help-33

[sturdy python]

?

[limber hearth]

Comment ça

[sturdy python]

le -

sa va pas dans le paranthese?

[limber hearth]

Si tu peux

[sturdy python]

de quoi si tu peux

[limber hearth]

?

Qu'est ce que tu veux savoir finalement ?

[sturdy python]

j'essaye de comprendre pourquoi tu dit que c'est (x+5)

quand c'est -(x+5)

[limber hearth]

C'est toi qui me l'a demandé

[limber hearth]

Même principe mais avec un - devant

Tu le mets dans la parenthese et tu fais la même chose

[sturdy python]

jvoulais comprendre quelque chose

desoler

[sturdy python]

ok donc -x -5 correct?

[limber hearth]

• x+5 est croissante ( / ) donc sqrt(x+5) est croissante ( / ) car la racine ne change pas les variations (si ça monte, ça monte encore)
• (-x-5) est décroissante ( \ ) donc sqrt(-x-5) est décroissante ( \ ) car la racine ne change pas les variations (si ça descend, ça descend encore)

[sturdy python]

ah ok

mais ici c'est quoi

(-x-5) ou (x+5)

[limber hearth]

-(x+5) de ce que je vois

Donc -x-5

[sturdy python]

donc decroissant

[limber hearth]

Oui voila

[sturdy python]

donc le c?

[limber hearth]

Non

[sturdy python]

d?

[limber hearth]

Mais pour le coup les variations n'ont pas trop de rapport avec la question

[limber hearth]

Oui

Car on résoud -x-5 >= 0

[sturdy python]

ah ok

mais decroissant c'est des nombres qui vont de plus en plus petits correct/

[limber hearth]

Oui

[sturdy python]

ok donc

comment est ce que

• infini a -5 c'est aller de un nombre plus grand a plus petits

-infini c'est plus petit

que - 5

[limber hearth]

-infini oui est plus petit que -5

[sturdy python]

donc c'est croissant

[limber hearth]

Non

[limber hearth]

.

[sturdy python]

ah

ok donc comment on fait la question

[limber hearth]

Bah la racine doit avoir qqchose de positif en dessous

Donc tu regardes quand est ce que la quantité en dessous est positive

[sturdy python]

ah ok

la racine est negative

en dessour

[limber hearth]

Non justement il ne faut pas

[sturdy python]

faut pas quoi

[limber hearth]

Que la racine est une quantité négative en dessous

[sturdy python]

mais dans ce cas c'est une negative en dessous

[limber hearth]

Pas tout le temps

[sturdy python]

ton ragebait est bonne

.close

[marsh citrus]

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[marsh citrus]

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[candid temple]

How do I factorize.... It's so confusing

[paper fiber]

Please elaborate

What are you trying to factorise

[candid temple]

I'll send it

Like 2x+6y+8

And 4y-8+4x

[paper fiber]

first you need to find the common factor

for example do you see anything common between 2x and 6y

[candid temple]

2?

[paper fiber]

exactly

what about 2x and 8

[candid temple]

2

[paper fiber]

so they all have a common factor of 2

we can write it like 2*x + 2*3y + 2*4

[candid temple]

Mhm

[paper fiber]

right now we multiply each term by 2

and then add them up

but we could also add them up and multiply the whole thing by 2

[candid temple]

Cause it is the common factor?

[paper fiber]

to get the same result

[paper fiber]

its being done regardless, but yes in this case its common

so we can do x + 3y + 4

and multiply that entire thing by 2

2(x + 3y + 4)

[candid temple]

Wait wait so

Huh

😭

[paper fiber]

do you know how to expand this?

[candid temple]

No..

[paper fiber]

do you understand what that notation even means?

[candid temple]

No

I just started year 7

[paper fiber]

2(x + 3y + 4) is shorthand for 2 * (x + 3y + 4)

[candid temple]

Ohhh

[paper fiber]

its 2 times the entire thing in parenthesis

[candid temple]

OHHH

[candid temple]

So that's how u got this

[paper fiber]

but we just write it like that because its easier

yes

[candid temple]

I get it now

[paper fiber]

and when you expand its the opposite of factorising

so if you expand that

[candid temple]

U get the original question

?

[paper fiber]

2*x + 2*3y + 2*4

yes

and then you simplify it to the original

[candid temple]

I get it now

Thank you!

[paper fiber]

no problem!

you can use .close if you have no more questions for now

[marsh citrus]

@candid temple Has your question been resolved?

[marsh citrus]

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[supple anchor]

im watching a video on this question i get everything until x^2 -2kx + 3k = 0, i dont understand why the discriminant is being set to equal 0?

[stable hemlock]

Is this a calculus question?

or algebra

[supple anchor]

algebra

the guy in the video just solves for k using b^2-4ac = 0

and then k is a non zero constant so k = 3

but i dont know why we are setting b^2-4ac = 0

[gritty vault]

ok good

so

you know discriminant right

[supple anchor]

ye

[gritty vault]

so

since we always do +- discriminant

we get two solutions for discrimant positive right

[supple anchor]

idk what this means 😭

i know if discriminant elss than 0

no real roots

greater than 0 means 2 roots

[gritty vault]

yes

[supple anchor]

equals 0 one root

[gritty vault]

we want only 1 root

[supple anchor]

why?

[gritty vault]

cuz tangent means they touch 1

[supple anchor]

OHHHhHHHHH

[gritty vault]

once

yes

[supple anchor]

there is only once solution when they touch each other

iccc

[gritty vault]

yea lol

[supple anchor]

ok i know how to do the rest of the question

tysm

[gritty vault]

np

[supple anchor]

.close

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[strong dagger]

I just dont know

[strong dagger]

ive been thinking

but its just blank in there

I started with the integers mod n

tried to create a subset thats a subgroup but it was also just a subring

probably because its integers mod n are a group under multiplication also?

[indigo nest]

That's not true in general

[strong dagger]

hmm its not?

[indigo nest]

Nope

ℤ_n is a group under multiplication iff n is prime

Well Z_n - {0}

But the idea of integers mod n is close

How about think about integers themselves

[strong dagger]

hm ok ill try that

[indigo nest]

Feel free to tag me

[static quarry]

do your subrings (and rings for that matter) have to contain 1?
definitions vary

[indigo nest]

Hmm true, in that case you have to come up with more clever examples

If you allow non-unital sets to be subrings

[static quarry]

try polynomials if that's the case

[strong dagger]

this book defines ring as an abelian group with these properties

so I dont think it needs to contain 1

[indigo nest]

Wait no

So we need to fail multiplicative closure

[static quarry]

yea

[indigo nest]

Yeah this works

I was thinking ℂ

[static quarry]

polynomials or matrices would be a good bet

[indigo nest]

I don't think any natural examples exist in matrices ngl

Product of upper triangle is upper triangle

Same with diagonal and lower triangle

Would have to be a contrived example

[static quarry]

anything whose square is not of the same form should work

[indigo nest]

Right

[strong dagger]

oh

so would

polynomials of degree n work?

since we multiplied them their degree would no longer be n

but addition doesnt change their degree

[static quarry]

yep that works (as long as you include the zero polynomial as well)

oh wait

[indigo nest]

Yup that works

Wait why are we waiting

[static quarry]

no not quite, the sum of two polynomials of degree n might have degree less than n

(if their leading coefficients are 1 and -1, for example)

[strong dagger]

oh right

[static quarry]

degree <= n works though

[indigo nest]

Oh we both just assumed smth else initially huh

[indigo nest]

Yeah that's what I thought he was saying lol

[static quarry]

or, polynomials that only have the degree n term

e.g. all scalar multiples of x^n

[indigo nest]

(n ≥ 1)

[strong dagger]

quick question on polynomials, when my book mentions Z[x], does this mean x takes only positive integer values and coefficients are only integers?

[indigo nest]

x doesn't take any values

[strong dagger]

oh

[indigo nest]

A polynomial ring is just that, polynomials

[static quarry]

btw (side note), for matrices, my first thought was the additive group generated by $$\begin{pmatrix}0 & 1 \ 1 & 0\end{pmatrix}$$

[elfin berry]

Bungo

[indigo nest]

You can evaluate them but that's a different thing

[static quarry]

did you mean the powers are positive integers?

[indigo nest]

ℤ [x] is just polynomials in x with integer coefficients

[indigo nest]

That's definition of polynomials no

[strong dagger]

no I just mean the x values because I thought polynomials were functions that you input values in for x

[static quarry]

yea i was just trying to guess what he might have intended

[indigo nest]

There are no x values, it's just a symbol

We can substitute from any ring as x values as needed

But those are evaluation maps that you'll probably come to later

[static quarry]

yea x is just a placeholder, you can evaluate (plug in some value for x) but the polynomial itself is defined with x as a distinct object that isn't in Z

[indigo nest]

Think of it similar to how ℂ is ℝ [i] with i² = -1

i is a symbol we are adding to ℝ with a certain property

For ℤ [x], x is a symbol we are adding to ℤ with no properties about its behaviour

[static quarry]

some authors give a concrete definition like: a polynomial is a sequence (a_0, a_1, a_2, ...) where only finitely many of the a_i's are nonzero, and the polynomial (0, 1, 0, 0, ...) is "x"

[indigo nest]

I'd like to meet those authors cuz why

[proud ice]

Leave no ambiguity in your definitions

[indigo nest]

Like sure polynomial ring is isomorphic to a countable direct sum but wtf bro

[static quarry]

because otherwise you have to be kind of vague about what x is

[indigo nest]

Fair enough ig, but that feels more confusing to me

[strong dagger]

oh ok I think I see the idea of polynomial now

[static quarry]

yea i mean they don't use that notation after the initial definition

[indigo nest]

Although ig it does make talking about modules slightly more natural

[static quarry]

(generally)

[indigo nest]

Also another example is purely imaginary numbers in ℂ

It's not closed under multiplication

[static quarry]

yea that's a nice one

[strong dagger]

oo yeah thats true

[indigo nest]

It also doesn't contain multiplicative identity

[static quarry]

in the reals, something like all integer multiples of sqrt(2) would work

[indigo nest]

Yeah

All integer multiples of any irrational would work I believe

[static quarry]

yep

[indigo nest]

Similarly, any line works in complex numbers

[static quarry]

(any line through the origin)

[indigo nest]

Right yeah

In ℚ anything of the form p/n for fixed n works

[strong dagger]

can you explain the line through the origin one

[static quarry]

well any line through the origin that isn't the real axis

[indigo nest]

Right yeah

[indigo nest]

It's the same idea as purely imaginary

Complex number addition behaves somewhat like vector addition

So if you take two complex numbers on the same line, their sum stays on the same line

[fervent rampart]

all the numbers on a line thru the origin have argument θ or θ + π. multiplying them together would give you argument 2θ (possibly adding a multiple of π) which is not generally on the same line (excepting the real line)

[strong dagger]

hmm so when we are talking about lines does that mean vectors?

I dont think im understanding like what the structure is of the set when we say line through the origin

[fervent rampart]

well complex numbers are basically vectors in R^2 equipped with multiplication

the set is really just all points on the line, interpreting that as a line thru the origin. you could set that up as like {x + iy | ax + by = 0} for some fixed reals a,b (a not equal to 0)

[strong dagger]

oh ok I see, and so adding those points gives you a point on the same line but multiplying them could take you out of that set

though I cant say I see why adding would give you a point on the same line and why multiplying could take you off but I think ill try proving it as an exercise

thanks for the help guys

.solved

[marsh citrus]

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[buoyant nacelle]

This is my answer and the key. How did the key get this answer, and are the answers the same?

[buoyant nacelle]

Im mostly curious about the 5(4x+3)^4 and cos((4x+3)^5)

[stoic saddle]

the answers aren't the same bc you misread where the ^5 was in the composition order

though you did mostly apply the chain rule correctly aside from that

you misread the ^5 as being to the left of sin in the chain

[buoyant nacelle]

Ah I see

Also, I'm unsure of what to plug in to find equation, is it just plugging in x into derivative to find the slope, and then solve for the equation?

[marsh citrus]

@buoyant nacelle Has your question been resolved?

[strange jungle]

Disintegrate

1. Tan
2. Sin
3. Algebra

So its tan(x)×dy/dx

[strange jungle]

F'x is the slope as you know f x is y and x is just x

Find intercept using that

[buoyant nacelle]

.close

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[torpid pulsar]

log(a+b)=loga +logb but how did they solve here?

[stoic saddle]

no, log(a+b) ≠ log(a) + log(b).

[torpid pulsar]

oh yesyes forgot to mention but yeahh which property did they use there

[stoic saddle]

wait, is that your solution or somebody else's

bc right from the get-go it looks like high-quality bollocks to me

[torpid pulsar]

cuz my teacher made us solve in this method

[torpid pulsar]

its someones else like from a yt lecture i solved

[stoic saddle]

well i mean breaking that log up the way that was shown is already just

pure nonsense

you cant do that

so the rest of it kinda goes down the window

are you required to do this without chain rule somehow or is that just more bs

[torpid pulsar]

nahh he mentioned he uses chain rule for tht

[stoic saddle]

aight

[torpid pulsar]

ohh thnk goddd

[stoic saddle]

well the sticky note says to use the factorization of x^3+y^3+z^3-3xyz

so you get $u = \log(x+y+z) + \log(x^2+y^2+z^2-xy-yz-zx)$

[elfin berry]

Ann

[stoic saddle]

which IS legit bc you break a product inside the log

[torpid pulsar]

no nooo its like in the end near to simplifying thts used

so wait the method he used is wrong?

[stoic saddle]

completely so, yes.

[torpid pulsar]

damnnn

thenksss

but wait sometimes even if we use wrong method we can get the ans??

[marsh citrus]

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[pearl haven]

So i was solving a question and got stuck here

I have expression (3+tan⁵A-tan²A)(3+tan⁵B-tan²B)(3+tan⁵C-tan²C) = (tan³A+2)(tan³B+2)(tan³C+2)

I can see that tanA = tanB = tanC = 1 satisfies it but that was a guess. How would i solve it the correct way?

[oak oxide]

eh?

[pearl haven]

I mean, you just cannot guess the solution all the time

[oak oxide]

No I meant 3 variables but only 1 equation

It makes me confused

[pearl haven]

Yeah this was it

Do you want me to send the complete question?

[still temple]

please do

[oak oxide]

No If you say so, I'll try it out maybe there's smth hidden here

[pearl haven]

a comes out to be 5

V(t) = 3+tan⁵t-tan²t

[still temple]

i don't think i can help here

[pearl haven]

It's alright, thanks anyway

[fervent wind]

What course is this from?

[pearl haven]

It's kind of a practice question for competitive exams

Attempted by 12th graders

[still temple]

what country are you from if you mind me asking

[pearl haven]

India

[still temple]

oh i see

[fervent wind]

crazy lol

[still temple]

i can send you our exams

if you want to practice

i think they would be similar

[pearl haven]

Nah, i have passed the exam this year. Just came across this question randomly

[fervent wind]

I was happy with solving calc 2 integrals rn in 12th grade

I hope i reach this level

[pearl haven]

The integrals used in this questions were easy

The data they have given is in like a format to scare you

[fervent wind]

Yea it looks crazy

[still temple]

this is from ours a it's a demo of Bertrand's postulate

give you everything you need and walks you through it

[pearl haven]

Is this attempted by high schoolers?

Looks crazy

[still temple]

yep

only 4 a year gets the scholarship

they're all researchers now lol either mit or other big names

[pearl haven]

How many people attempt?

[still temple]

we have a national exam in math the top 100 from the country are only the ones qualified to even pass this

it's mostly olympiads that win

[pearl haven]

Crazy man

.close

[marsh citrus]

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[waxen dust]

Do you know the solution

[pearl haven]

Final answer is 27

[waxen dust]

Yes

Okay

Have you noticed anything about the equation

[pearl haven]

Like

[waxen dust]

Symmetry

[pearl haven]

Yea like it's similar 3 expression

Multiplied together

[waxen dust]

Let's say x = tanA, y = tanB, z = tanC

$(3 + x^5 - x^2) \cdot (3 + y^5 - y^2) \cdot (3 + z^5 - z^2) = (x^3 + 2) \cdot (y^3 + 2) \cdot (z^3 + 2)$

[elfin berry]

USS-Enterprise

[waxen dust]

We can switch the variables around

And it doesn't change the equation

Right

[pearl haven]

Correct

[waxen dust]

So we can look for a solution where all the variables are equal

It's the easiest way

Say x = y = z = t

[pearl haven]

Similar to like what we do when a function is equal to it's inverse?

We put it equal to x

Because they are symmetric around y=x line

[waxen dust]

Not exactly

[pearl haven]

Then it's just 3+t⁵-t² = t³+2

[waxen dust]

I mean, in a sense yes

There we swap function inputs and outputs

Here we swap variables

[pearl haven]

Got it

[pearl haven]

After that i guess it's just easy factorisation

[waxen dust]

Correct

[marsh citrus]

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[waxen dust]

And you can now solve for t

[waxen dust]

Oops

[pearl haven]

Thank you man

[waxen dust]

No problem

.close

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[chrome ether]

I have a couple questions about the first image. First he defines B as a bounded plane measurable set. Why are we seeing it is measurable is this not implied from the defintion of bounded plane and the fact that we took as an axiom that all rectanglelsl are measurable? Next he says we have a area function a_s which takes a constant value a(B). When he says that b is a bounded plane set does he mean that b is the points that make these or b is a collection of planes? If its the first thing wouldn't just saying a bounded plane suffice. If it is the second how can you plug in B? B is a collection of sets? Woulldn't we need to say a(x) | x in B. Because each x is a cross sectional plane and that is the area that we want. Also is a_s supposed to be a step function where each interval represents an area of a cross sectional area from the beginning y coordinate of the cross sectional plane to the end y coordinate of the cross sectional plane?

[leaden monolith]

Does the word measurable mean anything to you?

[chrome ether]

well it is kind of a circular definition because I just see measurable as something with dimension that we can assign a real number to represent how much space it takes

[leaden monolith]

Bounded sets need not be measurable

Not every bounded plane is a rectangle

[leaden monolith]

This is a very technical thing that has a whole field of study dedicated to it (measure theory)

[leaden monolith]

Bounded planes only are contained in some rectangles, they need not be themselves a rectangle

[chrome ether]

but don't we define it as a subset of a square and if something is a subset of something measurable shouldn't it also be measurable?

[leaden monolith]

Well unfortunately, no

[chrome ether]

how can this be?

[leaden monolith]

Well, one can use the axiom of choice to construct a set that isn’t measurable

One construction are the vitali sets

But now we are delving into measure theory

[chrome ether]

so can these sets be constructed as a by product of the defintion of the axiom of choice? Because I just looked up what the axiom of choice is and it says that it allows to construct a new set by taking one of the elements from the original set no matter the size of the original set

[leaden monolith]

Another construction is also called the banach-tarski paradox if you’ve heard of that one

[chrome ether]

but isn't that a pretty intuitive idea?

[chrome ether]

no will look into it

[leaden monolith]

No this means you can make some weird ass sets

And these weird ass sets can be very weird and break your intuition

[chrome ether]

ok so then that answers the part of measurability.

What about what he means by when he says B is a set

the set of points that make up a plane or a collection of elements where each element is a plane?

[leaden monolith]

B is a bounded plane that’s measurable what’s the problem in that

[chrome ether]

but he is saying that it is a set and I am unclear if by this he means the set is the points that make that specific plane or by set he means each element in the set is a distinct plane

[leaden monolith]

The first one

Each point in S is a triplet (x, y, z), where every (x, y) happens to lie in B

[leaden monolith]

What is mathcal A in here

Because if we’re gonna be picky, then no B can be a subset of ℝ³ anyway

So it wouldn’t make sense to say B is a bounded plane in ℝ³

Unless we say B ⊆ ℝ²

But then it wouldn’t have zero volume in ℝ³ because it doesn’t even belong in ℝ³

[chrome ether]

im confused on the last statement. wouldnt the fact that it isn't in r3 mean that it has zero volume

also this means that we have an infinite amount of sets B right? one for each (x,yz)

[leaden monolith]

Well things that aren’t in ℝ³ don’t have volume

That’s not the same as having zero volume

The word volume here is doing the heavy lifting as well

[leaden monolith]

For each z

But no

There is not an infinite number of B’s

There is only 1 B

[chrome ether]

coudnt' we keep dividing the height into smaller and smaller planes?

[leaden monolith]

Can you imagine ${(x, y)| x = 2, 0 \leq y \leq 4}$

[elfin berry]

frosst

[chrome ether]

a line?

[leaden monolith]

Yes

[chrome ether]

ok

[leaden monolith]

Do you agree that there is only 1 2’s

[chrome ether]

one y coordinate with y=2?

[leaden monolith]

${(x, y)|x \in {2}, 0\leq y \leq 4}$

[elfin berry]

frosst

[leaden monolith]

This is the same thing yeah?

[chrome ether]

yes

[leaden monolith]

${(x, y, z)|x, y \in {2}, 0\leq z \leq 4}$

[elfin berry]

frosst

[leaden monolith]

A but different bit same idea yeah?

This is a vertical line in 3D

A line segment at least

[chrome ether]

ok I see same point (2,0) but where are moving it up and down

[chrome ether]

so we then take this idea to the plane right? Plane has the same x and y coordinates we just move it up and down

[leaden monolith]

Not same

But now let B be some bounded plane measurable set

[chrome ether]

ok

[leaden monolith]

B lies flat on the x-y plane

[chrome ether]

yes and we just move it up then

and down

[leaden monolith]

Now we say ${(x, y, z)| (x, y)\in B, a\leq z \leq b}$

[elfin berry]

frosst

[leaden monolith]

See this is a little bit cheating

[leaden monolith]

Here B is 2D

[leaden monolith]

Here B is a subset of 3D so it’s still 3D

[chrome ether]

wait how is b a subset of 3d?

[leaden monolith]

This should really say ${(x, y, z)| (x, y, 0)\in B, a\leq z \leq b}$

[elfin berry]

frosst

[chrome ether]

dont' we define it as a subset of a square which is 2d?

[leaden monolith]

(x, y) ∉ ℝ³

For any x and y

[leaden monolith]

That’s what I was being picky about

[chrome ether]

oh wait so your saying that we coulnd't have it as a condition for x,y,z?

[leaden monolith]

You can’t even describe that B is a measurable set

[chrome ether]

are you saying that by saying (x,y) in B in that set we are implyining B is a subset of R^3

[leaden monolith]

I’m saying B needs to be a subset of ℝ³ if you want to talk about it being a measurable bounded plane set

[leaden monolith]

I’m saying this implies B IS NOT a subset of ℝ³

Which is a problem

[leaden monolith]

Because then you can’t talk about this

So somewhere along the way your book has done some goofy shit

[chrome ether]

Im missing how this doesn't cover that case?

a plane is 3d and we say that this square is in the plane and that are bounded set is a subset of this square

[leaden monolith]

We said (x, y) ∈ B

If B were 3D there’d be nothing in B of the form (a, b)

[leaden monolith]

.

[chrome ether]

I see

and the a_s we have becomes a step function right?

[leaden monolith]

Sure

It’s just an indicator of if we are inside the region of z or not

[chrome ether]

it doesn't matter that it always has a constant value in a ceritain interval?

[marsh citrus]

@chrome ether Has your question been resolved?

[leaden monolith]

That part just says the area isn’t changing

Aka it’s a cylinder

[broken hearth]

guys i need a binomial help

anyone

[marsh citrus]

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[cold idol]

do i let z=x+yi?

[desert dirge]

It's worth a go

Give it a try

[cold idol]

ive got an answer but i dont htink its right

can u check

[stoic saddle]

this is quite gruesome

you should actually leave |2z-1| as |2z-1| for a while and not expand it

rather understand that since rhs is real, imaginary part of lhs will be 0

this will make your life a lot easier

[cold idol]

how do you kniow the rhs is real?

oh cuz its the modulus

[stoic saddle]

yeah exactly

[cold idol]

.close

[marsh citrus]

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[sand bear]

part a answer: f(x) > 1
part b answer: f^1(x) = 2 + 3/x-1 for x > 1

i was confused for part c on the composition of that function and how its domain and range are working together.
gf(x) means that we are inputting the values of f(x) into g(x).
now if the input of g(x) has to be greater than 0 (x > 0) that means thats the values of f(x) > 0.
but then the range of f(x) itself is f(x) > 1. so how is it working inside the composite function.

[marsh citrus]

@sand bear Has your question been resolved?

[brisk blade]

Dude the condition for composite functions is , range of inner function should be a subset of domain of outer function

x>1 is always a subset of x>0

[marsh citrus]

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[distant abyss]

Is there anyone who is good at maths?

[stone pelican]

Yes, but if you need help just send the problem

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

@distant abyss Has your question been resolved?

[lucid kelp]

<@&286206848099549185>

Pls remove that

[small berry]

<@&268886789983436800>

[lucid kelp]

Oh mb didn't know you could ping them

[echo river]

Theese spammers are everywhere

<@&268886789983436800>

[small berry]

Helpers can't delete messages, you can ping mods instead

[pallid jolt]

Has already been dealt with

[lucid kelp]

Ty for letting me knos

[pallid jolt]

We can't do much more than just jump on 'em when they happen

[echo river]

Whatchu mean

[echo river]

Sup giyu guy

[lucid kelp]

Yes?

[echo river]

Oh ok I didn't read yo msg

I thought u needed help with som problem

[echo river]

Was talkin' bout' this

[small berry]

@distant abyss You asked if someone can help you with math, someone told you to ask your question, and you haven't (but have said no to the "has your question been resolved" popup). Do you have a question you want to ask? If not, close the channel with .close.

[marsh citrus]

@distant abyss Has your question been resolved?

[marsh citrus]

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[halcyon stratus]

Hi Guys! I wanted to know if f(x) = x - pi or x ?
I know it's a periodical function of 2pi but I though if I wanted to know the graph i need to get something like f(x) = ax + b

[nimble prairie]

Depends on where on the graph you are

[halcyon stratus]

bcs if f(x) = x i would get 0 if x= 0 which doesnt' correspond at my graph

[nimble prairie]

A function doesn’t have to be of the form f(x) = ax + b

[halcyon stratus]

That's my issue I don't really know

[nimble prairie]

It could be defined by pieces

[halcyon stratus]

From what I saw it's a affine fonction in my head
(Yea it's a function defined by pieces

[nimble prairie]

Say when x is between 0 and 2pi

Then f(x) = x - pi

[halcyon stratus]

else ?

It would change by doing + 2pi ?

[nimble prairie]

You keep doing this for other intervals,

[halcyon stratus]

so f(x + k) = x - pi + 2kpi

Oh

[nimble prairie]

E.g. f(x) = x - 3pi if x is inbetween 2pi and 4pi

you can write this more concise

If you really wanted

[halcyon stratus]

I'm sorry I'm kinda confused

There's a lot of way of writing f then ?

[nimble prairie]

As f(x) = remainder(x, 2pi) - pi

Yes there’s different ways of writing the formula/rule for the (still same) function f

[halcyon stratus]

hmm

The easiest for me would be a function with a variable k € Z

so I can play with it and find the general definition of f

[nimble prairie]

In desmos you could write it like this

[halcyon stratus]

Wut's mod for you ?
x / 2pi but the rest of it ?

(Just for verifiyng if I get wut u mean or not)

[nimble prairie]

It’s almost like modular arithmetic but as an operator, I.e mod(1,2) = 1, mod(2,2) = 0 and mod(3,2) = 1 etc

[halcyon stratus]

The triple = thing in arithmetic

a === b [c] kind of ?

3 line equal

Either way we didn't saw it in this particuliar class so I can just translate what you said by doing
f(x) = x - pi(2k + 1)

[nimble prairie]

Yeah for suitable intervals

[halcyon stratus]

hmm okay..

Thanks for your help

.close

[marsh citrus]

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[marsh citrus]

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[oak oxide]

wassup mate, still grinding for AMC10?

What have you tried

[abstract musk]

ye

the lengths of 2 of the sectors are 4pi and 9pi

[oak oxide]

Hmmm what's your argument for that

[abstract musk]

well

let me draw

[oak oxide]

How is that 4

[abstract musk]

its in the question

oh shoot

wait

[oak oxide]

the other side clearly look bigger

[odd parrot]

what is that like 7th grade?

[abstract musk]

its 3 mp

[abstract musk]

?

[oak oxide]

[odd parrot]

anyways I need help guys

[oak oxide]

asking too much

[oak oxide]

!help

[marsh citrus]

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

[odd parrot]

cant u just help me

[oak oxide]

I dunno, but you can't get help here in this channel that's what I know

Open a new ticket mate

[serene nacelle]

shush pre algebra 8th grader

[serene nacelle]

go another channel

[abstract musk]

i read question wrong lol

but still stuck

[odd parrot]

how u do tat

[serene nacelle]

Buddy shut your mouth this isnt your channel

[abstract musk]

paste question there

[odd parrot]

k

[abstract musk]

ye i was thinking of that

[oak oxide]

I want you to compare the cyan line and lime line

[abstract musk]

they same

with pythag

[oak oxide]

and the angle between them?

[abstract musk]

i think 90

wait

[oak oxide]

could you prove that

[abstract musk]

no

oh wait

wait ik

180-90 = 90

[oak oxide]

Congrats

[abstract musk]

we know that angle x and y add to 90 because of triangle angle sun thrm.

so it is 25pi/4 the yellow part

[oak oxide]

Everything should be clear now

[oak oxide]

eh...

[abstract musk]

wiat

i made a mistake

i need to use circumference

not radius

[odd parrot]

ban @serene nacelle

[abstract musk]

???

bro get outta here

talk in ur own channel

[oak oxide]

[abstract musk]

am i done

[oak oxide]

You're gonna get yourself in so much trouble if keep doing this

[oak oxide]

anyway that looks right

[abstract musk]

ty

.close

[marsh citrus]

Channel closed

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[abstract musk]

i wil take break

[marsh citrus]

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[limber depot]

Hi

[limber depot]

Need help

[oak oxide]

Sure

Just send the problem

[small berry]

!da2a

[marsh citrus]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[limber depot]

I don’t get squeeeze theorem

[small berry]

What part of it don't you get?

[limber depot]

Whole thing

[small berry]

Can you define it in layman's terms?

[limber depot]

Idk when or how to apply it

[whole thorn]

also called sandwich theorem

[limber depot]

Layman’s?

[small berry]

Start by stating the squeeze theorem in your words

[whole thorn]

just look at the theorem

and see what part u don't get

[limber depot]

F (x) = g (x) = h (x)

That’s what I understand

But can’t solve problems

[small berry]

So if someone asks you "what's the squeeze theorem?" you'll respond with "F (x) = g (x) = h (x)"?

[limber depot]

Yup

[small berry]

here you go then. read this and then come back with the definition of the squeeze theorem https://www.sfu.ca/math-coursenotes/Math 157 Course Notes/sec_TrigLimits.html

[limber depot]

Alright

Ahaaaaa

So if g(x) n h(x) had the same value then f(x) would be forced to go there too

[small berry]

yeah exactly

formally, if g(x) is less than f(x) and h(x) is more than f(x), and both g(x) and h(x) approach the same point at some value, f(x) will get forced in between the two and have the same limit

so if g(x) approaches 1 (from below) as x approaches 0, and h(x) approaches 1 (from above) as x approaches 0, f(x) will get squeezed in between the two and approach 1 as well

Your f(x) is cosx/x^2, you want to find some function that is always less than f(x) and another that is always more than f(x), in order to squeeze f(x) at x -> 0

[limber depot]

g(x) would be -1 n h(x) 1

Yeah understood

[small berry]

not exactly

g(x) would be -1/x^2, and h(x) would be 1/x^2

[limber depot]

Oh

[small berry]

This is because -1 is always less than or equal to cosx

so -1/x^2 will always be less than or equal to cosx/x^2

and same thing with h(x)

[limber depot]

Hm

I guess I got the hang of it

[marsh citrus]

@limber depot Has your question been resolved?

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[finite forum]

for part b

dh = (1/50) * h(2h - 1) * cos(t/10) dt

1/h(2h - 1) dh = (1/50) * cos(t/10) dt

(2/2h - 1) - (1/h) dh = (1/50) * cos(t/10) dt

integrate both sides:

ln(2h - 1) - ln(h) = (1/5) * sin(t/10) + c

raise both sides to the power of e:

2h - 1 / h = Ae^(sin(t/10) / 5)

...

until i get:

h = -1 / Ae^(sin(t/10) / 5) - 2

even before subbing in values, this is wrong?

[tired knoll]

It looks good, you can continue

By multiplying by h on each side, and then you can solve for h

[finite forum]

i sub in h = 2.5 at t = 0

because of this

and i get -1 / A - 2 = 2.5

so 4 = 2.5A -> A = 1.6

so my final equation is

h = -1 / 1.6A^(sin(t/10) / 5) - 2

which is not this

nvm

they are the same

[tired knoll]

You mean \begin{equation*}
h = \frac{-1}{1.6e^{k \sin(t/10)} - 2}
\end{equation*}

right?

[elfin berry]

all matrices are invertible

[finite forum]

yeah

you just * -5/-5 on rhs

and its in the correct format

[tired knoll]

Yes

[finite forum]

nice

ty

.close

[marsh citrus]

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[marsh citrus]

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[sturdy python]

having troubles finding the reciprocal equation for

[warped basalt]

is that a decimal or a comma?

[sturdy python]

h(x)^-1

[plucky gust]

you better check if it's monotonic first

[limber hearth]

Comma, meant as a decimal

[warped basalt]

no after h(x)=0

[warped basalt]

ah

[sturdy python]

whats that

[plucky gust]

if it's always increasing or always decreasing because if not then it doesn't have a reciprocal

[sturdy python]

its increasing

[plucky gust]

hmm wait

[warped basalt]

where specifically?

[sturdy python]

i did h(x) = 0,5 sqrt x+4 -7 right

then i inversed x and y

x = 0,5 sqrt (y+4) -7

[plucky gust]

nooooo

[warped basalt]

i dont think thats right

[sturdy python]

then x + 7 = 0,5 sqrt (y+4)

[warped basalt]

im pretty sure thats precisesly where you went wrong

[plucky gust]

u can't swap them

[sturdy python]

what then?

[warped basalt]

inverse of h isnt that

[warped basalt]

imma call h(x) = y for now

[plucky gust]

remember what i sent u?

[sturdy python]

yes

[plucky gust]

do the same thing

[warped basalt]

you want H(y) = x

[sturdy python]

thats what im doing

[warped basalt]

this is the inverse of h(x)

[sturdy python]

replace H with x

[warped basalt]

????

[plucky gust]

no ur swaping them what you should do is put x in terms of y

there is a difference

example: f(x)=x +2 then x = f(x)-2 what ur saying is that x=f(x)+2

[warped basalt]

basically

you want to find 'x'

by algebraic manipulation

[sturdy python]

do the equation

and im gonna understand

[warped basalt]

technically against rules

[sturdy python]

no

i need to find the equation of the reciprocal

[warped basalt]

which is equal to x

[sturdy python]

so u want to find a number of x?

[warped basalt]

nono

you have

h(x) = y

= GIVEN

manipulate this to find x

find x wrt y

[steep garden]

I Don't Understand Anything About Math

[warped basalt]

$h(x)=y=\frac{\sqrt{x+4}}{2}-7$

[elfin berry]

Supernova

[warped basalt]

im afraid youre in the wrong channel then

[steep garden]

Sorry for it i write to the wrong channel

[warped basalt]

its ok

$h^{-1}(y)=x$

[elfin berry]

Supernova

[warped basalt]

its right but youre not supposed to send full solutions here.

[sturdy python]

why is the 0,5 0,25

[plucky gust]

oh i didn't know

[plucky gust]

i squared both sides

[warped basalt]

its in the rules, your supposed to guide the person asking to the answer without explicitly giving them it

[plucky gust]

sorry i genuinely didn't know

[warped basalt]

its ok

[sturdy python]

(y+7) ^2 = 0,25(x+4)

why is the ^2 on the left side still there

[plucky gust]

because why would you split the ()

[warped basalt]

because you squared from

$2(y+7)= \sqrt{x+4}
\Rightarrow 4(y+7)^{2}=x+4
\Rightarrow 4(y^{2}+14y+48) = x
\Rightarrow 4(x^{2}+14x+48)=h^{-1}(x)$

[sturdy python]

wheres the 0,5

[warped basalt]

multiplied 2 with both sides

[sturdy python]

u multiplied 0,5 by 2?

[plucky gust]

keep in mind 0,5=1/2

[plucky gust]

yes he did

[sturdy python]

ohhhhh 1/2

ok ill write that

[plucky gust]

use fractions if u feel lost with decimals

[sturdy python]

ive obtained

[elfin berry]

Supernova

[marsh citrus]

@sturdy python Has your question been resolved?

[marsh citrus]

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[quaint arrow]

i was wondering if it's possible to find a piecewise gradient for this function:

[quaint arrow]

since the derivative of -yxw is -yx, would the gradient be
-yx if -yxw > 0
0; otherwise

[marsh citrus]

@quaint arrow Has your question been resolved?

[quaint arrow]

<@&286206848099549185>

[marsh citrus]

@quaint arrow Has your question been resolved?

[marsh citrus]

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[proud sky]

Number 4

[elfin cairn]

,rotate

[elfin cairn]

it looks like you don't have any progress

why don't you draw the points given in the question

just drawing ABCD

[dire ravine]

Sorry to interrupt, was there an accidental modping here? Please don't delete modpings as this creates phantom pings

[proud sky]

Wait

[elfin cairn]

there was

sorry

[dire ravine]

No problem. Please do just remember not to delete 'em in the future

[proud sky]

Here’s an updated one

[elfin cairn]

looks good

so you need help to find the perimeter?

[proud sky]

Yep

[elfin cairn]

Pythagoras's theorem can help

do you see how?

[proud sky]

Yeah

[elfin cairn]

great, give it a try

and really I don't see a reason to use the other quadrilateral, so stick with ABCD

[proud sky]

How can you apply Pythagoras theorem here

[elfin cairn]

you know the two sides, and these are 90 degree triangles

so pythagoras applies

[proud sky]

So what would be there lengths

[elfin cairn]

what do you think

lets say the pink triangle

[proud sky]

educated guess I’ll say 5-6

[elfin cairn]

no need to guess

you know the exact side lengths

you have the grid, just count the squares

[proud sky]

So isn’t it 5-6

[elfin cairn]

it is in that range

but your objective is not to guess, but to find an exact solution

only if it weren't feasible then you would resort to approximations

[proud sky]

So make a box around the same

And count the squares

[elfin cairn]

nono

you are looking for the perimeter

[proud sky]

No I meant like finding the side length

[elfin cairn]

which is the sum of the side lengths, so if you can find the length of each side of ABCD you are done

yeah, try to do that

(count the horizontal and vertical lines, you can't count the other side accurately)

[proud sky]

Seems like it’s 6

[elfin cairn]

its not 6

[proud sky]

5 for all?

[elfin cairn]

no, use pythagoras

[proud sky]

Elaborate

[elfin cairn]

you know two sides of the pink triangle

pythagoras will give you the third one

[proud sky]

Oh

Do 5 squared by 2, 5 squared by 2

So 50?

[elfin cairn]

close

the side lengths are 1 and 5

so its 1^2 + 5^2

and remember that its equal to the square of the length of the third side

[proud sky]

Wait how’s it 1

[elfin cairn]

look in the diagram

one side has length of 1 square

[proud sky]

Ok so 26

[proud sky]

Does that apply to all shapes

[elfin cairn]

no, you have to count the length here for every vertical/horizontal line

[elfin cairn]

26 is the length of the third side squared

so the length of AB is sqrt(26)

[proud sky]

Is CD 17?

[elfin cairn]

sqrt(17)!!!!

important

[proud sky]

Hm

Then isn’t DA 29

[elfin cairn]

sqrt!!!!!

pythogras tells a^2 + b^2 = c^2

and you know a,b so you get c^2

so c^2 = 29/17/26 in each of these cases and to get c you need to take a square root

[proud sky]

I see

Square them all?

[elfin cairn]

square root

[marsh citrus]

@proud sky Has your question been resolved?

[proud sky]

Ok

Do I add them all up

[sand bear]

yeah im a dumbass

[marsh citrus]

@proud sky Has your question been resolved?

[marsh citrus]

@proud sky Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

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[wraith marsh]

Question and answer

[wraith marsh]

Answer seems right except for the fact that it contradicts the property of triangle which states sum of two sides of triangle is always greater than third side

[wraith marsh]

So is this triangle even possible?

[marsh citrus]

@wraith marsh Has your question been resolved?

[marsh citrus]

@wraith marsh Has your question been resolved?

[cunning fiber]

It’s not possible

Honestly this is a bad question

On so many levels

[wraith marsh]

Thank you so much. I thought, I was making some calculation mistake or something.

[cunning fiber]

[wraith marsh]

.close

[marsh citrus]

Channel closed

Closed by @wraith marsh

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[scarlet raft]

So on top would it still be one?

And then on the bottom would it be x (11-5?

[dry prawn]

top and bottom of what?

[scarlet raft]

So obv x +3 would stay on the bottom. So I am trying to figure out how to simplify the top

I know I gotta combine fractions in the numerator

[dry prawn]

ok so you're just trying to add $\frac{1}{x-5} + \frac{1}{x+11}$

[elfin berry]

Steakanator

[scarlet raft]

YES

[dry prawn]

aight cool

[dry prawn]

then no

[scarlet raft]

Am I right that a - (a+h) would be on top and a (a+h) would be on the bottom?

If that is the case, I am stuck trying to figure out what is a and what is h in this problem

[dry prawn]

i'm not immediately sure where a and h are coming from

[scarlet raft]

From my textbook:

[dry prawn]

[scarlet raft]

I am trying to figure out what a and h is. Then I am not sure if I know what to do next. On a side note, thank you for your time and helping me

[dry prawn]

their bits with a and h are really only useful in the context of that specific problem, let's instead focus on their general method

the denominator becomes a(a+h) because they need both fractions to have a common denominator

you've been introduced to this concept when adding/subtracting fractions, yes?

[scarlet raft]

When adding/subtracting fractions, yes. However compound fractional expressions are new to me.

[dry prawn]

we'll deal with the compound bit later

[dry prawn]

right now we're just adding fractions

[scarlet raft]

Here is the problem again so we don't gotta keep scrolling

[scarlet raft]

Ok. So we are just adding (x-5) (x+11)?

Wait a minute

Do we take a x out? So it's x(x-5+11) ?

[dry prawn]

no

[dry prawn]

if you insist on referencing this example:
the denominators of the original fractions are a and a+h
the denominator of the new fraction is a(a+h)

[scarlet raft]

How would I get to the new fraction then?

[dry prawn]

how were you taught to add fractions with different denominators in the past?

[scarlet raft]

Find the least common denominator?

But there is no common denominator for 5 and 11

[dry prawn]

your denominators aren't 5 and 11

[scarlet raft]

Least common factor I mean

[dry prawn]

least common multiple

[scarlet raft]

On top they are I thought. 1 over x-5

[dry prawn]

x-5 is not 5

[scarlet raft]

Oh

Maybe the least common multiple is x then

Because they both have x

[dry prawn]

this isn't the right direction

[scarlet raft]

Oh

I thought we had to find a least common multiple

[dry prawn]

i'm not sure you can really intuit what the least common multiple is

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:rational-functions/x9e81a4f98389efdf:adding-and-subtracting-rational-expressions/a/adding-subtracting-rational-expressions-advanced

perhaps following this would be easier

[scarlet raft]

Thank you

Wait a minute

Is it this for the 1st step

Don't I just take x + 11 and x - 5 in place for 1

[dry prawn]

what does "in place for 1" mean?

[scarlet raft]

Meaning (x-5) (x+11) / (x-5) (x+11) ?

[dry prawn]

sure

[scarlet raft]

Then you do the foil method right?

WAIT

[dry prawn]

don't do it before you need to

[scarlet raft]

I think you just straight add them

Right?

2x + 6?

[dry prawn]

add what?

[scarlet raft]

Add x-5 and x+11

[dry prawn]

well you have two fractions, so where is this adding being done?

[scarlet raft]

And im assuming why I can just add them is because I have a common denominator

[scarlet raft]

Up top

[dry prawn]

and this common denominator is?

[scarlet raft]

(x-5) (x+11) ?