help-4

[errant gyro]

you don't need to paint behind the mural

[quick pewter]

oh uh

[hazy pivot]

You have mural and frame+ mural

How do you find frame

[quick pewter]

(x +4)(x+9)

idk

do we subtract

[errant gyro]

yes

[quick pewter]

mmmm

okay

yep and im stuck hold on

x^2+13x+36 -(x^2 + 5x)

[errant gyro]

you have A. you have A+B. you want B.

[quick pewter]

sorry i couldnt find my numbers

[hazy pivot]

Wrong sign

[errant gyro]

check your signs!

[hazy pivot]

Bracket things when subtracting

[quick pewter]

heh

[errant gyro]

we really ought to have a "Mind the Signs" signboard

[quick pewter]

8x+36?

[errant gyro]

btw, when resending an answer, please avoid editing

[hazy pivot]

Yes

[errant gyro]

editing does not give a new notification and often we don't even know you edited something unless we were particularly looking for it

[hazy pivot]

And how much is this equal to?

[quick pewter]

oh okay sorry

[quick pewter]

uh wym

[hazy pivot]

Read the question again

[quick pewter]

100 ft squared

[hazy pivot]

Good

Now find x

[errant gyro]

(minor pedantic note: square feet for area)

[quick pewter]

10 😎

[hazy pivot]

No...

[errant gyro]

no way

[quick pewter]

uhm

scratch that

lmnao

16/7?

[hazy pivot]

No

Take it slow

8x + 64 is 100

+36*

[quick pewter]

oh whoops

i had 18x

x=8

[hazy pivot]

Good

We're done here then

[quick pewter]

dont we have to find how big it can be

[hazy pivot]

Well you have x

Find the area of the mural

[quick pewter]

.

64+40

104

ft^2

[hazy pivot]

Good

[quick pewter]

cant it only be 100 ft^2

not 104

[hazy pivot]

That's the frame

[errant gyro]

no. read the question a little bit more carefully

the area covered by the frame is 100sq. ft.

your mural technically has no size limit, except for the fact that it's bound by the size of the frame around it

[quick pewter]

man this is hard 🥹 i think i get it though

so C is just 104

ft ^2

[errant gyro]

yes. never forget units

[quick pewter]

okay can we do one last one

[errant gyro]

send it, don't ask to ask please.

[quick pewter]

ik revenue is like cost - profit

uhm

[errant gyro]

cost - profit?

[quick pewter]

i think i have a note on it somewhere

ill fact check

[errant gyro]

you mean profit - cost?

[quick pewter]

yes

[errant gyro]

you got it the other way round

profit is revenue - cost

revenue is raw earnings

[quick pewter]

do i even need that for this wuestion

[errant gyro]

not really

[quick pewter]

crud

[errant gyro]

you brought it up first so just thought to correct it

[quick pewter]

how do i start with A

[hazy pivot]

Well you're selling x products for y amount each

What's the total money you make

[quick pewter]

revenue is Profit-cost right

would i use that

[hazy pivot]

Don't worry about that yet

[errant gyro]

don't care about that

you're not even given cost

[hazy pivot]

You have 10 mangoes and you sell them for 5 each

How much do you make

[quick pewter]

50

[errant gyro]

also, once again, profit = revenue - cost

[quick pewter]

damn

(45-5x)(1.25x)

[errant gyro]

but ignore that for this question entirely

[quick pewter]

do i do that

[hazy pivot]

Yes

[quick pewter]

booya

[hazy pivot]

Now the next two are simple substitutions

[quick pewter]

so revenue is
56.25x - 6.25x^2

thats A

easy

[hazy pivot]

I do not feel like verifying that

So I'm just gonna trust you

[errant gyro]

,calc 45*1.25

[rocky lotus]

Result:

56.25

[errant gyro]

ok you're good

[quick pewter]

for C i just plug in 4.5 into this

which is 126.56? then for B i plug 4.5 into 1.25(X)?

[errant gyro]

units!

[quick pewter]

AHH

[errant gyro]

but the ideas are correct

[hazy pivot]

I keep forgetting I can just wolfram alpha to verify shit

[quick pewter]

126.56 dollars

[errant gyro]

$126.56 is faster

but yes

[quick pewter]

and B is $5.63

[hazy pivot]

,w 1.25*4.5

[rocky lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xavier_dragon

[errant gyro]

,calc 1.25*4.5

[rocky lotus]

Result:

5.625

[errant gyro]

ok, cool

[quick pewter]

oh heck yes thank you guys so much 🥹 have a good night

.close

[vale dock]

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[kind wave]

sorry just a quick question, if I have these six colors, pretend like each one is a different value, if I can have a mix of 10, or 20 total of these 6, how would I calculate that

(Please scroll down for a more mathematical question)

[kind wave]

basically whats the total combination of 6 different thing in a specified number that each unique thing can have multiples or none in

I honestly dont know how to word it

Im just tryna find an equation that could work for this so i can see what a reasonable cap would be for this

<@&286206848099549185> sorry for the ping lol but it does tell me to do it

[primal sluice]

What are you trying to do?

[kind wave]

I am planning a mixing paint feature for a game, Im using these six colors for it, how can i find out how many different colors you can make based off a certain set cap of how many total you can mix together

its kinda like a word problem in a sense

[primal sluice]

Ok

Is there a quantity option

Like how much of each paint

[kind wave]

any

[primal sluice]

Bc if there is then it’s basically infinite

[kind wave]

hmm ok how about 3

[primal sluice]

U can make any shade with a specific amount of each

[kind wave]

each

[primal sluice]

What I mean is

[kind wave]

I can work with 3 max of each paint

[primal sluice]

If you made a paint with 70% blue compared to 71% blue

You’d have a tiny difference

[kind wave]

yeah ik, its still a cap of 18 for all of them total though, so like 1 red and 19 blue without a 3 cap max per

but id prob do a 3 cap max

18 since thats the most possible

[primal sluice]

So you can add 18 portions of paintin total

[kind wave]

yes

and max 3 per

[primal sluice]

And no minimum requirement?

[kind wave]

atleast 1

but I mean obviously lol

[primal sluice]

Oh uea

[kind wave]

sorry but I really didnt expect it to take this long lol, its 11pm, do you know of maybe an online calculater I can use, or a resource

[primal sluice]

You’d have to calculate all possibilities for each quantity of 1-18 paints

[primal sluice]

Chat gpt can def help u with this one

[kind wave]

im not saying it will be exactly that, but im hoping there can be an equation

[kind wave]

id rather not use generative ai

thats the whole reason for this server no? if everyone used ai this server wouldnt have a use, well aslong as the ai is right

[primal sluice]

Can use combinatorics for each of the quantities

I think that’s the best way tbh

Still requires a big chunk of work

[kind wave]

im cooked bro😭

[primal sluice]

So for example

1 paint

6 possibilities

Oh wait it might not be that hard

Cuz then 2 is 6x6 possibilities

However if u pick the same paint again it won’t do anything

So 6x5 ig

[kind wave]

but you can also have the same mix just a higher value

how would I factor those out

[primal sluice]

Oh ok

Like as in 1 blue and 2 blues are different results?

[kind wave]

no

I mean as in

like how you said you could just have 3 red or 1 red and its still the same color but higher amount, im talking about in the sense you could also have 1 blue and 1 red, yet your still able to do 2 blue 2 red, how can i factor out repeating values

[primal sluice]

You’d have to calculate them separately

[kind wave]

https://tenor.com/view/am-i-cookedd-am-i-cooked-gif-7478847585523279700

this is gonna take a while huh

im just giving up ngl

[primal sluice]

There’s no shame in using ai for tedious tasks

[kind wave]

should I just reword it and see if someone can solve by the morning

[primal sluice]

Sure

[kind wave]

uhm btw will this be closed by the time I wake, if so then how will i see it

[primal sluice]

Idk if auto close is a thing

U can just request the response to be DMd to u

[kind wave]

alr ima reword then sleep

Lets say I have 6 different values, A, B, C, D, E, and F. If I were to combine these, with of a cap of K max repetition for each, and L for the max amount allowed in one group, how would I calculate the total ammount of different combinations, while excluding repetitive combinations, such as ABD, and AABBDD, or AAAA, and AA, so on and so forth

[next snow]

waht

if you want to exclude repetition you have to choose a different number of letters

[kind wave]

ok what should I write then

[next snow]

what's your total number of letters

[kind wave]

im tryna write something understandable before I sleep

[kind wave]

6

[next snow]

if you don't want repetition just use the multiplication principle

and how of those letters do you want to choose

[kind wave]

what

[next snow]

if it's 1 letter it'll just be 6

2 letters is 6x5

3 letters is 6x5x4

4 letters is 6x5x4x3

5 letters is 6x5x4x3x2

all 6 letters is 6x5x4x3x2x1

or 6!

5 letters is also 6! because 1 is a trivial element

this works because you want to choose (n - 1) letters

so that choosing all is avoided

[kind wave]

im trying to find an equation for the problem, I need to see what cap, and max is best for what im planning

[next snow]

wouldn't this just be equivalent to a letters and numbers combination problem

if you have a max to bound the number of shades iirc you can just do

(original color + its shades)

then proceed to multiplication principle

or you could interpret this differently

bound your number of colors so that it doesn't end up brown

and just say that a combination of colors leads to a new color

[next snow]

you can do this the same way i did above

by choosing (n - 1) of whatever

[kind wave]

dude im sorry but im too tired to understand this rn, if someone is able to write me an equation, it doesnt even have to be for 6 values, I might do 3 instead, I just need a way to calculate this, like I said I have an amount of input values, I apply a total amount cap to each of these for example cap 2 of each, and then a total max of them all combined. For example ABBC would be 4, B is capped out and cant have anymore inputted, this also had to ignore thing such as AABBCC, since in this scenario it would be equal to ABC, which means its the same value repeated, and wouldnt count towards the final count. These numbers are just examples, so im hoping there is an equation where I can input my own values for this to see which amount of each is best for me. Its 11:31 pm, gn, please dm if you figure this out or just ping me aslong as the channel stays open, ill be back on in less that 7 hours

[vale dock]

@kind wave Has your question been resolved?

[next snow]

then i think it would need to be a multiset

that is:

you want aabbcc

because normally abc does not repeat in sets

but in multisets they can

[vale dock]

@kind wave Has your question been resolved?

[vale dock]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[kind wave]

.reopen

[vale dock]

✅

[kind wave]

<@&286206848099549185> if someone is able to come up with an equation please do, sorry for the ping but its obviously being ignored

[ebon glade]

so you have 6 colors, for each color you have 4 options (dont use, use once, twice, thrice), so 4^6 total options

is that what you mean?

[vale dock]

@kind wave Has your question been resolved?

[vale dock]

@kind wave Has your question been resolved?

[vale dock]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[wet tundra]

hello

[flint phoenix]

yoo

Any questions today?

[jovial edge]

wassup

[analog herald]

hello

[hazy pivot]

Lmao

[wet tundra]

why do they use integral of a pdf to find the median for part D

[hazy pivot]

Just jump the guy before he can ask

[wet tundra]

cuz i wouldve thought it would be area of a cumulative distributiuon function

until m

[jovial edge]

I mean why not, they say line x=m devide f(x) into 2 equal areas

[wet tundra]

yes but

it doeznt make sense for me

[jovial edge]

what doesn't make sense exactly

[wet tundra]

is this wrong

[dark pewter]

it doesn't make sense to look at the area of the CDF

[wet tundra]

why

[dark pewter]

CDF is already the area of the PDF

why would we require finding the area of the area of the PDF?

remind you, that the PDF is given by f(x)

[wet tundra]

wait so wtf is a cdf for

i think i understand it now tho

ohh i get it now

[vale dock]

@wet tundra Has your question been resolved?

[vale dock]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[hoary salmon]

Please help with this question asap 😭🙏. How many ways can 4 identical croissants and 4 identical bagels be placed on a rotating tray?

[hoary salmon]

I also need help with a few more problems of a similar nature

[woeful rover]

Can the rotating tray hold all the bagels and croissants?

[hoary salmon]

Yes

[woeful rover]

Alright

You have 4 croissants and 4 bagels

8 items total

[hoary salmon]

Yes

[woeful rover]

We can choose 1 item from those 8 items to place as the first item

How many choices are we left with to place the second item?

[hoary salmon]

7

[woeful rover]

Yep

And after we place the second item, we're left with how many items for the third item?

[hoary salmon]

So 7! until the end, but how do I deal with the fact that they're identical?

[woeful rover]

8! actually

[hoary salmon]

Why is it 8? Isn't it rotating?

[woeful rover]

1 rotating tray can't carry all the 8 items?

[hoary salmon]

It can carry all 8, but wouldnt it rotate so that 8 combinations are the same???

So we'd be overcounting???

I've barely learned this and I'm still expected to solve this 😭

[woeful rover]

we're arranging on 1 tray

[hoary salmon]

Yes

[woeful rover]

So we're seeing the different permutations

Of that tray

[hoary salmon]

Ok

[woeful rover]

We have 8 items to arrange, so it'll arrange out to be 8! ways to arrange them on the tray

[hoary salmon]

Ok

[woeful rover]

Hmm

That's if we treat all the items as distinct choices

If we need to account for them being identical, then we'll consider 2 choices when we have an empty tray: either a croissant or a bagel

[hoary salmon]

Ok

[woeful rover]

Hmm lemme try drawing a tree

[hoary salmon]

Ok tysm

And the worksheet just gets harder from this question. We haven't even learned the definition of a combination 😭

[woeful rover]

Hmm

Alright so

Suppose we have 2 croissants and 2 bagels for now

[hoary salmon]

Ok

[woeful rover]

If we wanted to arrange them all as distinct objects, the total permutations is 4!

Now we have to account for the repeated croissants and bagels

[hoary salmon]

Yes

[woeful rover]

So we divide 4! by the permutations possible with the repeats

So 4!/(2! * 2!)

[hoary salmon]

Ok

[woeful rover]

It will be 6 different ways to arrange them

CCBB, CBBC, BBCC, CBCB, BCCB, BCBC

Now let's apply this to our bigger question

We have 8 items

4 bagels and 4 croissants

Total permutations = 8!

Permutations of croissants = 4!
Permutations of bagels = 4!

Different permutations = 8!/(4! * 4!) = 70 different ways to arrange the 4 croissants and 4 bagels

[hoary salmon]

I don't think that equals 1680

[woeful rover]

Why not?

[hardy coral]

on a rotating tray

[hoary salmon]

YES PLEASE HELP WITH THE ROTATING TRAY PART

[hardy coral]

Chill

[hoary salmon]

Also, 8765/43*2 is 1680? Lemme do the magh

[woeful rover]

Wait

[honest stone]

,w 8765/86

[rocky lotus]

Results provided by WolframAlpha

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Requested by finestperfectionism

[woeful rover]

My apologies.

[honest stone]

doesn’t look like it does

unless you meant (8765/43)*2

[woeful rover]

8!/(4!^2) = 70

[hoary salmon]

Ok. Also, I meant 8x7x6x5 I just used asterisks 😭

[honest stone]

oh… uh

maybe ‘x’s or × symbols would’ve been better

[woeful rover]

8 * 7 * 6 * 5

[honest stone]

,w (8•7•6•5)/86

[woeful rover]

Next time use spaces. It won't use italics this way

[rocky lotus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by finestperfectionism

[hardy coral]

I don't remember if there's a nicer way to do this but you can start by placing a bagel followed by two croissants, count it all with 5C2 = 10, remove one single repeated way, and then count the one last way that is alternating bagels and croissants

Total: 10

[hoary salmon]

Ok

[hardy coral]

There probably is a nicer way, I just can't think of it right now

[hoary salmon]

I have one last question. How many 4 letter words with no repeated letters and exactly 1 vowel can you make from the word "computer"

[woeful rover]

Wait so there's a gimmick to the rotating tray? It's not just a tray you can place the items on?

[hardy coral]

It just means rotations count as the same

[woeful rover]

So this example counts as a rotating tray?

[hardy coral]

CCBB and CBBC are the same

[woeful rover]

Ah got it

[hardy coral]

There's only two ways if you have 2 bagels and 2 croissants: either the bagels are next to each other, or they are not

[hardy coral]

There's no repeated letter in that word, so just choose 1 vowel out of 3, 3 consonants out of 5, and arrange them

[hoary salmon]

Ok ty

.close

[vale dock]

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[restive void]

Hello, I noticed that I am recurrently facing issues with representing equations of a variable in terms of another variable when the equation involved absolute values, square roots, and complex operations that restrict the set with which we are working - turning the given relation into one that is piecewise . . .

One question I am currently working on is the following (which I got from ChatGPT to practice such problems) . . . How'd I go about solving it as having had attempted solving it . . . I went through problems regarding restrictions and how to deal with getting rid of the operations and isolating y . . . Help much appreciated! (The question is like how to express y in terms of x)

[leaden kraken]

You can first try graphing it using Desmos to see what it looks like

Since you got it from ChatGPT, there’s no guarantee that the problems are correct, for example, it may not be possible to express y in terms of x here

[hardy coral]

This has at least two branches

[restive void]

I have another analogous problem where expressing it in terms of y is guaranteed, but it still has me confused

[hardy coral]

Also the absolute value is useless

[restive void]

This one

[turbid summit]

!nogpt

[vale dock]

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

[turbid summit]

but i advise you

get something like

|f(y)|=g(x) so that f(y)=-g(x) or +g(x) this gives you two equations of the similar kind to solve

or you could just square the whole thing

[hardy coral]

This also has two branches

[safe fulcrum]

yes, $\sqrt{3y + 9} - 3 = x - 5$ and $\sqrt{3y + 9} - 3 = -(x-5)$

[rocky lotus]

south

[safe fulcrum]

for $\sqrt{3y+9}-3\ge0$ and $\sqrt{3y+9}-3\le 0$ respectively

[rocky lotus]

south

[restive void]

And you're right, I asked ChatGPT to solve this one and it couldn't

That explains why I wasn't able to solve it

[lyric sundial]

!nogpt

[vale dock]

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

[turbid summit]

if you really have to use gpt, just use the web search function to gather links for stuff rather than trying to use it directly

[restive void]

I see, maybe using a solve function to make sure if y can be written in terms of x would be a better approach

the ti nspire has a solve function, for example

[restive void]

So in this case, why do we write the inequalities in terms of y?

Why not in terms of x?

[hardy coral]

It's just whatever is in the absolute value

[turbid summit]

its just saying it can either be positive, negative, or zero

[restive void]

Both have the same outcome, right? Like for the left branch, x-5>=0 and for the right branch -(x-5)>=0, right?

[hardy coral]

If you're just looking for exercises on inverse functions, it's easy to find online, just search "inverse function exercises with answers"

[restive void]

I was trying to work on inverse relations too, like finding the inverse relation of a non one-to-one function for example

[hardy coral]

Still easy to find online

It's not that different, you just need to find all the branches of the inverse

Like the inverse of x^2 has two branches, sqrt(x) and -sqrt(x)

[restive void]

All right, I'll try practicing on those,
thanks all!

.close

[vale dock]

Channel closed

Closed by @restive void

Use .reopen if this was a mistake.

[safe fulcrum]

no, both are x >= 5

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

the way I think about it is, recall that if you have y = |f(x)|, there's going to be a left branch and a right branch

so if you swap x and y to get x = |f(y)|

there's going to be a top branch and a bottom branch

[vale dock]

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• Show your work, and if possible, explain where you are stuck.
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[tiny torrent]

I just need a tiny hit to get started here

[tiny torrent]

What I know is that I need to show a bijection with N

My thought was to make a set A of the coefficients of all the terms in the polynomial which would have at most n + 1 elements

then somehow show theres a function to map those elements to N?

[keen tundra]

You should not build a bijection with N explicitly

It will be a mess

Use some known countable sets

[tiny torrent]

oh ok so as long as I show its a bijection with a countable set its countable?

[keen tundra]

This is the definiton

My point is that any bijection you will write will be a mess

So use other countable sets instead of N

[tiny torrent]

hmm

[ashen prawn]

hence we all must not rely on ai

!noai

[tiny torrent]

I just want to make sure, but I will be creating a function to show a bijection right?

or do you mean to not show a bijection formally since it'll be messy

[hardy coral]

Have you constructed a bijection between N and Q before?

[tiny torrent]

no I havent

this is my first time proving something is countable

my idea was that since the set of polynomails contain the set of integers

maybe I can map those integers to the integers in the rationals

and then for the next degree I map it to elements in between those integers???

im lost tbh

[hardy coral]

Well, are you allowed to assume both N and Q are countable, and that the set of pairs of elements from two countable sets is also countable?

[tiny torrent]

yeah we are we discuessed the countability of N and Q

but um

idk about the set of pairs of elements

[hardy coral]

The cartesian product

[tiny torrent]

no we havent discuessed that but we did discuss that the countable union of countable sets is countable

thats the only thing we discuessed besides countability of N, Q and R

[keen tundra]

This is very important here

[tiny torrent]

oo

[hardy coral]

Yeah that can work

[tiny torrent]

I think I see

but um

tbh I didnt really understand the proof to that theorem he did in class

maybe I should go take a look at that again

[leaden kraken]

even weaker, you can use an injection into a countable set

[leaden kraken]

you may need to use the fact that a finite product of countable sets is countable

[tiny torrent]

What does countable union mean

countable union of countable sets hmm

ive never heard countable union before

the union is countable im assuming?

is this the same as saying "Prove that the union of countable sets is countable?"

[tiny torrent]

oh ok i see, he did prove example 3 only showing injectivity

I was wondering why he didnt show onto

[hardy coral]

I interpret that as "the union of countably many countable sets is countable"

[tiny torrent]

ah ok that makes sense

[hardy coral]

(which implies that the cartesian product of two countable sets is countable, but you don't need to use that)

[leaden kraken]

This is not true. For example, any uncountable set is the union of its singleton subsets, which are each countable (in fact, finite)

[tiny torrent]

oh ok that makes sense

[hardy coral]

I think the point here is to use prime numbers, not just 2 and 3

If you have countably many E_j, you can assign a prime number to each one, and make the naturals that way

[vale dock]

@tiny torrent Has your question been resolved?

[tiny torrent]

Ok so I got somewhere I think
So if we let the polynomial with rational coefficients be represented as the union of a set D for each term's degree and a set R for each rational coefficient then if we can prove R and D are countable then the rational polynomials are countable by example 3. Since the degree of a polynomial is a subset of N then D is countable, since the set of rational coefficients is a subset of Q, then R is countable. By example 3 D U R is countable

Though I am not sure how I should formally go about representing the polynomial with ratinal coefficients as the union of these sets

there should be a more explicit formula right

feels wrong to just say it can be represented in this way

[hardy coral]

You need a set per coefficient

Your idea of D U R doesn't work; it would work with a cartesian product, but you don't have that

[tiny torrent]

hm ok let me think about that

[tiny torrent]

hm im not really understanding why I need a set per coefficient

If I have R be the set of singletons where each singeton contains a coefficient

then I cant really say R is a subset of Q so i'd have to show R is countable but if I just let R be the set of coefficients then R is a subset of Q and I can just say its countable

whats not allowing me to let R be the set of coefficients

hmmm

let me think about it some more

I think im still a bit far off on my proof, I tihnk im not understanding something fully

[hardy coral]

Perhaps I misunderstood

Think about constructing a polynomial

First you can take one natural number for its degree; that's also the number of coefficients, including ones that are zero

Then for each coefficient, you choose a rational number

You can't just have one set of rationals for every coefficients; you need one per coefficient

[tiny torrent]

just one natural number for its degree?

what if the degree is infinite, that wouldnt be in the natural numbers right

[hardy coral]

A polynomial has finite degree

[tiny torrent]

oh

[hardy coral]

A "polynomial" with degree infinity is a power series

Not really considered the same object

[tiny torrent]

oh ok

[hardy coral]

If you've heard of Taylor series or Maclaurin series, these are power series

[tiny torrent]

yeah I have in calc 2 a bit

how do we know that the degree is the number of coefficients

oh nvm I see

I was thinking of like x^4 + x^1 missing coefficients

buts its just 0

[hardy coral]

Ye

[tiny torrent]

ok ill try going at the proof again I think I can get it this time

[tiny torrent]

came to another problem but ill see if I can figure it out first

[worn sparrow]

ntw

btw*

Cantor-Bernstein-Schroeder Theorem

if you show theres an injection from X to Y and an injection from Y to X

that proves theres a bijection between X and Y

often easier to do that

make sense?

[tiny torrent]

yeah that theorem makes sense

[worn sparrow]

very relevant to you:

theorem:

$\mathbb N \sim \mathbb N^2$

[rocky lotus]

gfauxpas

[worn sparrow]

and you can get higher powers by induction

want a proof?

[tiny torrent]

is that true for N^N

because I think im stuck

[worn sparrow]

every finite power*

[tiny torrent]

since I have N^N many elements or im not really sure

[worn sparrow]

okay

[tiny torrent]

im just kind of stuck ill post what I have so far

[worn sparrow]

injection from N to N2:

$m \mapsto (m,0)$

[rocky lotus]

gfauxpas

[worn sparrow]

we're using cbs

agree this is an injection?

[tiny torrent]

whats m

[worn sparrow]

input

[tiny torrent]

ive never seen this notation so im a bit confused

[worn sparrow]

it means a function f(m)=(m,0)

it just avoids giving the function a name

[tiny torrent]

ok and m is just any natural number?

[worn sparrow]

yes

[tiny torrent]

ok I see

[worn sparrow]

okay for the other direction

we're gonna prove an injection

Q2->Q

agree thats good enough?

because Q ~ N, right

[tiny torrent]

hmmm

[worn sparrow]

sorry i meant

a bijection from positive rationals and N2

[tiny torrent]

ok makes sense

[worn sparrow]

https://en.m.wikipedia.org/wiki/Calkin–Wilf_tree

easier to see this is an injection than a surjection

welll you can argue

its triantle shaped not square

fine so its only an injection

good enough, with cbs

do you know what i mean by triangle and square shaped?

i mean N2 can be viewed as an infinite battleship game grid

starting from
(0,0),(0,1),(0,2),... first row

(1,0),(1,1),(1,2),... second row

(2,0),(2,1),(2,2),... third row

[tiny torrent]

yeah that makes sense

[worn sparrow]

this is a triangle so

just dont map anything to the numbers off the triangle

you need

one pair from the first row

2 pairs from the second row

4 pairs from the third

8 from the 4th

2^(n-1) ordered pairs on row n

i maybe didnt set up the proof right

but hear me out

you can just

take one from the first row, 2 from the secondrow , etc

and get a sequence, right?

so this shows at least that Q ~ N

Q positive that is

heres how we finish

if n =p/q a rational number in lowest terms

then f(p,q)=(p/q) is an injection from

N2 to Q

avoiding division by zero but that doeant change cardinality of an infinite set

[vale dock]

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[fading saddle]

q5

[hardy coral]

X isn't defined

[fading saddle]

it's solvable tho idk how

[hardy coral]

Without knowing exactly where X is? No it's not

[fading saddle]

it's from Cambridge textbook it has a answer

[hardy coral]

Suit yourself

[fading saddle]

leave it

u did the same thing last time

[distant galleon]

I do need to back up Nel here - the question isn't solvable as is. If they put a condition like "X divides ZY internally in the ratio 3:1", then it would be solvable. But as of now, the only the thing that be inferred about X (since nothing is explicitly stated about it) is that X lies on ZY, which leaves too many degrees of freedom to uniquely restrict its position.

[fading saddle]

ik but there's an answer ik the answer but not how to get it

OX = 1/2a+1/4c

[distant galleon]

If they're giving that an answer, then that isn't consistent with them asking you to give a proof (the end goals are fundamentally different)

I think the intention was "given O,X,M are collinear, determine the vector OX"

in that case, I'd use the fact that parallel vectors are scalar multiples of each other

think of what equations you can set up from that

[fading saddle]

the reasoning is scaler multiples

i got 1/2 a

not 1/4c

[distant galleon]

ah so you've already done some stuff so far

can you show what you've done so far then

it could just be a calculation error

in which case it's faster to look through what you've done than start from scratch

[fading saddle]

OM =OA+AM

OA =a

AM=1/2c

OM =1/2 C

+a

CZ=1/8a

YA=3/8A

a-1/8a-3/8a

=1/2a

i need to know how X is in straight line of midpoint of MA

so CB is parallel to OA which is parallel to the midpoint of MA and X

[distant galleon]

What I was going for here was something along the lines of $$\overrightarrow{ZX}=\lambda \overrightarrow{ZY}$$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

you can find $\overrightarrow{ZY}$ explicitly, so now you have $\overrightarrow{ZX}$ in terms of a parameter

[rocky lotus]

Civil Service Pigeon

[distant galleon]

you can do the same logic with $\overrightarrow{OX}$ and $\overrightarrow{OM}$

[rocky lotus]

Civil Service Pigeon

[fading saddle]

i thought the same thing and did that ,trying to find XY

but I couldnt

[distant galleon]

did you actually follow through with it

if you do it correctly, you should end up with some kind of system

[fading saddle]

i found YZ to find YX

but then there's no info to find YX

[distant galleon]

yes the best you can say with just $\overrightarrow{YZ}$ is that $\overrightarrow{YX}=-\frac{\lambda}{2} \mathbf{a}+\lambda \mathbf{c}$ for some constant $\lambda$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

but you don't just have that X,Y,Z are collinear

you also have that O,X,M are collinear

so you can make another equation with that

[fading saddle]

how will I solve the equation

[distant galleon]

if you do it correctly, you should end up with some kind of system

you can't do anything with just one equation

but with the two equations (X,Y,Z and O,Z,M), you can make a system that can be solved

which is why I keep asking you to do that

[fading saddle]

ok so what will be the second step after ZX = §ZY

[distant galleon]

you also have that O,X,M are collinear
so you can make another equation with that

[fading saddle]

k lemme try

@distant galleon i got OX as μ(a+c/2)

[distant galleon]

yup

so now we have two things

$$\overrightarrow{YX}=-\frac{\lambda}{2} \mathbf{a}+\lambda \mathbf{c}$$ $$\overrightarrow{OX}=\mu \mathbf{a}+\frac{\mu}{2} \mathbf{c}$$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

but these vectors don't match

YX and OX aren't the same

since the final result is with OX, we should try to get both equations with OX

I'll let you figure out how to do that

[fading saddle]

XO= λ(a/2-c)-5a/8
OX=λ(-a/2+c)+5a/8

we have 2 equations now right?@distant galleon

[distant galleon]

mhm

that's what I was talking about

[fading saddle]

but how do we solve that

[distant galleon]

it's a linear system

https://thirdspacelearning.com/gcse-maths/algebra/simultaneous-equations/

you have your standard methods like here

[fading saddle]

oo

@distant galleon but there isn't mew and lambda in both equations

how will we solve i can't figure it out

[distant galleon]

compare coefficients of $\mathbf{a}$ and $\mathbf{c}$

[rocky lotus]

Civil Service Pigeon

[fading saddle]

But they r multiplied by some constant so they won't matter ?is it wrong

[distant galleon]

what?

the whole point is that those constants are $\lambda$ and $\mu$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

the system gives you the values of those constants that work to satisfy both conditions

namely O,X,M and X,Y,Z being collinear

[fading saddle]

OX=μ(a+c/2)
OX=λ(+a/2+c)+5a/8

[distant galleon]

$\mu \mathbf{a}+\frac{\mu}{2} \mathbf{c}=\left(-\frac{\lambda}{2}+\frac{5}{8} \right) \mathbf{a}+\lambda \mathbf{c}$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

so $\mu=-\frac{\lambda}{2}+\frac{5}{8}$ and $\frac{\mu}{2}=\lambda$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

that's what I meant by comparing coefficients

[fading saddle]

ooo got it

so we take a and c out

now we substitude

right? @distant galleon

[distant galleon]

if that's how you want to solve the system

go ahead

[fading saddle]

.close

ty

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[sullen willow]

Can I do this?

[next snow]

do waht

[sullen willow]

Procede like this

Is it wrong or not

[next snow]

i can't see the rest of it

a bit confused on what you're trying to do also

you can pull out 3s and 2s to make like terms then add them together

[next snow]

nice simplification

[sullen willow]

So it's right?

[next snow]

the operation inside the parentheses should be addition

but you can still simplify it

it'll just take a bit more work

[sullen willow]

Ok thanks

[vale dock]

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[worn gyro]

guys i need help with logarithm.

[honest stone]

Post your question!

[proven wasp]

!da2a

[vale dock]

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

[proven wasp]

This is the final warning.

[honest stone]

That’s— not really applicable here?

[proven wasp]

||JK||

[honest stone]

dude you’re not a moderator

ANYWAY

@worn gyro feel free to post the question

[worn gyro]

it's actually not a question i just can't understand logarithm

[honest stone]

Well

you know what an exponent is right?

They’re inverses

[worn gyro]

oh i know it

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

a logarithm in 𝑎^𝑏 = 𝑐 aks “what value of 𝑏 when raising 𝑎 to it causes the value of the power to be 𝑐>”

[vale dock]

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[quasi glacier]

Currently struggling with these two challenges, also translation for the first one "Ahmed and Aicha have two identical bottles of water.
Ahmed drank 3/5
Aicha drank 7/10
Who drank the most water?"

[quasi glacier]

Again, help appreciated, but don't outright answer it for me please, just give me examples and explain it

[hasty knot]

Hmm how do u explain this😭

It’s like Ahmed drank 3/5 which means u cut it into 5 and he took 3 parts out of it

[quasi glacier]

Ts one as well, ts one is easy, just calculate, but I'm dumb :/

[hasty knot]

It’s ok 👌

An easy tip in this case u can make the denominator the same as the other one

It makes it easier

[quasi glacier]

That's what I did, but the end number would be negative, -1/9

[hasty knot]

Nope haha

It’s positive

If u want to make the 3 equal to 9 then u have to multiply by 3

So if u do that u also have to multiply the numerator by 3

So 2/3 is equal 6/9

[quasi glacier]

That's what I did

Ts it?

[hasty knot]

Ohh u flipped the negative sign for some reason

u turned it into 6/9 - 7/9

when it should be 7/9 - 6/9

[quasi glacier]

Oh, so I should put that first, then 6/9?

[hasty knot]

Yes, why would your 7/9 suddenly become negative even tho u didn’t touch it?

[quasi glacier]

Hmmm, makes sense

[hasty knot]

Right

And the 6/9 doesn’t become positive either

It stays negative

(- )x (+) = -

[quasi glacier]

Understood, back to this, what about the second image? I'm completely lost on that one

[hasty knot]

Same thing again

Make both denominators the same so that it’s easier to compare

[quasi glacier]

Oh now I understand

[hasty knot]

Are u French?😭

[quasi glacier]

Nope, but my school (Arabic school) has a lot of french, math, science, physics, all that

[hasty knot]

Ohhh

[quasi glacier]

I literally don't understand anything as well, lol, just use Google translate

[hasty knot]

Du coup tu comprends ce que je te dit là?

[hasty knot]

Ohhh

Hahaha ok ok

[quasi glacier]

Somewhat understand what that means "do you understand what I'm saying" right?

[hasty knot]

Yesss lmao

Ight so nothing else right?

[quasi glacier]

Amma do these by myself, see if there are any other problems that I don't understand before I close this

[hasty knot]

Ight ight

[quasi glacier]

Thx btw for the help

[quasi glacier]

Ts it?

<@&286206848099549185>

[dire cloud]

It is right

So 7/10>3/5

[quasi glacier]

Alrighty then. Thanks you and bagotto :3

Have a great day

.close

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[last star]

the blue below is correction information. This is graphing tan and cot functions

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[dull cipher]

A printing house must print a novel in two formats. This novel will be produced in at least 5000 copies but at most 8000. The number of pocket format novels must be less than the number of standard format novels, but must be at least 2000 copies. The pocket format novels cost $8 to print and will sell for $13, while the standard format novels cost $11 to print and will sell for $15. Knowing that the production costs amount to $1350 for printing, what maximum profit can this printing house expect?

[dull cipher]

x: number of pocket format novels
y: number of standard novels

x+y >= 5000
x + y =< 800

x =< y
x >= 2000

i got troubles with this part
A printing house must print a novel in two formats. This novel will be produced in at least 5000 copies but at most 8000. The number of pocket format novels must be less than the number of standard format novels, but must be at least 2000 copies. The pocket format novels cost $8 to print and will sell for $13, while the standard format novels cost $11 to print and will sell for $15. Knowing that the production costs amount to $1350 for printing, what maximum profit can this printing house expect?

[gritty sinew]

The pocket format novels cost $8 to print and will sell for $13
how much profit do you get if you make and sell one pocket book ?

[dull cipher]

5

then is it 5x + 4y?

[gritty sinew]

well almost

[dull cipher]

Z = 5x + 4y

[gritty sinew]

wait production costs = 1350

[gritty sinew]

so yeah this is the profit

[dull cipher]

alright ty

[gritty sinew]

but you have one more constraint

[dull cipher]

what is it

[gritty sinew]

printing cost = 1350

[dull cipher]

oh

i dont think thats constraint i think ill just have to subscract 1350 from the total amount at the end

?

[gritty sinew]

well idk

do you have the OG question ?

[dull cipher]

Knowing that the production costs amount to $1350 for printing, what maximum profit can this printing house expect?

[gritty sinew]

yeah ok at this point idk at all

[dull cipher]

ok

.close

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[oak bramble]

My strategy:

turn on derivative plot on calc

find the max/min of the derivative plot

I found the max/min to be a $\pm0.71$

[rocky lotus]

UCYT5040

[oak bramble]

as the x val

then i found area from that

however i had some error

of 0.0001

i guess i could've just gone further than 0.71

but it was annoying

[latent panther]

Can someone please help me with math I have a test tomorrow and I don’t understand anything

[oak bramble]

is there not a better way?

[oak bramble]

get out of my channel bro

[latent panther]

My fault

[oak bramble]

#❓how-to-get-help

[zealous pendant]

!occupied

[vale dock]

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).

[zealous pendant]

Dw not a big issue it’s happened like 3 times in the last hour

[woeful rover]

!occupied

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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).

[woeful rover]

Ah I'm a bit late

[zealous pendant]

Too slow

[woeful rover]

Perhaps

[woeful rover]

If you write down a formula for the area of that rectangle and use the derivative on that, then you might find something

[oak bramble]

I did that and got an answer, yeah

oh wait

nvm i dont think i understood you

the area of a rect is width * height

[woeful rover]

Yea

[oak bramble]

so i did 0.71*0.6040489749)

[woeful rover]

You can't use 0.71 without basis

First we write down the general form of the area

[oak bramble]

i'm sorry i mean 2*0.71

for the left and right side

I got 0.8577

correct is 0.8578

obviously i just need a more precise derivative

but

thats annoying to find

isnt there a better way?

im just using the graph, manually finding

[woeful rover]

Hmm

What was the original function you used to find the derivative of?

[oak bramble]

the one given from the problem

[woeful rover]

That alone isn't the function you need to find the derivative of

Sure, it happens to give you the result x value but you need to know how to do it properly

[oak bramble]

what do you mean?

[woeful rover]

Rectangle area = length * width right?

[oak bramble]

yeah

i guess prob height * width is better since y is height

[woeful rover]

Alright

Height * width

What is the height?

What is the y value?

[oak bramble]

height is f(x) where x is the x of the max/min of the derivative

so like f(0.71)

or f(-0.71)

[woeful rover]

Sure

[oak bramble]

its the same value

[woeful rover]

What is the width?

[oak bramble]

distance from the x of the max of deriv to the x of the min of deriv

or just 2* one of them

[woeful rover]

So you agree that the area is 2x * f(x)?

[wintry oxide]

Follow Vulcanone steps

[oak bramble]

yes

[woeful rover]

Alright, and the question wants the maximum of that area right?

[oak bramble]

mhm

oh wait so i just graph that?

haha yes i can just use the maximum calculation

although is there a way to do it algebraically?

what were you going to say?

[woeful rover]

I was saying that you could find the derivative of that by hand and find the x value then plug it in the original function

[oak bramble]

i dont know how to derive a function with eulers number in it

i only just learned how to do basic derivations

so i think i will learn this later then?

and if i ever see this question again before then ill just use calc

but your strategy here was genius

[woeful rover]

We follow the question's steps

[oak bramble]

alright, thank you!

.close

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Use .reopen if this was a mistake.

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[gaunt fog]

just quick questioners

[gaunt fog]

i got an equation called 65cm²

in like equation of variations

i didnt had time to write the full notes

this is the only equation i got

65cm² = k(10)(13)

so ofc, 10 x 13 = 130

therefore it would be 65cm² = k130

so now in order to solve for k i have to divide 130 on both sides

so 65 divided by 130 is

0.5

but but!

i have a feeling my process is wrong

since 65cm² has raise to 2 in it

does this mean i have to square it

[muted berry]

no that's right

the difference is that cm^2 is a measure of area while cm is a measure of length

they're just different kinds of measurement

[gaunt fog]

oh so i dont have to square it?

alright, thats cool that is cook

cool*

can u guys give me videos or tips

on how to divide faster

i'm having trouble dividing long numbers stuff lately

i usually just do the long process which i add the number up

I wanna master mental math

would drive me crazy

but would be cool too

[muted berry]

Probably just do division in 5s/10s/100s and subtract there

At least that's a good first step imo

[flint phoenix]

65cm² = k(10)(13)
Okay so the only way this works is when the RHS also has the unit cm^2

[muted berry]

Say like 710÷7 you can do 710-700 - 7 = 101r3

[gaunt fog]

oh because

7 times 100 is

700

o

so u made like a shortcut

now 710 - 700 = 10

10 divided by 7 is

approximately

1.428

so i guess

101.428?

[flint phoenix]

I don't quite get it

[gaunt fog]

LEZ GOOOO

i get it

[gaunt fog]

i wanto explain

[gaunt fog]

can i explain

[gaunt fog]

so you are dividing 710 to 7 yes?

[flint phoenix]

yes

[gaunt fog]

now since its a big nubmer it would be long process

to divide it 1 by 1

[hazy pivot]

It's just getting rid of the obvious multiples

[gaunt fog]

wats a multipule

sorry guys im like a 9th grade student, only developed my conscious of learning just like last year

rest of my school year grade i was unable to process simple info lmao

so i basically forgot all basic essentials of math

oh i get it

i researched multipule

[gaunt fog]

but ye he is right

its like finding the obvious multipules

in order for u to find the shortest way

[flint phoenix]

You do not understand what he writes lol. The approach you used is called distribution law, which is far from what he was talking about.

[gaunt fog]

o h

so u know how to do it now?

das cool

[flint phoenix]

The way you do it is correct, but I'd also like him to explain that if he feels like to

[muted berry]

yeah exactly

[gaunt fog]

yes ples explain

[gaunt fog]

how do u solve equations liek this though

36 divided by 1300

yes

36 divided by 1300

its hard for me

[muted berry]

well my mind starts with 36*10=360

[gaunt fog]

ok ok im listening close

[muted berry]

360x3=1080, and 360x4 is too large

so then you have 30*36=1080

and a remainder to solve of 1300-1080=220

so then you can do the same idea but this time 36 * 5 = 180 (again quick because you just half 36 and add a 0), 36*6 = 216

so that would just give you 36*36 r4

I mean there's definitely quicker ways but that's how I personally solve it mentally

[opal atlas]

what’s the problem?

[gaunt fog]

im just trying to ask for tips on how to divide quick

[opal atlas]

with or without calculator?

[gaunt fog]

without

im 9th grade

[opal atlas]

O

[gaunt fog]

alright

let me tell u what i understand

correct me if im wrong

36 x 10 = 360 (simple start)

then to get 1300

i divide 360 with 3

i mean

multiply 360 with 3

so it would be 1080

now since back at this part, 360 divided by 36 is equal to 10

since u multiplied 360 with 3,

therefore 10 x 3 is 30

which is 30 x 36

[muted berry]

yes

[gaunt fog]

so 30 x 36 = 1080

1300 - 1080 is

220

so i do with again

so i find the obvious multipule and easiest multipule which is 6

which equals to 216

[gaunt fog]

so basing on this divide count which is 30

[gaunt fog]

and on this divide count whic his 6

its 36 x 36?

which equals to