help-38

[fierce lake]

Yeah

But you know magnitude of a right

[mystic pilot]

√41

[fierce lake]

Finding cos theta should be your priority

Then find sin theta using cos theta = root(1 - sin^2(theta))

[mystic pilot]

I have to find a_c so i can put in R = 49/a_c where a_c is asintheta

[fierce lake]

So?

[mystic pilot]

If a_c is found R will be found

[fierce lake]

But what don't you understand when I say find cos theta first

[mystic pilot]

cos theta = 28/41

[fierce lake]

No?

First of all 28/root(41) is greater than 1

What you are saying is v cos theta

[mystic pilot]

Mb mb

cos theta = 4/7

[fierce lake]

?????? Where'd you get that from

[mystic pilot]

Let me send you

[fierce lake]

Ok

[mystic pilot]

I got something else this time

,rotate

[mystic pilot]

,rotate

[fierce lake]

Yeah this is correct

[mystic pilot]

Now

[fierce lake]

Find sin theta nwow

Now*

[mystic pilot]

It's is 5/√41

[trim jolt]

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[golden jay]

did I do anything wrong here with my steps, or the chat gpt is wrong? https://chatgpt.com/share/683193c9-f68c-800a-97e3-e1873fcdb24b , refer to the respond I asked the lastest.

[mystic pilot]

What are the following roots given in the question?

[trim lichen]

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[golden jay]

ah yea, that is why I sometimes need people to double check with my answer if I ever use chat gpt since I don't have the answer sheet with me 😅

@trim lichen any suggestion on how I could double check my answer without answer sheet or ai tools? 🫠

[golden jay]

oh my bad forget to sent this as well

need to reaffirm my answer for this as well

[trim lichen]

no.

[golden jay]

[trim lichen]

this one looks correct to me though

[trim jolt]

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[limpid shoal]

I need help with qn 12c

[little glen]

simple application of newton's equations of motion

[limpid shoal]

what is v?

velocity?

[limpid shoal]

I've calculated the acceleration of the system

[little glen]

ofc

it is the first equation of motion

v = u + at

[limpid shoal]

arent we finding speed and not velocity?

[little glen]

i give up

[limpid shoal]

wtf

[storm flame]

They're almost the same

[limpid shoal]

oh, so in that case v= u+a(t)
which is v=0+2(2)

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[limpid shoal]

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[smoky marsh]

question: find the first partial derivatives of the function

[smoky marsh]

i know how to find it but im not sure with respect to what

[heavy lion]

show what you have tried

[main sigil]

if they want partial derivativeS, then you should probably compute all of them

probably dw/dx, dw/dy and dw/dz

[smoky marsh]

so with respect to x y and z?

bruh

[heavy lion]

yes

[smoky marsh]

ok

[heavy lion]

its not even that difficult dude

[smoky marsh]

i mean

maybe for u

ok thanks thats all i needed

.close

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[forest stone]

For number 11 I divided it into 4 cases either n is prime so then we want to imply that n+1 is either prime or a product of primes or n is a product of primes and then n+1 is a prime or a product of primes

So n=cd n+1=c'd'

Add 1 and we have n+1=cd+1

What I did before was c'd'-1 = n

[trim lichen]

you're going to need strong induction here

[forest stone]

I don't know what that is

[trim lichen]

it is not really possible in any meaningful way to connect the factorizations of n and n+1

[trim lichen]

it's when instead of assuming P(n) you assume P(1)&P(2)&...&P(n)

[forest stone]

So in our case we assume that 1,2,3...n is a prime or a product of primes?

[forest stone]

?

[trim lichen]

well, not 1, you start from 2.

but yes, you assume that every number from 2 to n inclusive is either a prime itself or a product of primes.

and then you look at n+1

[forest stone]

And is it the same thing as before? I want to prove that n+1 is also going to be either a prime or a product of primes then

[trim lichen]

yes your goal's the same

so n+1 is either prime or not.

if it's prime, we win.

and if not...

[forest stone]

Ok will try thank you

Ok will try thank you

How should we set up the right hand side though since it can be two things?

[forest stone]

.

[trim lichen]

i would prefer if you @ mentioned me instead of pulling a random message and reply-pinging to that

it confuses me this way because it looks as if you're trying to ask about a specific thing i said but you don't

anyway you break down into 2 cases

the first of which is trivial

and then n+1 = cd

c and d are both greater than 1 and less than n+1 so they are both ≤ n

therefore by the strong inductive hypothesis...?

[forest stone]

Oooh but that proves both cases then that 2,3,4...n can be either prime or product or primes no?

@trim lichen

[trim lichen]

sure they can

you can either laboriously examine all 4 possibilities

or lump them both in as "product of 1 or more primes"

[forest stone]

Ye but doesn't it just prove that n+1 is either prime or product of primes in one case

Since we say that 2,3...n is prime or not prime

Then we look at n+1 =cd

So we know that cd is less than n meaning cd-1 is less than n. But how can we be sure that c or d is less than n. We know c and d are less than n+1 but not n

[trim jolt]

@forest stone Has your question been resolved?

[forest stone]

For this we know that c or d < n+1. But if we have n+1=cd then subtract cd-1 then how can we draw that c or d less than n conclusion in a rigourous way?

Oh I see proof by contradiction assume cd-1>n then cd>n+1 which is false

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[winter pasture]

Question number 5

[stoic garden]

,rccw

[stoic garden]

@winter pasture
What are you trying to do? Prove/Disprove this statement, or get an answer checked?

[winter pasture]

Get my answer checked

[stoic garden]

x + 1 = 1 is wrong, sadly.

[winter pasture]

Can you explain what’s the correct answer

I have been staring at it for like 10 minutes

[stoic garden]

No. I can explain the problem to you to help you get the solution.

[winter pasture]

Sure

[stoic garden]

First, since both exponents are the same, I recommend setting some new variable u = x - 1.

This gives us: $7^u = 3^u$

[solid kiln]

@stoic garden

[stoic garden]

Do you understand this?

[winter pasture]

No

[stoic garden]

Which part do you not understand?

[winter pasture]

Power = power?

[winter pasture]

What are we trying to prove in the question

[stoic garden]

We are simply trying to solve for a value of x that makes the equation true.

[winter pasture]

0?

[stoic garden]

Are you saying $7^{0 - 1} = 3^{0 - 1}$?

[solid kiln]

@stoic garden

[winter pasture]

Then it’s -1 +1

[stoic garden]

$$7^{1 - 1} = 3^{1 - 1} \implies x = \boxed ?$$

[solid kiln]

@stoic garden

[winter pasture]

X = -1

[stoic garden]

This is what we get if we say x = 0.

[stoic garden]

Then you're saying: $7^{-1 - 1} = 3^{-1 - 1}$.

[solid kiln]

@stoic garden

[winter pasture]

Oh it’s - so it’s 1-1

Therefore x = 1

Aright thanks

.close

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[dapper oxide]

when do we draw x = a or y = b as the coordinates of a point and when do we draw them as lines? what's the general rule? (I don't have a specific example)

[marsh forum]

context generally

[timid imp]

I would say if you're told to draw x = a or y = b you should draw them as lines unless otherwise specified

[simple rock]

do you mean you want to know difference between plotting point (a, b) and plotting lines x=a and y=b?

[dapper oxide]

when do we do which

[simple rock]

Depends on what you are finding. Depends on your question

If you could provide any example to clarify what exactly you want to know..

[dapper oxide]

if I have something like (just a random example):

Find the solution of:

y = 10
y < x + 5

[boreal shuttle]

then just find the solution ?

[dapper oxide]

do we draw y = 10 as a line?

[boreal shuttle]

so your trying to get the solutions but with a graph

geometrical solution , or w/e its called

[boreal shuttle]

i would just give an interval of x

10 < x + 5 => x > 5

as stated above , an example or an image of something you drew is going to help

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[kind island]

hi need help studyinf for my upcoming exam

[kind island]

and this question is annoying me

so ik that

f'(0) is 3

[wraith hinge]

do you know whats x in math?

[kind island]

and f(0) is 4

yes

[wraith hinge]

can u tell me please

[kind island]

is this a serious question

[wraith hinge]

yes

[wraith hinge]

im dumb and i need to fix 2 of my grades to pass math

[kind island]

thats the x

are you serious bro

[supple copper]

Do you know what a tangent line is

[wraith hinge]

no

[stark bison]

@wraith hinge This channel is occupied. If you need help, open your own one

#❓how-to-get-help

[kind island]

a straight line that touches a curve

of a point?

listen i

[wraith hinge]

fuck you

[kind island]

wrote

y - y1 = m(x-x1)

for the formula

[supple copper]

Like this?

[kind island]

and i got y - 4 = 3(x -1)

but i feel like its wrong

or im missing something else

[kind island]

uh

isnt it more like this

[supple copper]

No that’s correct

The tangent line is a straight line

[supple copper]

You need only the slope and 1 point on the line to find the equation of the line using

You have the slope via the derivative and the point (1, 4) you get from the original function

You’ve done the right steps

[kind island]

this is the answer

but i got a different answer

so im not sure if im right

[supple copper]

What do you mean

[kind island]

this and this

[supple copper]

Simplify it!

[kind island]

ohh

and when i simplify that

i get

[supple copper]

Yes

[kind island]

this i suppose

wait

wrong image

[supple copper]

Just move the 4 to the other side

[kind island]

oh well i gotta start thinking more

[supple copper]

You’ve got the method right have more faith in yourself :D

[kind island]

yeah about that

so its y - 4

and if i put it onthe other side

it becomes positive yeah

im not wrong on that

right?

[supple copper]

Yep

[kind island]

alright

[kind island]

yeah i should lol thanks

wait

i already know this isnt right lol

[supple copper]

wtf

Add 4 to both sides!!

[kind island]

oh yeah

y - 4 + 4 = 3x -3 + 4

y = 3x + 1

.close

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[woeful pasture]

I'm trying to find ways to speed up how fast I expand two or more brackets that include trinomials or higher, such as (x^2-4x-4)(x^2+6x+9).

I tend to just distribute each term in the first bracket with each term in the second in consecutive order, which works out to be the foil method when dealing with binomials but I'm wondering if there's a faster way of doing it for expressions with more terms

[dusky thunder]

i really like the method of adding coefficients

ask urself "what possible products yield an x^3 term" for example

and add the product of all the coefficients

let me see if i can find u a video on it

[woeful pasture]

oo

that'd be great thanks

[dusky thunder]

https://youtu.be/uiSUP_biejg?t=153

ignore the taylor series stuff

i time stamped the link

so it only looks at polynomial multiplication

[woeful pasture]

Okay thanks

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[void dew]

how would i do this assuming with the equations that i had to do to kmake the first one and then transform that all

[dusky thunder]

have u learnt function transformations?

f(x-h)+g means
move the graph to the right by h units and up by g units

so f(x+2)+4 would mean moving the graph to the left by 2 units and up by 4 units

[void dew]

no bruh like how would i do that becuase i need to groah the origanl one first buu cant do that

thats what i need help with and then using those to transform i need help with that

@dusky thunder

[tidal schooner]

the original one is given

[void dew]

yea but i have to grpah that and then apply the rules for each pone to trasnform that oringal one

[tidal schooner]

okay so which bit are you stuck on

[void dew]

applying the rules to the functions that make the parent function

[tidal schooner]

okay so for f(x+2) + 4 move the original graph given 2 to the left and 4 up

u get it?

[void dew]

yes

i know the rules but i need help applying to the parent function becuase its not 1 equation

[tidal schooner]

how do u mean?

just apply the rules to the entire graph given

[void dew]

yes

one second

i will share my groah i made

[tidal schooner]

okay

[void dew]

@tidal schooner

there

[tidal schooner]

yes that’s the original graph

now just apply the rules

[void dew]

yes but how for the whole groah thats what i need help with

i dont get how to do that

@tidal schooner

so like this

@tidal schooner

[placid radish]

do you know what transformation of f does f(x+2)+4 represent?

[void dew]

yes but i need the lines made

[placid radish]

can you first tell me what f(x+2)+4 reprensent?

just to make sure we are on the right track

[void dew]

yes

it means that the grpah we have now movess -2 and up 4

to make the new graph

[placid radish]

where are you moving for the -2?

[void dew]

to the left

becuase its the other way inside

[placid radish]

so move left for 2 and move up for 4?

[void dew]

yes

accorind to the rules

[placid radish]

alright, now remember, translation does not change the shape of the graph: if you have a straight line, after the translation, it will still be stright lines

[void dew]

yes i know i just dont know how to apply them like this

[placid radish]

It is better if you break the graph into 4 line segments

[void dew]

i idd

[placid radish]

translate each line segment left by 2 and up by 4

[void dew]

i idid to make the parent function

[void dew]

i need help with that

[placid radish]

then you should get the desired function

[void dew]

but i need help to trnasofrm them

and do it

i never done it like this

without just drawing it myslef

[placid radish]

do you know how to translate line segments?

[void dew]

yes

[placid radish]

then try and translate each line segment

there are 4 line segments in the graph of f

[void dew]

i already tried it doesnt work

[placid radish]

show me

[void dew]

i know how to but i cant draw them by hand right now

[placid radish]

because it should work

[void dew]

h\left(x\right)=f\left(x+2\right)+4

[placid radish]

$ signs around the equation

[void dew]

$h\left(x\right)=f\left(x+2\right)+4$

[solid kiln]

ppq#7826

[placid radish]

right but translation is geometric you dont need to worry about the function

[void dew]

$f\left(x\right)=3x+13$

[placid radish]

its simply "move the shape to the left by 2 and up by 4"

[solid kiln]

ppq#7826

[void dew]

i did but the line doesnt work

[placid radish]

it should work you are moving line segments

[void dew]

but it doesnt thats the issue

can you give me the equations

[placid radish]

like here is a triangle

here is the trangle translated to the left by 4 units

you very literally just move it to the left

here is a graph of a function, can you guess what "move to the left by 4" does?

[void dew]

yes

[placid radish]

then I think you messed up the computation, can you show me your work so I can pinpoint where went wrong?

[placid radish]

right, it should just be

[void dew]

one second

[placid radish]

so with this intuition, you can traslate the graph of f

its as simple as moving it two unit to the left first, and then moving up by four units

[void dew]

can you send me the images

i cant draw them so

[placid radish]

here is an example: red arrow moves left by 2, blue arrow moves up by 4

[void dew]

yes i know how it works but i just cant do it

[placid radish]

you can just draw it there no?

[void dew]

no like i cant draw it

[placid radish]

can you elaborate?

[void dew]

https://www.desmos.com/calculator/cqv56kklb3

no drawing tool

[placid radish]

https://www.geogebra.org/calculator

this has drawing tools

you can also just screenshot and draw with paint app

[placid radish]

also for piecewise functions, do this https://www.desmos.com/calculator/brcorvbc0f

[void dew]

now it works

$h(x)=f(x+2)+4$

[solid kiln]

ppq#7826

[placid radish]

for piecewise function, if you are doing it like that you should name your function f_1, f_2, f_3, f_4

dont use the same name

it confuses desmos if you define 4 functions for f

[placid radish]

and also they should allow you to draw on the paper if this question appears

maybe it was because there were difficulties drawing on the screen

[void dew]

i got all 4 now thanks

[placid radish]

yeah but i think the question was asking you to hand-draw it without assistance

[void dew]

it said graph they dont speicfy you can do both

[placid radish]

like print out the paper and draw it on the grid

[void dew]

naw thats to much work i only did that at school

[placid radish]

yeah but as long as you know how to do it without tools its fine

[void dew]

i do

[placid radish]

cuz if it is tool assisted this question is kinda pointless, it just test how well you know how to use desmos

[void dew]

.close

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[earnest oak]

a

[earnest oak]

someone go through A

[fierce lake]

!status

[trim jolt]

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

[earnest oak]

3 tbh

[fierce lake]

What did you get

[earnest oak]

so i did y-1 - m1/ m2-y2 right?

i did 2-5/5-2

and i got -1

[fierce lake]

What is m1 and m2

[earnest oak]

idk where to go from there on

[fierce lake]

X coordinates?

[earnest oak]

yes

[earnest oak]

yes

what country r u

maybe its different in ur country math

[fierce lake]

Not where you are from probably

[earnest oak]

our formula is y-1-m1/m2-y2

im australia lol

[fierce lake]

Generally we write x1 and x2

Anyways

[earnest oak]

or

wait

[fierce lake]

Why is one of you signs inverted

[earnest oak]

we can do y2-y1/x2-x1

[earnest oak]

wdym?

[fierce lake]

Why is it y1-m1/m2-y2

What even is this first off

Point slopes form?

[earnest oak]

well

its rise/run isnt it

m=y-x?

i think i messed something up

[fierce lake]

Yes

[earnest oak]

let me do rise/run rq

should i do this

m= y2-y1

x2-x1

?

[fierce lake]

Yeah

Idk where you got y2-x2 from

[earnest oak]

i got

2-5/5- -2

[earnest oak]

same

ok so i re did it

y2-y1/x2-x1

2-5/5- -2

is that good so far

so now i find the difference between y2-y1

and x2-x1?

i got -3/7 is that correct

[trim jolt]

@earnest oak Has your question been resolved?

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[supple copper]

I’m trying to understand ℚ[x]/p(x), when is this a field?

[supple copper]

Does p(x) need to be irreducible? I feel like it makes sense that if p(x) is reducible then quotienting by all the roots of p would identify all the “0”’s in the ring, turning it into a field

[vale gull]

Yes

[amber python]

wow frosst does algebra

[vale gull]

Its basically R/I is a field iff I is maximal

[marble wharf]

if p is not irreducible, then eg p=f*g. now what is [f]*[g] in the factor ring?

[supple copper]

It’s 0?

[marble wharf]

yes. why

(of course this only shows the easy direction of the equivalence)

[supple copper]

Ngl I’m lost

[amber python]

uh oh

[marble wharf]

how does the multiplication in the factor ring work

or a step before that

what are the elements of the factor ring

[supple copper]

Polynomials that look like k(x)p(x)?

[marble wharf]

no

have you seen the construction of modulo n using a factor ring?

[supple copper]

From ℤ you take the nℤ as the ideal then quotient ℤ by it?

[marble wharf]

yes

what are the elements of Z/nZ

[supple copper]

{0, 1, …, n-1}

[marble wharf]

no

[supple copper]

wat

[marble wharf]

the element of quotient rings are sets

[supple copper]

Oh

[marble wharf]

equivalence classes, residue classes, cosets, however you wanna call them

[supple copper]

{{0, n, 2n, …}, 1+{0, n, 2n, …}, …}?

And the minus ones too

[marble wharf]

yes

the first one we write as [0], the next as [1] and so on

later we leave the [] away but early on those are helpful

[supple copper]

Okay so what about with the polynomials then

[marble wharf]

lets stay with mod for a second

what does [a]*[b] mean

[supple copper]

a * b mod n?

[marble wharf]

yes but no

in this factor ring notation etc

[supple copper]

[a*b]?

[marble wharf]

yes

you take the representatives of the cosets and multiply them to get your new coset

lets say n=6

what is [2]*[3] ?

[supple copper]

[0]

[marble wharf]

because [0] and [6] are the same set

[supple copper]

Yeah

[supple copper]

That’s not the actual class but an element of Z I see the issue

[marble wharf]

well modulo notation is all around a bit of a clusterfuck

[amber python]

depends on what you mean by mod n

[marble wharf]

sometimes it means take remainder, sometimes it means "interpret everything as the relevant class"

anyway

back to polynomials

Q[x]/(p(x))

[amber python]

might help to figure out when Z/nZ is a field

[supple copper]

That I figured was when n is prime

When nothing in Z/nZ is a zero divisor

[marble wharf]

well field and not having a zero divisor are not the same thing in general

only because the ring is finite

[supple copper]

Oh

[marble wharf]

(the classic example of course being Z)

[amber python]

or if you're a finite dimensional vector space over Q

[supple copper]

Well I was just thinking that you don’t wanna multiply a bunch of stuff together that aren’t 0 and somehow end up with a 0 (or n for Z/nZ) as a result

[amber python]

you also need inverses to exist

[supple copper]

Otherwise 0 would have a multiplicative inverse

[marble wharf]

0 never has a multiplicative inverse

that has nothing to do with zero divisors

the problem is that other elements might not have an inverse

[supple copper]

Oh

[amber python]

0 = 1 moment

[supple copper]

If it multiplies to 0 then it doesn’t have an inverse

Okay I see the problem there

[marble wharf]

0*x=1 doesnt have a solution

except in the trivial ring

[amber python]

you cant multiply nonzero stuff together in Z and get 0, but you also cant divide by 2 for example

[supple copper]

Ohhh

So what I’m thinking isn’t the issue

It’s something else for not finite fields

[amber python]

what are you thinking?

[supple copper]

I suppose I had a sufficient condition which was that non zero stuff can’t multiply to be 0

But with Z that doesn’t happen and it’s not a field

[amber python]

yeah when your ring has finitely many things, your hand is forced

[supple copper]

Okay so what’s the equivalent of this for non finite things

[amber python]

well if you're a finite dimensional vector space over a field contained in your ring

and you have no zero divisors

the same conclusion holds

[marble wharf]

finite dim vector spaces are about the closest thing you can get to finite sets

[supple copper]

I’m sorry where did vector spaces come from

I thought we only had rings

[marble wharf]

one of the key points being that for finite sets injective iff surjective and for finite dim vector spaces that same thing is true for linear maps

[amber python]

just as you can think of R as an R vector space

you can think of Q[x] as a Q vector space

similarly for Q[x]/(p(x))

[supple copper]

So…Q[x] has more structure than just a ring?

Oh I suppose that’s true

[marble wharf]

its an algebra

[amber python]

vector spaces have less structure than a ring

[supple copper]

Wait what

[marble wharf]

well, different structure

[amber python]

yeah different

[marble wharf]

both have the addition. and then different kinds of multiplications

[marble wharf]

(awful name btw imo)

[supple copper]

So you mean to consider say the polynomials of degree n or less, but restrict that to only rational coefficients, and that’s what ℚ[x] looks like?

[amber python]

[marble wharf]

who said anything about restricting the degree

Q[x] is not finite dim

[supple copper]

Oh whoops

[marble wharf]

Q[x]/(p) is

[supple copper]

So it’s like the vector space of rational coefficient polynomials but I can multiply them together

[marble wharf]

yes

[supple copper]

And the scaling by rational numbers as a vector space can be described by scaling by rational constant functions

[amber python]

sure

[supple copper]

So this is not helpful for ℚ[x]

[amber python]

its helpful for Q[x]/(p(x))

[supple copper]

But that right

[amber python]

but that makes sense because Q[x] doesn't have zerodivisors

it also aint a field

[supple copper]

sorta like Z?

[amber python]

ye

[supple copper]

so that's why we want the maximum ideal I to quotient by, which is what the other guy said at the beginning

if we turn Q[x] into finite dim at the same time as removing all the 0 divisors then it becomes a field

[amber python]

ye

[supple copper]

okay so certain polynomials act like primes and non primes

and they do so in the way of reducibility

okay because if you're reducible then you're like, not a prime or something

[amber python]

i mean you can just call them primes

[supple copper]

you can? o.o

[amber python]

prime = irreducible in Q[x]

[supple copper]

well i guess from the structure that is what it means to be prime

[amber python]

theyre different in other rings

[supple copper]

so Q[x]/p(x) is not a field if p(x) is reducible?

or we don't know yet

[amber python]

you do know

[vale gull]

Yes

[supple copper]

i feel like it was an iff for the Z/nZ case

[amber python]

.

[vale gull]

Yes

It is in principal ideal domains

[supple copper]

so if p(x) = k(x)h(x) where k and h are linear polynomials, then Q[x]/p(x) can't be a field because k(x) and h(x) are zero divisors

[amber python]

they don't have to be linear

they can be any nonconstant polynomial

[supple copper]

wait what

doesn't 1 of them need to be a linear polynomial at least

[amber python]

no

[supple copper]

ohhh

they just need to have coefficients in Q

and that's enough

so even something like p(x) = (x^2+1)(x^2-2) if i do Q[x]/p(x) is not a field?

[vale gull]

Yes

The polynomial has to be irreducible

[trim jolt]

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[lunar stirrup]

so far what ive been doing is manually calculating the cases

is there a better way?

[cobalt drift]

f(x) = f(8 - x) tells you the function is symmetric about x = 4

f(x + 10) = f(24 - (x + 10)) tells you its also symmetric about x = 12

Is that what you've been exploiting?

[lunar stirrup]

uh no i didnt think of that

[cobalt drift]

Hmm hold on

[lunar stirrup]

could u tell me how to exploit that symmetry

[solid jewel]

the values will be in AP

start writing down some of them

you can find the CD then

then its just a question of no. of terms in the AP

[lunar stirrup]

im not getting an AP

0,3,5,8...

[cobalt drift]

f(x + 10) = f(14 - x) ==> f(x) = f(8 - x)
f(14 - x) = f (8 - (14 - x)) = f(x - 6)

So now, f(x + 10) = f(x - 6) ==> f(x) = f(x + 16)

[solid jewel]

there will be 2 aps

[worldly wing]

No they won’t

[solid jewel]

one for each mirror

[cobalt drift]

So f is 16-periodic

So you have 0, 16, 32, 48, 64, 80, 96, and, all of that + 3

[lunar stirrup]

huh i didnt clearly understand this..what did u do here

[solid jewel]

used f(x) = f(8-x)

[lunar stirrup]

accha and replaced x with 14-x

[solid jewel]

yes

[cobalt drift]

Also use the reflections

f(x) = f(8 - x) ==> f(8) = 6 too

Similarly, f(5) = 6 as well

[cobalt drift]

Beware of repeats when using the 16-periodicity for this one

[lunar stirrup]

ok got till here

[cobalt drift]

We established that f(x) = f(8 - x), so plug in 0 and 3 for x

You have f(8) = 6 = f(5)

[lunar stirrup]

yeah

[cobalt drift]

So 16-periodicity goes back into action

[lunar stirrup]

oh so we have to do 8+16.... and 5+16.....so on?

[cobalt drift]

Yes

You don't need to actually add them

[cobalt drift]

Because everything else is fine except for 96 since 96 + 5 > 100

Should be fine for 8 as well I think

[lunar stirrup]

oh why not

[cobalt drift]

So you have 6 + 6 + 7 + 7 = 26

[cobalt drift]

You need only the number

[lunar stirrup]

ohk so ur saying we just need to know like the +3 +5 +8 part for the number of terms

[cobalt drift]

Yes

[lunar stirrup]

ohk got it thanks a lot

.close

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[balmy basin]

Why must you take the min of the eps bound with 1

I understand the eps bound, don't get 1

[jagged phoenix]

What if I gave you $\epsilon$ really big, $|f|_\infty$ really small

[solid kiln]

Element118

[jagged phoenix]

yeah on the next line we assume $\delta\leq1$ replacing $\delta+2|f|\infty\leq1+2|f|\infty$

[solid kiln]

Element118

[balmy basin]

I see that it's necessary for the replacement. There's no intuitive reason beyond that then

[jagged phoenix]

the norm is that min is required

it's kinda exceptional when there is no min

that the same bound works everywhere even for super large epsilons

[balmy basin]

Hmm I don't understand here. It's always "for all epsilon", meaning you can always get super large epsilons

[jagged phoenix]

to have the same bound work everywhere iirc you need some sort of lipschitz condition

[trim jolt]

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[wispy salmon]

What is the minimum spanning tree for this network using prims algorithm?

[trim lichen]

you came here 40 minutes ago with the exact same question but you disappeared and it timed out

@wispy salmon what are the edge weights?

[wispy salmon]

here

ignore how it looks like it came from an insane aslyum

and the edge weight between B and A is 7km

[trim lichen]

ok so weights are distances

[wispy salmon]

yes

[trim lichen]

ok

[wispy salmon]

its a networks assignment due today

[trim lichen]

is a starting vertex mandated?

[wispy salmon]

idk what that means

Do u want me to just send u the whole booklet

[trim lichen]

yes i do

[wispy salmon]

okay

[trim lichen]

are you instructed to start from vertex H for example?

[wispy salmon]

let me take photos hold on

[trim lichen]

ok right

[trim lichen]

so the graph is made by you and doesn't come with any special conditions, got it

ok

so let's see

do you know how prim's algorithm works generally

[wispy salmon]

Yes i have instuctions

[lunar stirrup]

dam

[wispy salmon]

what

[lunar stirrup]

nothing just the lighting is kinda low

[wispy salmon]

yeah its unfortunately 2am

[wispy salmon]

where did bro go

[trim lichen]

please do not call me "bro"

[wispy salmon]

Oh sorry.

wait my bad i just realised that could be mis gendering

[trim lichen]

i am still here -- i was just reconstructing your graph in a graph editor to get a cleaner picture

[wispy salmon]

Okay thank you

[trim lichen]

in this case yes it is misgendering

anyway

[wispy salmon]

I appreciate it alot

Yes sorry I am not very educated on that

[trim lichen]

ok so it looks like you're required to start with the cheapest edge

[wispy salmon]

But I am learning

Okay

[trim lichen]

in your case that's edge AB

[wispy salmon]

Yes

I understand that part

[trim lichen]

at each step, as instructed, you look at all the edges connecting a node in your tree to a node not in your tree,

and pick the cheapest of those

[wispy salmon]

But with the prim thing you cant make a circuit

[trim lichen]

would you like to be taken through the algo step by step

[wispy salmon]

yes please

[trim lichen]

gimme a couple mins to actually recreate your graph so i don't have to open your photo every time

[wispy salmon]

Okay

[wispy salmon]

mine doesnt look like that bc

It needs to look like this

[jagged phoenix]

it's fine to distort the graph after we already placed the edge weights down

[wispy salmon]

Wait I don't want to leak my address

My assignment is to choose

10 locations in maybe like Melbourne

[trim lichen]

i am not making a photorealistic reproduction of your graph here

that is not my goal and you should not concern yourself with that

[wispy salmon]

So for example I chose 10 different airports in victoria

[trim lichen]

im just treating it as a graph

and keeping your naming convention and all edge weights

bc that is all that matters to us

[wispy salmon]

Okay that makes sense

[trim lichen]

are you OK with that

ok

[wispy salmon]

Yes

[trim lichen]

so here is our proto spanning tree so far

i will highlight nodes and edges belonging to it in red

so everything black is not in our tree

[wispy salmon]

okay i will draw a red line on mine now

Wait let me find my pen

[trim lichen]

now what i want you to do for me is execute, here in discord, the following instruction:
List all edges connecting a red node to a black node. Write down their names, not their weights.

[wispy salmon]

Okay ill just use a highlighter

What 😓

I dont understand

I understand the words but not what you want me to do

[trim lichen]

i want you to tell me the names of all such edges

based on my pic

an edge is named by its endpoints, for example edge AB which is currently the only red edge

[wispy salmon]

Ohhhhh

But why

[trim lichen]

this is what prim's algo does

i am taking you through prim's algo step by step

[wispy salmon]

Im having difficulty understanding how to not create a circle

Okay

[trim lichen]

don't worry about it

just do as i say

you have my assurance that no cycle will happen

[wispy salmon]

This is the instructions i have

[trim lichen]

,rccw

[wispy salmon]

Okay I understand

[trim lichen]

yes i'm taking you through step 2 here

do what i say please?

[wispy salmon]

Okay I will try

[trim lichen]

also DON'T JUMP AHEAD

i will say this upfront

[wispy salmon]

Okay I wont

[trim lichen]

i will repeat the instruction here:

List all edges connecting a red node to a black node. Write down their names.

and the graph so far

[wispy salmon]

BC?

like I am confused

[trim lichen]

is that the only such edge?

[wispy salmon]

Nope

[trim lichen]

well i want them all.

[wispy salmon]

There are lots of options

Okay okay

[trim lichen]

there are not "lots" of options.

remember, the edges we're looking for are those which join a red node and a black node.

(red means in the tree and black means not yet in the tree)

[wispy salmon]

okay let me try again

BC, AE, AH, AF

[trim lichen]

ok

[wispy salmon]

Is that right 🥳

[trim lichen]

now tell me: which of these edges is the cheapest (i.e. has the lowest weight, or in your case is the shortest)?

answer as "The shortest edge is ___, with weight ___."

[wispy salmon]

Umm

AH

I think

[trim lichen]

answer as "The shortest edge is ___, with weight ___."

[wispy salmon]

Wait no

I don't know how this is helping me 💔

[trim lichen]

i am taking you through this by giving you direct instructions to follow

[wispy salmon]

Like it is but I just want to get my booklet 90% complete

Okay

[trim lichen]

don't try to think ahead, just do as i say here

you are just executing an algorithm

[wispy salmon]

Okay

The shortest edge is BC with weight 37km

Wait no

AH with 35km

[trim lichen]

yes

[wispy salmon]

😎

[trim lichen]

now we take this shortest edge we just found, and add it to the tree

[wispy salmon]

I think I might have to pull an all nighter

Okay

[wispy salmon]

This is making sense

[trim lichen]

now we do the exact same process all over again.

List all edges, by name, that connect a red node with a black node.

[wispy salmon]

Okay 👍🏻

HI, HG, AF, AE, BC

[trim lichen]

good

now tell me what the shortest one among those is.

and its weight

[wispy salmon]

BC and 37km

[trim lichen]

ok, good.

[wispy salmon]

Yayaya

Thanks 🙏🏻

[trim lichen]

ok, i take it you can continue by yourself from here?

[wispy salmon]

Okay

If i need more help though

With other questions do I just come back here

Or will u not be available to assist

Wait!

Wont it complete a full circuit

[trim lichen]

i specifically may be available or not but i cannot predict that

and no, the algorithm will never complete a full circuit.

[wispy salmon]

Okay thats okay

Okay 👍🏻

[trim lichen]

to do that, the last edge in the circuit would have had to connect two red nodes.

but we only ever connect red with black.

so that never happens.

[wispy salmon]

Okay that makes sense

Okay thank you very much

I have an issue

[trim lichen]

so far so good but you're one step short of completion

you need to connect the tree to node E and you need to do it with one edge only

[wispy salmon]

what does that mean

Ohh

Would it be EA then?

[trim jolt]

@wispy salmon Has your question been resolved?

[wispy salmon]

Hello

[trim lichen]

yes

[wispy salmon]

okay tysm u have helped me sm

[wispy salmon]

Ann

How do you answer the first image

[trim jolt]

@wispy salmon Has your question been resolved?

[trim jolt]

@wispy salmon Has your question been resolved?

[trim jolt]

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[near hearth]

can anyone help me with this question pls (find the unknowns in the figures)

[zinc ginkgo]

Use Pythagorean's theorem to find FH

[near hearth]

but how do I find x after that

[zinc ginkgo]

Get a couple equations for EG, angle H, theta

[little glen]

,calc 169 - 64

[solid kiln]

Result:

105

[near hearth]

so it's like this

[boreal shuttle]

YEP

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[boreal shuttle]

@near hearth how did you get it

[trim jolt]

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[real geyser]

i’m confused why the number of objects bought would replace the variable? i thought the cost replaces the variable

[unique minnow]

Apart from the fact that they somewhat give you the variables in parentheses (sweaters are denoted by s, and so on), practically speaking, you wouldn't want to know the price of your order in terms of the price of the items, but rather in terms of the number of items you buy. Prices tend to be "set.".

However I do agree that they could have been more explicit in defining that s denotes the number of sweaters

[real geyser]

ohhh i see

would it usually stand for the number of objects in these types of questions?

or does it vary

[trim lichen]

you should always explicitly say what it stands for

[unique minnow]

I think in general it'll stand for the number of objects.

[dapper swift]

It really depends on the context

The first line indicates that the prices of each garment are the known quantities, but the number of each garment are the unknowns

And then in the 2nd line of the question you sub in the unknowns

[trim lichen]

sometimes it stands for the quantity and sometimes for the price and that is why "let s be the sweaters" is unacceptable as a variable introduction

[dapper swift]

It's just an exercise in reading comprehension but yes it's not good style

[real geyser]

wait do you mean i need to state it when i ask my question or the question on the homework needs to state it

[trim lichen]

both

[real geyser]

wait what did i do wrong with asking my question 😭 i don’t wanna be confusing

[trim lichen]

it's not you it's them

[real geyser]

ohhhhh okay!! well thanks for clearing it up everyone

.close

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[tawny cypress]

Hi, it might be a bit of a silly question but how do I correctly format a proof, and is mine correct?
The screenshot is the question and I will write my answer below.

Answer:
Let n be a positive integer. Let X = {1,2,...,n −1}.
Let x ∈ X.
Prove that if there are integers a, b such that ax + bn = r, then d | r, where d = gcd(x, n).
Assume if r = ax + bn, then d = gcd(x, n) is divisible by integers x and n.
In d = gcd(x, n)
x = d * k1 or dk1
n = d * k2 or dk2
In r = ax + bn
r = ax + bn
r = a(dk1) + b(dk2)
r = d(ak1 + bk2)
Making ak1 + bk2 a product of integers,
Meaning r is divisible by d.

Is it correct and am I structuring it correctly?
Thank you!

[trim jolt]

@tawny cypress Has your question been resolved?

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@tawny cypress Has your question been resolved?

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@tawny cypress Has your question been resolved?

[simple rock]

Let n be a positive integer. Let X = {1,2,...,n −1}.
Let x ∈ X.
Prove that if there are integers a, b such that ax + bn = r, then d | r, where d = gcd(x, n).
Assume if r = ax + bn, then d = gcd(x, n) is divisible by integers x and n.
Wrong! gcd(x,n) is not divisible by x and n, instead x and n are divisible by gcd(x,n). You did things Correctly, just phrased it wrong.

In d = gcd(x, n)
x = d * k1 or dk1
n = d * k2 or dk2
In r = ax + bn
r = ax + bn
r = a(dk1) + b(dk2)
r = d(ak1 + bk2)
Making ak1 + bk2 a product of integers,
No! It should be 'making r a product of integers d and ak1+ak2'. Again what you did was right but you phrased it wrong.

Meaning r is divisible by d.

[tawny cypress]

Ooooh

Yeah, I get that now

[simple rock]

Your proof is right

[tawny cypress]

And how do I structure it?

Is the way I did fine or is there a better way?

[simple rock]

It's correct

[tawny cypress]

If it's correct, then amazing

Thanks so much!

.close

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[swift oriole]

i need help with Area and Volume calculations ;-;
i did it for GCSE's but i forgot everything the moment i finished the exam and now i need it for A-Level 😭
i'm googling questions and stuff but i still don't get it

[astral orchid]

MOPA

make obama president again

[swift oriole]

not related but..

[native shuttle]

If you have a particular question everyone can help

[swift oriole]

like just understanding basic area and volume shapes is what i'm stuck at

idk where to begin either

[round imp]

!occupied

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[round imp]

go preach elsewhere

@swift oriole which shapes you need help

[swift oriole]

the one with the

4

i think it's a cuboid..

[round imp]

wat

wait

This?

[swift oriole]

yup

there is smt to do with 1/2

and idk where u get that from ;-;

and pyramids

[round imp]

what

theres no 1/2

formula is base x width x height

[swift oriole]

..

what am i seeing on google then..

jeez imma fail 😭

[round imp]

Oh surface area

Not volume

yea theres a factor of 2

[swift oriole]

idk i have to learn both

[round imp]

but think of it like this

theres 6 faces

and then the three facing you is the same as the three back facing you

so its x2

[swift oriole]

hm

[round imp]

and the area facing you is 3 rectangles

which have area formula base x height

[swift oriole]

i see

thank you

[round imp]

np

[swift oriole]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[woven sigil]

HELP MEEEE

Its simple but i forgot how

Dont mind my bad writing

[clear cloud]

,rccw

[boreal shuttle]

what are you stuck on exactly ?

[woven sigil]

Everything bro idk how to start and like make it simpler so i can dind x

Find*

[boreal shuttle]

you'd need to gather them all in one denominator first

[woven sigil]

How

[boreal shuttle]

,, \text{i.e } \frac{.....}{(9x^2+2x+4)(2x^-4)(3x-2)}

[solid kiln]

<rajel />

[woven sigil]

How did u do it tho

What do i do to get to this

[boreal shuttle]

,, \frac{a}{b} \pm \frac{c}{d}=\frac{ad \pm bc}{bd}

[solid kiln]

<rajel />

[woven sigil]

So i do a X d and b X c?

X as multiplay

[boreal shuttle]

yep

you should already know this

[woven sigil]

how would it look like in this question tho

I do know but idk how it would look like if i multiplayed 9x^2 and those things

[boreal shuttle]

do it fort the first 2 , and then do it with the result of the first 2 and the last one

[woven sigil]

Leme try

[boreal shuttle]

well just start , you'll figure it out along the way

[woven sigil]

Right?

[steep knot]

all the denominators can be factorised

they all have common factors

[boreal shuttle]

why did you develop the denominator tho ?

your trying to solve for x , no need to touch it

[steep knot]

so you can just multiply both sides by their LCM to get rid

[boreal shuttle]

just solve the numertaor

i.e $27x-2=0$ if you didnt do any mistake

[solid kiln]

<rajel />

[steep knot]

I think it makes the calculations easier