You will never be one of us 🥀

[torn jolt]

dam outcasted

[slate glacier]

[torn jolt]

https://tenor.com/view/storm-alone-sad-gif-22428000

[torn jolt]

feels like start of a movie, icl they never end good

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[visual turret]

I have a question about differentiating

[slate glacier]

just ask the question

[visual turret]

yes sorry im busy drawing it in paint to show

[slate glacier]

no problem

[visual turret]

sorry for my awful handwriting i just decided to send my book instead

you see where i drew a star on my page, when i move the exponent back under after differentiating and you are making it positive again do you drop the negative?

[slate glacier]

you drop the negative for the power, but keep it for the term

like -x^(-2) becomes -1/x^(2)

[visual turret]

but id get x squared = -162000/20 no?

or x squared = -8100 simplifed

[slate glacier]

no

-162000/x^2 + 20 yes?

[visual turret]

yes

[main vault]

You made 20 positive when bringing it over by accident

[slate glacier]

just send the minus term to the other side and it becomes positive

[visual turret]

ohh yes and then i cancel the negative

sorry guys super silly mistake

[slate glacier]

no problem man

[visual turret]

thanks for the help!

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[knotty grail]

uh umm this is a weird one

[knotty grail]

lhopital abandoned me on this one

[round sparrow]

Lhopital is generally not a good method anyway

[knotty grail]

propaganda

ok yeah true though

[round sparrow]

Do you know Taylor series ?

[knotty grail]

sure

[round sparrow]

That should work

[torn jolt]

which taylor series workshere

[round sparrow]

You develop each 1/f(kx) in taylor series

[torn jolt]

,, \frac{1}{1- (kx)^{p}}

[glossy valve]

Hi

[round sparrow]

Yup

[torn jolt]

hum..

oh this is the generating function one maybe

idk what do they call it $\frac{1}{1-x} = 1 + x + x^2 ...$

[glossy valve]

Hi

[round sparrow]

Kind of, but stop at first term

[torn jolt]

at 1?

[knotty grail]

what the hell bruv

why did that work

thank you

.close

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[torn jolt]

lmao

[round sparrow]

At x

[torn jolt]

ic

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[queen pilot]

Can anyone help me understand how to do synthetic division

[round gust]

I mean there s plenty of videos

On youtube

https://m.youtube.com/watch?v=FxHWoUOq2iQ&pp=ygUSc3ludGhldGljIGRpdmlzaW9u

[limpid viper]

is this the horner method?

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[full forum]

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[dark wigeon]

I'm trying to find the equivalence class of the function

This is my work

[ember quartz]

Hello, what does the "o" do in ⁨f o theta⁩?

[narrow ermine]

It’s composition.

How are F and S defined?

[dark wigeon]

F is defined as : E -> E

I think S just means it's bijective

[narrow ermine]

Like functions from E to E?

[dark wigeon]

Yep

[narrow ermine]

Okok

[ember quartz]

Ah ok, it's not written as $f \circ \theta$, they have used word for that

[glossy valve]

Mor Bras

[dark wigeon]

And R is a binary relation

[dark wigeon]

Yep , skill issue from our teacher

[ember quartz]

Je suis désolé pour toi

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[craggy saddle]

My friend said it's like "one move theatre popcorn is 4 dollars and a ticket is 3 dollars another theatre popcorn is 3 dollars and the ticket is 4 dollars and we have to rewrite it graph it and write an equation

[craggy saddle]

Can you help with this type of question

[thorny bluff]

wdym

this is just 4+3=3+4

[craggy saddle]

How to rewrite it graph and write an equation for it tho

[void nova]

Graph what??

[thorny bluff]

there arent any variables here

[craggy saddle]

Exactly

[void nova]

Equation?? What even is the variable???

[craggy saddle]

Bags of popcorn ig

[thorny bluff]

could you give an example of an actual problem like this that you need help with? i dont understand what you are trying to ask

[void nova]

Are you sure your friend is not trolling you? 🙈

[craggy saddle]

Idk I wasn't there so idk

[void nova]

There where??

Lol

[craggy saddle]

My school during the test

[void nova]

Ohh so this was an exercise on the test and you were absent?

Are you 100% sure the question was that one?

[craggy saddle]

She said it was smth like that but not the actual problem

[void nova]

There we go, as I expected

[craggy saddle]

No like not the exact numbers but that's how it was set up

[void nova]

Without the original we can't help you
Not because we don't want, but because what you reported doesn't make sense

[craggy saddle]

I understand

[cold lagoon]

it was probably that they price it a certain way and then you're calculating profit based off of customers

[craggy saddle]

I was thinking maybe it wanted to make a graph to see like when they'd cost the same pric

[cold lagoon]

and the variable is money earned

[craggy saddle]

Price*

[cold lagoon]

well theres two sets of prices and it doesnt give a rate of whether they're increasing them or decreasing them

[craggy saddle]

Yea

[cold lagoon]

so i think your friend explained it badly

[craggy saddle]

Alr ty

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[brittle spade]

i dont really understand

the solution

can someone help plz

thx

if you message feel free to ping me i dont mind

im just doing some hw while i wait

[neon basin]

Hello

What's

[brittle spade]

LETS GO

ITS THE GOAT

[neon basin]

sqrt(108306)

[brittle spade]

idk what i did

[neon basin]

And 234

[brittle spade]

thats in an alternate question

the one im looking for is

In the figure, what is the area of triangle ABD? Express your answer as a common fraction.

[neon basin]

Right

Can I see

What you did

for the slightly high number

[brittle spade]

So I first found the area of certain triangles and found the length of certain lines

So ACB = 14, ACF = 6, ECB = 7

[neon basin]

👍

[brittle spade]

and AF = 5, EB = sqrt(53) and AB = sqrt(65)

[neon basin]

Okay

[brittle spade]

and i dont know what to do

I thought AED and DFB had same heights but no

and if i try to subtract ACF, i cant get efb and vice versa

also for context i dont know coord geometry

so thats all i know

[neon basin]

Right

What about ABD and ABF

[brittle spade]

what about them?

they add to make 8 ig

[neon basin]

Right

But their height from B onto AF is the same

[brittle spade]

wait what?

can you elaborate

[neon basin]

If we look at ABF

The height to B is 4 right

[brittle spade]

yes

abf height is 4

[neon basin]

wait no it's not

that's not a right triangle

holy

I am way too tired

Sorry but I need to go to sleep it seems 😭

Can't even see angles right

[brittle spade]

oof

gn

what do i do

do i wait for another helper?

or ping??

[neon basin]

yeah someone will come

You can mention yeah

Sorry again 🙁

😴

[brittle spade]

Its ok

gn gng

<@&286206848099549185> the person who was helping me is asleep sry

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[thin scroll]

this the question provided

[thin scroll]

and this is my oslution

is work?

[rapid rain]

correction, this is 40 + 0 + 0

dot product adds all the coefficient products together

[thin scroll]

ic

[rapid rain]

and the formula in general is $|a\cdot(b\times c)|$, but since here you got a positive result it didn't matter

[glossy valve]

Rafilouyear2026

[thin scroll]

m

ok thank you

.c;pse

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[rapid rain]

also quick fix to the b x c computation

it's (8,-4,-5) I think

1 * 1 - 3 * 2

[thin scroll]

oh yea

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[plush egret]

https://en.wikipedia.org/wiki/Leibniz_integral_rule

.

[pseudo trout]

Ty

.close

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[drifting summit]

hi

[drifting summit]

(d) There can be at most one maximum element

I'm trying to prove this

Let $x$ be the maximum of a set $X$ with a partial order $\leq$. Let $x'$ be another maximum of $X$. Because $x$ is a maximum, $x' \leq x$. Because $x'$ is a maximum, $x \leq x'$. Because of the Antisymmetric Axiom of partial orders, $x = x'$ so there is one distinct maximum element.

[glossy valve]

toast

[drifting summit]

And here is asimple proof i wrote up

I think it's right

but why aren't maximal elements unique?

like wouldnt both y and x be maximals, but theyre the same element?

[stiff musk]

how do you know that x and x' are comparable?

[drifting summit]

hm

[drifting summit]

well i would assume x' and x are both in X

but ig that doesnt mean theyre comparable

[drifting summit]

wait i see

you can only assume x and x' are comparable if $\leq$ is linear?

[glossy valve]

toast

[drifting summit]

Let $x$ be a maximal element for a linear partial order $\leq$ on a set $X$. By definition of linear, $X$ is a chain, that is any two elements are pairwise comparable. Let $x'$ be an arbitrary element of $X$. Note because $X$ is a chain, $x$ and $x'$ are comparable, so either $x' \leq x$ or $x \leq x'$. If $x' \leq x$, we are done. If $x \leq x'$, by definition of maximal element, $x = x'$. So $x' \leq x$, thus $x$ is a maximum.

[glossy valve]

toast

[drifting summit]

here is my proof for f (a maximal element in a linear order is a maximum)

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@drifting summit Has your question been resolved?

[slender onyx]

looks good

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[mental pewter]

hey uhm im wondering how to find The cross-section of the plane (α) passes through A and is perpendicular to SC.

[mental pewter]

with ABCD is a square, O is the centroid of ABCD, and SO ⟂ (ABCD)

the only thing that i can do is draw AM ⟂ SC

but thats not enough

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[mental pewter]

<@&286206848099549185>

[round gust]

can you reformulate your question

whats plane alpha

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[civic dust]

Can someone help me with composite functions?

[grave elm]

What's your question?

[civic dust]

Wait Im gonna screenshot give me a second

I am trying to get the meaning of this, im reading some tutorial on composite functions but I don't really understand it

[grave elm]

Composite function is basically when you put one function as an input to the other

(f ◦ g)(4)

(f ◦ g)(x) means f(g(x))

[civic dust]

What is the symbol between f and g?

[grave elm]

It's a symbol for composition

it means that we put g as an input to f

[torn jolt]

it is called f of g(x) sometimes

[civic dust]

Ok so In this question that I sent, (f o g)(x) will be f is the input function for g function?

[grave elm]

Hmm, i think you reversed it

[civic dust]

output of function g for x will be the input for f function?

[grave elm]

(f o g)(x) means that we first calculate g(x), and then plug that as an input to f(x)

So e.g. to compute (f ◦ g)(4), we need to find f(g(4)). We start by finding g(4), that's 5 * 4 = 20. Now we put that as an input to f, so f(g(4)) = f(20)

now we need to compute f(20)

[civic dust]

ok? Let me try? It's weird I didn't know you cando that in math

[grave elm]

which would be -30

[civic dust]

Nested functions

[grave elm]

Are you familiar with programming?

[civic dust]

Yes, that's why I mentioned this

[grave elm]

def f(x):
    return 10-2*x

def g(x):
    return 5*x

print(f(g(4)))```

this is computing (f o g)(4)

[civic dust]

Which lang is that

[grave elm]

python

But it should be pretty much the same in almost every lang

[civic dust]

I don't think that going to programming would help this

But ok I got the main idea

So I take the value of g(x) which is = 5x, and put this 5x into f(5x) = 10 - 2(5x)?

[grave elm]

Exactly that

[civic dust]

Wow that is complicated

How do I read this notation?

Or understand it

[grave elm]

(f o g)?

f of g

[civic dust]

Yeah

[grave elm]

or f composed with g

[civic dust]

So this is a composite of 2 functions?

How do I know in which order should I solve them?

[grave elm]

f o g means f(g(x))

g o f means g(f(x))

[onyx glen]

aight one moment

i literally drew a graphic for another helpee about this exact thing

[civic dust]

No I got this, I get the main idea

[onyx glen]

composition of two functions

[civic dust]

Ok so can there be more than 2 functions in a composite?

[onyx glen]

composition*

and yes you can chain as many as you want

nothing stops you from taking like 8 functions and composing them in one big chain

[civic dust]

So it's basically putting output of 1 function to anotehr function, and it's called composition of functions

[onyx glen]

like $f_1(f_2(f_3(f_4(f_5(f_6(f_7(f_8(x))))))))$

[glossy valve]

Ann

[onyx glen]

that is exactly what it is

[civic dust]

And it's literally g of f

[onyx glen]

other way around

f of g

[civic dust]

function f for the input of function g

[grave elm]

f o g is f of g
g o f is g of f

[civic dust]

"f o g is f of g
g o f is g of f"

How can I note that big composition into the symbols?

[onyx glen]

$f_1 \circ f_2 \circ f_3 \circ \dots \circ f_8$

[glossy valve]

Ann

[grave elm]

And to read it, you can just keep chaining the "of"s

f1 of f2 of f3 ...

[civic dust]

So f8 is the first function, f7 takes f8's output and put it's into itself

[onyx glen]

yes

f8 is the only function that sees the "raw" value of x directly

[civic dust]

Great man, thank you a lot

I love the help

[torn jolt]

its like concentric layers of onion

[onyx glen]

im not a "man" but you're welcome

[civic dust]

Man, in the sense of a human

[onyx glen]

i had a much more gruesome comparison in mind tbh

[civic dust]

What

[onyx glen]

more gruesome than this i mean

[civic dust]

[onyx glen]

but i will not say it unless someone specifically says they want to hear it

[civic dust]

I want to hear it

[grave elm]

I kinda want to hear it now..

[torn jolt]

edging

[civic dust]

[torn jolt]

yh yap it out

[onyx glen]

aight since 2 people said it

||human centipede||

[civic dust]

I don't even know what is this

[torn jolt]

https://tenor.com/view/why-cry-why-pepe-why-why-pepe-cry-pepe-pepe-cry-gif-5847161834541812219

[civic dust]

holy shit ok

[torn jolt]

@modulators

[civic dust]

MODULATORS

[torn jolt]

so if you dont have any doubts you can close this

[civic dust]

Yeah thank you

[torn jolt]

by typing .close

[civic dust]

A lot, didn't expect to solve it that fast

.close

[full forum]

Channel closed

Closed by @civic dust

Use .reopen if this was a mistake.

[torn jolt]

do come back if you have more doubts

and help me join green gng

[civic dust]

I don't know what's green

I don't smoke zaza

[grave elm]

im green

[civic dust]

Ok then solve more tickets i guess?

[torn jolt]

like the username becomes green

[torn jolt]

yh ig

[torn jolt]

https://tenor.com/view/hulk-smash-gif-12677792749566644516

[onyx glen]

!redir

[full forum]

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

[torn jolt]

there is no channel they closed it

[full forum]

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[random kettle]

Foundations

Are these both correct or is one correct and the other is or are they both incorrect

[grave elm]

T <=> x in A?

Did you mean x in X?

in the first one

[random kettle]

I'll ask him

[grave elm]

And how do you define complement?

[grave elm]

Is this someone else's proof?

Why don't they come here instead?

[random kettle]

They don't speak english

[grave elm]

Do they speak arabic, or what language?

[random kettle]

He said if x in A' then x ~in A

Like that's the def.

[random kettle]

Yes

[grave elm]

And what exactly is X?

[random kettle]

Universal set

In both cases

[grave elm]

Which I suppose is defined in a way that every x belongs to it

[random kettle]

Exactly

[grave elm]

Yeah, so the first proof is alright, just replace T <-> x in A with T <-> x in X

i suppose that's a typo

[random kettle]

He said yes

[grave elm]

in the second proof, this part is kinda weird

I can understand the -> direction, that follows just by A subset X

[random kettle]

The left side of <=> is the same from the first line to the last so we both didn't write it in all the others

X in (AUA')

In all the other lines

[grave elm]

The only small issue is that the <- direction isnt justfied just by A subset X here

[random kettle]

Excuse me sir

Wait a min

[grave elm]

(x in X) v (x in X) -> (x in X) -> (x in A) v (x in X)

it's not too difficult to justify the <- direction, but it isnt justified just by A subset X

[random kettle]

This further explains what I mean

However the professor doesn't write it to save time

[grave elm]

yeah, i understand that part

[random kettle]

Ok so what's the issue

[grave elm]

But what you're claiming here (this is the 2nd proof) is
(x in A) v (x in X)
<-> (x in X) v (x in X)
and you justify it by A subset X

which works for the -> direction

but <-> consists of 2 directions

the other direction, <-, is justified a bit differently

[random kettle]

A subset of x is given

Which means every x in A is x in X

[random kettle]

Wdym

[gray quarry]

In my opinion …..the denominator could be 2 in the last line…..Is it correct…..?

[grave elm]

For the <- direction, you'd have to prove that if
(x in X) v (x in X), then (x in A) v (x in X)

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[random kettle]

This help channel is reserved

[random kettle]

But the other direction is x is in AUA'

We used <=> and the def on union since p<=>p = T

[grave elm]

My point is that you need to justify both why the next line follows from the current one and why the current one follows from the next line

A subset X only justified why (x in X) v (x in X) follows from (x in A) v (x in X)

it doesnt justify the other direction

[random kettle]

Ohhhh

Our proffesor didn't tell us that

[cyan river]

yes x ∈ X doesn't imply x ∈ A

[grave elm]

The reverse direction can actually still be justified

x in X v x in X -> x in X -> x in A v x in X

thanks to the or

[random kettle]

Unfortunately the professor didn't teach us none of this

She used contradiction in her solution

[grave elm]

Does your prof use words?

Or does she write purely symbolic proofs just like you?

[random kettle]

The students asked if they could use something other than contradiction

[random kettle]

Want me to show you her full proof?

Using C!

[grave elm]

sure

[random kettle]

Bruh

Number 3

[grave elm]

Ic, so she does use at least some words

I'd probably learn to do that as well

you can use words in your proofs

[random kettle]

Yeah thx

[grave elm]

Anyway, other than that minor nitpick, your proof is correct

just keep in mind that when you write <->, you need to justify both directions

in this case, you were kinda lucky (or maybe you knew of it, idk)

[random kettle]

So no.2 is correct

[random kettle]

Lol I didn't please tell me why

[grave elm]

To prove my point, here is a proof that x in A <-> x in X:

x in A <-> x in X (A subset X)

I did practically the same thing as you, I replaced x in A with x in X, because A is a subset of X

but in this case, it is completely wrong to do that with <->

what I could say is
x in A -> x in X (A subset X)
This would be valid and justified

[random kettle]

without justification

[grave elm]

Yeah, but I cant really justify that in general

[random kettle]

Ok

How can I justify mine

[grave elm]

If x in X is just T for you, you can do:
x in A v x in X
<-> x in A v T (x in X is T)
<-> T (P v T = T)
<-> x in X

[random kettle]

Based on the info I gave you about what our proffesor taught us

Could it have been possible that no.1 used AI

[grave elm]

AI as in artificial intelligence?

or what

[random kettle]

Because when I asked him to explain he mentioned none of what you explained here

[random kettle]

Yes

[grave elm]

Like you're asking whether this proof used AI?

I dont think so

Not necessarily

but it's possible, i cant say

[random kettle]

Noted

All together, do you recommend the students sticking to C! To prove since our teacher didn't teach us how to prove using other methods?

[grave elm]

I think that both ways are fine tbh

[random kettle]

And ofc use words lmao

[random kettle]

Ok

[grave elm]

Yeah, especially as your proofs will get more complicated, words will become more and more important

[random kettle]

Thanks a lot I really appreciate your patience

[grave elm]

np

[random kettle]

.close

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[torn jolt]

these conditionals in general

[torn jolt]

is there a general pattern here

I did A/2 = ...
B/2= ...
C/2 = ...

Looked messy

I am doing number a rn

[shy gull]

anyone here vietnamese?

i need help with this

[nimble ridge]

!occupied

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[white badge]

!occupied, unfortunately. #help-29 is open though.

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[torn jolt]

I am going to sleep

.close

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[patent hill]

so I don't understand how i got part b wrong. I found the line that arg π/6 would be and then substituted it into the equation for the circle to find where they intersect. And then tried finding the modulus of that. but its not right, i checked my workings so i dont think i made any mistakes there, im not sure tho

[languid talon]

you might have not gotten credit because you didn't write z explicitly, and/or you never specified what x and y were

unless if the green was you checking your work?

[patent hill]

yea it was my own marking

this isnt exactly hw this is just practice questions

[patent hill]

ill keep that in mind when i do other questions ig

[languid talon]

the problem is actually somewhere in your self-check

[patent hill]

ohhh

wait what line exactly

or are you expecting me to check everything again

[languid talon]

well all of the work is omitted so I can't say, but the expression containing the radical is just not (sqrt(6)-1) / 2

you should find that $|x+iy|^2=\left(\frac{3\sqrt2-\sqrt3}{2}\right)^2+\left(\frac{\sqrt6-1}{2}\right)^2=7-2\sqrt6=(\sqrt6-1)^2$

[glossy valve]

Flip

[patent hill]

wait why do you not root that? i thought to find modulus you had to square root (x^2+y^2)

[languid talon]

you do, but you don't need the radical until the very end so I just work out |x+iy|^2 and then take the square root when I'm done

like in the end I write $|x+iy|=\sqrt{|x+iy|^2}=\sqrt{(\sqrt6-1)^2}=|\sqrt6-1|=\sqrt6-1$

[patent hill]

ohh sorry i forgot that the ans here is squares

[glossy valve]

Flip

[patent hill]

thanks, i think i know what i did wrong

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[proper tapir]

Why can’t 5x1 + 5x2 +6 =0 ?

[viral jasper]

,rccw

[viral jasper]

Because $x_1$ and $x_2$ are in the domain $\bR_+$

[glossy valve]

SWR

[proper tapir]

Yeah it just hit me

Thanks i should’ve read the question more carefully

[viral jasper]

Happens to the best of us

[proper tapir]

.close

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[still current]

f(x) = 10ln(x-4) + 1

[still current]

10 log_e (b)(x-h) + k

?

[foggy vapor]

Err

What are you asking

[still current]

i have to find the domaine and variation so i need b

[foggy vapor]

The domain and range?

[still current]

variation is increasing ]?,?[ and decreasing ]?,?[

[foggy vapor]

Okay i see

Do you know the domain of log(x) by itself

[still current]

its ]-infinite,4[

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[sinful robin]

How do i start with this proof? Our current unit is using double derivatives to find inflexion points

[sinful robin]

9a i mean

[woeful pasture]

what have u tried

[sinful robin]

I took the first and second derivative to get f'(x) = 3ax^2+2bx+c and f"(x)=6ax+2b

But i dont really know how to start

[woeful pasture]

ok

from f'(x) = 3ax^2+2bx+c

how do u generally find the stationary points

[sinful robin]

f'(x)=0

[woeful pasture]

cool

so what would be the condition to have two distinct stationary points

[sinful robin]

Uhhh there are a few?

Be a polynomial of degree 3 or higher?

[woeful pasture]

3ax^2 + 2bx + c = 0

[sinful robin]

Have a single point of inflexion?

[woeful pasture]

stationary points are the ____ of this quadratic eq

[sinful robin]

Roots

[woeful pasture]

good

[sinful robin]

Wait

Is this like

I throw it in the quadratic formula and show that discriminant has to be positice

Oh wait yeah that works

[woeful pasture]

bravo

👍

[sinful robin]

Thank you!

Lemme just solve this on paper before closing the channel to know i did it correctly

[woeful pasture]

aight

[sinful robin]

Great! Thank you very much

.close

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[grave elm]

<@&268886789983436800>

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[bitter comet]

<@&286206848099549185>

[onyx glen]

you should send your question(s)

[gritty rose]

!15m

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[bitter comet]

I dont understand q no 4, 6 , 10 ,11

[ocean sluice]

for number 4 do you know the rule of rule of exponents

when it’s a negative

[bitter comet]

Yes

[ocean sluice]

okay what is 81^-2

wait are you allowed to use a calculator

[bitter comet]

No

[oak ruin]

Do you know that $(a^b)^c = a^{bc}$

[glossy valve]

Ari

[oak ruin]

It'd be easier to just multiply all the exponents instead of doing 3 computations

[ocean sluice]

yeah true

[bitter comet]

No

[oak ruin]

well that's a good one to know

You have exponents on the 81:

1. -2
2. 1/2 (from the square root)
3. 1/4

multiply thoe

[bitter comet]

So should i just apply a^(-n)=1/a^n

[onyx glen]

you can but it isn't essential here

[bitter comet]

Alr

Is the answer 1/3

[onyx glen]

let's check

,w (sqrt(81^-2))^(1/4)

[glossy valve]

Results provided by WolframAlpha

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

Requested by ann_dec

[onyx glen]

yup

[bitter comet]

Thx

Oh ya and in the 9th question is it SAS or RHS

[onyx glen]

what data have you got

[bitter comet]

.

[onyx glen]

yes, i can see the questions

im asking you to read the diagram and tell me what congruence data you've got

[bitter comet]

I think RHS

[onyx glen]

i wanted you to list it out in more detail:

• a pair of right angles
• a pair of equal hypotenuses
• and a shared side

so yes RHS. because the right angle isnt between the sides involved so SAS can't apply

[bitter comet]

Ok tysmm

Tht itt

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[fading creek]

https://www.youtube.com/watch?v=HfACrKJ_Y2w
so i am currently watching this calculus course by freecodecamp.org, but on 9:09:09, it skipped to integrals without teaching anything about them previously. both in their pre-calc course and the calculus course

can anyone tell me just enough knowledge about integrals for me to understand what she is saying for topic "proof of mean value theorem"? also isnt she supposed to +c? thanks

[full forum]

@fading creek Has your question been resolved?

[fading creek]

<@&286206848099549185>

[prime pier]

this is a weird order to put the sections in, especially considering that the mean value theorem is stated in terms of integrals rather than derivatives.
you could consider skipping to 10:27:45 (Antiderivatives) and watching until 11:22:17, end of The Fundamental Theorem of Calculus II to get an idea of how these two theorems are equivalent

wait disregard. they posted the wrong section during that interval, they link to the correct video in the description

@fading creek here's what you need

[fading creek]

oh thanks! didnt see that

[prime pier]

i almost didnt see it either :D

[fading creek]

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[spiral vigil]

$W = \int \vec F\cdot \dd{\vec s}$

[umbral dome]

the "dx" in the integral is the displacement

[glossy valve]

hayliänus austrǎlis

[spiral vigil]

well yes, you're summing up a bunch of force * very tiny displacements, over the path of the object

in this case it's made easier by the fact that the object is traveling in a straight line, and your force is being applied along that line; so the fact that it's a vector isn't that important

but the fact that the force changes as you pull it is important

to make it more concrete:

• you first have to stretch it from 10cm to 11cm, this takes 2N of force over 1cm so 2cJ of work
• then you stretch it from 11cm to 12cm, this takes 2.2N of force over 1cm so 2.2cJ of work
• then you stretch it from 12cm to 13cm, this takes 2.4N of force over 1cm so 2.4cJ of work
• ...
• then you stretch it from 19cm to 20cm, this takes 3.8N of force over 1cm so 3.8cJ of work

but that's not really precise is it? you could work with mm instead of cm and get a more accurate result with more stages

an integral is just "what if we worked with infinitely small stages and infinitely many of them "

also ty for the excuse to use centijoules as a unit that was fun

well i did multiply by the displacement in that example

yes this

i assume the question here is like how much water is required to fill this to a certain height

like lifting the whole container?

wouldn't that just be the weight of the whole thing * 1m?

ohhh i see

generally for these you figure out how much energy it takes to lift one "slice" of water ("infinitely thin") up and out

so you'll have a factor associated with the height you have to lift it to, and a factor associated with how big that rectangle is

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[stone finch]

I think here we can use bernoulli,s theorem

[stone finch]

to find out the height

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@stone finch Has your question been resolved?

[warped frost]

do you have a question

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@stone finch Has your question been resolved?

[stone finch]

yeah

can you tell me how to solve that problem

[stone finch]

this one

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[mental pewter]

can ya help me with this geometry problem

[mental pewter]

we have a pyramid S.ABCD with ABCD is a square

Find the cross-section cut by the plane (α) passing through point A and (α) perpendicular to SC

the only thing that ive done is draw AM ⟂ SC

but that's not enough

[onyx glen]

draw lines in planes SDC and SBC passing thru M and perp to SC

[mental pewter]

oh

im dont think thats that easy...

.appreciate ya a lot

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[thick hedge]

what;s this even supposed to be asking me

[spiral vigil]

which sentence do you not understand?

[split hatch]

There are basically three questions

• Let E be any closed subset of R. Is it possible to find a differentiable f with zero set E?
• Let E be any closed subset of R. Is it possible to find an n-times differentiable f with zero set E?
• Let E be any closed subset of R. Is it possible to find a smooth f with zero set E?

in case the wording was confusing

[thick hedge]

The wording was confusing

[thick hedge]

Thansk a lot!

[thick hedge]

I suppose if yes, I have to give an example of a function for each

or is proving existance possible

[split hatch]

yes

[thick hedge]

yes to the 2nd ?

[split hatch]

I think all of them are true

So if you just prove no3 then you're done

[thick hedge]

I don't think coming up with a function is possible here though is it

the cantor function will wreck havoc

[split hatch]

have you seen the exp(-1/x) thing

[thick hedge]

$e^{-1/x}$?

[glossy valve]

wai

[split hatch]

yes you basically use this as the building block for smooth bump functions

[thick hedge]

hmm

I'll think a bit more I guess

thanks!

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[split hatch]

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[grave elm]

<@&268886789983436800>

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[torn jolt]

Does anyone have practice resources for 10th grade math(general) aka 16 year old

[torn jolt]

Non-video is preferred, thanks

[fast peak]

khanacademy

[full forum]

@torn jolt Has your question been resolved?

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[civic dust]

What means invertible here?

[civic dust]

I can't solve c

[torn jolt]

invertible means that g is a one to one function

[civic dust]

What means one to one

[torn jolt]

different values are never the same

they have their own image

[civic dust]

Yeah but isn't that how all the functions work?

[torn jolt]

example:

f(a) != f(b) ever

nope

[civic dust]

But no function can give the same result twice

Other wise it's incorrect

[torn jolt]

nope

[civic dust]

What

[torn jolt]

its not incorrect

there are functions when inputted different values for x give same output

take for example

f(x)=x^2

[civic dust]

The vertical line test shows this function is correct

[torn jolt]

for a inverse theres a horizontal line test

[civic dust]

Never heard about horizontal line test

[torn jolt]

https://matterofmath.com/algebra/horizontal-line-test/

[civic dust]

So If the function is one to one, there is no inverse for the function?

[torn jolt]

if the function is one to one, there is a inverse for the function

[civic dust]

Isn't this when a function is incorrect?

[torn jolt]

how is this an incorrect function

[civic dust]

different x's give the same y

Wait I am confused

[torn jolt]

this is just for invertibility

Does not mean the function is incorrect

[civic dust]

"Remember that each input must have only one output to be an official function."

So one-to-one function is not a official function?

[torn jolt]

You interpreted that incorrect

one function cannot give more than one outputs

not that the outputs can be equal

cant*

[civic dust]

In one-to-one function input is same as output?

[delicate torrent]

i can make $f(x) = 1$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

then many inputs correspond to 1 output

it's still a valid function

[torn jolt]

no

[civic dust]

So what is the connection to this?

[torn jolt]

it just means that the function all have unique images

there will be no overlap

[civic dust]

Ok

[torn jolt]

I think visualizing with a arrow head diagram will be easier

[civic dust]

I just did plot it on desmos

One-to-one doesn't give out same y outputs

for any x

[torn jolt]

exactly

[civic dust]

Ok this is the invertible?

[torn jolt]

yup

[civic dust]

So If I put -1 into the g function, Ill get 4 in y?

[torn jolt]

if you put 1 you'll get 4

[civic dust]

What about -1 for x

[torn jolt]

not given

[civic dust]

So how can I determine what is g(2)?

It's also not given

[torn jolt]

yep

you cant

[civic dust]

So I can't solve it?

[torn jolt]

you mean c?

[civic dust]

Yes

[torn jolt]

qn c?

Inverse and non inverse cancel each other out

so 2 = x

[civic dust]

What

But there is no g(2)

So how can I know what to put into g^-1

Oh ok because they cancel each other out

Ill get back X

[torn jolt]

I am looking if I can find an example

are you ok with an AI generated example?

[civic dust]

Ok I did calculations and it appears right

[civic dust]

yes

I hate AI explaining math, that's why I am here

How can I proove that x = 2?

[torn jolt]

it all nicely cancels each other

[civic dust]

$f^{-1}(x)=2 <-> f(2)=x

Great, it's not putting up

[glossy valve]

firestepper

[civic dust]

So this is the definition

But it doesn't explicitly say that x is equal to 2 here

It just shows the definition of inverse function

[torn jolt]

Its by definition

unless you know the function you can't really prove it

you just have the table

[civic dust]

ok thank you

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[crude tiger]

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2317090 uhh the aops links for sols isn't working where should i read the sols

[crude tiger]

Prob 52

[delicate torrent]

!status

[full forum]

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 tiger]

1

[delicate torrent]

okay

[crude tiger]

Well not really I've been case working abit

[delicate torrent]

it's a congruency exercise

[crude tiger]

I just wanted to know where should i read the sols

[rapid rain]

You could start thinking about the prime factorization of p^q + p^r if it were a square number

[delicate torrent]

-# ||WLOG q < r||

[rapid rain]

Not sure we have q and r different necessarily

[delicate torrent]

-# ||since if q = r then it ain't square if p is odd||

[crude tiger]

Yrp

[crude tiger]

Hmm can u prove it?

[rapid rain]

Let's start thinking about p = 2, it was hinted at a bit

[crude tiger]

Wait

[delicate torrent]

-# ||the following = 2p^r if q = r, use the prime factorization||

[crude tiger]

Yeah if it was 2 it isn't square

[crude tiger]

But how bout >2

[rapid rain]

Not sure about that statement

[rapid rain]

We can start talking about p > 2 if you want

[delicate torrent]

WLOG, q < r

as i said above, q can't be equal to r if p is odd

[crude tiger]

Ah yeah if the exponent is odd it indeed square

[crude tiger]

Yes please

[delicate torrent]

to prove, consider the prime factorization if p is odd

[rapid rain]

The prime factorization of p^q + p^r*

If q = r, then this is 2p^q = 2 * odd

So can't be a perfect square

So if q < r

You can maybe find out what q must be?

[delicate torrent]

-# q and r are prime btw

[crude tiger]

Hmm

[rapid rain]

Hint: ||biggest common factor||

[crude tiger]

Uhh

[rapid rain]

What is the biggest common factor between p^q and p^r?

[crude tiger]

P

My bad

P^q

[delicate torrent]

assuming that q < r

cool

so factorize it?

[crude tiger]

Factor what?

[rapid rain]

p^q + p^r

[delicate torrent]

the expression?

[crude tiger]

Hmm interesting

[delicate torrent]

what do you get?

[crude tiger]

P^q-r)+1)p^q

[delicate torrent]

$p^q(p^{r-q} + 1)$

[glossy valve]

1 divided by 0 equals Infinity

[crude tiger]

He need more latex

[delicate torrent]

lmao

[crude tiger]

We case bash with this?

Wait yeah it need to be square

I think I'll take it myself after this

. Close

+close

Uhh idk how to close

[rapid rain]

it is .close

No caps

[crude tiger]

.close

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[broken sage]

help please

[broken sage]

i took x-a as a, x-b as b and x-c as c

idk how to water it down to one of these 4 forms

[torn jolt]

tarun sir

i remember doing this one

what have you tried so far

[torn jolt]

multiply first column by a second by b and so on, then you see 3rd row is abc