help-23

[lean otter]

radians?

[glass sonnet]

There's a nice formula for finding sector area

Radians express angles in terms of pi

[lean otter]

But I’m looking for the volume not area

[glass sonnet]

Rather than degrees

[lean otter]

Does it still apply?

[glass sonnet]

I was thinking about going down the volumes of revolution route

[lean otter]

yes that route

[glass sonnet]

So rotation the circle around the y axis to produce a sphere

So you must have come across radians?

You can't integrate without them

[lean otter]

There are no values given or angles

Just expressions

[glass sonnet]

Correct

You create an angle

Theta

[lean otter]

ohh

[glass sonnet]

Its just an arbitrary angle used for the formula to derive

[lean otter]

okayy I think I can solve it now thanks for the tip ^^

.close

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[kind dove]

how to prove if these groups are isomorphic ?

[safe radish]

@kind dove Has your question been resolved?

[feral linden]

They are not

An isomorphism is a bijection, and there isn’t bijection between them because the former is uncountable while the latter is countable

[kind dove]

I exactly thought about this

do you think this reason is enough ?

[feral linden]

Yes

[kind dove]

ok thanks.

.close

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[lean otter]

For a function g: A $ \rightarrow $ B where a = {a,b,c} and B = {0,1} which of the possible options for the function g is 1 to 1.

[lean otter]

For a function g: A $ \rightarrow $ B where a = {a,b,c} and B = {0,1} which of the possible options for the function g is 1 to 1.

$For$ a function g: A $\rightarrow$ B where a = {a,b,c} and B = {0,1} which of the possible options for the function g is 1 to 1.

[flat frigate]

Γιάννης

[lean otter]

I don’t know how to do this

[feral linden]

|A|>|B|

[lean otter]

This question is so confusing

[feral linden]

Told you

|A|>|B|

[lean otter]

I know this

[feral linden]

So?

[lean otter]

It cannot be 1 to 1?

A is domain

[feral linden]

Yeah

[lean otter]

B is codomain

What if a=0

b=1

And c=0

[feral linden]

You just said it

[lean otter]

Yes

[feral linden]

So?

[lean otter]

But it doesn’t make sense for the codomain to be smaller than the domain

[feral linden]

Problem solved

[lean otter]

No

The question makes no sense

[feral linden]

It’s the answer

No, it makes perfect sense

[lean otter]

Pls explain how this works

[feral linden]

I don’t know where you got the idea that |codomain| can’t <|domain|

[lean otter]

Give example please

[feral linden]

What?

[feral linden]

You just gave one

It’s not injective

[lean otter]

Oh

Simple

But how does that stop it from being 1 to 1

[feral linden]

…

[lean otter]

Oh

Sorry

[feral linden]

Because |A|>|B|

[lean otter]

I underst

[feral linden]

Good

[lean otter]

I’m now seeing textbook pages in my mind

I’m actually stupid to make this mistake

But then again

This is a very strange question?

[feral linden]

No

It’s a perfectly normal question

[lean otter]

But why is it 3 marks

?

[feral linden]

Idk

[lean otter]

It seems strange

[feral linden]

No

Again, it’s perfectly normal

[lean otter]

That it is 3 marks

Because the solution is so simple

[feral linden]

Strange to you. No one ever said |codomain| can’t <|domain|

You made that up

[lean otter]

No strange that it’s 3 marks

[lean otter]

This makes plenty sense

I was confusing myself

[feral linden]

I don’t know what I can say, good for you? Easy to make 3 marks?

[lean otter]

Hopefully that is what he wanted

I also feel that it’s strange wording to ask which of the possible options for the function, without listing any options

The main problem here is that I forgot what codomain was

[feral linden]

List them if you want

All 8

[lean otter]

Ok

What are they?

[feral linden]

Show that none of them is injective. Just so meaning less to me

[lean otter]

Oh wait

[feral linden]

I wouldn’t make such list

[lean otter]

I know what they are

[feral linden]

You can make it if you insist

[lean otter]

Thx

I think he wants the list

That explains why it’s 3 marks

Thx so much

[feral linden]

Np

[lean otter]

.close

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[stark grove]

if i integrate from A to D what area am i finding

[stark grove]

.close

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[cold quarry]

I honestly do not understand how to go on with this natural deduction

[cold quarry]

A is deducible from A in the OR ofc

Damn, I suppose there aren't a lot of people working with this ND system

<@&286206848099549185> Don't rly like to ping, but I've tried 3 times before 🙃

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[cold quarry]

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[elfin willow]

is independence the same as association

[elfin willow]

e.g for a chi squared hypothesis test null and alternative hypothesis

instead of no association and some association

can i write independent and dependent?

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@elfin willow Has your question been resolved?

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@elfin willow Has your question been resolved?

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@elfin willow Has your question been resolved?

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@elfin willow Has your question been resolved?

[ember stratus]

that's fine. as long as you elaborate on your results.

[safe radish]

@elfin willow Has your question been resolved?

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@elfin willow Has your question been resolved?

[elfin willow]

.close

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[gleaming yacht]

How to proof:
$$AP^2+PD^2+BP^2+PC^2=4R^2$$

[flat frigate]

Tanjiha

[thin token]

you said that you got it

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[supple ore]

@thin token im curious how did you prove it?

[thin token]

well i gave her a lead

even i wasn't sure if it will end up being proved

https://discordapp.com/channels/268882317391429632/486844875632017408/951517797421310003

see this is what we discussed

at the end ig she found out with the lead, but now idk

[supple ore]

I see

Well if they come back I think I've got a valid proof

[thin token]

uno reverse, now I'm curious @supple ore

[supple ore]

Damn you uno reversed me

So if we think about the circle in the form X^2 + Y^2 = R^2 then we can get the lengths of our parts like so

AP^2 = (Y + Sqrt(R^2 - X^2))^2
PD^2 = (Sqrt(R^2 - X^2) - Y)^2
PC^2 = (X + Sqrt(R^2 - Y^2))^2
BP^2 = (Sqrt(R^2 - Y^2) - X)^2

If you expand these and add them all together you will find it simplifies to 4R^2

Y being the y coordinate of BC and X being the x coordinate of AD

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[teal cape]

heya, function question:
f: R->R, f(x) = x-[x]

[winter pivot]

What is []

And what is your question

[feral linden]

Gauss floor function

[teal cape]

a) show that f is periodic with the function T0=1

i’m not sure if 0 is the correct answer

[winter pivot]

So [-3.3] = -4

[feral linden]

Yeah

[winter pivot]

Not a fan

[feral linden]

it’s periodic and the minimal period I can find is 1

[safe radish]

@teal cape Has your question been resolved?

[teal cape]

so the answer is [-3,.3] = -4?

oh forgot the other question
b) show all the possible sets

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[wind quest]

@fading kayak sorry for ping, can u see what i did wrong for a

[fading kayak]

You forget to apply f to the first answer

That is, you only wrote p(x) + q(x), not f(p(x) + q(x))

Same goes for second answer with p(x) and f(p(x)); q(x) and f(q(x))

Same in b

Well, you forgot to apply f everywhere 😅

[wind quest]

ohh bc of the lin transfo from P2 to P3?

[fading kayak]

Because of the application, yes

You don't know yet if it's a linear transformation

[wind quest]

i should only apply f for p(x) because of P2 to P3 right

[fading kayak]

No, you should apply it because that's what is asked

You're asked to write the result of f(p(x))

e.g. for the first question :
f(p(x) + q(x)) = x(p(x) + q(x)) = (a2+b2)x³ + (a1+b1)x² + (a0+b0)x

[wind quest]

i think i did that and then only applied f for p(x) in part 2

[fading kayak]

1. In a. second answer you forgot to update the second part (i.e. f(q(x))).
2. In b. first answer you forgot to mutiply a0 by c

And, finally, yes in fact f(cp(x)) = cf(p(x)) and yes f is a linear transformation

[wind quest]

applying f works for any function?

[fading kayak]

What do you mean ?

[wind quest]

nvm but i think im getting it, just confusing myself a bit

i think theres a with a that would make it a lin transformation

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[lean otter]

hey

[lean otter]

The probability of being a vegetarian and a smoker is .24. the probability of being a smoker alone is .4. given that someone is a smoker, what’s the probability that they’re also vegetarian?

not sure how i’d do this

.4 / .24 = .06

[broken yew]

draw a venn diagram perhaps

[lean otter]

‘let’

“let”

my logic i guess. don’t know if i’m even properly utilizing the vent diagram

*venn

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@lean otter Has your question been resolved?

[lean otter]

i think so yeah

.6

.close

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[granite cape]

Hey can someone help me with experimental and theoretical probability, I have a question I’m stuck on where it’s 12 snails racing, two dice are spun and whatever the sum of the dice the corresponding number of the snail will move forward, for example if it rolls a 9, the 9th snail will move forward. However I’m not sure how to find the experimental and theoretical probability for this

[past hearth]

whats the actual question

[granite cape]

?

[grizzled fossil]

“Experimental” prob just count ways each were obtained and divivide by total

For theoretical do a table for example

[granite cape]

What do you mean?

[grizzled fossil]

I mean what I said

[granite cape]

Can you clarify what you said?

[grizzled fossil]

Seeing how you aren’t clarifying on what you don’t understand, not really no

[granite cape]

.close

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[native jay]

How/why does the x appear/disappear in the numerators of these equations?

[marsh walrus]

? its just partial fractions

so nothing super magical is happening

although partial fractions are kind of magical

[native jay]

wait what even are partial fractions

[marsh walrus]

you know how to do this?:

$\frac{1}{x+3} + \frac{2}{x-4}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

make this into a single fraction

so its $\frac{(x-4)+2(x+3)}{(x+3)(x-4)} = \frac{3x+2}{x^2-x-12}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

so an x "appeared" in the numerator

partial fractions is the reverse process from the right hand side here

we know this can be split into a sum of fractions

each fraction will have a linear factor in the bottom

well

they wont necessarily be linear

but theyll be simpler

usually

[native jay]

I believe I'm just working with linear fractions for now

idk though my professor did not explain this

[marsh walrus]

so yea heres the question

[marsh walrus]

it sounds like magic, its just algebra

start here $\frac{3x+2}{x^2-x-12}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

heres how you start the process

ill get u started then u try on your own, if you want

and confirm it equal to what we started with

[native jay]

yeah that sounds good

[marsh walrus]

okay

so step number 1

you factor the bottom

its (x+4)(x-3)

[supple ore]

It will split into linear factors as long as the degree is no greater than 4

[marsh walrus]

you can create uhh

i mean it might not split at all

you can create a fourth degree with a quadratic denominator tho

anyways

thats not important

[supple ore]

You can always split it into linear factors

[marsh walrus]

you start here:

[supple ore]

As long as its less than 5th degree

[marsh walrus]

$\frac{3x+2}{x^2-x-12} = \frac{A}{x+3} + \frac{B}{x-4}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

now solve for A and B

consider re writing it this way

$\frac{3x+2}{(x+3)(x-4)} = \frac{A}{x+3} + \frac{B}{x-4}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

and multiplying through the denominator on the left-hand side (you get many may cancellations)

[native jay]

Is it going to need a little guesswork or is there a consistent formula or something?

I think i can do it with guesswork but if there's a consistent way to do it I think i'm a little confused

[marsh walrus]

so id do like

you multiply through by (x+3)(x-4), yea?

so we got to this

$3x+2 = A(x-4) + B(x+3)$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

@native jay your real tool to utilize here is that you have two functions of X

they are always equal

so for any x, this equality must be true

lets pick convenient values for x, then

since it must always be true

what happens at x=4?

[native jay]

B = 2

right?

[marsh walrus]

lets see here

so itd be

14=7B

sounds good 😄

whats the other convenient x value?

[native jay]

and then you would do the same thing to solve for A

so x = -3

A = 1

[marsh walrus]

alright

[native jay]

Ok I think i've got it

[marsh walrus]

so now you go back just a bit

and back to a million years ago when we started

https://media.discordapp.net/attachments/903481106819612714/952709508889706516/331558346333224964.png

[native jay]

Ok this is not as hard as I was trying to make it lol

ty for the help!!

[marsh walrus]

nah

np

[native jay]

.close

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[lean otter]

https://i.imgur.com/qkbQnQx.png

[lean otter]

not very sure what to do here

i know this

https://i.imgur.com/v9w2HT3.png

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[opal mason]

Hello Basically my teacher taught this math problem step by step with my classmates solution but I don’t really get the problem it’s self and would want help to solve this

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[tall prawn]

why do you have 2 rooms

[coarse plover]

Oh so

I made a mistake

And sent the wrong message

And it was pinned

So i tried to close it

But it didnt

So it was bugged or something

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[coarse plover]

.close

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[proud igloo]

how to solve for x?

[analog forum]

logarithms

just as the inverse of addition is subtraction, and the inverse of multiplication is division, the inverse of exponents is a logarithm

I'd recommend you do some reading yourself since they're a complicated topic but generally the way I convert it is that $base^{exponent} = solution <==> log_{base}(solution)=exponent$

[flat frigate]

FeO

[proud igloo]

oh okay thanks for the explanation, i think I understand it a bit more now

.close

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[sturdy aspen]

https://math.stackexchange.com/questions/2826123/odds-of-coming-out-ahead-in-roulette
I understood the problem, but my answer seems to be incorrect for the first part of the problem

[proud cipher]

hmm page 100

@sturdy aspen ping helper

its been 15 min

[safe radish]

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[gentle stirrup]

I wanna calculate the distance between a point and a line SEGMENT, i use vector projection for that and it works unless the projected point isnt on the line segment anymore. Since i want the distance between the Segment it should then calculate the distance between the Point and the end of the line segment

[gentle stirrup]

here is an example

the white point is the point i want the distance too

and the red point is the projected point

normally i calculate the distance between the white point and the red point

but here the distance would be to the left end to the line Segment

[safe radish]

@gentle stirrup Has your question been resolved?

[gentle stirrup]

ok figured it out

.close

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[compact parrot]

hi

[compact parrot]

my qn wasn't answered before can i ask again

[pulsar condor]

dont ask to ask

[compact parrot]

no one seems to be helping

[pulsar condor]

Well no one can help if there's no question.

[compact parrot]

I did ask before but yea

can u help me with this

[pulsar condor]

Yeah, that doesn't matter

help with what?

You've posted a random expression

[compact parrot]

the graph between Aul and lambda is a straight line given in my textbook

flu ,gu,gl are constants

how's it possible tho cuz it has a 1/x^2 dependency

[pulsar condor]

yeah it's not a line

[compact parrot]

textbook says otherwise tho and i asked my professor

he said it's the right graph

[pulsar condor]

Oh yeah, I can totally see your textbook

[compact parrot]

and asked me to find the answer on my own

[pulsar condor]

If you want to give the entire context of your question, go ahead

as for rn, I can only say it's an inverse square law, not a line

[compact parrot]

well this is the graph

beats me idk

the constants dont have an effect obviously do they

[pulsar condor]

[compact parrot]

nvm

thanks for ur time

[safe radish]

@compact parrot Has your question been resolved?

[safe radish]

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[rare vale]

I need help with this one

[rare vale]

im getting both +∞ and 0

[idle grove]

It's 5

Sqrt (5) i mean

[rare vale]

how?

[idle grove]

What I'm gonna say is not rigorous but helps

When x gets bigger, the one that has bigger exponent wins, in this case 5x^4+2x is 'dominated' by 5x^4 so when the x gets larger, 5x^4+2x can be approximated as 5x^4

[rare vale]

we dont have to think abt the 2x?

[idle grove]

No

That's why I'm saying this isn't rigorous

The formal way is to take commo factor x^4 and so 2x is only left with 2/x^2 and it's limit goes to 0

[rare vale]

I did something like this. I have no idea of what im doing

[rare vale]

what did I do wrong here?

[idle grove]

The common factor

Is wrong

You take common factor x^4 and not its inverse

[feral linden]

Divide by x^2 above and below

What?

No

Sqrt(5+2x^-3)

x^-2=sqrt(x^-4)

[rare vale]

Divide them by x^2 or x^4?

[feral linden]

I told you,x^2

But don’t you understand that x^-2=sqrt(x^-4)?

[rare vale]

Oh wait sorry. Ill send a pic

I saw this one. How did they factor out the x²?

[feral linden]

It didn’t exactly what I told you

[feral linden]

Can’t you just do this?

Why are you so stubborn

[rare vale]

im really trying but I can't get it right

sorry our teacher did not teach this to us

[feral linden]

You don’t understand that a=sqrt(a^2) when a is positive?

[rare vale]

I understand that

[feral linden]

$\sqrt{5x^4+2x}\frac{1}{x^2}=\sqrt{(5x^4+2x)\frac{1}{x^4}}=\sqrt{5+\frac{2}{x^3}}$

[flat frigate]

Cogwheels of the mind

[feral linden]

Jesus

[rare vale]

OOOOOOHHHHHH I SEE

[feral linden]

Good

[rare vale]

is this correct?

[rare vale]

thanks btw!!

.close

[safe radish]

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[supple nimbus]

How is riemann sum used in probability theory?
I know that Riemann Sum is used to approximate the area under the curve, I don't understand how it is used in probability theory

[pulsar condor]

Finding probabilities of continuous rv

Or approximating with a truncated sum

[safe radish]

@supple nimbus Has your question been resolved?

[supple nimbus]

Okay thanks! I get it now

[safe radish]

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[candid depot]

what are the steps of solving this derivative?

[opaque prism]

Differentiate ln(x^4 +1) and multiply with differentiation of x^4 +1

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[potent geyser]

There already has been some discussion regarding this previously, but no conclusion came, we basically had a false start.

[spice grove]

So um

the sum will be

$\frac{n(2m+n-1)}{2}$

[flat frigate]

What the hell am I doing here?

[spice grove]

This should be divisible by

$\frac{n(n+1)}{2}$

[flat frigate]

What the hell am I doing here?

[feral linden]

No

[spice grove]

What?

[feral linden]

It isn’t

IT ISN’T

[spice grove]

What isn't?

[feral linden]

Your statement

It isn’t divisible by that

…

[spice grove]

Oh it isn't I know.

We have to prove that.

[feral linden]

n=4,this 4 numbers:5,6,7,8

[spice grove]

There is some possible value.

[feral linden]

10 doesn’t divide 26

[spice grove]

I never told it did?

[feral linden]

You are saying I am blind?

[spice grove]

When did I tell 10 divides 26?

...

[spice grove]

Ah

[feral linden]

Look at what you said bro

[spice grove]

Light mode.

Remove.

Right should is not something I should have said.

Relax, I agree on that.

I never said 10 26 thingy.

[feral linden]

Just pointed out a mistake

[spice grove]

Also, 100% bruh wtf?

[feral linden]

I never said you shouldn’t say anything

Just found a mistake and I pointed it out, that’s all

[spice grove]

I said, I should not have said should. I should have tried the problem though, lol.

[spice grove]

Exactly, that's completely alright.

What we have to prove is

There are

$m \in Z and n \in N$ so that it is divisible, now better?

[flat frigate]

What the hell am I doing here?

[feral linden]

No

m and n are given

We prove the existence of a subset whose sum is divisible by that number

[spice grove]

For all m?

[feral linden]

I believe m should >=1

[spice grove]

Well in any case the first sum is

[spice grove]

[feral linden]

Because it’s false when m=0, a little error in the problem, we can fix that, just by assuming m>=1

[feral linden]

Like I said, it isn’t always that subset

I told you a counterexample already

[spice grove]

Forget the should be divisible thingy now.

[safe radish]

@potent geyser Has your question been resolved?

[clever sorrel]

I think I’ve got the answer

I’ve tried it through induction

Here’s the pattern

Can you do a proof by induction based on this pattern?

[potent geyser]

lemme see

[potent geyser]

bro you should read the chats above by the way, we had a light mode user and a guy that for a moment said that 10|26 argue with each other. They are very smart and kind individuals tho, which makes this funnier.

[feral linden]

Tried, didn’t work. those n numbers , there exists one divisible by n, say kn, if I use induction I can find a subset of the remaining n-1 numbers whose sum is divisible by 1+2+…+n-1, but sadly that sum divided by 1+2+…+n-1 isn’t necessarily k. Like {5,6,7,8,9} you have a subset of {6,7,8,9} whose sum is divisible by 10, which is {6,7,8,9}. Sum is 30, 30/10=3 doesn’t equal 5/5=1

[potent geyser]

you just rewrote the question tho

[feral linden]

Induction failed

[potent geyser]

This problem is very difficult.

after all

[feral linden]

Yeah of course they are

Competition questions

[potent geyser]

Yep

The competitors had 1 week to solve these questions tho

there were 8 of them

[spice grove]

Wait what really?

[potent geyser]

Yeah

[spice grove]

Woah.

[potent geyser]

I have already solved 6 of them

One is above my grade to attempt

This is the last one

This is a previous year prob

Not ongoing

[feral linden]

Maybe just send it too. one question we can’t solve or two questions we can’t solve. That makes no difference😂

[potent geyser]

Yeah, but this is not ongoing, so I have infinite time(theoretically) to solve it

But practically, I have a test in a few days

[feral linden]

What’s the another question you didn’t solve I mean

[potent geyser]

The circle one

[feral linden]

?

I answered you that one

[potent geyser]

You regarding two points and a 90 degree angle

[potent geyser]

Yes, and thank you again for that

But I was not supposed to, it was for the grade above

[feral linden]

I see

[potent geyser]

I am in 10th, it was for 11th grade students

[potent geyser]

I have no clue how to solve this

still

[feral linden]

Didn’t realize competitions are classified by grades, I thought all high school students attend together

[clever sorrel]

I'll try and elaborate on what I wrote

[potent geyser]

I am part of an 'elite' society of math enthusiast students in India, it is exclusive because they make us take an IQ test and only 145+ scorers can enter. It is shitty and discriminatory this practice. But they give us really cool problems and resources though. Part of this society is solving these questions on a weekly basis. So these are grade based

[feral linden]

I see

[potent geyser]

It is still a shitty society though

[potent geyser]

thnx a lot

[clever sorrel]

I'm so hungry I can't solve math anymore

bye temporarily

[potent geyser]

poverty is a prevalent issue in mathematicians after all

¯_(ツ)_/¯

[clever sorrel]

Yeah, I think I'll be giving up on this problem permanently for a while, good luck though

[potent geyser]

bro, pls dont

[clever sorrel]

Try doing some data crunching and try and find patterns though

[hushed mirage]

not tea

[safe radish]

@potent geyser Has your question been resolved?

[feral linden]

I am forming an idea, first we can assume that 1<=m<=(n-2)(n-1)/2+n-1 right? Because WLOG We let that sequence be contained in [1,n(n+1)/2-1]. It derives the general result.
Then we actually have the range of those sums, union of [(2m+k-1)k/2,(2m+2n-1-k)k/2]. Where k from 1 to n.So we want to find integers k and s such that (2m+k-1)k/2<=sn(n+1)/2<=(2m+2n-1-k)k/2. So we want to find an integer between ((2m+k-1)k)/n(n+1) and ((2m+2n-1-k)k)/(n(n+1)). Notice that the whole thing is contained in [(2m)/(n(n+1)),(2m+n-1)/(n+1)] except some gaps: [(2m+2n-1-k)k)/(n(n+1),((2m+k)(k+1))/(n(n+1))] , where k from 1 to n-1. So I need to show that number of integers [(2m)/(n(n+1)),(2m+n-1)/(n+1)] contains is greater than sum of numbers of integers [2m+2n-1-k)k)/(n(n+1),((2m+k)(k+1))/(n(n+1))] contain, where k from 1 to n-1. So I need to prove an inequality:gauss floor function [(n^2+(2m-1)n-2m)/(n(n+1)] is greater than Σgauss floor function [(2k^2+(2-2n)k+2m))/(n(n+1)]. Since the right hand side, those n-1 terms , the greatest terms appear when k=1 or n-1, so I am thinking proving
[(n^2+(2m-1)n-2m))/(n(n+1)]>(n-1)max{[(2m-2n+4)/(n(n+1)],[(2n+2m)/(n(n+1)]}

[tame raft]

this is trivial

the sum you want it to be divisible by is n(n+1)/2

so it's just gonna be a pigeonhole thing

[feral linden]

How?

[tame raft]

it's like

if you can't find one that's divisible by n(n+1)/2

then all the possible subsets of that set will sum to something that isn't n(n+1)/2 divisble

and that's gonna be a contradiction somehow bc of pigeonhole

like, there's gonna be 'too many' possible subsets somehow

[feral linden]

Why?

[tame raft]

if there exist no subsets such that the sum of the subset is divisible

then all the possible sums of the subsets are not divisible

[feral linden]

3 and 17 aren’t divisible by 10, but their sum 20 is.

[tame raft]

no

say you have subsets A, B, C, etc.

and say the sum of the elements in A is a

and the sum of the elements in B is b

etc.

then, if none of a, b, c are divisible by n(n+1)/2

then all of a, b, c are not divisible by n(n+1)/2

right

it's just logic

[feral linden]

Yeah

[tame raft]

i gtg, i'll take a shot at writing something up when i get back in a bit

[feral linden]

But you said sum to

What does it mean , sum to, like a+b+c?

[safe radish]

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[potent geyser]

.reopen

[safe radish]

✅

[safe radish]

@potent geyser Has your question been resolved?

[feral linden]

Need help

[tame raft]

oh yeah i forgot about this

i'll try and write something up

[safe radish]

@potent geyser Has your question been resolved?

[tame raft]

so considering modulo n(n+1)/2

suppose m is a modulo n(n+1)/2

then let us consider the possible sums of the subsets

the 1 element sums can be from a to a + n - 1, and all the intermediate values are possible

the 2 element sums can be from 2a + 1 to 2a + 2n - 3, and all the intermediate values are possible

the 3 element sums can be from 3a + 3 to 3a + 3n - 6

the 4 element sums can be from 4a + 6 to 4a + 4n - 10

etc.

and the single possible n element sum must be n(2m+n+1)/2

ok

[feral linden]

Yeah

[tame raft]

so assuming maximum overlap

like

merge all the intervals, assuming they all overlap

well hmm they don't have to all overlap but

well for a = 1, let's simplify it, there's probs a WLOG argument somewhere

then we have it's just:
from 1 to n
from 2 to 2n - 1
from 3 to 3n - 3
etc.

so clearly these overlap

so in the end it'll be from 1 to n(n+1)/2

so that covers all n(n+1)/2 possible values

bc it's modulo n(n+1)/2

[feral linden]

What do you mean merge?

[tame raft]

wait i think i screwed up a little

[tame raft]

union

ok fixed

[feral linden]

Why they all overlap? Some cases they don’t intersect at all

[tame raft]

ok sure

but you can probs WLOG or transform it

[feral linden]

No I can’t

[tame raft]

bet you can, let's see...

[feral linden]

And what is the special case you considered? When a=1?

[tame raft]

yeah

for a = 2:

from 1 to n + 1
from 2 to 2n + 1
from 3 to 3n
etc.
up to something

[feral linden]

That’s trivial right

Sum of all

[tame raft]

i wouldn't expect it to be especially trivial

it illustrates the general point

it turns out that indeed the sum of all of them is the one that works

[feral linden]

? Find a subset of {1,2,3,…,n} whose sum divides 1+2+…+n

[tame raft]

but overall you're covering n(n+1)/2 different values

[feral linden]

Don’t you think that subset is just the whole set?

How is that WLOG?

[tame raft]

look

i'm showing the general technique

this is what i mean by pigeonhole

it's gonna be a long finnicky argument

essentially you cover all the n(n+1)/2 'holes' modulo n(n+1)/2

[feral linden]

I still don’t know why you said it covers all values

[tame raft]

it's clearly going to turn out that it does

this is my intuition, i'll follow it through

gimme a sec

[feral linden]

Those closed intervals, they can have empty intersection

[tame raft]

they can't all have empty intersection

there's too many of them

they're too big

[feral linden]

{5,6,7,8}
Then [5,8],[11,15],[18,21]{26}

All empty intersection

[tame raft]

??

[feral linden]

?

Those 4 intervals

Each two have empty intersection

[tame raft]

but modulo, what, 12?

10?

[feral linden]

Oh I see

[tame raft]

it'll work out

i haven't worked it all the way through yet but this just smelled like pigeonhole from the start

gimme a sec

[feral linden]

You still need to tell me why it covers 0 , they are big, they overlap, but they can still be contained in [1,n(n+1)/2-1]

[tame raft]

you want to think in terms of maximum overlap and minimum unique values modulo n(n+1)/2

if there are at least n(n+1)/2 different unique values for the sums modulo n(n+1)/2, then by pigeonhole you're done

so how do we show this

that's the question

[feral linden]

😂 n(n+1)/2 values mod n(n+1)/2 are just all of them, 0 included

[tame raft]

yes, exactly

pigeonhole

[feral linden]

Then why even bother😂pigeonhole

You have all values then sure you have 0

[tame raft]

we need to show that we have all values lol

that's all

so

the 1-element subset sums cover n different unique values

[feral linden]

That’s what I was trying to ask you from the beginning

How?

[tame raft]

not sure lol

we have to find a lower bound on like total overlap? idk

ok so

[feral linden]

Seems doesn’t work to me

They overlap

Some values appear in several intervals

[tame raft]

well that has to be the case

[feral linden]

How will you rule out the worse case

[tame raft]

there's way too many of them for them not to do that

[tame raft]

something about the structure

[feral linden]

That is they are all contained in the longest one

[tame raft]

not sure yet

[feral linden]

And even the longest one is not long enough

(2n-2k)k/2

[tame raft]

well yeah ofc the longest one isn't long enough

the longest interval is just n different values right

[feral linden]

No

[tame raft]

if they're all contained within the first one

no wait i'm a fool

it'll be midpoint

yes

ok but

[feral linden]

Yeah k=[n/2]

[tame raft]

we can deliberately go from the 1-element sums

and add another element and break out of that interval, more deliberately

then add another one, break out of those

[feral linden]

Don’t understand

[tame raft]

we can approach it more systematically

ok lemme

fuck

gtg

5 mins

[feral linden]

Gl

[safe radish]

@potent geyser Has your question been resolved?

[safe radish]

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[slim pawn]

Not rlly sure how to differentiate c

[austere peak]

Do you know how the derivative of inverse trig function are found?

It’s done using implicit differention.

[slim pawn]

ya i just dont know what to do next

how can i put the csc on the other side

[austere peak]

So you know the derivative of arccsc(x)? If so then apply chain rule.

[slim pawn]

yes i know the derivative of csc

but not arccsc

is it the same

[austere peak]

No, I derive it for you.

Just realized it would be better to watch this instead. https://youtu.be/WFCWtycjKAk

[slim pawn]

thank you!!!!!

[austere peak]

Sorry, the explanation by me would have taken too long and wouldn’t be clear enough.

[slim pawn]

but this video doesnt have the natural log

that is what was confusing me

[austere peak]

This video is about finding the derivative of arccsc(x). After knowing that, you can do the above problem.

Using chain rule.

All that is left for you problem is multiply it by the derivative of arccsc(x).

[slim pawn]

oh sorry

.close

[safe radish]

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[lean otter]

need help with b

[pallid nymph]

did you try to evaluate the integral?

[lean otter]

yeah

i got it down to

integral from 0 to 2pi of sqrt(t^2 + 4)

using trig identities

@pallid nymph

[pallid nymph]

ok

[lean otter]

hnot sure what to do from here though

oh

i can just put it in calculator

nvm

[pallid nymph]

yea that just looks like itll be hard

[lean otter]

ok

i just plugged into calc

got 24.44

[pallid nymph]

square root + ln something

[pallid nymph]

looks good

[lean otter]

seem reasonable?

ok thanks

[pallid nymph]

yea

[safe radish]

@lean otter Has your question been resolved?

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[woeful vector]

How fo i work thid out

[woeful vector]

,rotate

[woeful vector]

5

[safe radish]

@woeful vector Has your question been resolved?

[fleet condor]

is the problem to find the x,y,z angles?

[woeful vector]

yep

and give reasons

@fleet condor

[fleet condor]

try finding the angles in this order

1 2 3 can be found using a line is 180 degrees, 4 uses some theorem i forgot the name of with the parallel lines, 5 is 180 degrees, 6 is the same idea as 4

[safe radish]

@woeful vector Has your question been resolved?

[woeful vector]

@fleet condor

only need to find

x y and z

[fleet condor]

yeah i rather gave some help for that

restating the problem doesnt help

[iron solar]

you can start with this

i forgot the theorem name but it makes those two congruent

then with reflexive you get this

and then all angles on a triangle =180 so 180-75-55 gets you 50

all angles on a line = 180 so 180-50-55 = 75 (x answer)

i dont remember this theorem either but you get this

again all angles on a line =180 so 180-30 is 150 (z answer)

then the final answers are x= 75° y= 55° and z= 150°

[hollow skiff]

Corresponding

That’s the name I believe

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[thin wraith]

.close

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[lean otter]

bruh

[slate galleon]

It's by 20%*2/5 = 8%

[lean otter]

hello is this room free

[slate galleon]

#❓how-to-get-help

[lean otter]

what

[slate galleon]

20% weighted with it being worth 2 of the 5 tests

[lean otter]

it was a 20$ increase

[slate galleon]

As an example, test scores: 9, 10, 11, and on the final test 10*1.2. The average of the test exams is 10, the final test is 20% more. The average of all tests is (9+10+11+2*12)/5 = 10.8, an 8% increase overall compared to the average of 10 from the tests

[lean otter]

2*12?

oh nvm

so the answer is 8%?

[slate galleon]

Yes

[lean otter]

hmm alright

so you just like plugged in numbers and see how it went?

[slate galleon]

I just did 20%*2/5 as I knew that the 20% was to be effectively spread across 5 exams where the 20% came from 2 of them/exam weighted as 2 exams. I knew that it would also be 0%<x<20% so if I tried 20%/(2/5) I'd know I was wrong too

[lean otter]

hmm alright

.close

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[crude shoal]

evaluate 1 * 100 + 2 * 99 + 3 * 98 + ... + 99 * 2 + 100 * 1.

I tried to do something like this: 2( 1^2 + 1 * 99 + 2^2 + 2 * 97 + ... + 50^2 + 50 * 1 ) but how do i continue?

[crude shoal]

<@&286206848099549185>

nvm

.close

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[lean otter]

What is root mean squared value?
RMS whatever.
How is it useful?
What's the major difference between that and the regular avg?

[lean otter]

Ping me please.

[opaque prism]

Average is used to get the central tendency of a given data set while RMS is used when random variables given in the data are negative and positive such as sinusoids

[lean otter]

Ah, makes sense.

[lean otter]

So, RMS is like a straight line which goes from the middle of a sine wave?

[opaque prism]

Kinda

[lean otter]

Yeah, as expected.

[lean otter]

Thank you so much!

[opaque prism]

you are welcome

[lean otter]

.close

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[vernal lark]

Help #2

[vernal lark]

Sry it sideway

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@vernal lark Has your question been resolved?

[vernal lark]

<@&286206848099549185>

Sry I couldn’t find the original

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@vernal lark Has your question been resolved?

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[raw jewel]

i have the first part

just dont know how to do the rest

[grizzled halo]

what is problem asking for

[raw jewel]

and on the top its 24x above the 6y on the first portion

combine like-terms

[grizzled halo]

you can just add all the same term then

[raw jewel]

alright thanks

.close

.close

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[elfin bough]

hey

[elfin bough]

quick question, what's the difference between scientific notation and e notation?

are they exactly the same thing?

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@elfin bough Has your question been resolved?

[safe radish]

@elfin bough Has your question been resolved?

[lean otter]

They are the same but in e notation e is used to condense the equation. E is equal to "x10 (times ten)
E.g: 5.94 e 19 (e notation) = 5.94x1019 (scientific notation)

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[sand pivot]

anyone help?

[safe radish]

Please don't occupy multiple help channels.

[scarlet sage]

you have to post the question

[sand pivot]

Sorry

[scarlet sage]

what do you think you have to do

[sand pivot]

idk

[scarlet sage]

any ideas?

[umbral meteor]

You have to solve it like a was an x

So how you solve this
x + 1 = 2

you subtract both sides by -1

x = 1

now what do we do for a + 5 >= 9?

[sand pivot]

Idk

I’m so confused

[umbral meteor]

What confuses you?

[sand pivot]

How to do it

[umbral meteor]

do you know how to solve x+3 = 2

[sand pivot]

5

Now it’s changed

[pulsar condor]

5+3=2?

[lean otter]

1+1=2 quick maths

[umbral meteor]

No it’s not 5 because if you move the 3 to the right side, its sign changes from +3 to -3

x + 3 = 2
x = 2-3
x = -1

—
So how you do you solve x + 5 >= 9?

[sand pivot]

I understand Now ty

Sorry wheat i meant is The whole wuedtion changed

Like it restarted my Thing

and gave me a new set of question

[umbral meteor]

Can you solve it?

[sand pivot]

I could

But now i can’t

<@&286206848099549185>

[scarlet sage]

minus 8 from bof sides

u want x on its own

so that u have x >= something

the way to do that is minus 8 from the left side

and if u do that, u have to minus 8 from the right side as well

[sand pivot]

Thank u

[scarlet sage]

no problem

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@sand pivot Has your question been resolved?

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[rigid kernel]

https://media.discordapp.net/attachments/490999623339474944/953861028909842462/20220316_234300.jpg?width=507&height=676

[rigid kernel]

How do I do number 8

I get a different answer, and since I can look at the answers in the back what i get is not true but I dont understand why

[light shoal]

,rotate

[drowsy moss]

what did you try?

[rigid kernel]

WYM

sry caps*

wym

[drowsy moss]

you said you got a different answer, what did you do to get your answer?

[rigid kernel]

Oh

so first i got P=4x+4y

then p is 600

600 = 4x+4y

150 = x + y

y = -x +150

used x=-b/2a

x= 75

but the answer says it shoud be 100

[drowsy moss]

why 4x and 4y?

[rigid kernel]

cause divided in two sections

4x and 4y

[drowsy moss]

ok, but that middle fence is shared by both areas, it isn't there twice

[rigid kernel]

oh ye I was doingthat as well

But I still dont get 100

I got 133.3

[drowsy moss]

draw a rectangle

[rigid kernel]

i made it P=4x+3y

[rigid kernel]

ok

[drowsy moss]

and divide it down the middle

[drowsy moss]

don't split it up, just cut it down the middle. Yeah!

so, ok, the perimeter is 4x+3y. Got it

[rigid kernel]

I get 600 = 4x +3y

3y = -4x+600

y = -3/4x + 200

but when i put that into x= -b/2a

like when i put it into area formula

[drowsy moss]

that's not a quadratic

[rigid kernel]

a = x(-3/4x+200)

[drowsy moss]

x=-b/2a finds the vertex of a quadratic

[rigid kernel]

ye

it is a quadratic

a(x)= x(-3/4x +200)

and i get a(x) = -3/4x^2 + 200x