help-10

[outer moat]

And apple equals to x+5

What can we say about the apple?

Its the same thing as here

[hasty snow]

apple equals positive or 0?

[outer moat]

apple is positive or 0

We say that x+5 is positive or 0, it cant be equal to positive or 0

Positive or 0 is a list of infinite numbers, yk

Back to your problem

y^2=x+5

And the domain for x is [-5, +infinity)

Which means that x+5 is either positive or 0

What can we say about y^2?

[hasty snow]

y^2 is + or 0

[outer moat]

Exactly

And what would the range be?

You need to show the range the same way as we did for the domain. That means (something, something)

[hasty snow]

positive real numbers [0,+infi)

[outer moat]

Cant it be 0?

You excluded 0, why?

y^2=x+5. And the domain is [-5, +infinity). Let x=-5 then y^2=-5+5=0

[hasty snow]

mb zero can be included

[outer moat]

So the domain is [-5,+infinity) and the range is [0, +infinity)

[hasty snow]

thank you so much i learned alot

[outer moat]

Dont let y^2 confuse you. The square is there to help you find the domain

Glad to help :>

[hasty snow]

bye have a good day

.close

[obtuse pebble]

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[crude comet]

yes

[obtuse pebble]

@timid silo Has your question been resolved?

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✅

[warm shale]

Apertyx

[brazen gorge]

it's correct

just multiply the paratheses

$\frac{6}{x^3} \cdot \left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right) = \frac{6}{x^3} \cdot \left(\frac{x+1}{x}\right)\left(\frac{x+2}{x}\right)$

[warm shale]

artemetra

[brazen gorge]

like this

i just showed you

oh

factor

but in any case you are correct

it doesn't matter in what form you represent the answer

[crude comet]

the answer is y''

🙂

jk

[brazen gorge]

ong

x+1 takes one x, x+2 takes one x

but fair, without knowing the answer you wouldn't be able to spot that

i just don't know why you concerned about that

your answer is correct, period

[obtuse pebble]

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[obtuse pebble]

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[undone lotus]

Can somebody check these two questions for me? Whether they’re right or wrong

[obtuse pebble]

@undone lotus Has your question been resolved?

[undone lotus]

<@&286206848099549185>

[wooden delta]

Ok

The last step is wrong

It should be an integral of tanx sinx

Not tanx secx

[undone lotus]

Oh

Yeah

Ur correct

How do i go from there

@wooden delta well?

[wooden delta]

Add a constant term

You are done

[obtuse pebble]

@undone lotus Has your question been resolved?

[undone lotus]

Thanks a lot man

[obtuse pebble]

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[tough cape]

a) Task 1: Make a business plan for each group

Agree on the following tasks:

• Select 20 products (items to be traded) so that the total value of those 20 products (calculated based on the input price of each product type) does not exceed the assumed 100 USD;

• Choose a business form, discuss business strategies;

• Assign work to each team member; Each individual plans his or her way of doing things and the whole group exchanges ideas.

[obtuse pebble]

@tough cape Has your question been resolved?

[latent walrus]

interesting

what are you expecting us to do with this?

its not a question, theres not really any information

[tough cape]

issues that i cant even solve... so.. what products do u think is good to sell?

[latent walrus]

what products, literally anything?

[icy granite]

discord.gg/business

[latent walrus]

this doesnt seem like a maths question from what youve provided

^

[tough cape]

litterally this is for grade 7 math..

[obtuse pebble]

@tough cape Has your question been resolved?

[obtuse pebble]

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[heavy belfry]

I don't understand how to do this at all

[heavy belfry]

I know A = 1/2 * d_1 * d_2

and I know Perimeter is 4S = K

Could say S = K/4

I have a right angle, so I could use the pythagorean theorem to say

K/4 or S = (a^2 + b^2)^(1/2)

I don't know where to go from there

Maybe I could find a or b using the pythagorean theorem too?

(s^2- a^2)^(1/2) = b, then substitute?

[obtuse pebble]

@heavy belfry Has your question been resolved?

[heavy belfry]

<@&286206848099549185>

[timid silo]

Use that 2ab<=a^2+b^2

[heavy belfry]

How did we reach that

[timid silo]

AM-GM or (a-b)^2>=0 because squares are non -negative

[heavy belfry]

wait one sec

is 2ab from

b x h

2a * 2b

[heavy belfry]

AM GM?

I don't know what the abbreviations mean, sorry

Mind explaining

[timid silo]

It's an inequality of the arithmetic and geometric mean

But do you understand that (a-b)^2>=0 and that we can get 2ab<=a^2+b^2 from there

[heavy belfry]

I have no clue what (a-b)^2 is tbh

How did you reach it

This is what I've got written down

[timid silo]

$(a-b)^2\geq 0$

[warm shale]

Ryanstaal2006

[timid silo]

This can be rewritten as $2ab\leq a^2+b^2$

[warm shale]

Ryanstaal2006

[heavy belfry]

Okay well

My questions are

How did you reach (a-b)^2

a^2 + b^2 I get

But also second question is, why can you say it's bigger than 2ab

[timid silo]

I didn't reacht it or something but it is just to show you how to get that a^2+b^2>=2ab which is what we are going to use

[heavy belfry]

Alright

So is this a given for any rhombus?

With those half-diagonals

Actually it's fine keep going

[heavy belfry]

Where do we go from here

OH

Ok

[timid silo]

I'm going to draw something and send a picture

[heavy belfry]

Got it now

[heavy belfry]

yea

[timid silo]

So what is the area of the rhombus?

[heavy belfry]

it should be

1/2 * 2a * 2b

instead of 1/2 * a * b

[timid silo]

Yes so the are is 2ab right?

[heavy belfry]

yes

[timid silo]

Now recall that 2ab is at most a^2+b^2

And what is a^2+b^2 equal to?

[heavy belfry]

k/4 or s

[timid silo]

So $2ab\leq a^2+b^2=\frac{k}{4}$

[warm shale]

Ryanstaal2006

[heavy belfry]

yeah

can I get a value for b^2 with that

[timid silo]

The questions asks for a, b when the area is maximal

[heavy belfry]

b = sqrt(k^2/4 - a^2)

[timid silo]

So obviously the area is maximal when 2ab=k/4 right

[heavy belfry]

yeah

[timid silo]

So 2ab should be equal to a^2+b^2

[heavy belfry]

got it yeah

[timid silo]

Now this leads to (a-b)^2=0

[heavy belfry]

a = b

right?

[timid silo]

Exactly

[heavy belfry]

ah..

[timid silo]

And now calculate a in terms of k with Pythagoras

[heavy belfry]

so is this a square then?

[timid silo]

Exactly

[heavy belfry]

oohh

well that was helpful

tysm

[timid silo]

You're welcome

[heavy belfry]

🫡

.close

[obtuse pebble]

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[timid silo]

the power in the nominator and the denominator are equal

try doing the euclidean division first

[royal basin]

numerator*

but yes this is a mistake that's non trivial to spot but is not terribly hard to recover from

[obtuse pebble]

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[halcyon dust]

anyone free here?

[cloud belfry]

i am

[halcyon dust]

can u help me with the question above, my deadline coming soon?

[halcyon dust]

u still here dude?

[cloud belfry]

angle subtended by arc at centre

is half

that subtended at circle

something like that right?

@halcyon dust familiar with this?

[halcyon dust]

maybe

im not so sure so i join here to claridy my doubt

[halcyon dust]

never heard of this?

[cloud belfry]

double

https://byjus.com/maths/angle-subtended-by-an-arc-of-a-circle/#:~:text=The angle subtended by an arc at the centre is,remaining part of the circle.

go thru theorem quick

[halcyon dust]

is this like a different type of the arrow one?

[cloud belfry]

angle o is twice angle c

[cloud belfry]

what

[halcyon dust]

so c is o/2 ?

[halcyon dust]

this one

[cloud belfry]

yes

[halcyon dust]

if i am correct

ah alright thx dude

[cloud belfry]

y=2x ryt

[cloud belfry]

and then oab is isosceles

[halcyon dust]

ye ik that

[cloud belfry]

n stuff...angle sum n stuff

[halcyon dust]

the o is 76

but thx anyways :DDD

.close

[obtuse pebble]

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[cloud belfry]

wlc

good luck with math

[obtuse pebble]

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[shy hawk]

does anyone know how to find the inverse of y=xsin(x) so I can integrate it from the y-axis

[cloud belfry]

standard process doesnt work?

show work

[zealous haven]

why do you want to integrate over the y-axis

[shy hawk]

because i know the limits for it for the bounded area

$$ but the with the x-axis, i tried doing pi/2=xsin(x) $$

[warm shale]

Hector

[zealous haven]

yeah

you can't solve it algebraically but you can at the very least approximate it

whereas it's just plain impossible to get the inverse of y=xsinx

[zealous haven]

i'm assuming you're allowed to use a graphing calculator to help with this or something

[obtuse pebble]

@shy hawk Has your question been resolved?

[shy hawk]

no, were not

[obtuse pebble]

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[hollow parrot]

I need help proving the following: Let $A$ be a commutative noetherian ring, $M$ a maximal ideal and $Q$ a $M$-primary ideal ($\sqrt{Q} = M$). Prove that there exists $n\ \in \mathbb N$ such that $M^n\subseteq Q \subseteq M$.

[warm shale]

Eduardo

[obtuse pebble]

Please don't occupy multiple help channels.

[hollow parrot]

.close

[obtuse pebble]

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[haughty coyote]

#groups-rings-fields

[obtuse pebble]

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[steady rose]

hello

[steady rose]

So i have a problem

i really like maths

its like poetry of logical reasoning

quote by einestine

but the thing is

i always end up scoring less

and feeling bored, and tired of repetetive problems

ending up having a feeling of not even wanting to do it

any tips?

yes i do use slow tempo music while doing maths, otherwise my mind is too empty and i can't focus

one of my problems

for eg. just today i had an exam on compound interest

on of the easiest topics right

but i scored 75% ONLY

i got all the calculations wrong

and when time was running out i got even MORE nervous and

ultimately just scored badly

same happens in like school exams

this was a private tutor's exam btw

i loose my nerve in the hall, and stuff i have done 100s of times seem new

[obtuse pebble]

@steady rose Has your question been resolved?

[steady rose]

<@&286206848099549185>

ik it isn't really maths but can you guys help me still

[obtuse pebble]

@steady rose Has your question been resolved?

[obtuse pebble]

@steady rose Has your question been resolved?

[obtuse pebble]

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[unreal vessel]

can someone check if this is correct?

[unreal vessel]

because when I put it in my calculator it gives me 4.934

nvm

.close

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[obtuse pebble]

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[stoic shell]

apply chain rule and plug in x=2

[high lily]

g'(x) isn't 2

[stoic shell]

Well, the chain rule application is correct and you have all the other values

[high lily]

you have y' = f'(g(x))g'(x)
to find the value when x = 2
sub in x = 2

[stoic shell]

nice

[obtuse pebble]

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[west willow]

Hi

[west willow]

I want to find the coordinates of C. Wich way is the correct one?

If we only loot at C

[stoic shell]

Use the rectuangular prism they give you. With points H, A, and B, you should be able to find the dimensions of the prism

[west willow]

So there is no way to get c without the other coordinates?

[stoic shell]

No, not by looking at the picture directly. It's 3d on a 2d page

[west willow]

Okey ty

.close

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[novel ermine]

i need some help

[muted shuttle]

is this a assignment/test?

[novel ermine]

assignment

[timid silo]

$since theyre similar its \frac{12}{7}=\frac{9}{w}$

[warm shale]

Prime Minister

[timid silo]

what the fuck

[dark stirrup]

Only surround dollar signs around the actual math expressions, not the plain text

[timid silo]

sorry im a little slow

[novel ermine]

so is that the answwer

[timid silo]

you get $12w=63$ therefore $w=\frac{21}{4}$

[warm shale]

Prime Minister

[timid silo]

because they want simplest form

[novel ermine]

.close

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[ebon sand]

how do i do this?

wait i forgot

i have to differentiate

erm

so i got 16x*-3+2

do i differentiate again?

[gusty tartan]

No need, just equate it to 0 and solve

[ebon sand]

oh

thats what i did

i got x = 2

but i have no idea what to do with that info

do i plug it back into the equation

the original equation?

to get y?

[bold bane]

Yes.

[ebon sand]

sweet

is that the entire answer?

also when do u have to differentiate twice

[bold bane]

When determing concavity.

[ebon sand]

thx ok

i expanded the brackets and differentiated each term individually

is it weird that its 5 marks or am i missing somehting?

like then i simplified fractions

idk how to do this tho

[snow dawn]

What have you got down so far?

[gusty tartan]

It asks for the COORDS, you have the x value , sub it into the original equation and get a y-value

Ahh didn’t see it was already answered , dumbass phone

[ebon sand]

hehe xd thx anyway

[ebon sand]

3/2x^2-3x+2

is the gradient function

[snow dawn]

Ooo, I think you're close.

Let me double check, after deriving the equation I got this answer:

$\frac{dy}{dx}\left(\frac{(x^2+4)(x-3)}{2x}\right)=\frac{6}{x^2}+x-\frac{3}{2}$

[warm shale]

@snow dawn

[snow dawn]

I can guide you through the steps?

[ebon sand]

wait which one r u doing

a or b

[snow dawn]

6(a)

[ebon sand]

oh yh ofc

erm yh

pls

i thought u its j basic differntiation

[snow dawn]

Okay... bare with me

[ebon sand]

ohhh

wait sry

acc

i see my silly mistake

i forgot to differentiate

3x^2

no

thats still not right 🤦‍♂️

[ebon sand]

erm yh ok dyu mind going on? xd

[snow dawn]

One sec this is going to get messy

I'm going to work backwards as opposed to work forwards

[warm shale]

@snow dawn

@snow dawn

[snow dawn]

Are you able to work from that?

[warm shale]

@snow dawn

[ebon sand]

woah what

quotient rule

ive never learnt that before

[ebon sand]

this line looks crazy haha

[snow dawn]

Okay

You haven't been introduced to the quotient rule yet...

Generally speaking,

The rule:
$\frac{d}{dx}\left(\frac{u}{v}\right)\equiv
\frac{-\frac{dv}{dx}u+\frac{du}{dx}v}
{v^2}$

[warm shale]

@snow dawn

[snow dawn]

It looks a bit intimidating at first, but don't fret too much.

Do you know where I obtained u and v from?

[ebon sand]

erm nope

my teacher dosent expect me to use

quotient rule

like we only learn that next yr

im rly tired so am gna stop maths but thx

[snow dawn]

It's really useful the quotient rule as it allows for the differentiation of fractions.

But just note for future reference, in a simpler form:

$\left(\frac{f}{g}\right)'=\frac{f'\cdot g-g'\cdot f}{g^2}$

[warm shale]

@snow dawn

[snow dawn]

Sorry I can't be of much further help

[obtuse pebble]

@ebon sand Has your question been resolved?

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[ebon sand]

.close

[obtuse pebble]

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[west willow]

Why is that wrong

[fossil crag]

Because you have to do - after - before +

[west willow]

I know the right way

bur

[west willow]

why have to?

[fossil crag]

Same reason why 2-1+1 is 2 and not 0

Order of operations

[west willow]

ok

[fossil crag]

When you only have - and +, do left to right

Easiest way to never screw up

[stone spoke]

you can change all of the (-)'s to (+)'s by distributing the negative to each component and then just add all together in any order

[west willow]

I see

[fossil crag]

Works too

[west willow]

ty

.close

[obtuse pebble]

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[obtuse pebble]

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[scenic fjord]

May I know how to differentiate this? The answer given is at the below

[fervent meadow]

The stat mech experience lol

It's nasty but it is just product/chain rule in the end

Using properties of ln will make things easier tho. Since there's a fraction in the equation for Z you can split it into two separate terms when you take the natural log since ln(a/b) = ln(a) - ln(b)

[scenic fjord]

Like this?

[scenic fjord]

But why my exp(-Bhw), is the answer given lacking out of negative ?

[fervent meadow]

Exactly

[fervent meadow]

The energy is defined as -d(lnZ)/dB

[scenic fjord]

I mean the negative in the bracket

[fervent meadow]

You're missing a step when differentiating the second term

[scenic fjord]

Which step?

[fervent meadow]

When you're doing the chain rule for the second term, you'll have an extra factor of exp(-Bhw)

$$\frac{-1}{1 - \exp(-\beta \hbar \omega)} \times \hbar \omega \exp(-\beta \hbar \omega) $$

[warm shale]

thedude365

[scenic fjord]

Yea, I just realised it. Let me try with others

Is this correct?

[scenic fjord]

Okay thanks

[obtuse pebble]

@scenic fjord Has your question been resolved?

[fervent meadow]

Yeah that looks right

[obtuse pebble]

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[scenic fjord]

.reopen

[obtuse pebble]

✅

[scenic fjord]

Thanks

.close

[obtuse pebble]

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[hearty pendant]

does taylor expansion works over finite limit for which curve almost resemble the integrand given to us

[hearty pendant]

and thus i can expand that function in polynomial

so can i do like this?

[tardy epoch]

Do you have an actual problem

[obtuse pebble]

@hearty pendant Has your question been resolved?

[hearty pendant]

no

its mind brew

[fallow pasture]

One edge of a cube coincides with a radius of a sphere. If the radius of
the sphere is R, find the volume of that portion of the cube which lies outside
the sphere.

is my drawing right?

[hearty pendant]

i think one face must be slightly outside

[fallow pasture]

can you draw it?

[hearty pendant]

i can but do u agree with me

i made that prposition by ur asking only

[fallow pasture]

yeah

[hearty pendant]

ohk wait i will try

[fallow pasture]

thank you

[hearty pendant]

idk how use 3d drawing software

my claim is the point as centre of sphere has to giving us three radii of circle and at same time also those side will be sides of cube

and we know all side are mutually perpendicular

[fallow pasture]

I get the right answer now

thank you

[hearty pendant]

so the problem can also be simplified as finding the volume in first octant

[hearty pendant]

ohk

[hearty pendant]

i see u used that one octant analogy before me saying

by didviding by 8

nice question tho

was this in ur course book tho?

[obtuse pebble]

@hearty pendant Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

What does this question mark mean, is there a symbol there which isn't formatted for my laptop?

[royal basin]

yeah it's a technical fuckup of some kind

[sage geode]

Unless the question mark is actually defined somewhere, then, yeah, probably the original symbol isn't formatted

[royal basin]

you should tell your teacher that the symbol is not displaying properly

and abandon this problem in the meantime

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

@royal basin@sage geodeThanks

.close

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

Wait

Why do they use ln rather than log base 10

Does it matter?

[sage geode]

Not really

I would have picked logarithm base 2 actually

[high lily]

unless your calc has log base 2
you'd have to use change of base anyway

[timid silo]

Alrght thanks. I must've made a mistake because I used log base 10 and got something different

[high lily]

show your work / what you're putting into the calc

[timid silo]

I should send a picture?

[timid silo]

bcs thats what theyve done for 100s of years

before they had calculators

[high lily]

yeh, helps us spot your issue, so you can avoid that in the future

[timid silo]

it just became standard practice

[timid silo]

Gotchu

Part a

@high lily

[high lily]

you had 1700 instead of 17000

[timid silo]

Oh my

Ok thanks

Silly me

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

just want some clarification, but for the following schematic

[timid silo]

would you describe the inputs of the primitive gate G3 to be that of w1 and C, or w1 and E?

[royal basin]

w1 and E obviously

[timid silo]

is the white dot meant to represent the wire connecting E to G3 or G2 to G3?

unsure on how you're meant to determine that

[obtuse pebble]

@timid silo Has your question been resolved?

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[timid silo]

Guys? How do I show that g 1-1 just by knowing this : f(g(x))= -3x

[royal basin]

!xy

[obtuse pebble]

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

[frigid venture]

i need help in differential equations

[royal basin]

!occupied

[obtuse pebble]

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

[spice citrus]

g doesn't have to be surjective though if f(g(x)) = -3x

It does have to be injective

[timid silo]

It's says : given the functions f,g: R->R for which this is true (f°g)(x)=-3x

And it asks to prove that g is 1-1

[spice citrus]

nvm, that's 1-1, got confused with bijective

You can prove it by contradiction, what happens if g is not 1-1

[timid silo]

Hm

Yeh I suck at those

Idk how U can do that

[royal basin]

suppose there existed two different x and y such that g(x)=g(y)

then f(g(x)) would also be equal to f(g(y))

[timid silo]

G(x)= G(y)
=> F(g(x))= F(g(y)) how do you do that if you don't know that f is 1-1

[royal basin]

you don't need to know f is 1-1 for that

you just need to know it's a function

it's for the REVERSE implication that you would need to know f is 1-1

also by the way don't mix upper and lower case

[timid silo]

My phone does this on its own

Sorry

Let me think this through for a bit

Ok nvm I got it

Yea it works with contradiction thank you

How do I close this channel

[tepid sapphire]

.close

[timid silo]

.close

[obtuse pebble]

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[hoary hatch]

what are the values of n so it satisfies : 2010^n +2011^n + 1432^n = 0(mod 7 )

[obtuse pebble]

@hoary hatch Has your question been resolved?

[spice citrus]

You should first reduce the bases mod 7

[hoary hatch]

yup it becomes ; 1 + 2^n + 2^2n = 0 (mod 7 )

[spice citrus]

Have you heard of fermat's little theorem?

[hoary hatch]

no i havnt yet ?

whats that ?

[spice citrus]

In this context it says that 2^6 = 1 mod 7

(you can verify this by hand as well)

because 7 is prime and 6 = 7 - 1

[hoary hatch]

well yea u mean it becomes : 2^n + 2^2n = 6 (mod 7 ) ?

[spice citrus]

That's true, but not what I meant, what happens if you replace n with n + 6 and simplify

[hoary hatch]

after expanding i cant see anything that can help me

[spice citrus]

maybe try plugging in some values for n as well, you will see a pattern if you plug in enough (0, 1, 2, ..., 8 should definitely be enough)

[hoary hatch]

okkk thank you for the hints brother

[obtuse pebble]

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[obtuse pebble]

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[maiden bone]

How the heck is 2pi(radians) = 360 degrees

i mean aint pie covers whole circle and circle == 360 degrees?

[rocky shale]

no

pi = 180 degrees

[final thunder]

One radian is the angle subtended by an arc that is equal in length to the radius, and the circumference is 2pi*radius.

[warm canopy]

https://tenor.com/view/radians-math-angles-circle-pi-gif-20830003

[maiden bone]

wait
how is that small portion is pi ?
i mean aint pi a ratio of diameter of any circle to its circumference
see this image

[maiden bone]

the 2nd image

[royal basin]

pi isn't the small red portion

it's the entire green stuff

half of the circle

the red is pi-3

[maiden bone]

oh ohk but still in this picture its clear that pi is whole circle

i guess i am mixing up 2 diff topics?

[royal basin]

pi times the DIAMETER is the whole circle

pi times the RADIUS is half

[obsidian isle]

Tau moment

[maiden bone]

really >?

its confusing
oh okay see
it takes 3.14......... diameters to cover whole circle , right ?
and whole circle (rotation) equals 360degress , right
then pi should be 360 degrees
hence proved

[royal basin]

"hence proved" feels like you are trying to attack the people trying to help you.

maybe you don't intend it but that's how it reads tone-wise.

anyway

radians are about covering the circle with RADII and not with diameters!

[maiden bone]

nah sorry i didnt mean that

[maiden bone]

ahh they didn't tell me this

[royal basin]

the gif showed it clearly.

[maiden bone]

okay suppose if the idea of radians is to cover the circle with diameter the pi = 360 degree
, right ?

[royal basin]

...

yes technically, but this is only relevant if you really really want pi to be equal to 360°

[maiden bone]

oh yeah lol
i didn't notice that

[royal basin]

if you do that, NOBODY will understand you!

maybe you like it when nobody understands you. idk

[maiden bone]

yeah cause there is no definite term for pi

[royal basin]

...

💀

[maiden bone]

oh lol

i mean no definite value

idk what i am saying

[polar fossil]

there is though

[balmy mortar]

No, what are you saying

[maiden bone]

anyways my doubt is cleared

[maiden bone]

i need to take a deep breathe

[balmy mortar]

This is nonsense.

[wooden cipher]

Are you talking about rational value?

[maiden bone]

please explain

[balmy mortar]

It is better if you do. About what you meant

[maiden bone]

@royal basin Thank you 🍻

[balmy mortar]

"definite value", what does this mean

[maiden bone]

i just meant pi is irrational so there is not definite value

idk if definite is a right word to use

[balmy mortar]

it is not

irrational is.

[maiden bone]

oh lol

my bad

[royal basin]

it is not the right word to use

[maiden bone]

yeah sorry
my bad

.close

[obtuse pebble]

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[restive narwhal]

(sets), is Z+ the same as N - {0}

[kind hawk]

depends on who you ask

[restive narwhal]

damn

cos i was tryna do this question and was confused on why they're saying N - 0 in the answer

[timid silo]

(personal opinion) always use $\Z_{\ge0}$ and $\Z_{>0}$ instead of $\N$ to avoid that ambugity crap

[restive narwhal]

aight

[brave bramble]

"Positive integers" never includes 0

We write "non-negative integers" when we want 0 and all positive integers

[obtuse pebble]

@restive narwhal Has your question been resolved?

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[urban drum]

is (7) a prime ideal in Z/15Z?

[royal basin]

what do you think

especially given that 7*13 = 1

[urban drum]

yes because (7) {0,7,14} and since every a*b in (7) has either a or b in (7) but my solution says it's not a prime ideal...

[robust bloom]

Yes Yes Yes
YesYes Yes Yes Yes
Yes YesYes Yes Yes
Yes Yes Yes

[royal basin]

(7) is not {0, 7, 14} !

[urban drum]

what is it then?

[royal basin]

what do you think it is, given 7 is a unit?

what ideal does a unit generate?

[urban drum]

7 is the set {...-23, -8, 7, 22, ...}?

[royal basin]

no, that is not what i am talking about.

forget about Z. we are in Z/15Z.

first do you know what the word "unit" means?

Y/N

[robust bloom]

||he will prolly say 1||

[royal basin]

shush.

if he says the wrong thing, i will correct him.

[robust bloom]

You are so great at explaining teach me too. I am mesmerized... not joking, fr fr, no cap

[urban drum]

no sorry im german speaking..

[royal basin]

a unit is an element which has a multiplicative inverse.

[urban drum]

ah yes

[royal basin]

what ideal is generated by such an element?

[urban drum]

the whole ring?

[royal basin]

yup

[urban drum]

but I don't really get the idea of how subsets generate an ideal. We just have the definition with the intersection over every ideal containing the subset. Is there an easier way to think about this?

[royal basin]

given a ring R and a subset S in it, the ideal (S) is the set of all elements of R that can be expressed as r_1 s_1 + r_2 s_2 + ... + r_n s_n, where the r_i belong to R and the s_i to S

if S = {a}, i.e. if S is a singleton, then (S) = (a) = {ra | r in R}

i.e. (a) is the set of all multiples of a

[robust bloom]

pls welp me ann when u free

I am not very good at this art

[royal basin]

!noadvert

[obtuse pebble]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[robust bloom]

?

[robust bloom]

huh?

just asked for help anyways ._.

i wont disturb bye

[urban drum]

so like (3) in Z/15 would the be {0,3,6,9,12}? and then be prime ideal

[royal basin]

yes

[urban drum]

so for rings of the form Z/nZ we have that every element a with gcd(a,n) = 1 is a unit and therefore (a) = Z/nZ and if gcd(a,n) is not 1 then it is possible that it is an not trivial ideal. Is that a way to classify this?

[obtuse pebble]

@urban drum Has your question been resolved?

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[fallow widget]

I can't remember what next steps to take

I'm trying to prove that my 2nd step can become the induction step, right?

[dense badger]

Yes, but be careful with the wording. You could put something like "suppose it is true for n=p, and let's prove that it is also true for n=p+1". Then, organize the last part a little so that the process is clear.

[tepid sapphire]

You seem to be doing well so far, you replaced $(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots 1-\frac{1}{p^2}$ for $\frac{p+1}{2p}$, which is valid since it is your inductive hypotheses. The next step should be to derive an apparent equality from $\frac{p+1}{2p}\cdot(1-\frac{1}{(p+1)^2})=\frac{p+2}{2p+2}$.

[warm shale]

Crystopher

[obtuse pebble]

@fallow widget Has your question been resolved?

[fallow widget]

It's not going well..

It keep going wrong somewhere

[dense badger]

from the second step you have to:

$\frac{p+1}{2p} \cdot \frac{p^2+2p}{(p+1)^2}=\frac{p+1}{2p} \cdot \frac{p(p+2)}{(p+1)^2}=\frac{p+2}{2(p+1)}$

[warm shale]

José

[dense badger]

remember that you must use equalities, not implications (these are used only with propositions)

[fallow widget]

I can’t get it right

Please..

@past sand Are you here ?

[dense badger]

check that

[fallow widget]

I have

But you aren't fully showing how you solve it

Why are you factoring out p for example?

[warm shale]

José

[fallow widget]

NO WAY IS IT

:p

But what is the purpose of factoring out p?

[plucky sentinel]

I can help

[dense badger]

is for simplify with the other denominator

[plucky sentinel]

This is really easy broskie

[fallow widget]

help me siski

[plucky sentinel]

Ok

First do the bracket one

[fallow widget]

WAIT

i see it

i cancel out 2

and p+1

[dense badger]

yep

[plucky sentinel]

No

[fallow widget]

Jesus i seriously have to go to an optician

[plucky sentinel]

(P^2 + 2P = P(P + 2)\

(P^2 + 2P - P(P + 2) = 0)

Factor out a common factor of (P):

(P(P + 2) - P(P + 2) = 0)

Simplify:

(0 = 0)

So, any value of (P) satisfies the equation. My bad for the confusion earlier.

[warm shale]

adrita

\(P^2 + 2P = P(P + 2)\

\(P^2 + 2P - P(P + 2) = 0\)

Factor out a common factor of \(P\):

\(P(P + 2) - P(P + 2) = 0\)

Simplify:

\(0 = 0\)

So, any value of \(P\) satisfies the equation. My bad for the confusion earlier.
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.53 
     
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l
ocal/texlive/2023/texmf-dist/fonts/enc/dvips/lm/lm-mathit.enc}{/usr/local/texli```

[fallow widget]

tysm that makes sense

.close

[obtuse pebble]

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[graceful marten]

I dont think the resistances are drawn properly?

Also how do I calcualate resitance at TA and TB
(Ik it will be the same answer but I want to calcualate it ty < 3 ]

[obtuse pebble]

@graceful marten Has your question been resolved?

[graceful marten]

<@&286206848099549185>

[graceful marten]

https://media.discordapp.net/attachments/782492147713572878/865482024100691998/4552-sus-dance.gif

[brisk oracle]

I'd write newton's second law separately for the lorry and trailer

50000 - 4000 - Tb = 1800 * 5 => Tb = 37000 N
Ta - 10000 = 1800 * 3 * 5 => Ta = 37000 N

[graceful marten]

Ah this makes a lot of sense @brisk oracle

Thank you

[graceful marten]

The lorry and trailer experiences resistances of 4000N and 10,000N respectively

Oh

Nvm im an idiot

Ty elya

.close

[obtuse pebble]

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[wheat juniper]

Hey, I have to solve this complex equation in C: 2z - 3iz̄ = -5 -i

[wheat juniper]

after I replace z with it's algebrical form z = a+ib, and develop I end up with 2a+2ib-3ia+3b = -5 - i

And now I don't know what to do anymore

[dark stirrup]

a and b are real numbers

So in the end, real components must match and imaginary components must match

[wheat juniper]

So 2a + 3b + 5 = -2ib+3ia-i?

[dark stirrup]

Not quite.

[wheat juniper]

hmm

Aren't they matching correctly?

[dark stirrup]

If I wrote a+bi=2-3i, I think you would know how to find a and b very easily

[wheat juniper]

a=2 and b = -3

[dark stirrup]

Yup. And you did that by first looking only at the real components and solving for a, then looking only at the imaginary components and solving for b

[wheat juniper]

Don't quite follow you

I get that we identified a and b

But how can I apply it on the other equation seeing there's so much stuff mixed in

[dark stirrup]

It may be easier if you wrote this in vector form $$(2a+3b, 2b-3a)=(-5,-1)$$

[warm shale]

SWR

[wheat juniper]

Vector forms?

[dark stirrup]

Complex numbers can equivalenty be written as ordered pairs

$a+bi\equiv(a,b)$

[warm shale]

SWR

[wheat juniper]

Still don't get it

Will look into it tomorrow with my teacher

I have a bit of a language barrier since I never use english with math

Thanks for the help though

[dark stirrup]

You are trying to solve the simultaneous equations $2a+3b=-5$ and $2b-3a=-1$.

[warm shale]

SWR

[wheat juniper]

let me see

hold on I don't see where the -1 comes from

[dark stirrup]

From the -i at the -5-i spot

[wheat juniper]

Right 2 real numbers can't equal a complex

well an imaginary one

Wait

So in this case b and a = 1

[dark stirrup]

Yup

Good job

[wheat juniper]

thanks man

👍

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[fallow widget]

Does anyone know of a good reference where I can learn about permutations and combinatorics ?

[civic sapphire]

I always recommend khanacademy.org

[obtuse pebble]

@fallow widget Has your question been resolved?

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[tardy epoch]

,tex .log rules

[warm shale]

riemann

[tardy epoch]

Memorize that table

Yes

[tardy epoch]

Use product

[tardy tusk]

Nah

See

When there is + between logs you put multiplication inside log

[mortal garden]

Yes

[tardy tusk]

Lol

[mortal garden]

LMAO

[tardy tusk]

no difference

exactly no difference

[obtuse pebble]

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[blazing bramble]

Do i multiply with a³-9

[blazing bramble]

Or only -9

[timid silo]

only -9

[blazing bramble]

Okk

Nase kl kai pali

[obtuse pebble]

@blazing bramble Has your question been resolved?

[blazing bramble]

Yes

[obtuse pebble]

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[strange oyster]

Where is ∞~?(complex infinity)
It`s like ∞i or what?
In google very small information about this

[forest sinew]

well theres multiple right

i mean, many

in what context do you mean?

[forest sinew]

sometimes certain functions shoot off in certain directions so you can talk about a certain infinity something goes towards

[forest sinew]

theres an infinity in every direction

[strange oyster]

hmmm

Like this?

[forest sinew]

think like

[strange oyster]

Complex numbers is a real with imaginary, but ∞ is like complex numbers, have a real numbers and imaginary? Or ∞~ is a imaginary infinity?

[forest sinew]

if i have $f(t) = 5 + it$

[warm shale]

jan Niku

[forest sinew]

where does this go as $t \to \infty$

[warm shale]

jan Niku

[forest sinew]

you could say that $|f| \to \infty$ as $t \to \infty$

[warm shale]

jan Niku

[forest sinew]

just think of infinity as a place really far away from the origin

if you go infinitely far away from the origin

in any direction

you get to infinity

[strange oyster]

yes but

∞~ like complex number or just imaginary infinity? Or complex and imaginary at one time?

[forest sinew]

infinity here means distance

distance from the origin

it could be $0 + \infty i$ or $\infty + 0 i$ or whatever

[warm shale]

jan Niku

[forest sinew]

someone is gonna have a problem with that notation lol

[strange oyster]

hmmmm

.close

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[chilly swan]

yo
if we have a^k = x mod t
and we have a and x and t
when do we know that this will cycle
for example(every fifth k)?

[obtuse pebble]

@chilly swan Has your question been resolved?

[tranquil sonnet]

Eulers theorem

[obtuse pebble]

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.close

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[regal jolt]

I need help with finding the area, my answer was 96 ab, but I’m not sure if it’s correct

[high lily]

is incorrect, show your work

[regal jolt]

I’m sorry, but I have no idea where to begin

[blazing forge]

hint: the shape is made up of shapes that you do know how to find the area of

[regal jolt]

So I divide into 2 shapes then just find the area by multiplying the length and width?

I’m not sure where to go afterwards

Could someone explain how

<@&286206848099549185>

[stoic shell]

what shapes did you get

[regal jolt]

Square and rectangle

[timid silo]

are a and b not given?

[stoic shell]

so calculate the areas

[hazy elk]

this isn't a square btw

[stoic shell]

not sure how you constructed a square and rectangle but go for it and i'll help if it goes off the rails

[hazy elk]

unless 3b = 2

if 3b = 2 then it is a square but at that point that's a bad question

[regal jolt]

Then is there other shapes I should use that are correct

[blazing forge]

yeah. something that is (except in really horrifying cases) true is that if you decompose the shape into disjoint smaller shapes and find the areas of the smaller shapes, the sum of all of the areas is the area of the big shape

[stoic shell]

2 rectangles would be a better way to put it

[regal jolt]

Would you illustrate the 2 rectangles since I’m not really seeing them properly ig

[stoic shell]

this work

[regal jolt]

So is the answer something like 16a + 6b

Because you multiply 2 and 3b(6b), then the other side you multiply 4 and 4a(16a). Am I wrong or did I misunderstand something?

@stoic shell

[stoic shell]

show calculations

[regal jolt]

@stoic shell <@&286206848099549185>

<@&286206848099549185>

[meager laurel]

It is $4(3a+3b)+6b=12a+18b$

[warm shale]

Natural7

[regal jolt]

Ah, I see

Ty

[obtuse pebble]

@regal jolt Has your question been resolved?

[obtuse pebble]

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[terse flame]

Given A = [ 3 1 ] and B = [ 3 5 ], find a matrix P such that AP = B
[ 2 1 ] [ 1 2 ]
The answer is [ 3 8 ]
[ -7 -19 ]
but im getting P = [ -8 -13 ]
[ 5 -8 ]

[terse flame]

my bad

[brave bramble]

What's your strategy?

[terse flame]

B is actually [ 2 5 ]
[ -1 -3 ]

my strat is to let P = B times the inverse o A

which leads me to the answer i got above

[brave bramble]

Which order?

BA^-1? Or A^(-1)B?

[terse flame]

but appaently

the first one

[brave bramble]

There's the problem

[terse flame]

aparently the correct ne is obatined by the second one

why?

[brave bramble]

You have:
AP = B

In order to isolate P, we want to multiply both sides by something. Note we have to choose to left-multiply or right-multiply

If you're right-multiplying by A^(-1), that gives:
APA^(-1) = BA^(-1)

Which doesn't isolate P like we'd want

[terse flame]

why though?

oh nvm

i see

.close

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[opaque lily]

h(x) = f(x) + g(x)
assuing f(x) and g(x) are periodic, would that mean that the period of h(x) is the period of g(x) * period of f(x) or can it be smalelr

[native inlet]

let f(x)=sinx and g(x)=cosx then h(x)=sinx+cosx, however the period of h(x) is 2pi which is not 2pi*2pi

[opaque lily]

ok i'll ad that f(x) and g(x) are both specifically the sin function too

oh wait that still applies ok

[native inlet]

If f(x) is period on k, and g(x) is periodic on n, then f(x)+g(x) is periodic on lcm(n,k) iirc :)

cos(x) is periodic on 2pi

sin(x) is periodic on 2pi

sin(x)+cos(x) is periodic on lcm(2pi,2pi)=2pi

[opaque lily]

oh yeah that makes sen

ok thanks

[native inlet]

[opaque lily]

.close

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[vivid lintel]

what i have

[hybrid swan]

compare the hypotenuse of both the smallest triangle and the biggest triangke

[vivid lintel]

but its in variable form

[hybrid swan]

do you see the hypotenuse of the big triangle as 7?

[vivid lintel]

yes

now what

[hybrid swan]

$\frac y7 = \frac2y$

[warm shale]

onion

[vivid lintel]

what is this

y+7

y=7

@hybrid swan

what is this

[hybrid swan]

these are the two triangles

because they’re similar, we can compare the values of the sides and the hypotenuses

[vivid lintel]

ok

[hybrid swan]

the first one is the bigger triangle, just to make sure

[vivid lintel]

ye

so is y=7

[hybrid swan]

no, not really

[vivid lintel]

ok now what d i do

[hybrid swan]

$\frac y7 = \frac2y$

[warm shale]

onion

[hybrid swan]

we have this

[vivid lintel]

yes

@hybrid swan

now waht

pls reply faster

[hybrid swan]

my apologies

we multiply crossing the numbers

[vivid lintel]

2y=14

y=7

[hybrid swan]

y.y = y^2

[vivid lintel]

oh yeah

y^2=14

[hybrid swan]

$y=\sqrt{14$

[warm shale]

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

[vivid lintel]

yes

i agre

[marsh geyser]

No

[vivid lintel]

what

[hybrid swan]

what’s wrong?

[marsh geyser]

|y|

Sqrt(y^2) = |y|