help-0

[arctic wren]

@flat tundra i think it's 1/2, because you have 2 possibilities or is head ou is tails

But i'm not sure, i'm really bad at probabilities

[flat tundra]

Basically you have 2 coins , one is head head (hh) , and one is head tails (ht). If one coin is picked ,and one of the sides showed it's a H , this is one of 3 possible scenarios , because the scenario in which the side showed is tails is eliminated

[flat tundra]

Yeah but this would be true if we didn't see that the shown side is heads

[arctic wren]

The other dide must be head or tail

So it,s 1/2

[arctic wren]

But, if the shown side is head, and we have 2 coins or the other side is heads or the other side is tails

[flat tundra]

If the side is not showns there are 4 scenarios:
Pick coin HH , flip and get h .
Same scenario as above but with the other side.
Pick coin HT and the side up is H so it gives T , or the other way around , so it gives H

But since we know that the shown side is Head the possibility in which tails was the side facing up is not there anymore

And from this we get to 2/3 probability of getting head again

[flat tundra]

in the text it says that one coin has head on both sides , and one has head on one and tail on the other

[alpine sable]

Here's what I'm thinking, if you care for my input (I am also a bit terrible at probability):
So, we grab one coin. We know there are only two coins, HH and HT.
We pick one coin with the coin on H. We know there is one more coin in the box, but the one you're holding is either HH or HT. We won't know unless it's flipped.
When flipped, it can either be heads or tails, since there are two coins with different outcomes on one side, and I conclude it's 1/2. (Take this response with a grain of salt)

What would the scenario be with three coins, two of which are HH, one HT?

[flat tundra]

The point that is missing is that you are shown one side , and that side is head

This changes things

Because you change the parameters

On of all the 4 possibile scenarios , the one in which tail is facing up , is not possible anymore

[velvet pelican]

Hint: ||use the conditional probability formula||

[flat tundra]

So there are not 4 scenarios

[civic rivet]

Can anyone come Discord and help us by our problem? We think it is simple but too hard for us to solve xd

[arctic wren]

I dont remember

[velvet pelican]

@civic rivet please post in an unoccupied channel

[flat tundra]

I don't even know what it is , I didn't use mathematical formulas to solve it , that's why I'm asking it here

[alpine sable]

The way your explanation sounds with the three different scenarios is making it sound like you pick the HH coin twice as much than the HT coin.

We need a probability expert here lol

[velvet pelican]

Basically $P(\text{HH coin | we see H}) = \frac{P(\text{HH coin AND we see H})}{P(\text{we see H})}$

[arctic wren]

@velvet pelican that's it?

[velvet pelican]

yes

[arctic wren]

So, it must be 1/2 cause the numbers of intersection between A and B is 1 and the number of A is 2?

[flat tundra]

The possibile scenarios are: taking the hh coin on one side , and that side is h , doing it again with the other side and it's H again , taking the ht coin and doing the same process we get one scenario with a h outcome and one with t outcome. But now were given an other parameter in the question, it says that the side shown is H , and this is 100% true, if the shown side is H the scenario in which the shown side is T is impossible, so it's out of the data of the problem,

[velvet pelican]

no, it's not

$P(\text{HH coin }| \text{ we see H}}) = \frac{P(\text{HH coin AND we see H})}{P(\text{we see H})}$

[ocean seal]

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

[velvet pelican]

ok this works

[flat tundra]

Does my logic work or is it flawed ?

[velvet pelican]

what's your final answer?

[flat tundra]

66.66666...%

Of getting H

2/3

[velvet pelican]

the question asks for the probability that you have the HH coin given that you flipped a H

so I believe that is incorrect

[arctic wren]

So the numbers of intersection between coin 1 and coin 2 is 2, because we have 2 H left, and at total we have 3 Heads, is that correct?

[velvet pelican]

you can't intersect the number of H

that is the intersection of probabilities

[woeful pulsar]

there are two hh coins and one ht coin?

[arctic wren]

No

[velvet pelican]

no 1 of each

[woeful pulsar]

yeah then what's wrong with 2/3

[velvet pelican]

wait actually nvm

[flat tundra]

But the problem says that you get the coin on one specific side , so there is the probability of getting HH on both sides ,so 2 scenarios in which you get H , but if you got the HT coin you get T

[velvet pelican]

it's (1/2)/(1/2 + 1/4) = 2/3

[arctic wren]

I still didn't get why is 2/3 and not 1/2

[flat tundra]

Because we change the parameters of the problem by saying that the shown side is H

[woeful pulsar]

because if you have no evidence it would be 1/2, the conditional probability changes based on the evidence

[flat tundra]

So 2/3 is correct?

[velvet pelican]

yes

sorry

I put in the wrong numbers

earlier

[velvet pelican]

Very similar to the monty hall problem

[flat tundra]

No , I'm the one who's sorry haha , I know very little about math , so I had to verify if my process was right

[flat tundra]

What problem?

[velvet pelican]

Monty hall problem, it was a gameshow some decades ago I think

you had 3 doors, 1 with a car and 2 with goats

[arctic wren]

[flat tundra]

It makes me think of those riddles in which an evil guy you tells you which door it's better to take to don't die lol

[velvet pelican]

you pick one of the doors

then the host (who knows what's behind which door) opens a different door with a goat

[alpine sable]

oh HI

[flat tundra]

Ok , yes it has the same logic

[alpine sable]

is anybdoy free right now

[velvet pelican]

now you get the choice to switch or keep your door

and the question asks whether you should switch or not

to maximise the chances to get the car

[flat tundra]

Ok wait

Let me think

We say that the car is 1

And the goats are 0

[candid torrent]

@alpine sable i can help you

[flat tundra]

001

[candid torrent]

but this channel is occupied

go to #help-1

[alpine sable]

oh lets move down yea

[flat tundra]

If you choose 0 it shows you the other 0 , if you switch you get 1 , of you choose one , he shows one of the 0s you switch and get the other 0

So if you accept to switch choosing the car would lead to goat and choosing the goat would lead to car

So yes ? @velvet pelican

[velvet pelican]

yes

[arctic wren]

||you have to maintain?||

[velvet pelican]

||no, the answer is you have to switch||

[flat tundra]

The coin one was easier, it think this one would have taken me a lot longer without knowing it worked similarly

[arctic wren]

Why??? I still didn't get this and the coin

[flat tundra]

I can draw a diagram

[velvet pelican]

it's similar in the way that knowing that when the host shows a goat in the other door, the probabilities change

[arctic wren]

Yes

[velvet pelican]

btw if you switch the door, you have a 2/3 chance of getting the car

and 1/3 chance of getting the goat

so it's much better switching

[arctic wren]

I'll try to understand

[flat tundra]

Sorry if it is a bit messy

I may have made some errors too , I'm tired and not a native speaker

[arctic wren]

I get it now

Thankks

But the goat and car , no

[flat tundra]

It has the same process , try to make a diagram , keep in mind that the variable that changes is that you are shown one of the doors with a goat that is not the door you choosed

[ocean seal]

TheMane3

[warped phoenix]

If I vertically stretch the above function by a factor of 2 and then reflect it across the x-axis, the function would become

[ocean seal]

TheMane3

[warped phoenix]

right?

[fluid kite]

63>77

[warped phoenix]

?

[flat tundra]

I think it does , but I'm not the right person to trust for an answer with this

[warped phoenix]

hmm...the answer was $f\left(x\right)=-2\sqrt[3]{x+3}+14$

[ocean seal]

TheMane3

[warped phoenix]

and i have absolutely no clue where the -14 came from

anyone know??????????? D:

[flat tundra]

The 7 was out of the square root

You had to multiply it by 2

[warped phoenix]

ah, what's the concept/reasoning behind that, if any? Or is it just a rule for transformations

[flat tundra]

It's a rule for transformations but as everything in math there's a concept behind it

[fluid kite]

16>33

[warped phoenix]

u good? @fluid kite

yeah makes sense

[fluid kite]

it's a riddle

if anyone gets it dm me

[warped phoenix]

...

[fluid kite]

44>33

[flat tundra]

Square root of 16 , then plus 4 it's 8 , square root of 16, times 2 then plus 4×2 it's 16

[warped phoenix]

this channel is in use so do it elsewhere please

[fluid kite]

which?

[warped phoenix]

@fluid kite well if you need help solving the riddle, than any of the question channels, if not, then go to #chill or #math-discussion

[fluid kite]

k

[warped phoenix]

and yeah shanti makes sense because if I put $f\left(x\right)=-2\sqrt[3]{x+3}-7$ into desmos then its not reflected across the x axis

[ocean seal]

TheMane3

[arctic wren]

How can i demonstrates that ? $lim [sin(\theta)/\theta]; \theta \to 0$ I've seen the demonstration but didn't get it

[ocean seal]

Guilhotina

[sage jacinth]

squeeze theorem

also

that goes to 1

not 0

if you want it to go to 0 you have to take the limit as $\theta$ approaches infinity

[ocean seal]

(バカ) baka

[alpine sable]

someone help me with this

Pls

[lethal stump]

perms and combs is probably my weakest subject but it should be 1/2 right?

[rich basin]

Yeah, but i was wondering why is my answer wrong

this is what I have right now $\frac{3C1\cdot 5C1 + 3C1\cdot 5P2+ 3C1\cdot 5P3 + 3C1\cdot 5P3 + 3C1\cdot 5P4 + 3C1\cdot 5P5}{
6P2 + 6P3 + 6P4 + 6P5 + 6P6}$

[ocean seal]

BorutoEyePower

[rich basin]

$\frac{3C1\cdot 5C1 + 3C1\cdot 5P2+ 3C1\cdot 5P3 + 3C1\cdot 5P3 + 3C1\cdot 5P4 + 3C1\cdot 5P5}{
6P2 + 6P3 + 6P4 + 6P5 + 6P6}$

[ocean seal]

BorutoEyePower

[rich basin]

I could explain how i managed to get this location if you want?

<@&286206848099549185>

I understand that it is 1/2 based on that either one having 50 percent of being chosen even or odd

Butt i just want clarification on why this won't work

[arctic wren]

is there a chat to talk about another thigs besides math ?

[lilac lantern]

Why is the integral of 1/u = ln|u| and not ln(u) , like why is it || ? does this mean it is the ln of the absolute value of u?

[rich basin]

@lilac lantern I thought this wasn't an integral question?

[lilac lantern]

Yes its just a general question lmao

Ok I found the answer for myself

[rich basin]

Please can anyone help me?

[nocturne mirage]

whats the question sir ?

[lethal stump]

the main reason for the absolute value sign is that x has to be > 0 if you want real values for ln(x)

[rich basin]

this

[lethal stump]

omg

forget what i said

[rich basin]

this is what i have right now

$\frac{3C1\cdot 5C1 + 3C1\cdot 5P2+ 3C1\cdot 5P3 + 3C1\cdot 5P3 + 3C1\cdot 5P4 + 3C1\cdot 5P5}{
6P2 + 6P3 + 6P4 + 6P5 + 6P6}$

[ocean seal]

BorutoEyePower

[lethal stump]

do you have answers?

[rich basin]

Yeah

[lethal stump]

it isn't 1/2?

[nocturne mirage]

I would say 1/2 since only the last digit matters

[rich basin]

It is 1/2

[lethal stump]

oh ok

thank god

[rich basin]

But i'm confused how come my working out does not give out 1/2

Like i've taken all taken all the cases when it is even, and numbers are even when their first digits are divisible by 2

[nocturne mirage]

last digit*

[rich basin]

Yeah last digit

So basically in my numerator, I take cases of various amount digits i want to make

Like forexample $3C1\cdot 5C1$

[ocean seal]

BorutoEyePower

[rich basin]

I choose 1 one of the 3 even numbers and then times it with the rest of the numbers left

It doesn't matter which is picked, as long as the last digit is even. It will still work

[rich basin]

I finally figured why i got it wrong

@crisp iron

right now i have $\frac{3^{13} \cdot 48^2}{52^5}$

[ocean seal]

BorutoEyePower

[hazy dome]

I’m pretty sure you’re supposed to find each part separately

[rich basin]

Isn't this with replacements

[hazy dome]

It doesn’t specify

Which leads me to believe that each part is an independent event from the previous part

Asking as if nothing had happened before the draw for each part

[rich basin]

I am confused

[hazy dome]

Basically treat each part as it’s own separate entity

Don’t even consider the other parts when solving for any particular part

[rich basin]

I am confused with what you mean by treating what as a separate entity

[hazy dome]

Just solve each problem separately

As their own separate event

[rich basin]

this is part (a) we are talking about

[hazy dome]

This is all parts

Pretend that you’re given a full deck in part a, you draw the cards and calculate the probability, then you reset the deck and put in all needed cards then you do the same procedure for part b and c

[rich basin]

How can we solve part (a) then in your view when separating each of the entity

[hazy dome]

Just calculate the probability of drawing 3 hearts from a deck of cards given non replacement each time

Because once you draw those 3 hearts, it won’t matter what the next 2 cards are

You’ve already satisfied the requirements

[rich basin]

But then once you draw 3 cards, there are many other options you can draw it with

[hazy dome]

But that doesn’t matter

[rich basin]

Like forexample 3 hearts and 2 spades is one option and 3 hearts and 2 aces are another

[hazy dome]

You’re asked for 3 hearts

That’s it

You could draw any other 2 cards you wanted, you still got 3 hearts

Part a simply asks for 3 hearts

[rich basin]

But then there are many ways you can get the 3 hearts though

[hazy dome]

But exactly 3 hearts

So no more no less

[rich basin]

That is why it should be 39^2

I removed all the hearts

[hazy dome]

When dealing with a situation such as this, you’re gonna multiply every single time, and multiplication is commutative, so order does not matter

[rich basin]

What equation do you think it will be

[hazy dome]

Basically, solve as if you draw 3 hearts in the first 3 cards, and then you draw anything but a heart for the last 2

[rich basin]

Based on what you have interpretated and just doing it non replacement it would be 13C3 over 52^5

[hazy dome]

(13/52)(12/51)(11/50)(39/49)(38/48)

The first 3 are hearts, and the last 2 are anything but hearts

I’ll have exactly 3 hearts at the end

I’m sorry I couldn’t explain it in more detail

[rich basin]

@hazy dome That is the right answer, but I am still confused

Like isn't it supposed to be with replacement

[hazy dome]

The problem is that the question doesn’t specify

[rich basin]

Like you can have HHHAA

[hazy dome]

It’s kind of assumed that if you draw 5 successive cards that you’re not gonna replace them

[rich basin]

Doesn't it say with replacement

[hazy dome]

But that’s just me

...

I’m dumb XD

[hazy dome]

so then it’s (13/52)^3 times (39/52)^2

[rich basin]

Wait but this is wrong

And your other answer is correct

[hazy dome]

Ok now I’m confused XD

[rich basin]

,w solve (13/52)(12/51)(11/50)(39/49)(38/48)

[ocean seal]

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[rich basin]

Wait this is still wrong

The only difference with the answer is with the denominator being an extra 0

[hazy dome]

Wut lol

The cubed one is with replacement

[rich basin]

That is the answr i had before i'm pretty sure

[wary stream]

Why are you decrementing the values?

You're replacing the cards back into the deck

[rich basin]

@wary stream Well we now have a different answer

right now we have $\frac{3^{13} \cdot 39^2}{52^5}$

[ocean seal]

BorutoEyePower

[rich basin]

@hazy dome I think there is another issue i think

[hazy dome]

Idk what it is then

[rich basin]

I think it is because we have duplicatse

Like we can have AABPP

And we can have ABAPP

which would be the same when we select 5 cards

[wary stream]

Do you know what the exact answer is?

[rich basin]

,w solve (13/52)^3 times (39/52)^2

[ocean seal]

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[rich basin]

this is the answer

[wary stream]

,w 2717/33320

[ocean seal]

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[nocturne mirage]

the channel is already used, @warped phoenix

[wary stream]

This channel is busy

[rich basin]

,w solve (3^13 * 39^2) divide (52^5)

[ocean seal]

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[rich basin]

,w solve (13^2 * 2^39) divide (52^5)

[ocean seal]

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[rich basin]

,w solve (13^3 * 39^2 * 2) divide (52^5)

[ocean seal]

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[rich basin]

,w solve ( (13^3 times 39^2) divide 4!) divide (52^5 divide 5!

[ocean seal]

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[warped phoenix]

Why is it 1.9% and not 2.1% when 25/12 = 2.08?

anyone???

[rigid smelt]

yes

oh well idk about this

<@&268886789983436800>

[tall wing]

ah

yeah it looks like a test

banned

[woven flume]

in matrices

uh

I'm trying to get A inverse = (1/88) B

[alpine sable]

Hi how do i solve this?

[ember lava]

expand it, then find coefficients of p^2, p and the constant term, the values of coefficients will be a, b and c, respectively.

[sudden nimbus]

so opt 4

??

@ember lava

[ember lava]

yes

[sudden nimbus]

ok

[alpine sable]

<@&286206848099549185>

[ember lava]

ok so lets take food A first

[alpine sable]

Sure

[ember lava]

so you need 500 units of calcium

and food A gives 32 units of calcium every ounce

so how many ounces are needed?

[alpine sable]

15.625?

[ember lava]

yes

[alpine sable]

Oh so its division

[ember lava]

but they are telling you to round it off to the nearest whole number

so its 16

[ember lava]

yes

find out the others in the same way

[alpine sable]

So i add them up?

[ember lava]

yes

[alpine sable]

Like (500/32) + (470/24)

[ember lava]

yes

[alpine sable]

Thanks

[ember lava]

np

[alpine sable]

@ember lava is there any discord group for chemistry too

[ember lava]

yes there is. im in it.

if you want, i can send the link in dm

[alpine sable]

Yes please

@ember lava

[ember lava]

ok

[alpine sable]

Does anyone know about the card game Hand and Foot?

Because I have an idea about the game, but I need help with it.

[muted owl]

So how am I supposed to do the working for this

I know the answer but in general how do I show steps

[vale wigeon]

with words

perhaps a good intermediate step is to find the common difference of this AP

[muted owl]

its alright I got it

I need to put formula and then plaece values in the formula

[tired hamlet]

How would I start this one?

[plain dome]

1+1=

[tired hamlet]

Huh?

[vale wigeon]

"i didn't read #❓how-to-get-help and i think it's funny if i post a 1st grade math question hoping for someone to give out the answer, in violation of this server's purpose"

[quiet crown]

$2005=2183\cdot(1-r)^15$

[ocean seal]

Vex

[quiet crown]

Ok fuck latex lol

but i think you now just need to reverse to find the rate

Assuming in 1990 the population was 2,005

Which i think the reverse is

It will look like this

@tired hamlet

but that fifteen will be above the square root thingy

I'm probably way off

[tired hamlet]

Is this right or is the way you did it right?

Oh I think I need to switch the 2183 and 2005

[ionic jewel]

,w 2183 = 2005e^(15k)

[ocean seal]

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[ionic jewel]

0.0056 about

[heavy canopy]

What signs should we put in between so the result is 13?

[tired hamlet]

Oh 15 instead of 0?

[dark granite]

there is a question being asked here already. Please refrain from asking until this question is answered. or ask elsewhere

[heavy canopy]

Oh I didn't know that im sorry!

[dark granite]

it's all good; just thought I should bring it to your attention

[ionic jewel]

is there actually any combination?

you can't use digision, besides between the 4 and the 2 i don't think

and that's not useful

i don't see a way and i don't really want to try all 256

[coral cosmos]

Exactly what I'm thinking

[alpine sable]

Its $(\sqrt{9} 7) - (42)$

[ocean seal]

Quinnt

[alpine sable]

Which is (3 * 7)-(4 * 2)= 21-8=13

[coral cosmos]

but can we use sqrt though

Does it come under a sign? Because I've never heard of it

but your answer is correct, so maybe yea

[left echo]

someone help pls 🙏🏿

[ionic jewel]

i though square roots (and parenthesis although yours don't matter are illegal)

but fair solution then, id still call it a trick question

[paper oar]

this is a 3rd (odd) degree function and in odd degree functions, it rises at one side and falls to the other which means that we can eliminate options B and D. Now since the leading coefficient is negative (-4) it will rise to the left and fall to the right, and thus it will be option C.

[left echo]

thank u sm bro fr

[paper oar]

no problem, hope you understood it.

[left echo]

also

why does it have to rise from the left and decrease from the right? how does that correlate w the coefficient

why cant the end behaviour go both up or down

like they’re both positive or negative infinity

[paper oar]

because it is -4x^3 which is an odd degree and if there is an odd degree then both ends will not go to the same end behavior

[ionic jewel]

i don't feel like explaining but you should probably tl;dr limits and how to approximate infinite limits with big numbers

don't make him memorize all the cases

[paper oar]

yeah I guess.

[ionic jewel]

[left echo]

ohh that makes sense ok

[paper oar]

that is a nice emoji

[left echo]

thank u

[paper oar]

no problem

[tight locust]

if p and q are the roots of ax^2 + bx + c then what equation yields the roots 1/p and 1/q

[ionic jewel]

(x-1/p)(x-1/q)

[tight locust]

in terms of a, b, and c

[ionic jewel]

are you trying to solve abc in terms of pq or the other way around?

if you want a,b,c in terms of pq just multiply out how i sent it

if you want it the other way do the quadratic eqn

oh I'm illiterate sorry

[minor heath]

a good hint could be vieta's formulas

[tight locust]

i got it lol

ty

[dry thunder]

the sum of 15 repetative integers in 550 find the integers?

any help?

[ionic jewel]

what are repetitive integers?

the same one 15 times?

[dry thunder]

my bad consecutive*

i think you can keep it like x, x+1 and so on

[tight locust]

29 thru 43

[ionic jewel]

you could yes

that would be an annoying expression though

[tidal grotto]

Find integers of n if possible 🙂

[tight locust]

not if you're smart

[ionic jewel]

so the better way is just doing 550/15 then using the answer as your middle term and it'll probably work

,w (1+2^n+3^n)/5 = n!

[ocean seal]

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[tight locust]

that's not even an integer

[dry thunder]

thanks man!!

[minor heath]

exactly

[ionic jewel]

looks like no integer solutions

[tidal grotto]

😠

[minor heath]

,w (1+2^n+3^n)/5 = n! over the integers

[ionic jewel]

yeah just round it's close enough, you use it as your "guess" then change as needed

[minor heath]

bruh

[tidal grotto]

(╯°□°）╯︵ ┻━┻

[ionic jewel]

wherw is this problem from

[tidal grotto]

ME

[tight locust]

the sum of the integers from 1 to n is n(n+1)/2

[ionic jewel]

it's not a diophantine common form, so idk how you expected to do it

[tidal grotto]

I thought it had integers

[dry thunder]

ye thats why im asking

[tight locust]

so you have 15x+14*(15)/2 = 550

[dry thunder]

was asking

[ionic jewel]

it works, but it's easier to just approximate then edit as needed

[dry thunder]

ye

[dry thunder]

yeah

thanks guys!

[tight locust]

because the first term is x+0 so it's actually the sum from 1 to n-1

[ionic jewel]

nah doesn't look like it, and in general finding integer solutions to non-polynonial equations is not going to go well for you

[tidal grotto]

ok

[tight locust]

lmao

did you really expect that to have an integer solution?

[tidal grotto]

yes

wait

how did the computer calculate the non-integer value of factorials

it showed a graph

confusion

[jagged imp]

it certainly cant have positive integer solutions

since the numerator of that fraction is even

[ionic jewel]

haha gamma function go whoosh

[jagged imp]

wait no

lmfao im dumb

[ionic jewel]

gamma function is a continuous extension of factorials I suppose

,w (1/2)!

[ocean seal]

Results provided by WolframAlpha

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[ionic jewel]

this is a fun one @tidal grotto

[tidal grotto]

for you

why pi

whyy

[ionic jewel]

😆

[tidal grotto]

btw if 0! is 1

how can 0.5! can be lower than 1

[ionic jewel]

,w plot gamma function

[ocean seal]

Results provided by WolframAlpha

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

Requested by 兔子#6562

[tidal grotto]

oh

so It can get lower at some point

[ionic jewel]

except it's off by 1

[tidal grotto]

what does gamma function do btw

like

can I calculate by hand

[thorn kindle]

if cos(0) = 1 then how can cos(pi) be lower than 1

[tidal grotto]

laugh

[ionic jewel]

yes, it has an integral definition iirc

,w gamma function

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by 兔子#6562

[tidal grotto]

um

[ionic jewel]

not given on that page

but there is one

[tidal grotto]

oh

[tidal grotto]

O_

wow

that looks so

cool

[ionic jewel]

🤠

math is cool

[shrewd otter]

😎

[ivory otter]

im supposed to prove the above identities

but im not sure how to continue from here lol

[rich basin]

@ivory otter I can help you if you want? unless you help me?

[ivory otter]

help you with what

i dont understand your offer

[rich basin]

Combinatronics

[ivory otter]

xD

[rich basin]

I will help you with trig, however yuou must help me with combinatronics

[ivory otter]

lmfao i cant help

[rich basin]

Why?

I've been stuck on this one question all day

[vapid swift]

just use cos^2y=1-sin^2y

[rich basin]

Only the doctors on mathematics can help me cure this nightmare

This is the only question left until I'm up to the easy part of the combinatronics section, dealing with binomial expansion

This morning my equation that I had is

[ivory otter]

dont see how that would solve it

[lethal stump]

ah but it will

[vapid swift]

it will

[rich basin]

$\frac{13^{3} \cdot 39^2}{52^5}$

[lethal stump]

you will see when you try it

[ocean seal]

BorutoEyePower

[lethal stump]

also remember that cos^2x + sin^2x =1

[ivory otter]

ye

[rich basin]

@ivory otter Try subbing it in

[ivory otter]

i did

[rich basin]

@ivory otter To solve proof question in trig, you want to try to make the functions the same as the equation that you are trying prove

[ivory otter]

you dont say

[rich basin]

@ivory otter Havee you got it yet?

[ivory otter]

ye

i mean

i dont like this

i wouldnt have thought about this

unless you told me

how the fk am i gonna solve it like this during exams

i cant use discord in exams

[rich basin]

@ivory otter Don't worry, I was like you before 2 years ago. It is going to get extremely easy once you practice more

[ivory otter]

i have been practicing for days..........

[rich basin]

@ivory otter Just make sure to make the functions the same as the equation that you are trying to prove as well as the degrees

Degrees priority first and then the functions

[lethal stump]

also sometimes it helps to look at the question backwards

as in starting from the answer and thinking 'how could i make this look like the thing they gave me?'

[rich basin]

@ivory otter When is your exam?

[dark granite]

This question implicitly assumes that the order of selection doesn’t matter

[ivory otter]

in september

[dark granite]

So 52^5 shouldn’t be the denominator

[rich basin]

@dark granite I thought that might be the case, how can we remove the duplicates?

That is the only issue I have so

far

I was thinking of dividing 52^5 by 5!

[dark granite]

The denominator must be 52+5-1 choose 5

Because were selecting with replacement

[rich basin]

What is with minusing 1?

[dark granite]

You may have learned that as 52 multichoose 5.

It has to do with boxes and balls with dividers

[rich basin]

don't know what you mean by boxes and balls?

[dark granite]

So you have 52 distinguishable boxes

5 identical balls

So you only need 51 dividers to tell boxes apart

So 51+5 choose 5 places the dividers down and the balls fill the remaining slots

Also known as stars and bars

[rich basin]

What do you men by dividers?

[dark granite]

Literally just bars that divide the distribution of the 5 balls amongst the 52 boxes.

If all the balls are to the left of the 51 dividers, that means we picked all 5 of say aces

So line up the 5 identical balls in a row. Where these balls land, represents which cards you’re selecting

[lethal stump]

i have never seen this in perms and combs before and now i am scared

[dark granite]

Yeah it’s fun yo

[rich basin]

I am still rather confused how can just 51 bars can divide the distributin of the 5 balls amongst the 52 boxes

[dark granite]

You’ve learned selection with repetition where order matters right?

[rich basin]

Can you please give me a smaller example?

[dark granite]

Sure

[rich basin]

So I'm right now imaging like 6 boxes and there are at least 3 bars minimal to divide them

[dark granite]

From a store that sells only three varieties of coke, I wish to order exactly two cans of coke. How many ways can I order two cans of coke if I’m allowed to have cokes of the same variety?

[dark granite]

No, look at my example

[rich basin]

okay

[dark granite]

So I need three distinct boxes

[rich basin]

yes

[dark granite]

And 2 identical balls

And 3-1 dividers

[rich basin]

How would you say the bar represents?

Is it like 0 | 0 | 0

[dark granite]

Using this method, if I have n distinct boxes, I need n-1 dividers to distinguish between the distribution of balls into the boxes

[rich basin]

What about something like 6 boxes

wouldn't you need 3 bars to divide?

[dark granite]

Remember only two balls for this example

[upper pond]

how do i calculate probability

[dark granite]

You’d need 5 dividers

[rich basin]

Why is that the case for 6 ?

[dark granite]

A question is being answered rn. Please wait until this question is finished or ask elsewhere

Because 6-1=5

[rich basin]

Unless they are lined up linearly

[dark granite]

Read my comment bout n boxes

[rich basin]

yeah

[lethal stump]

wait so can you write the denominator as 5*52C1?

soz about interrupting anything

[dark granite]

So you’d be essentially arranging a ‘word’ with bars and balls where everything to the left of the left most bar represents one box, everything in between two consecutive bars represents another box, and everything to the right of the rightmost bar represents another box

[rich basin]

okay

[dark granite]

Make sense?

[rich basin]

yes

[dark granite]

0|0| is a different ordering of the cans of coke than |0|0 for instance

[rich basin]

Not quite make sense

[dark granite]

I’d say it’s (52+5-1) Choose 5

Well the first distribution there says I bought cherry coke and Diet Coke, but not sprite. The second distribution says I bought Diet Coke and sprite but not cherry coke

[rich basin]

Okay that now makes sense

[dark granite]

Nice

Where the 0 is represents that I ordered that item

If there’s more than one 0 in a box, that means I ordered more than one item of that variety

Make sense?

[rich basin]

yes

[dark granite]

Dope

So back to your card question

Essentially we’re choosing 5 cans of coke from a store that sells 52 varieties and were allowed to select cans of the same kind

Right?

[rich basin]

yes

[dark granite]

Hence the whole (52+5-1) choose 5 denominator

So what’d the numerator be for part c?

[rich basin]

is the 5 for the extra replacements that is necessary to keep up with the amount of choice the user is making?

[dark granite]

Essentially yeah, but it makes more sense to think in terms of the identical balls into distinguishable boxes idea

[rich basin]

so i think it will be like (13+3 -1 ) choose 3 times (39+2 -1 ) choose 2

[lethal stump]

yo do you have sols

for the qs

[dark granite]

That ain’t the answer for part c man

If we can’t allow hearts in the selection, just throw em away in the first place. So the numerator for part c should be (52-13+5-1) choose 5

[lethal stump]

do you guys know the answer for part a?

[rich basin]

yeah

[dark granite]

Were working on it

[lethal stump]

i was thinking about maybe using binomial probability?

since the chance of pulling a heart is 1/4

[rich basin]

I am still rather confused how the idea of 52 + 5 is related to the identical balls idea

[lethal stump]

and there is no replacement

could you do something like 5C3((1/4)^3)*(1)^2 = 5/32?

[dark granite]

This is the numerator for part a yes

[dark granite]

Rewrite 52+5-1 as 5+(52-1)

Now see it?

[rich basin]

wouldnt that be 56 chose 5

[plucky sun]

how to skip count by 6 and 7

[dark granite]

Yes

[plucky sun]

Help

[dark granite]

A question is being answered here at the moment

[plucky sun]

Well answer my question first mf

[rich basin]

Still confused with how you put it like that

[dark granite]

Like what?

[rich basin]

so you getting the divider and then you adding 5 more bars

[plucky sun]

Eo

E

Eaoa

Hdhhd&Hebburn

[dark granite]

Those 5 more things aren’t bars, those are the balls

[plucky sun]

Rested we've eggs seen wii eh ez u yum tea to tea to gx t yum tea yum tea

To ez 60 tea tea td up tea to dm tea cm f4

🖕

[rich basin]

can you please represent it in like a diagram out of text, maybe for a smaller example?

[plucky sun]

Answer my question first mf

[dark granite]

If you’re not going to wait your turn or help answer he current question previously asked, please leave and go to another chat room and ask your question if one isn’t already being answered.

[lethal stump]

what language is this

[dry thunder]

idk

[dark granite]

Bruh the can example is about as small as I’d go. Any smaller and the big picture becomes lost

[dry thunder]

i guess thats the language idiots speak!

yes

[lethal stump]

mfs really tryna cast a spell on me

[dry thunder]

lmfao

[rich basin]

@dark granite like something like this 0|0|0

[dark granite]

Yeah can someone get a mod here to throw that dude outta this chat

Okay I see it

[dry thunder]

i think your the one whoes out

watch your back @dark granite

[dark granite]

No I’m talking about @plucky sun

[dry thunder]

anything can happen 😉

[sly mantle]

@plucky sun knock it off

[dry thunder]

oh ye sowwy @dark granite

[rich basin]

So in the can example where you want two cans of the same type. It will be (3+2-1) choose 2 1||

[dark granite]

@dry thunder dude I literally have no prob with you. Im just talking about that other dude who flipped us off

[lethal stump]

mathematical beef

[dry thunder]

yeye im sowwy

didnt see shit

didnt mean too :))

[dark granite]

It’s all good just get that other dude outta here

[dark granite]

Yes!

(3+2-1) choose 2 is right

[rich basin]

what about the 1||

[dark granite]

Oh I thought that was a typo you made

[rich basin]

the | is the bars

[dark granite]

Okay, why is the 1 there?

Where are the balls?

[rich basin]

Would you be able to please represent that scenarioo in terms of | |

[dark granite]

Sure

0|0| is one distribution

Notice there is no zero to the right of the rightmost bar

[rich basin]

yes

[dark granite]

So that particular can of Coke isn’t selected in this distribution

Make sense?

[rich basin]

yes

[dark granite]

Kk well there you go! You should be gold now

[rich basin]

But i'm confused with how the + 2 relates to that?

[dark granite]

You’re talking about 3+2-1?

[rich basin]

yeah

[dark granite]

Okay if I give you the word aabb and ask you to arrange the letters in that word in a row, how many distinct arrangements are there?

[rich basin]

there will be 4! divide by 2! and 2!

[dark granite]

aa represents the 2 balls (cans of coke we’re selecting) and bb represents the 3-1 dividers

Also known as 4 choose 2

[rich basin]

why does bb represent 3-1 dividers?

[dark granite]

More quickly, 4 choose 2. Since we could select two of the 4 spots for the a’s and the remaining spots will be where the b’s will go

[woeful pulsar]

please just ask in one channel

[rich basin]

yeah

[dark granite]

Replace aa with 00 and bb with the two bars

Now does it make sense?

@rich basin

[rich basin]

Sorry, I am sitll confused with the bb being two bars

Like i get how you do 4C2 being you choose 2 aa and the rest being the bb

[dark granite]

Okay we must place bars and balls in a row

[rich basin]

Don't quite link this to the bar idea

[dark granite]

Aabb makes up four total spots

Of those 4 spots place the a’s (the balls) and then the b’s (the bars) go in the remaining spots

The distribution of balls around the bars represents the distribution of balls into boxes

Because the dividers help us distinguish between what balls are in what boxes

[rich basin]

How would the aa and the bb in the idea of | |

[dark granite]

Aa is two identical letters the bars are two identical symbols

Okay explain this whole boxes into balls business back to me the best you can and hopefully we can figure out where your confusion is

Or dm me if you don’t wanna do this publicly

[rich basin]

So basically the boxes have n total boxes - 1 divisors, and we know what the distribution balls are because of the divisors

[dark granite]

Should say n-1 instead of just -1 there

[rich basin]

Yeah, and forexample something liek 1 | | , there are 2 balls in that specific type

whereas those who are null, would not be selected

[dark granite]

What does 1 represent there?

[rich basin]

The 1 represents that there are 2 balls in that specific type forexample, blue red and green. There would be 2 blue balls

[dark granite]

Uh okay, so how would you represent that there is only one ball in that type?

[rich basin]

0 | |

or | 0 | or | | 0

3 - 1 + 1 C 1

[dark granite]

Okay, why not just put two zeros where that 1 is originally?

[rich basin]

I thought you meant before that 0 is suppose to mean 1 item

[dark granite]

Because the way you have the 1 there doesn’t allow us to choose one item of one kind and another of another kind

Yes

To two items being selected would be two zeros

3 items, three zeros

And so on

[rich basin]

so like 0 0 0 | |

or 0 | 0 | 0

[dark granite]

The number of balls is the number of things you’re selecting

The number of bars is one less than the number of distinct varieties you have to choose from

[feral apex]

Can someone dumb it down for ke

[dark granite]

A question is being answered rn. Please wait or ask elsewhere

Does that make sense @rich basin

[rich basin]

Yeah i get that

[dark granite]

Dope

Kk so the card thing follows pretty similarly from there

[rich basin]

but don't quite see the distinction of adding the amount of balls ontop of forexample 3 +2 -1 choose 2

[dark granite]

5 identical balls and 52-1 dividers

[rich basin]

That is the only confusion I have

[dark granite]

Well remember it’s (3+2-1) choose 2

That whole quantity In parenthesis choose 2

3+2-1 represents how many spots we can choose 2 from to place the 2 dividers

[rich basin]

is it written as (3+2) -1 or is it (3-1) + 2?

[dark granite]

Or the 2 balls

Those are the same

[rich basin]

Lile how is it seen?

[dark granite]

Addition is associative

[rich basin]

Because i see the fifrst one as +2 being that it will firstly add the 2 replacement balls and then getting the dividers

[dark granite]

$\binom{3+2-1}{2}$

[ocean seal]

logician_pdx

[dark granite]

Like that

[rich basin]

and then the second being finding the dividers first

[dark granite]

Okay yeah the way you just described doesn’t quite encapsulate what’s going on here

Do you mind if I chat with you bout this later today? It’s 2:24am where I live and I’m doing this from my phone so I’ll be able to provide better explanation once I’m on my laptop

Could ya dm me the card question?

[rich basin]

Yeath that is fine, but what would the equation be before you go?

Would it look something likw

,w solve ((13+3 -1 ) choose 3 times (39+2 -1 ) choose 2) divide ( 56 choose 5)

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by BorutoEyePower#4998

[dark granite]

Yeah that’s for part a

[rich basin]

But then the answer is different

[dark granite]

I’d personally right that 56 as 52+5-1

The answer is different from what?

[rich basin]

it is different frm the wolfrma

[dark granite]

Wym?

Wolfram just calculates what you give it

[rich basin]

This is apprantly the answer

[dark granite]

That first fraction that I saw you post. Did that fraction equal this one here that you say is the answer?

[rich basin]

unfortunately not

[dark granite]

Can this answer not be reduced?

[rich basin]

,w solve 2717/33320

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by BorutoEyePower#4998

[rich basin]

It has a different decimal approximation

[dark granite]

Hmm any explanations that came with that answer?

[rich basin]

They didn't say anything, this is one of the even type question

[dark granite]

That answer is wrong dude

And frankly textbook answers are sometimes wrong

[rich basin]

Yeah, the answer could be wrong

The textbook has wrong answers

[dark granite]

I’d email your prof of the answer we worked out

Wait really?

Why are you using that textbook then?

[rich basin]

This is the textbook we have to use

[dark granite]

Damn but you’ve noticed it’s had wrong answers before?

[rich basin]

I got a wrong answer question before,

[dark granite]

Wdym?

Your answer was wrong or theirs?

[rich basin]

Like there was a question with a wrong answer in the previous chapters

[dark granite]

Oh gotcha

[rich basin]

So there is a chance this one could be wrong

[dark granite]

Yeah def email your teacher

Yes! Definitely

[rich basin]

Okay thanks. I will try to DM you tomorrow

[dark granite]

Okay cool sounds good

When you email your teacher, I’d show em exactly what we worked out and provide explanation for our answer

[rich basin]

Good night

[dark granite]

Thanks

[rich basin]

My teacher barely reply to emails

[dark granite]

Wow

Maybe also Throw your Question in the competition math chat and see if other people get what I’ve got

Just In case your prof doesn’t answer

This chat is open for questions now

[rich basin]

Okay I sure will, thanks

[dark granite]

You’re welcome

[clever pewter]

This question is from a book and that book usually contains pretty less solutions (solutions to selected problems) and I am stuck on this since quite a while; I will be glad if somebody helps me prove this.

there's also a hint on the book and the hint says: ||use the argument used in IMO 1988 #6 or in other words use vieta jumping/ infinite descent||

I have made some progress as well

let a^2+b^2+1 != 3ab or a^2+b^2+1 = kab where k != 3
and there exists (a,b) belongs to positive integers such that a^2+b^2+1=kab
then rearranging it we get x^2-(kb)x+b^2+1=0
Let (a,b) be the smallest solution possible (a+b= minimal)
wlog assume a>=b
then let ab=l
then, a^2+b^2+1=kl

and I don't see how I can proceed

I tried to prove it using contradiction

<@&286206848099549185>

[ivory otter]

can i do this

$\int 5x^6$ $dx = \frac{1}{7} * 5x^7 = \frac{5x^7}{7} = 5x^6$

[ocean seal]

MEOWBRO 父

[ivory otter]

ah shit noooooooo

i cant lol idk why in my head i thought it was (5x^7)/(x)

[thorn kindle]

Huh?

Wtf are you on man

[ivory otter]

xD

ok so a serious question

shouldnt the integral of

$\int 5x$ = x^5

[ocean seal]

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

[coral cosmos]

Wait, its 5x^2/2

[ivory otter]

once again i realise my mess up

if it was 5x^4

it would be x^5 right?

[coral cosmos]

Oh yes then it would be x^5

[ivory otter]

ok

[coral cosmos]

$\int 5x^4$ = x^5 + c

[ivory otter]

ye

[ocean seal]

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

[flat flicker]

Hi guys, I'm trying to solve a task using gauss' method but I'm struggling a bit. Could you help me please?

[woeful pulsar]

maybe share more information?

[flat flicker]

Ops, yes sorry😅

[vale sapphire]

A horizontal tangent line simply means that the derivative equals 0 at that point

[dark granite]

Use the method of proof known as constant descent root flipping

[clever pewter]

that's what the hint says

[dark granite]

Really? Okay I’m on point

[clever pewter]

yes seems like it

#help-0 message

[dark granite]

Suppose a=b. Then what do we have?

Look at the given condition

[clever pewter]

Hmph well I tried it with small values like (1,1) and it worked but how will I prove it

I mean that's the thing right

I have to prove it

[dark granite]

Bro look at what I just said

[wheat hare]

how to get the alternate form of this n/(n+1) ?

[clever pewter]

right

[dark granite]

A question is being answered here. Please wait or ask elsewhere

[alpine sable]

@dire pilot that is a

[clever pewter]

ok so I can basically assume that a=b instead

[dark granite]

At first yes

[clever pewter]

and do the grunt work

[dark granite]

Then we’ll cover a>b

Wlog

[clever pewter]

oh

so you mean wlog a≥b

[dark granite]

No

I mean exactly what I wrote

Cover the equality case first

[clever pewter]

okay

[dark granite]

Then do a>b

So if a=b, what happens?

[clever pewter]

we can change the equation and make that a^2|2a^2+1=3

[dark granite]

No

Still no

Look at the left hand side of the divides symbol

There we go

But why did you set that equal to 3?

We can’t do that

[clever pewter]

so are we trying to contradict the statement ab|a^2+b^2+1=3

[dark granite]

No

We are directly proving the equality case holds

Then we will use contradiction later for a>b

Dude change everything back to what you had with the equality case, but delete =3

[clever pewter]

ok so

a^2|2a^2+1

[dark granite]

Yeah

[clever pewter]

which is only possible for a=1

[dark granite]

Nice

How do you know that

Prove it

[clever pewter]

prove it? 😳

[dark granite]

You can’t just jump from that and say it only works for a=1 without showing any work

[clever pewter]

I don't really see how I can prove my claim

[dark granite]

What else does a^2 divide?

Does a^2 divide 2a^2?

[clever pewter]

ok I get what you mean

so => a^2 must divide 1

[dark granite]

Right but why

[clever pewter]

which forces a=1

but honestly really we need that?

[dark granite]

Yeah it is a proof right?

[clever pewter]

because a^2 divides 2a^2 and it must divide 1 to make the expression true

[dark granite]

We have a^2 divides two things, so it must divide their positive difference

And that positive difference is 1

[clever pewter]

right

[dark granite]

So a^2 divides 2a^2+1-2a^2=1

Since a is a natural number. Only a=1 satisfies this

[clever pewter]

yes

[dark granite]

So now we’re still not done with this equality case

What do we have to show?

[clever pewter]

we need to show that this whole thing =3 right

[dark granite]

We need to show 3a^2=2a^2+1

Just plug in a=1

[clever pewter]

yes

[dark granite]

Kk that is the end of the equality case

Did that part make sense?

[clever pewter]

but did we really need to show that it works lol

[clever pewter]

ofcourse it did

[dark granite]

Yes you’ll see why later

Okay so is it okay to assume a>b now? Explain why/why not

[clever pewter]

it is because even if we change the places of a and b it'll be the same

[dark granite]

Perfect

[clever pewter]

basically we don't lose generality

[dark granite]

So let’s assume without loss of generality that a>b

[clever pewter]

yes

[dark granite]

Suppose the statement is false now

Then what happens

Think about what it means for the statement to be false

[clever pewter]

well it basically means what I said right

before

[dark granite]

Which was

[clever pewter]

a^2+b^2+1 != 3ab

[dark granite]

Right

[clever pewter]

or just a^2+b^2+1=kab with k != 3

[dark granite]

Exactly

Okay now let’s make that equation look like a quadratic in terms of a

So move kab to the left hand side

[clever pewter]

lol so my solution was not completely wrong

[vale sapphire]

Quickly jumping in to say that there is one case of inequality that can be very easily eliminated

[clever pewter]

attempt to sol*

[dark granite]

Yeah sounds like you weren’t too far off

[clever pewter]

x^2-(kb)x+b^2+1=0

with a as one of the 0es

[dark granite]

Nice

Yes

By Vietas formulas, what must the other root look like

[vale sapphire]

nevermind

[clever pewter]

umm

[dark granite]

Look up em formulas yo

[clever pewter]

something like b^2+1/a?

[dark granite]

Right

And that equals another thing too

In terms of k

Kb-a

[clever pewter]

right

[dark granite]

Don’t forget order of operations. We’re dividing the sum by a, not just 1

[clever pewter]

(b^2+1)/a

[dark granite]

So this other root c equals bk-a and that fraction (b^2+1)/a

Is c an integer?

[clever pewter]

hmph

how do I judge

[dark granite]

What does c equal

Bk-a

Is that an integer

[clever pewter]

ok so a is an integer

and so is b

and so is k

so it has to be right

[dark granite]

Yes.

So now show c must be positive

Look at the other thing c equals for this

[clever pewter]

oh

(b^2+1)/a this should be positive right

[dark granite]

Yes but why

A and b are both positive right?

[clever pewter]

because a>b and b^2<a unless b is a natural number b<=1

[dark granite]

Um all you need to show c is positive is the fact that a,b>0

[clever pewter]

ah

[dark granite]

So the numerator and denominator of that fraction must both be positive

[clever pewter]

yes and we proved that it should be an integer from the other part

[dark granite]

So c is a natural number

Right?

[clever pewter]

yes

[dark granite]

Kk

Remember when we assumed the statement was false?

[clever pewter]

yea

[dark granite]

Notice that we can also assume that there is a smallest pair of natural numbers a and b for which it’s false. We can define them to be the smallest pair by saying the product ab is smallest

We’re allowed to do this by the well ordering principle

We need to show know that cb <ab

This will contradict the minimality of ab and finish the proof

[clever pewter]

right

[dark granite]

So using the fact that a>b and all we know about c, how can we show cb <ab?

Hint:

Show a^2>b^2+1

[clever pewter]

ca=b^2+1

[dark granite]

Right

[clever pewter]

and a^2>b^2

[dark granite]

Yes

But that won’t help with showing a^2>b^2+1

Since a>b, how can we relate a and b?

Write a in terms of b

[clever pewter]

hmph