help-19

[modern gyro]

Since it is in 3rd quadrant now

Got it m

?

[acoustic grotto]

Yes yes

[acoustic grotto]

So for example, sin (270-x), and because it's an odd multiple of 90, it will become cos x, and then because it is in the 3rd quadrant it will stay positive?

Does that make sense 😭

[modern gyro]

hey

When you check the quadrant

Check it with respect to first function

Is sin + ve in 3 rd quadrant ?

[acoustic grotto]

no

[modern gyro]

@acoustic grotto then ?

[acoustic grotto]

-cos x ?

[modern gyro]

Sin 245 lets consider

So sin 245 is - ve

And how will its cos value be + ve ?

[acoustic grotto]

yes

[modern gyro]

So - cosx

[acoustic grotto]

ahhh okay

[acoustic grotto]

could you give me another example to try

[modern gyro]

Ok

conver sin (450 - 30)

[acoustic grotto]

One sec which quadrant is 450 degrees?

[modern gyro]

Its in 360 + 90

So 360 ° = 0 ° a full rotation

[acoustic grotto]

would it be cos 30?

I've never done a question past 360 degrees so im not really sure how this works

[modern gyro]

Yeah

Its cos 30

[acoustic grotto]

🥳

[modern gyro]

🙂

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[light hornet]

Can some read my argument in (4b) and see if it's good enough?

[bronze fox]

No, this is not a proof. To show the contra positive you need to show it for all non injective functions f. You gave an example of one function.

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[warped urchin]

This question is hopefully not too complicated but I don't understand parallelograms very well yet so hopefully someone can help?

[warped urchin]

So first of all

this line going across

wait I think I got it

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@warped urchin Has your question been resolved?

[warped urchin]

<@&286206848099549185>

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@warped urchin Has your question been resolved?

[warped urchin]

I was all excited but i'm so finished

[warped urchin]

WAIT FINALLY I GOT IT

And know i feel stupid

double stupid

*now

its like this

9*8 to get the area

which is 72

then you divide it by the base length

72/10

7.2

that is the length of h

took me way too long

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[fallen prism]

can someone explain

the cross multiplication

that hapepend here

[summer hollow]

They rearranged for sin(alpha)

im guessing the modulus values were already calculated before so have been plugged in

[fallen prism]

can u explain how

[summer hollow]

I cant see the value of the vector from the image

but calculating the modulus is essentially square root of the sum of the squares of each component of the vector- almost like pythagoras

[fallen prism]

this was the whole question

they essentially cross multiplied

but i duno how

[summer hollow]

if you have a vector $\textbf{v}$, then $\textbf{v}.\textbf{v} = |\textbf{v}|$

so for a vector if you take a dot product of itself- you get the modulus, and you can distribute over the dot product which is why you can cross multiply

[clever fjord]

rh4n

[fallen prism]

ok

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[tacit zinc]

I need help figure out where I went wrong on this bath question because I picked the first option and it said it was wrong

[thin violet]

First one should be right

[tacit zinc]

Thank you

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[silk bramble]

Hello! I need help with a notes question I had during my lecture on friday over differential equations and eigenvectors specifically.

At the bottom im unsure how to actually approach it and i just used a SoE (systems of equations) calculator and im unsure if these are correct

[silk bramble]

The eigen values I found were correct 👍
the given matrix is at the top, im just stuck on the eigenvectors @ the bottom :(

[forest sky]

the usual approach to solving linear systems of equations like this would be gaussian elimination

[silk bramble]

could you walk me through this for the first eigenvalue of lambda = 1?
then i can try it for the other values?

[forest sky]

https://www.youtube.com/watch?v=eYSASx8_nyg

[silk bramble]

thank you 👍

the goated organic chem tutor

i'll ping to check when im done to just check over my work

@forest sky so i tried it so far and im stuck

oop sorry for the upside down

,rotate

,rotate

[silk bramble]

i got here then im unsure what to do, is it just solving another SoE?

[forest sky]

that is the correct elimination. now you want to use that to find the general solution to the system of equations (which should be some free variable multiplied by a vector)

something like this: https://www.youtube.com/watch?v=HPE1yPYklh0

[silk bramble]

so something like this?

or would I replace X3 with C and just leave it at that

C(-1; 4; 1)

[forest sky]

that works

[silk bramble]

does the x3 thing work as well?

or do i have to replace it with C so i stay consistent

[forest sky]

they're both just variables. it's a style thing at that point

[silk bramble]

👍

tysm!

im p sure htese are correct from here

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[rigid crystal]

R^2 means that it has x and y only no z

but i dont get the point of the E

or is it just to identify it as a set ?

like any set

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[fresh ruin]

yea idk it's just a name they're giving it in that context

sometimes E stands for euclidean but I doubt that's it here

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[frigid vine]

How does this work?

[honest quest]

I think you need to consider central limit theorem. Applying that would suggest E as the answer

[runic swift]

size 2 sample size seems too small for central limit theorem

[honest quest]

What would you suggest then

[runic swift]

probably a

[honest quest]

A looks exactly the same as the original graph?

[runic swift]

actually yeah mb

might be b

idk really

[frigid vine]

if size 2 is too small then what do you suggest

[frigid vine]

<@&286206848099549185>

[honest quest]

go with B

Has some good diagrams at the end: https://en.wikipedia.org/wiki/Central_limit_theorem

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[runic tinsel]

asking for a friend she says its A or D she just doesnt know which one

[crisp wadi]

It's neither A nor D

[runic tinsel]

oh wow

so how should she go about it

[crisp wadi]

Given a rational function with equal degrees in the numerator and the denominator, the limit as x -> infinity is the ratio of the leading coefficients

[runic tinsel]

right

simplified

sorry

she said C or D

not A and D

i told her its not D

[crisp wadi]

Now she is closer

Wait

No it's not C either

[runic tinsel]

okay wait

she said she doesnt understand

[crisp wadi]

Tell her this

[runic tinsel]

shes redo-ing it

f(x) --> 2

?

she said she was confused cus its not a polynomial and she didnt remember the formula

[crisp wadi]

Yes

[runic tinsel]

okay thank you!

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[runic tinsel]

is c correct?

[fallen summit]

c should be correct...

[runic tinsel]

Thank you!

.close

.close

...

.close

[fallen summit]

.... its already closed i think...

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[undone inlet]

this is the solution. why does p have to be greater than 0? i understanding their working out but i don't really understand how they came to that conclusion

[summer cradle]

it says "positive values of p"

[ivory dock]

I assume it's because at no point is any limitation to p set during the solving process, nor in the solution so it can be any positive value of p

hence p > 0

[weary pelican]

you're already asked which positive values of p work, so p has to be positive since we ask it to be that way

while it's true that any real value of p will also allow convergence, but it isn't the question that was asked

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[mortal siren]

can someone please check my answer any help is appreciated

[cold sage]

your work confuses me

were you changing cos(2x) into a form of sin^2?

if so then the right is 2+1-2sin^2(x)

you missed a 2

but then you fixed it? eh, your work isnt consistent

this is just no

you need to make sure youre writing things properly

the end results are fine but your working is a mess

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[slim raft]

I have done limit questions but this concept is something I can't understand

[slim raft]

What is the approach behind this?

How did he equated that?

[wooden pasture]

plug in 0 for x

[slim raft]

Oh ok , I didn't saw that

I need one more help @wooden pasture

[wooden pasture]

sure

[slim raft]

How is it possible to do it

[wooden pasture]

do what?

[slim raft]

Image loading

It will be undefined

Why it has any solutions then?

[wooden pasture]

thats why we use limits

[wooden pasture]

the function is equal to 3 at pi/2

[slim raft]

Question was in cos, how can it become sin

[wooden pasture]

but if we take the limit as it approaches pi/2, we can equate the two

trig property

[slim raft]

Pls tell me , I frinking forgot everything

[wooden pasture]

$\cos x = \sin{(\frac{\pi}{2} - x)}$

[clever fjord]

math X meth ✓

[slim raft]

If it is continuous then can we equate both of the solutions?

alpha cos part... and 3

[wooden pasture]

yes

[slim raft]

We will get value of alpha then?

[wooden pasture]

yes

[slim raft]

Sry bro my battery was dead , wanted to ask that only SinX/X will be 1

Other can be or not?

[wooden pasture]

yes

[slim raft]

Other trigo functions cant?

So we need to change them to sin?

In this part can I take pie by 2 as comman and do it as 1 ?

Cos X by X = 1 ?

Last question bro @wooden pasture

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[tepid ingot]

Hello, I've been trying to solve this for about 3 hours.
So I understand because I'm looking for where the function is >5 it can't exist in -1 and -4 'cuz due to the denominator, in >-1 it is negative so it's not there either, same on >4.
Now I understand by looking at it and at the graph that on >72/13 although positive it can't be >5 because of the denominator being a higher degree.
My problem here is, how does one arrive to this conclussion without looking at the graph, 'cuz if it for example the inecuation said > 0,1 the range would be different.
So how can I get there?

[mystic saffron]

you wanna graph this?

[tepid ingot]

no I just want the range of the function

[mystic saffron]

ok so do you know how to graph a function

[tepid ingot]

I mean jsut do a table value right?
it is graphed in geogebra there.
I need to understand how to get to this:
which I am not being able to in paper

that reads Solve(a), sry for the language setting xd

[mystic saffron]

ok

so basically u want to find the range

without graphing the function

[tepid ingot]

yes

[mystic saffron]

ok

gimme a min

[tepid ingot]

💙

[mystic saffron]

ahhhh

long ahh calculations

basically take f(x)=y

write function in terms of y

[tepid ingot]

as you see here I already kind of have the solution with the study of the sign of the function butwhat confuses me is how do I identify that the function lies between -1 and 4 without knowing what it looks like.

[mystic saffron]

ull get a quadratic and then put discriminant>=0

[tepid ingot]

no idea what you're talking about

[mystic saffron]

gimme a min

see this

u get the quadratic equation in terms of y

now in this equation find the domain of x that will be equal to range of y

[tepid ingot]

now I'm even more confused

[mystic saffron]

mind sending what u have done

[mystic saffron]

this picture gives no clue

[onyx cloak]

basically what you do to find range of a function of this form is

you assume y = <the given function>

then you cross multiply and form a quadratic in x

[onyx cloak]

like here

[mystic saffron]

yes

i have no idea what that guy is doing

[onyx cloak]

what

[mystic saffron]

what?!

[mystic saffron]

bruv which grade u in

[onyx cloak]

oh wait

u arent the one who asked the question

sorry, my bad

[mystic saffron]

[onyx cloak]

yeah i have no idea what this guy is doing either

[mystic saffron]

bro just post the question properly

word to word

[tepid ingot]

I'm actually in uni by now but I was always really bad at math and I am now studying art soooo :p
I'm just doing this to help my mother through highschool :3

[mystic saffron]

help ur mother through highschool??! Im sorry?

[tepid ingot]

you asked to show what I had done so there it is.
What I'm trying to do here is find the range for which that function is >5

[mystic saffron]

ohk

i solved it for function >0

[tepid ingot]

yeah, you asked what grade am I in, so I answered

kind of hurtful tbh

[mystic saffron]

no worries

[tepid ingot]

so if you assume y to be a(x)... I'm not following sorry I'm dense

[mystic saffron]

leave it

it was for f(x)=0

[tepid ingot]

am I dum for not knowing any of this stuff after graduating if I didn't have a math class in like 5 years?

[mystic saffron]

no no chill out its alr

[tepid ingot]

would it be possible to just get rid of the fraction?
or does it having independent number change things, I don't even remember how to divide polinomials T.T

[mystic saffron]

what youve done here is

uve found domain

possible values of x

[tepid ingot]

yes

we agree on that

[mystic saffron]

yes

so now u hve to find the range

of a(x)

[tepid ingot]

yes

[mystic saffron]

see

first

cross multiply the denominator

and get a quadratic

did you get through?

[tepid ingot]

cross multiply with what?

[mystic saffron]

with 5

[tepid ingot]

I was trying that at some point so I get:
x²+28x-52 / x²-3x-4 right?
if I do my 13x-72 / x²-3x-4 - 5/1?
is that correct or am I misunderstanding something here?

[mystic saffron]

nono

[tepid ingot]

yeah thought so

[mystic saffron]

try to make it a single equation

[tepid ingot]

hold on

[mystic saffron]

i am still doubting the question

[tepid ingot]

what?

[mystic saffron]

yes

if we cross multiply

the roots turn out to be complex

[tepid ingot]

but it don't matter right?
you can always just leave the number written like a root right?

[mystic saffron]

so range is 143/10 to infnity

[mystic saffron]

i said complex not irrational

D<0

@onyx cloak take a look

[tepid ingot]

something like this are you saying?
then rearrange it and get the roots?

[mystic saffron]

yes

[tepid ingot]

right, I mixed it up

[mystic saffron]

just replace equal to sign with greater than

and find the range of that quadratic

but still its not making sense

[tepid ingot]

what isn't?

[mystic saffron]

2.8,14.3 to infinity

this should come ig

for a(x)>5

range

if a(x) were 0

it was straight forward

[tepid ingot]

I haven't even gotten the range of this xd

but it's complex right?

[mystic saffron]

roots are complex

range isnt

if roots are complex it simply means

the function is not intersecting the x axis

but it has its extent on y axis

[tepid ingot]

right

[mystic saffron]

and thats what we gotta find

[tepid ingot]

I don't think they taught me this in highschool actually

I notice my mother's actually getting better education than I did.

[mystic saffron]

wait

i got it

that was a long ass problem

after cross multiplying

u should get this

now do D>0

after that you get this

[tepid ingot]

where you get that y?

[mystic saffron]

i interpreted the question incorrectly my bad

let a(x)=y

[tepid ingot]

u told me to find the range and I found they never taught us that

[mystic saffron]

so

this step

forget the >5 part i messed up there

[tepid ingot]

back to square one then

I get you assumed the function as y but then I don't understand why you have multiplied the denominator by y and have the numerator being substracted

[mystic saffron]

arithmetic

[tepid ingot]

the one thing I am increadibly bad at

[mystic saffron]

i multipled both the sides by denominator

[tepid ingot]

which both sides

[mystic saffron]

sides opposite to equal sign

[tepid ingot]

sorry I'm the kind of guy who writes every single thing he does in the paper 'cuz I can't follow otherwise

[mystic saffron]

dawg this problem too hard for u

i recommend polishing the basics

dont mind me otherwise

[tepid ingot]

I don't even know what I should polish

[mystic saffron]

i suggest drop the question for now

for you this would be complex

[tepid ingot]

this is exactly why I was hesitant to ask in the first place

[mystic saffron]

and focus on basic domain and range related question of simple polynomials

[tepid ingot]

thank u for trying I guess

I don't have time for that anyway I've already spent too long on this and it's not my class I'm just trying to be helpful

[mystic saffron]

its like you jumped over here.. and the question wasnt clear too

[mystic saffron]

no problem

[tepid ingot]

it was

[mystic saffron]

if u feel

[tepid ingot]

how do I close this?

[mystic saffron]

u can ask me on my dms

[tepid ingot]

thank u <3
I might later for now I gotta go work

[mystic saffron]

ya maybe i wasnt focused enough nvm

[tepid ingot]

!close

!close

[odd edge]

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[wild plover]

How do I solve this?

[odd edge]

@wild plover Has your question been resolved?

[wild plover]

No

[odd edge]

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[wild plover]

.reopen

[odd edge]

✅

[wild plover]

<@&286206848099549185>

[open crown]

Pretty intuitive

If a 20 sided die is rolled, what the probability that the result is 4?

@wild plover

[wild plover]

I don’t know

[open crown]

How abt a 6 sided die (normal one), whats the probability that the result is 1?

[wild plover]

1/6

[wild plover]

1/20

[open crown]

Good

Now

[open crown]

Same question but how abt the result is 4 or 5?

What do u think?

[wild plover]

2/20

[open crown]

YES

[wild plover]

7/20

[open crown]

Now 4 to 10

[open crown]

Bingo

[wild plover]

7/20

I have to use a formula to answer tho

[open crown]

Use logic

[wild plover]

😐

[open crown]

Like

A die with n-th side

[wild plover]

It has to be a distribution formula

[open crown]

Then any 1 side should have 1/n-th chance to get it

[wild plover]

No it has to be like a binomial distribution formula

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[tender kindle]

5%

[mystic saffron]

💯

https://tenor.com/view/the-count-the-count-sesame-street-sesame-street-five-beta64-gif-23660183

[winged orchid]

wow i guess 5%=0.05 or 1/20 np you're welcome

[tender kindle]

Did I ask you

:]

[winged orchid]

no

[fallow tapir]

what is your question?

[tender kindle]

If a 20 sided dice is rolled what is the chance of getting the number 4

[winged orchid]

you are correct

[tender kindle]

1/20

[winged orchid]

5% you are correct 👍

[fallow tapir]

any other question?

[tender kindle]

Thanks

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[tender kindle]

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[rich fiber]

how do u know how to use what order of f(x)

the first one

is x^3-4x-5x

but the second

is 5x-x^3+4x

[onyx cloak]

when you are writing the area with respect to x axis then you do (right function - left function)

[odd edge]

@rich fiber Has your question been resolved?

[rich fiber]

wait

but how do u know

oh u just check the lines

on ur grapghing calculator or smthg?

[onyx cloak]

u have the graph given

use that

in your particular case you can determine which one is which because one of the equation is a straight line (y=5x)

and there's only 1 straight line in the graph

[rich fiber]

5x

but from -3 to 0

wouldnt 5x be the right function

[onyx cloak]

nvm, my bad

what i said was for with respect to y axis not x

youre writing with respect to x axis so its upper function - lower function

[rich fiber]

ook

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[high jay]

show that this function is continuous

{x} is the fractional part of a number

for example {7.2} is 0.2, because 7.2 - [7.2] (whole bit) = 7.2 - 7 = 0.2

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[fallow tapir]

ok

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[fair bane]

i know in the binomial thereom that n = n-k + k, but i dont know why that's the case. can someone explain why?

[nimble blaze]

just an algebraic property not specific to binomial theorem

n - k + k simplifies to n

[ember oak]

$(n-k)+k=(n+(-k))+k=n+((-k)+k)=n+0=n$

[clever fjord]

SWR

[nimble blaze]

additive inverse property if you wish to tack on a name

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[grizzled charm]

When connectivity implies arc connectivity ? and please can you provide me with a proof ?

[ember oak]

!original

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

[odd edge]

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

[ember oak]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[grizzled charm]

i was reading the proof a Corollary that uses Hurwitz's theorem, in order to use the theorem here D{w} a domain, and i am trying to proof that D{w} connected

[grizzled charm]

is it clear now ?

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[grizzled charm]

.reopen

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✅

[echo ginkgo]

Va plutot dans #real-complex-analysis

Et traduis ta preuve de préférence

@grizzled charm

[grizzled charm]

okieee merci

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[dreamy wasp]

What does Z[x]/(x^3) look like? And can someone explain quotient rings easily?

[odd edge]

@dreamy wasp Has your question been resolved?

[fresh ruin]

it's like all the remainders you get when you divide Z[x] by x^3

which is just quadratics ax^2+bx+c I think

another way to see it is that you can add or subtract x^3 over and over, so basically you can cross out any x^3, x^4, x^5, etc

[dreamy wasp]

Is that why they call it mod?

Same idea as something like Z/6Z or Z6

?

Is there anything that is different from this idea also? Or can I use that for all quotient rings?

[fresh ruin]

yea the same idea as mod

if you mod out by something different like x^2+1 though it can get really complicated

but for single terms like 6 and x^3 it's that simple

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[topaz trout]

Does the TI-84 have a productlog/Lambert W function?

[fresh ruin]

no

you just use graph intersection

[topaz trout]

Okie ty ❤️

[gilded vector]

!done

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[tribal relic]

guys someone help meeeeee

plsssss

plss

pls

[gilded vector]

ask your question in #help-34

[tribal relic]

ok

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[past ginkgo]

I just need help with part a

[past ginkgo]

I dont understand how to rewrite r in terms of h to suit dh/dt

[fresh ruin]

to go from r to h you use the ratio of r to h for the whole cone

like to get from 2.5 to 4 you do 4/2.5, and that works for any r because of how it's a triangle that shrinks at a steady rate

r= h*2.5/4

[past ginkgo]

as in to get from 2.5 to 4 we do
2.5 * 4/2.5

i understand that its a steady rate decrease
but i just find it a bit difficult to see the relationship between 2.5 to 4

because 4 is a height whilst 2.5 is a radius

[fresh ruin]

since it's a straight diagonal line, any depth you pick will have the same shape of cone, slightly taller than wide by a 4/2.5 ratio

like going halfway down means you have half the height and half the radius, so the ratio is the same

[past ginkgo]

mhmm the heights and the radius will differ

so do u mean the ratio is just
r:h
2.5:4

for this ive seen ppl write this as then
(r/2.5) = (h/4)

[fresh ruin]

yea that works

I just think of if I have to multiply by a big number or small number to decide the fraction

[past ginkgo]

oh so how did u know which way round for this one

So I have this

but the h is to the minus half power

when simplified

<@&286206848099549185> ;v;

[fresh ruin]

I can't tell where you get dh/dv but it's from taking the derivative of V=1/3pi*r^2*h = (2.5/4)^2*1/3pi*h^3

[past ginkgo]

I got dv by dh via the general formula for the cone

But it doesn't seem specific TvT

Oh why is h to the power of 3

[brittle beacon]

Substituting that r in will get you h^3

[past ginkgo]

Ohhhhhh

@fresh ruin
@brittle beacon
Thank you guys 😭 ✨️

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[wooden swan]

how do i find this

is it 1400?

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[tall lagoon]

Please brothers

[tall lagoon]

help

😦

[elfin zodiac]

What is your problem

[tall lagoon]

I don't know how to start, please 😦

If you could tell me which is the shortest method, please.

[elfin zodiac]

Do you have more context?

You will have to factor denominator

And do partial fractions

But looks like a mess unless im missing something

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[pearl birch]

Good evening, how are you all?

I have a question...

I consider the functions f and g in such a way that I know that f has at least one root and g has at least one real root.

The question is: does the function (f+g)(x) have at least one root?

What I have done is this:

Since f has at least one root, for example "a", then f(a)=0.
Since g has at least one root, I will call it "b" so g(b)=0.

Now, the function (f+g)(x) = f(x) + g(x).

The question is, can I find some x such that f(x)+g(x)=0?

If I calculate (f+g)(a)=f(a)+g(a)= 0 + g(a), I don't know the value of g(a).

If I do the same with b, I don't deduct anything.

That is, everything seems to indicate that it does not matter that both functions have at least one root, we cannot ensure that their sum also admits some root (different from or equal to the previous ones).

But, trying to look for an example where this happens, I can't find any... If I choose two functions with at least one root, I can't get the sum to have no root/roots.

Does anyone know if there are any examples or if I'm actually wrong and the sum of two functions with at least one root each is another function with at least one root?

The context of the exercise is actually about Linear Algebra, where I have to find out if a given set (the functions f such that there exists at least one "k" for which f(k)=0) is a vector subspace of (F, +, R, .)

f and g are functions of R in R and can be anything (polynomial, irrational, exponential, etc.).

[odd edge]

@pearl birch Has your question been resolved?

[brittle beacon]

e.g. f(x) = (x - 1)^2 and g(x) = (x + 1)^2 furnishes f + g as 2(x^2 + 1) which has no real roots

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[light jungle]

How do I solve for the shaded area

[blissful musk]

area of hexagon - pi*r^2

180(n-2)
180(6-2)
720 degree is total angle in hexagon

/6

120 degree each angle in hexagon

on the right

you have sid elength 10 and 10

and includd angle 120

wait

nuh

ignore me i messed up

[light jungle]

💀

[blissful musk]

actually

maybe....

u can get radius from that

i mean diameter

side length 10 and 10 and included angle 120... use cosine to find the other side

that sshshhould be your radius

diamter

[light jungle]

Idek what u be saying

[wise jasper]

bru

the hexagon is 6 equilateral triangles

with side length 10

so the radius of the circle is the height of the equilateral triangle

which is 5sqrt3

[blissful musk]

oh

haha

xd

[wise jasper]

so the circle has radius 5sqrt3

[blissful musk]

[light jungle]

What 😭

[wise jasper]

the hexagon area formula is $\frac{3a^2\sqrt{3}}{2}$

[clever fjord]

Dork9399

[light jungle]

Oh

[wise jasper]

so the hexagon has area 150sqrt3

[light jungle]

Wait is there a way to solve without that formula? Cuz like say I have a test and it’s a 10 sided one instead

Is there a way to solve with just trianfles

Triangles

[wise jasper]

do you have calculator?

[light jungle]

Yuh

[wise jasper]

are you supposed to use trig or smth

[light jungle]

Lowk idek my teacher left this unit

[wise jasper]

was it supposed to be trig?

[light jungle]

And We’ve had 3 diff subs

Trig was our last unit this is circles

O sure I can use anything. Tho

[wise jasper]

ight

trig is easiest

[light jungle]

💀

[stuck hare]

Hi...

[wise jasper]

so we can split the polygon with n sides into n triangles

[wise jasper]

hi!

[light jungle]

Like this right?

[wise jasper]

,rccw

[wise jasper]

it wont be 120

itll be 360/6

60

[light jungle]

Oh

Whoops

Then all angles are 60?

Dosnt that mean all sides are 10

[wise jasper]

in the hexagon

yes

which is why we can use simpler formulas for the hexagon

but for the 10-gon, we don't have that luxury

would you like to take that example?

[light jungle]

Uhh

What

[wise jasper]

in a polygon with 10 sides, the triangles won't be equilateral

so you won't know the area of the polygon or the circle

both of which you need to know

[light jungle]

I can’t just use law of sines?

[wise jasper]

exactly

you should use law of sines

so for a polygon with 10 sides, the vertex angle will be 36

[light jungle]

Wait back to my original question I think I did something wrong

[wise jasper]

and the other two angles will both be 72, because the triangles will all be isosceles

[wise jasper]

sure

[light jungle]

So

Since the inside triangle is 10 on all sides

Oh wait

Nvm I see what I did wrong

Nvm

I didn’t solve for height to find area of triangle

Nvm I did something wrong

I’m getting the area of the circle to be bigger than the hexagon

[wise jasper]

show your work

[light jungle]

I split the triangle in half

To solve for height

Once I got height I did b x h

To get triangle area

Than multiplied it by 6

To get 259

[wise jasper]

what did you get for triangle area?

[light jungle]

43.3

[wise jasper]

as a radical

[light jungle]

Oh

Uhh

25 root 3

[wise jasper]

what did you get for triangle height?

[light jungle]

5 root 3

[wise jasper]

so what did you got for the full area

of the hexagon

as a radical

[light jungle]

Oh uhh

150 root 3 I think

[wise jasper]

what did you get for the circle area?

[light jungle]

314

[wise jasper]

work

and keep it in terms of pi

just type it out

[light jungle]

I just thought that the radius is 10 so I did 10^2

To get 100

Then did 100pi

[wise jasper]

the radius will be the height of the triangle, not the side of the triangle

look at the diagram

[light jungle]

Oh

Oooh

So then it should be 5 root 3 squared?

[wise jasper]

yes

[light jungle]

Oooh

I got the answer now

Tysm

[wise jasper]

np

[light jungle]

.close

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[light jungle]

Can someone point out my probably stupid mistake

I solved for the shaded area

By doing arc area

60/360 x 25pi

Then got 13.1

Multiples it by 3 to get 39

[odd edge]

@light jungle Has your question been resolved?

[odd edge]

@light jungle Has your question been resolved?

[light jungle]

<@&286206848099549185>

[cobalt nacelle]

its an equilateral triangle

so find area of triangle

and then area of circle

then subtract circle from triangle

[light jungle]

How do I find the area of the circle

[eternal oyster]

The medians of an equilateral triangle intersect each other at the 1/3 of the height, maybe u could do something with that

Height of the triangle is 3√5

[cobalt nacelle]

pir^2

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[light jungle]

I give up

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[rotund dawn]

Could somebody proof-read my proof?

[rotund dawn]

proof-read by reading proof xd

in the kernel part I changed implies to iff

[brittle beacon]

What's a supposed to be, sorry?

[rotund dawn]

any element from the ring R

[weary pelican]

and the a here?

[brittle beacon]

isn't the kernel supposed to be the ideal I though, you needn't have a being zero

[rotund dawn]

fuck, that should be r

[brittle beacon]

[this was the a I was referring to]

[rotund dawn]

I didn't know how to really write that tbh

I just showed what it means for it to be a kernel which is $\ker(\phi) = {a\in R\ |\ \phi(a)=0}$

[clever fjord]

Juke | ping me if no response

[rotund dawn]

or at least, I tried to

not sure if I expressed that correctly though

[weary pelican]

just say $r+I = I \Longleftrightarrow r \in I$

[clever fjord]

rafilou2003

[rotund dawn]

wait, is that needed though

I thought it's fine even if r in R, outside of I

[brittle beacon]

[in particular, that 0 you're getting sent to is supposed to be the 0 of R/I, which is I]

[weary pelican]

yes the 0 of R/I is I

so pi(r) = 0 literally means r+I = I

which means that r has to be in I

[rotund dawn]

i think the fact that I haven't done the part of the class where we talk about kernels has caught up to me, fuck

bc afaik, if r in I, then isn't all of pi(r) = 0?

since I is the kernel of pi?

[weary pelican]

if r in I, then pi(r) = the 0 of R/I

[brittle beacon]

(you can have elements r in the ring R, that aren't in the ideal I)

[rotund dawn]

ah but r not in I then it is something else

I see

makes sense

pi isn't from I to R/I, but R to R/I, tru

[rotund dawn]

I'm still a bit confused about this though

you said "yes the 0 of R/I is I" but I can't quite make the connection between that and what I asked

[weary pelican]

you wanted to know if "r+I = I <=> r in I" was needed

I said if you want to come to any conclusions as to pi(r) = 0 means about r, then you have to write r+I = I

[rotund dawn]

oh sorry, when i asked, i meant to ask why the RHS "r in I` is needed

[summer cradle]

[rotund dawn]

?????? ^

yo chat, what's goin on above me xd

anyways

[brittle beacon]

You wanna show the kernel is I

[weary pelican]

why is it "needed"?

[rotund dawn]

yes

[rotund dawn]

yeah

like why must r be in I

[weary pelican]

r + I = I

so r + 0 must be in I

[rotund dawn]

uhh

[weary pelican]

what do you not get?

r+I = I, both being sets

[rotund dawn]

is the statement, r+I=I for any r, not necessarily in I, true?

[weary pelican]

It's true for r in ker(pi)

and we just SHOWED

[rotund dawn]

ahh I see

AHH

[weary pelican]

that r+I = I can ONLY be verified when r in I

[rotund dawn]

I was confused bc I thought r+I=I is true for any r

but like

u said it three times yet my brain somehow refuses to understand

okay

[weary pelican]

Okay, r in ker(pi) <=> r+I = I is ok for you right?

it's the definition

[rotund dawn]

this is a new definition for me kinda, but i think it makes sense

I'm thinking of I being like a 0

in addition

[weary pelican]

yes, it is the 0 in R/I

[rotund dawn]

okay, huge

[weary pelican]

so

[weary pelican]

we focus on this

suppose r+I = I

this is an equality between SETS, correct?

[rotund dawn]

yes

[weary pelican]

so, what is inside r+I?

[rotund dawn]

that's a good question...

lemme think

it'll be some subset of R..? I think

[weary pelican]

yes

[rotund dawn]

okay

[weary pelican]

write me in set notation {...} what r+I is

[rotund dawn]

Idk what each element would look like though

[rotund dawn]

sure

[weary pelican]

(recall how a+B is defined)

[rotund dawn]

hmmmm

$r+I = {r\in R \ |\ \pi(r) = I}$

[clever fjord]

Juke | ping me if no response

[rotund dawn]

i think..........?

[brittle beacon]

(for one, that's slightly circular, seeing that pi(r) uses r + I)

[rotund dawn]

mmmmmmmm

good point

lemme cook...

[summer cradle]

hahahaha that react

[rotund dawn]

$r+I = {r \ |\ r\in R/I}$

[brittle beacon]

I love it, pretty much my default response to anything related to cooking

[rotund dawn]

this looks silly

idk

[summer cradle]

isn't that just R/I

[rotund dawn]

ye

[clever fjord]

Juke | ping me if no response

[rotund dawn]

there i guess

[brittle beacon]

[notice what r + I kind of hints at...]

[weary pelican]

that's still R/I

[rotund dawn]

r+I is

[weary pelican]

what does it mean when we add an element with a set

what set do we get?

[rotund dawn]

it's added to each element in the set?

[weary pelican]

exactly

so write it in set notation

[rotund dawn]

the issue is like

ik 0 is in I

but that's pretty much all ik

idk what elements of I would look like

wait, what even is an ideal then..

an ideal is like when you have

$rI\subseteq I$

[clever fjord]

Juke | ping me if no response

[rotund dawn]

for all r in R

[weary pelican]

an ideal is a subgroup with this property yes

[rotund dawn]

well that's multiplication but similar for addition

but how does one write this hmmmm

[weary pelican]

no need to know what every element in I looks like

just call them i

[rotund dawn]

oh

$r+I = {r+i\ |\forall i\in I}$

[weary pelican]

no for all

[rotund dawn]

how come

[weary pelican]

in set notation, you don't write "for all"

[clever fjord]

Juke | ping me if no response

[weary pelican]

$r+I = {r+i\ |i\in I}$

[clever fjord]

rafilou2003

[rotund dawn]

o

yeah actually that-

ye

[weary pelican]

ok*

[rotund dawn]

LMFAO "no 'for all'" changes it to \forall

xd

[weary pelican]

so gimme one element in that set

[rotund dawn]

r+0

r

[weary pelican]

yes

so if r+I = I...

[rotund dawn]

then r=0

[weary pelican]

no

[rotund dawn]

shit

hmm

oh

r in I

OR RATHER

r is in the kernel of pi

which means r in I

I think

[weary pelican]

nono

r in I was it

[rotund dawn]

ah

[weary pelican]

you just said r was in r+I

r+I = I

r in I

[rotund dawn]

oh-

that's

that should have been so obvious

mb

I see

[weary pelican]

ok

and reciprocately, if r in I

then r+i in I for any i (because I is a subgroup)

and i = r + (i-r) in r+I for any i

so r+I = I, both inclusions being proven

[rotund dawn]

what's an inclusion

[weary pelican]

subset

[rotund dawn]

ah

ohh

I see

subset of one another literally means they're the same

I see

hmmm sooo, what does this have to do with kernels then

well let's see

the kernel is the set of any element from R that is the 0 in R/I

but...

is I the 0 in R/I?

[weary pelican]

yes

[rotund dawn]

I see

is that by definition of I? or would I need to somehow show this?

well wait, ig atp i'd be proving what an ideal is..?

[weary pelican]

R/I is a ring with neutral element I

this should be in your notes

[rotund dawn]

what is a neutral element though

[weary pelican]

the 0 element

[rotund dawn]

ahh I see

oh.. bc it's mod I

like that's--

yeah i see

[weary pelican]

so

now that r in ker(pi) <=> r in I...

[rotund dawn]

ker(pi) = I

[weary pelican]

yes

[rotund dawn]

okay great

and the other part of the proof was easy and should be fine so I think I'm good on this one

I should probably go back and catch up on kernel and image stuff for groups tho

Thanks a lot

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[low wagon]

help

[low wagon]

The sum of the weight is: 15+30lb+40lb=85 lb+

To hire the transfer of electrical equipment by means of a crane, $15,000.00 is paid, which added to the payment per kilometer of travel with the load makes a total of 55,800.00. What is the price of the transfer per kilometer if the equipment moves a distance 8.5 km

Incognita and plantation

cost of transfer per kilometer: x.

Helpme

It is first degree equations