help-26

[lapis lava]

Like say it was 15 though

Wouldn’t the number be even smaller than .01?

[verbal crater]

yep

but says nothing about being smaller than 0.001

[lapis lava]

Ohhhh so you’re saying that the numbers bigger than 13.227 are in the .01 category

And we don’t know at what number a .001 category starts

[verbal crater]

yes

because your table doesnt have it

maybe if you use a different table with more columns then you'll be sure

[lapis lava]

It it possible there is no .001 to consider because it’s not on the table

So we never pick a category that isn’t listed on the table

Like even if instead of 96 it was 1 million. We still don’t pick .001 because it’s not on the table

[pastel birch]

yeah

[lapis lava]

@pastel birch bro can you check if I did #1 and 2 correct?

[pastel birch]

Im pretty sure you did bro

but you will need to double check with someone

[lapis lava]

@verbal crater final answer .01?

[verbal crater]

sure

[lapis lava]

Is number 2 correct?

I did .61 x 1,000

Got 610

[verbal crater]

yeah

[lapis lava]

I guessed on question 1

It’s 50/50

Is there a correct way of deciding?

[verbal crater]

null hypothesis always assumes the thing claimed is true

alternative opposes it

[lapis lava]

Sounds good

Did I get this one right?

It’s a similar practice problem

They’re saying they want to see if the 5 teas have different sales

[verbal crater]

again like i said

null hypothesis supports the claim

and alternative opposes it

[lapis lava]

Did I do it correctly?

[verbal crater]

no

its the bottom one

[lapis lava]

They say : We will use the chi- square goodness-of-fit test to assess whether the five…

How come it’s the bottom one?

[verbal crater]

read the null hypothesis

does it agree or disagree

[lapis lava]

If I pick the bottom answer null hypothesis it says that all the sales are equal. That disagrees with the sentence of assessing if the 5 drinks have different sales.

[verbal crater]

ok i probably explained it poorly

the alternative hypothesis is basically someone like questioning the data

so they want to do a test

[lapis lava]

I submitted it and the .01 was the wrong answer

It doesn’t say what’s right though

[verbal crater]

then probably 0.05

[lapis lava]

Why?

[verbal crater]

0.05 is like the cutoff point for statistical tests

if p<0.05, you reject the null hypothesis

otherwise you accept the null hypothesis

[lapis lava]

So if p=.04 you reject the null hypothesis?

[verbal crater]

yes

[lapis lava]

How do you figure out what the null hypothesis is?

[verbal crater]

for example in this test, they want to test if the sales are different right

[lapis lava]

Yeah

But they also show you that the sales are different

[verbal crater]

thats the alternative hypothesis

[lapis lava]

In the problem statement

So the thing they tell you is the alternative hypothesis

[verbal crater]

the alternative hypothesis will be that the sales are indeed different

then the null hypothesis will be the opposite of that

so the null hypothesis is that the sales are not different

[verbal crater]

pretty much yeah

[lapis lava]

In the first one about the 1971 study it’s the opposite of that though

They show you the distribution fits and the null hypothesis is that the distribution fits

[verbal crater]

in that case, the test is about whether the distribution fits

so the alternative hypothesis will be that the distribution does not fit

[lapis lava]

But I thought right now we just agreed the alternative hypothesis is the one they tell you in the problem

Not the opposite

[verbal crater]

right now how do i explain this

[lapis lava]

Maybe give some random examples that would explain it better?

[verbal crater]

the null hypothesis is basically like what is claimed

so like if i said i earn 1000 dollars a month

thats the null hypothesis

now the alternative hypothesis challenges my claim

so it will be something like i dont earn 1000 dollars a month

[lapis lava]

I don’t think that explains the drink example problem

In that one they said the sales are different for each drink, but the null hypothesis is that they’re the same

[verbal crater]

because they are doing a test

doing a test means you're challenging their data correct?

hopefully this helps

im really bad at explaining this lmao

[lapis lava]

I’ll see if I can find a YouTube video or something

I have a final exam tomorrow so I’ll see if I can study other stuff to optimize the total points I get

If I get a mid C I can pass the class

[topaz sinew]

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[jolly glade]

how do I get the starting radian

[jolly glade]

my answer is correct except for the 4pi/3 part, I should've written 2pi/3. tbh I just randomly put 4pi/3 cuz the question before this had this answer but I don't know how to get it, where did 2pi/3 come from?

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@jolly glade Has your question been resolved?

[jolly glade]

COME ON

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@jolly glade Has your question been resolved?

[agile phoenix]

are u sure that its 2pi/3?

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[neon iron]

Question 3

[neon iron]

Pls welp

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@neon iron Has your question been resolved?

[weak cliff]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[weak cliff]

?

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[weak cliff]

.reopen

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[weak cliff]

This

[weak cliff]

(I failed in reopening the channel somehow)

@neon iron do you still need help?

[brave coral]

IIRC you need the "Helpful" to able to reopen channels that aren't your own

[weak cliff]

Ah thank you

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@weak cliff Has your question been resolved?

[neon iron]

No

[weak cliff]

👍

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[gray horizon]

im tbh not sure

[tawny moth]

the only things I can think of are:
do you usually decorate your unit vectors? i.e. $\hat{\imath}$ or $\vec{i}$?

[thorny flame]

robin goodfellow

[tawny moth]

Else maybe they want x, y, z instead of i, j, k

[gray horizon]

ima try

it worked

wow

[tawny moth]

very nice

[gray horizon]

thanks

[tawny moth]

all good! gl with your maths

[gray horizon]

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[slender bay]

How do I find I_3 & I_4?

[slender bay]

Damn scrolled up and nothing was answered for the past few hours

[neon iron]

not a physics server..

[slender bay]

Hm I thought I saw some physics stuff though

Eh Idk sorgy

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[pseudo pond]

Does c not exist because both one sided limits are different?

[pseudo pond]

Number 4

[restive inlet]

yes

[pseudo pond]

Thank

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[pseudo pond]

.reopen

[topaz sinew]

✅

[pseudo pond]

Just to make sure f(2) exists because there’s a solid point there right? 4.d

[restive inlet]

yes

[pseudo pond]

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[pseudo pond]

https://tenor.com/view/tank-m60t-turkish-tank-tsk-syria-gif-23298742

You

.reopen

[topaz sinew]

✅

[pseudo pond]

Is it possible for L to exist on this graph shown? Like can it approach 5 from the left if there’s another graph there

[weak cliff]

What do you mean with if theres another graph there?

[pseudo pond]

Like there’s a separate slope or line

I’m not sure what to call it

This is the question I’m answering

.rotate

,rotate

[weak cliff]

Only thing I can say is that the limit from the left DNE and the limit from the right does

[pseudo pond]

Could you explain why it is DND?

DNE

[weak cliff]

Because we have a sinusoid on the left

Which i think is divided by x or whatever it kinda doesnt matter

The point is

[pseudo pond]

It’s because like it’s a separate curve right

[weak cliff]

It can be any value between 4 and 2

We do not know

There is no limit

From the right it obviously goes to the value 3 though

Do you understand what Im saying? Im trying to explain it using intuition right now not really mathematical terms

[vocal void]

Just apply the definition of limit. For example, for epsilon = 1/2 you already can't find a delta such that.. blablabla

[weak cliff]

It could be 2 different curves if thats what your asking

[pseudo pond]

Is it because there’s too much sinusoid stuff happening

[weak cliff]

Yes

For example this limit here (blue curve i drew)

Also 2 separate curves

This limit does exist

Because we can find it from both the left and the right and it is the same

[pseudo pond]

I see

So what should be my explanation

[weak cliff]

Your explanation should just be that the limit from the left DNE

[pseudo pond]

I don’t think I could just write too much sinusoid stuff

[weak cliff]

Would be what I would do haha

But you could try saying it doesnt exist because the limit oscillates between 4 and 2

[vocal void]

Explanation is that in any neighborhood of the point, there are always two points a,b such that f(a) - f(b) is 2

Which is greater than epsilon

[pseudo pond]

Ok thanks

[weak cliff]

I dont think light is already at the epsilon delta explanation

Or are you?

[pseudo pond]

Yea

I don’t

I just said thanks though as a gesture

[weak cliff]

Yeah then it probably doesnt need to be that technical

[pseudo pond]

Ok cool

[weak cliff]

Saying too much sinusoidal stuff happening would probably be good enough

[pseudo pond]

Ok thank you

[weak cliff]

These kind of exercises are just to get a feel for limits

[pseudo pond]

alright cool

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[pseudo pond]

https://tenor.com/view/tank-m60t-turkish-tank-tsk-syria-gif-23298742

You

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[clear rivet]

Task: Assume that the points P, Q, and R lie on the sides AB, BC, and AC, respectively, of △ABC. Show that the circumcircles of △APR, △BPQ and △CQR meet at a single point.

I understand that I am suposed to use Miquels theorem, but I do not understand how I can prove that this is the case. I understand that if this is the case then the angles APR=AMR, BPQ=BMQ, CQR=CMR. But this does not prove anything if I just write it out.

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[clear rivet]

<@&286206848099549185>

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[rocky jewel]

"Using a convenient exponent law prove the logarithm law for log (x/y)"

[rocky jewel]

I really hate these kinds of tasks.. Any recommendations on how to attack them?

Like I know that log (x/y) = log(x) - log(y)

I also know that we can have a^b / a^c = a^b-c

[glass canyon]

$e^{a-b}=e^{a}/e^{b}$,

[rocky jewel]

yes

[thorny flame]

fish

[glass canyon]

perhaps, uh

[rocky jewel]

and whenever you look at the answers from the old exams they always just write "see book" and in the book they're like "oh we'll leave this as an assigment for you to figure out yourself"

so what would the "ideal" way be to show this? and similar problems?

I had someone tell me to assume that it's the opposite for another kind of question that was of the "prove"-type yesterday but i'm not sure how to apply it here 😢

[glass canyon]

well, if $e^{log(x/y)}=e^{log(x)-log(y)}$, then the law is true

[thorny flame]

fish

[glass canyon]

or alternatively, $log(x/y)=log(x)-log(y)$ gives x/y=x/y if exponential is taken

[thorny flame]

fish

[rocky jewel]

you think smth like this would work?

i feel like it's circular reasoning when i take out the ln e^b in the 2nd row on RHS

like im using what i want to prove ._-

[glass canyon]

i would say no

try having $\ln(x/y)=\ln(x)-\ln(y)$ as your initial statement

[thorny flame]

fish

[rocky jewel]

oh so show the transformation i want to prove

i like that

[glass canyon]

yeah

if $e^{ln(x/y)}=e^{ln(x)-ln(y)}$, then the initial statement is true

[thorny flame]

fish

[rocky jewel]

wat

so start like this

[glass canyon]

yes

[rocky jewel]

and then hit it with e

[glass canyon]

yes

[rocky jewel]

so like this then

[glass canyon]

yes

[rocky jewel]

and perhaps write some context like

"Using the fact that e^x - y is equivalent to e^x / e^y"

ok i'll try to remember this tactic

so essentially set up LHS to be where i start and RHS where i want to end assuming the rule is true

thank you a ton

legend

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[grim swift]

im stuck

did i good so far?

[wintry urchin]

I think you forgot to multiply the denomenator by sqrt(6+x)+3

[grim swift]

oh im stupid

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[severe lichen]

how do i get the derivative of (1-(6/(x+8)

1 is just 0

do i do -6 * 1/x+8

d/dx 1/(x+8) is 1

cause d/dx of x is just 1 and d/dx of 8 is 0

[brave coral]

Its 1/(x + 8) not (x + 8)

So the derivative wouldn't be 1

Rewrite it in index notation and apply the power rule

[severe lichen]

wdym

like

(x+8)^-1

-6 * (x+8)^-1

[brave coral]

Yes

[severe lichen]

how do i get the derivative of the last term

[fallow heart]

Using power rule

$\frac{d}{dx}\left(x^n\right) = n\cdot x^{n-1}$

[thorny flame]

Alberto Z.

[fallow heart]

(with n ≠ -1)

[severe lichen]

u mean chain rule

so g'(x) = 6(x+8)^-2

or 6/(x+8)^2 whichever u prefer

so i did
x^3 * 6(x+8)^-2 + (3x^2)(1-6/(x+8))

(-6x^3)/(x+8)^2 + (3x^2) - (18x/8)

[fallow heart]

Yep correct

[severe lichen]

am i done

[fallow heart]

I suppose so

If you want you can factor some stuff

[severe lichen]

this is a mess

[severe lichen]

hm how

[fallow heart]

The first minus (in front of the 6) should be a plus, now that I look carefully

[fallow heart]

And also the third term is -18x²/(x + 8)

[severe lichen]

(x-8)?

u mean x+8

[fallow heart]

Yeah

[severe lichen]

thanks

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[fallow heart]

You're welcome 🤗

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[fiery sigil]

how come I dont get the right answer

Not sure

[icy sky]

look at the top function again

[fiery sigil]

ryuh

[fiery sigil]

https://tenor.com/view/chinyk772-chinyk-fight-kratos-gif-13125980

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https://tenor.com/view/chinyk-chinyk772-fight-kratos-gif-13125977

[icy sky]

what

what did you plug in for the top function in desmos vs what is the top function

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[odd birch]

Each of the five quarters was marked using the same process, so it is possible to have duplicate letters within the set of five coins. No coin has the same letter written on both sides.

After marking the coins, they are placed in an opaque cloth bag. You reach into the bag and select a coin using a blind draw. Once selected, you flip the coin and place it on a table. You repeat this process for each coin, drawing it at random from the cloth bag, flipping it, and then placing it on the table. You do not have any special coin-flipping skill; the coin flip causes a random side of the coin to be selected. After the coins have all been drawn and flipped, the coins are arranged in a neat row in the order in which they were drawn.

Suppose you were hopeful that the result would spell the word "HELLO" after the selection and flipping process was completed. Furthermore, suppose that by random chance the marking of the coins coincidentally maximized the odds that you would get a favorable outcome. Keep in mind that the order in which the coins are drawn from the bag is still entirely random.

In this favorable scenario, what is the probability that the letters on the coins spell the word "HELLO"?

Could someone tell me if I am going in the right direction here?

My initial thinking was that the optimal markings for the coins to spell out "HELLO" would be for each coin to have one of the needed letters on either side, but not the same letter as the other side. So for example the coins might looks like this:

(H, O)
(E, L)
(L, E)
(L, H)
(O, L)

So all I would need is to calculate the probability of them being in the correct order using 5! or 1/120.

[odd birch]

The one thing I am not sure about is if there being 2 repeating L's makes getting "HELLO" more likely than just 1/5!

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@odd birch Has your question been resolved?

[odd birch]

<@&286206848099549185> 😳

[mint crescent]

Wouldn’t it be more favorable to have two double-sided L coins?

[odd birch]

I assume so but I have no idea how to calculate that

[mint crescent]

1. Brute force

2. Approach it one coin at a time. Ex. For the first coin, what’s the probability that it has an H? Then, what’s the probability you actually end up with an H? Rinse and repeat.

[grave girder]

can someone help me with calc2 problems?

[mint crescent]

!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).

[odd birch]

Wouldn't I have to like list out every single combination of heads/tails across the 4 letters for all 5 coins for this to work?

[mint crescent]

No

[odd birch]

Well how do I know the probability of the first coin being an H if I don't know each side of all the remaining coins?

What if the remaining 4 coins all have an H? What if none of the remaining coins have an H?

[mint crescent]

What happened to the “favorable scenario” part

[odd birch]

Well yeah obviously that is the part I have to figure out, the way that the 4 letters are spread across the 10 sides of all the coins, but I don't know how to figure that out

[mint crescent]

I’ll do this case since it’s the easiest to demonstrate my point with, but it’s the same for the other cases

Let’s say we have a (H,O), (H,E), (O,E) coin

[odd birch]

The same coin cannot have the same letter on each side

[mint crescent]

Nvm

[mint crescent]

Then, there’s a 2/3 chance we pick a coin with H and a 1/2 chance we actually land on H

Let’s say we pick the (H,O) coin

Then, we have (H,E) and (O,E) coin

And you can continue the logic from here - there’s a 1/2 chance of picking the coin with O and a 1/2 chance of actually flipping O

And then a 1/2 chance of picking an E

The other two cases are basically the same thing, so I’ll leave those to you

Anyway, I gtg now, but hopefully this helps 👋

[odd birch]

Does this include accounting for needing the coins to be in the correct order though?

The coins are drawn randomly

[mint crescent]

I forgot the two (L,L) coins

But yeah, this clearly accounts for order

[odd birch]

There cannot be an L,L coin

assuming you mean heads L, tails L

[mint crescent]

… I’m actually high

[odd birch]

I'm like more confused now lol

[mint crescent]

(H,
(E,
(L,
(L,
(O,

ok let's pair two of the letters with the L's

wlog, let's use H and E

(H,
(E,
(L, H)
(L, E)
(O,

so we just need to pair off L, L, O

so O pairs with L

and wlog, H pairs with L and E pairs with O

(H, L)
(E, O)
(L, H)
(L, E)
(O, L)

ok yeah your original config is the same thing, just with E and H switched

1. Flip H -> p = (2/5)(1/2) = 1/5

so then you have (E,O), (L,H), (L,E), (O,L) left

2. Flip an E -> two cases: (E,O) or (L,E)

I. (E,O) -> p = 1/8

(L, H), (L, E), (O, L) remaining

3. flip an L -> p = (2/3)(1/2) = 1/3

4. flip another L -> p = (1/2)(1/2)

5. flip an O -> p = 1/2

II. (L, E) -> p =1/8

(E, O), (L, H), (O, L)

3. flip an L

• (L,H) -> p=1/6

(E,O) and (O,L) left, the probability of L then O is 1/8

• (O,L)

(E,O) and (L,H) left, the probability of L then O is 1/8

yeah it's just nasty casework

[odd birch]

I cannot imagine that this is the answer considering that this question comes from like a very basic entry level CS thing where they ask me to explain my reasoning for the answer in a single sentence lol, but I will try to figure it out

[mint crescent]

hm

[odd birch]

wut

[mint crescent]

idk where that image came from

ignore that

Could someone tell me if I am going in the right direction here?

My initial thinking was that the optimal markings for the coins to spell out "HELLO" would be for each coin to have one of the needed letters on either side, but not the same letter as the other side. So for example the coins might looks like this:

(H, O)
(E, L)
(L, E)
(L, H)
(O, L)

So all I would need is to calculate the probability of them being in the correct order using 5! or 1/120.

The one thing I am not sure about is if there being 2 repeating L's makes getting "HELLO" more likely than just 1/5!

[mint crescent]

we could test it for an easier case

lemme do that rq

[odd birch]

There should be like an exact answer in the format of "N in M" according to the question

So I really think it is simple

[mint crescent]

how long do you have to do this

[mint crescent]

because I think the repetition is an issue

[odd birch]

The instructions for the thing say it shouldn't take long at all

[mint crescent]

eh maybe we are overthinking it

[odd birch]

It is the only question I am not 100% sure about

[mint crescent]

have you learned any fancy probability stuff

or just the definition and nothing else

[odd birch]

No

[mint crescent]

so no recursion, right?

What if we were to reorder "HELLO"?

Like if we did L, L, H, E, O

get the Ls out of the way first

(H, O)
(E, L)
(L, E)
(L, H)
(O, L)

ah wait

this forces us to pick (E, L) and (L, O or H)

so let's say we pick (E, L) and (L, O) in some order

then we're left with (H, O), (L, E), (L, H)

and then this part is smooth sailing

actually I think this is it

oh wait...

[odd birch]

I feel like the question is asking how to start a car and you are trying to like explain how to build a car from scratch

[mint crescent]

nah I'm using too many words to explain my point (mainly cause I was thinking out loud)

(H, O)
(E, L)
(L, E)
(L, H)
(O, L)

If we have this, then we must have (E, L) and either (L, H) or (L, O) for the L coins

and then the other three coins are fixed

that's it

._.

[odd birch]

1 in 3840 was my first answer where I assume each coin had one of HELO and one other random letter on the other side, then I would do (1/2)^5 multiplied by 1/5!

But that didn't seem as optimal as each coin having a needed letter on each side

[mint crescent]

wouldn't you need to multiply that by 4 for this part tho? (which of the L coins you pick first and which of the two other L coins you use for an L)

which gives you 1 in 960

[odd birch]

I cannot even keep up with all of the like strange formatting so I honestly have no clue

[mint crescent]

Do you want me to try to explain it differently?

[mint crescent]

cause I'll admit I did kinda gloss over this part

[odd birch]

yeah idk i was like maybe 30% confident when I posted and now i'm at 0% tbh lol

[mint crescent]

so is that a yes?

[odd birch]

i guess but i feel like you are just not understanding how simple of a question this is meant to be

[mint crescent]

Coins: (O, H), (L, E), (L, E), (L, H), (O, L)

Note that there's two coins with E's, so one of them must give us the E and one of them must give us the L

so that means we need to figure out how to order (O, H), (L, H), (O, L)

[mint crescent]

and at this point, there's 2 possibilities:

• (L, H) gives us the L, (O, H) gives the H, (O, L) gives the O
• (O, L) gives the L, (O, H) gives the O, (L, H) gives the H

tldr: with the optimal labelings, only two orderings of coins work

[odd birch]

How am I meant to explain this with two regular sentences lol

[mint crescent]

Take the coins to be labeled as (insert), observing that there are two coins with Es (meaning they must give an E and an L). This leaves two ways to order the other three coins (and thus all the coins) to spell HELLO, yielding the probability to be (2/(5!/2!))(1/2^5) = (whatever the hell that is).

like I said, I write a lot of stuff when I think out loud, so it appears way longer than it actually is

@odd birch are you good?

[odd birch]

I'm trying to like figure out how to write that as a fraction so I can turn it into N in M

but I dunno it just still seems convoluted given the context the question is in

[mint crescent]

,w (2/(5!/2!))(1/2^5)

[thorny flame]

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Requested by civil_service_pigeon

[mint crescent]

eh I'm not gonna force you to use it

but that's what I would say

[odd birch]

so I know that (1/2^5) is the chance that each of the coins gets the side you need, yes?

[mint crescent]

yes

[odd birch]

and the 2/(5!/2!)) is just accounting for there being 5 possible permutations since the set is has 5 things in it, but there is a repitition in the correct permutation?

[mint crescent]

2 is the number of favorable permutations, 5!/2! is the number of permutations, and 1/2^5 is the probability you do all the flips properly

[odd birch]

is the 5!/2! part in any way related to this? https://www.ck12.org/Probability/Permutations-with-Repetition/lesson/Permutations-with-Repetition-BSC-PST/

[mint crescent]

based on the title, yeah

ok yes it is

I just skimmed through it

[odd birch]

And this works for like either of the two possible sets of coins?

[mint crescent]

what does "two possible sets of coins" refer to here

the two favored ones or two sets of coins that both yield the optimal result

[odd birch]

Coins: (O, H), (L, E), (L, E), (L, H), (O, L)
Note that there's two coins with E's, so one of them must give us the E and one of them must give us the L
so that means we need to figure out how to order (O, H), (L, H), (O, L)

whatever this was about

[mint crescent]

5!/2! is the total number of permutations given that the coins are labeled optimally

[odd birch]

so why is that under 2 then in the first fraction?

[mint crescent]

it's the exact same logic as this for the most part

[odd birch]

So there are 60 possible arrangements that can be made from HELLO as well? Because both Apple and Hello are 5 letters with a letter that repeats twice?

[mint crescent]

yes

we care about how to order the coins and not the letters themselves

but like I said, it's the same thing

we have 5 coins and two of them are the same

[odd birch]

so if you know how, can you explain why the 5!/2! goes under the two? the diagram doesn't make a whole of of sense to me

I can see where 5!/2! is coming from but not the 2 above it

[mint crescent]

tldr: with the optimal labelings, only two orderings of coins work
two orderings (permutations) of the total 5!/2! work

[odd birch]

So am I like wording this right then?

I think I understand now

[mint crescent]

I would provide an example of an optimal arrangement for extra clarity

I would also say "there are 5!/2!=60 possible orderings of coins" rather than "we can use (5!/2!) or 60"

and I'd say "... land on the correct side, yielding an answer of 1/960"

basically cut out the last sentence of pure arithmetic

[odd birch]

(O, H), (L, E), (L, E), (L, H), (O, L) so this is one of the 2 optimal sets of coins?

[mint crescent]

you could double up (L, O), (L, H), or (L, E)

I wouldn't mention the total number of optimal labelings

it's not particularly relevant

[odd birch]

The optimal arrangement of letters on the coins would allow for each coin to have a different needed letter on both sides (ex: (O, H), (L, E), (L, E), (L, H), (O, L)). Since the bag contains exactly 5 coins but with a repetition (letter L) in the correct permutation, there are 5!/2!=60 possible permutations. Because there are 2 permutations out of the total 60 that give us the correct letters, we can use (2/60) and multiply it by (1/(2^5)) since we need all 5 coins to land on the correct side, yielding an answer of 1/960.
So this?

I understand it but like don't know how to like word it so they know what I am thinking lol

[mint crescent]

The optimal arrangement of letters on the An optimal labeling of the coins ~~would allow for each coin to have a different needed letter on both sides (ex: ~~ is (O, H), (L, E), (L, E), (L, H), (O, L). Since the bag contains exactly 5 coins but with a repetition (letter L) in the correct permutation but two of the same coin, there are 5!/2!=60 possible permutations orders to draw the coins. Because There are 2 permutations out of the total 60 that give us the correct letters, __possible orders to draw the coins in where it is possible to obtain the order "HELLO", making the probability of drawing such an order we can use (2/60). Since each coin must land on the correct side, and multiply it by we multiply this by (1/(2^5)) since we need all 5 coins to land on the correct side, yielding an answer of 1/960.

@odd birch ?

anything else

cause I'm going to sleep soon

[odd birch]

lol i was trying to get rid of all the markdown

but i think thats it

thx for helping

[mint crescent]

👍

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[mint crescent]

you could've just asked me for the version without the markdown too 😭

[topaz sinew]

@odd birch Has your question been resolved?

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[pastel birch]

<@&286206848099549185>

[pseudo sonnet]

this is not a question

do you have an actual question?

if not, please close this channel and stop trolling

[sharp dew]

Are you a pilot

He’s running out of oxygen I suppose

[pseudo sonnet]

well, I'll give OP one more minute

if nothing, then I'm closing this

a tragedy.

.close

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[thorny flame]

Lizzy

[median pivot]

i cant seem to find it

but since u can factorize the denominator

u can try partial fraction

yeah ppl only use it in integrals

for sum

reason

everyone is

same here

but ur math seems more advanced

so what we study

is like baby play to u

i dont do calc

im still in analysis

i js learn it for fun by myself

eh

ty

HELL YES

some minecraft to ease the pain

from education

[pearl fog]

play modded :3

[median pivot]

still w bud

[pearl fog]

non create modded still underrated

[median pivot]

yes

hard mode minecraft

[pearl fog]

gregtech

[median pivot]

wut

[knotty path]

you’re learning math the way that it developed historically

[median pivot]

if i learn triangle similarity before thales theorem

am i learning math in a historically developped way

[pearl fog]

mod

[median pivot]

did u not figure out partial fraction

[median pivot]

gregtech

ill js tell some bot to do it rq hold

[pearl fog]

greg tech 🔥

[median pivot]

im tucked in bed

[knotty path]

What does it mean by finding a p series of a function?

@neon iron

[median pivot]

here it is

partially fractioned

if thats a word

1/(x-1) btw can be deemed to be -1/(1-x)

[knotty path]

Yeah that should do it

[median pivot]

sup jasono

u want it in 1/(1-x) form kinda?

well

u can add up two terms

in one sum

[knotty path]

So turn the two fractions into individual p series

[median pivot]

well would it work if the second term is in 1/1-x form?

[knotty path]

Then add them together?

[median pivot]

what i thought

to add the sums?

if they have same boundaries

u can add them

[knotty path]

Im confused, to me the phrase p series means a series of the form 1/x^p

Lizzy do you have any example work I can see

[median pivot]

so like

yk how 1/(1-x) is the sum from 0 to infinity of x^n

yes?

[thorny flame]

Lizzy

[median pivot]

ahem

so whats the p series of -1/(1-x)

no u add -

[knotty path]

Perhaps we can get them into a geometric series? Is that what you mean

[median pivot]

lizzy in confusion

[knotty path]

Thanks

[thorny flame]

Lizzy

[median pivot]

this is where i go sleep

goodnight guys

[knotty path]

Lizzy, just for future reference, a power series and p series is not the same thing

Anyways first the fraction decomposition should give you

$\frac{3x}{(x-1)(x+2)} = \frac{1}{x-1} + \frac{2}{x+2}$

[thorny flame]

°Jason Parker°

°Jason Parker°

[knotty path]

Oops

$-\frac{1}{1-x} = -\sum_{n=0}^{\infty} x^n$

Also $\frac{2}{x+2} = \frac{2}{2(1 + \frac{x}{2})} = \frac{1}{1 + \frac{x}{2}}$

[thorny flame]

°Jason Parker°

°Jason Parker°

[knotty path]

So $\frac{1}{1 + \frac{x}{2}} = \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{2}\right)^n$

[thorny flame]

°Jason Parker°

[knotty path]

Thats it

Then you combine the two series

$\frac{3x}{(x-1)(x+2)} = \sum_{n=0}^{\infty} \left( (-1)^n - 1 \right) x^n$

[thorny flame]

°Jason Parker°

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[olive nexus]

hi

is p prime or positive integer or real number or whatever

[neon iron]

$\frac{p^4-1}{3p^2+3} = \frac{p^2-1}{3}$

[thorny flame]

Tchaikovsky

[olive nexus]

Oh wait this is true for any p, is the question to prove it is true for any p or something

[icy sky]

yeah this is true for any p

[neon iron]

im wondering how they got the answer

[icy sky]

use some factoring

[neon iron]

what steps were taken

[olive nexus]

oh okay

Do you know difference of squares

[neon iron]

i know binomial formulae

[olive nexus]

(a-b)(a+b) equals (a^2-b^2)

that

[neon iron]

yea

[olive nexus]

okay p^4 equals (p^2)^2 right

If you take square and then take square again you get 4th power

[neon iron]

ok

[olive nexus]

Then we can do difference of squares on p^4-1

p^4-1 = (p^2-1)(p^2+1)

[neon iron]

is this easy to do mentally for you

or will you have to write it down to solve it

[olive nexus]

Oh why you ask

[neon iron]

because the book says i should be able to solve that mentally

but im not that good yet

[olive nexus]

Who cares what the book says

[neon iron]

at least i dont think

[olive nexus]

Focus on understanding the problem at hand

you can write down if you want

[neon iron]

ok

[olive nexus]

Are we good with this

[neon iron]

yea

[olive nexus]

3p^2+3 = 3(p^2+1) are we okay with this

[neon iron]

$\frac{(p^2-1)(p^2+1)}{3(p^2+1)}$

this

[olive nexus]

no 3(p^2+1)

[thorny flame]

Tchaikovsky

[olive nexus]

Yeah, p^2+1’s cancel each other

we multiplied by that and then divided by that cancelled it out

[neon iron]

okay i see now thank you

[olive nexus]

no problem

[neon iron]

i will send a simular problem here im having trouble with but you dont have to help me if you dont want to

[olive nexus]

You can send it

[neon iron]

thank you

$\frac{15u^2-24uv}{12u^2}=\frac{3u(5u-8v)}{12u^2}=\frac{5u-8v}{4u}$

[thorny flame]

Tchaikovsky

[neon iron]

$\frac{15u^2-24uv}{12u^2}=\frac{\frac{5u^2-8uv}{3u}}{\frac{12u^2}{3u}}=\frac{5u-8v}{4u}$

which way is more ideal

when the book gave me this problem i solved it the bottom way

but it told me the solution with the top way

[olive nexus]

Both gibe you the answer

[olive nexus]

there isn’t any problem then?

[thorny flame]

Tchaikovsky

[neon iron]

okay

i was just wondering if my way was not the best

since the book gave a different approach

[olive nexus]

you got the answer doesn’t matter

[neon iron]

okay

[olive nexus]

In both you just simplify everything by 3u

I personally would’ve thought of the first but doesn’t matter they are doing the same thing

[neon iron]

okay thank you very much

you are very helpful

.close

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[elfin niche]

What is the Zero of the function?

[odd pagoda]

a zero of a function f is a number x with f(x)=0

[elfin niche]

Can it be interpreted as, what numbers that can make the function 0?

[restive inlet]

yes

[elfin niche]

What if theres none?

[odd pagoda]

well then there is none

[elfin niche]

Then no numbers can make it 0?

I see

$\frac{-2}{x-10} < 0$

[thorny flame]

Beersathought

[elfin niche]

Then its something like

Zero of the function : None
Restriction x = 10?

[odd pagoda]

well thats not a function

do you only mean the left side?

but yes

[elfin niche]

Yes

[elfin niche]

It's a Rational Inequality

Though I have no idea what that means

[odd pagoda]

a rational function is a polynomial divided by a polynomial

[elfin niche]

Is Rational function same thing as Rational equation?

[odd pagoda]

no

[elfin niche]

Ok

Rational functions are with domain and range so to speak?

[odd pagoda]

they are connected obviously but a function is not an equation

[elfin niche]

Smth like $f(x) = -\sqrt{x-2} + 2$

[thorny flame]

Beersathought

[elfin niche]

I see

[odd pagoda]

well thats not a rational function but yes

[elfin niche]

Oh wait this is Radical functions right

[odd pagoda]

yes

[elfin niche]

Ohh are Rational functions the one with, Asymptotes?

The Vertical, Horizontal, Oblique, x integers and y integers?

[odd pagoda]

well lots of functions have asymptotes

[odd pagoda]

.

so something like f(x)=(x-3)/(x^2+17x-1)

[elfin niche]

I see

$f(x) = \frac{4x-2}{3x+1}$

[thorny flame]

Beersathought

[elfin niche]

This is a very dumb question, how do you know if a term? is a polynomial?

[odd pagoda]

a polynomial is a sum of terms of the form number*x^k where k is some natural number

[elfin niche]

Something like 4x , and -2 is a single term right?

[odd pagoda]

yes

[elfin niche]

But if combined, 4x-2 this is a polynomial, correct?

[odd pagoda]

even on their own they would already be polynomials

[elfin niche]

I see

Random question, but why solve for Asymptotes? Like irl, how could it help?

[odd pagoda]

well for example x could be time in some physical experiment. then the horizontal asymptote would tell you what happens to whatever f represents when the experiment runs for very long

[elfin niche]

What's "f"

[odd pagoda]

f(x)

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@elfin niche Has your question been resolved?

[elfin niche]

I see

Thanks a lot

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[long magnet]

alright, I understand all the sin/cos/tan stuff I just struggle with the way of solving the square roots, like why did the sqrt(65) go up and is multiplied by the b(4)

[strange kindle]

When a square root is in the denominator, we multiply and divide the square root by itself

So the denominator becomes rational

[long magnet]

I see, ok I have another question. is there a simple way of finding 4sqrt(26)?

[strange kindle]

The closest perfect square number close to 26 is?

[long magnet]

hmm I know that, but is there a quick way of finding it? or do I have to keep dividing by numbers until I find it's perfect square number

[strange kindle]

You could remember some perfect square numbers like upto 15 or something

But yes there is a way to calculate

Square roots

You can look it up on YouTube or wait for someone else as I last used it years ago

[long magnet]

alright, I will go do some searching. thank you so much

[strange kindle]

No problem.

[long magnet]

.close

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[vale hemlock]

nevermind i figured it out

.close

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[sturdy forge]

How can I solve it without using polar form?

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@sturdy forge Has your question been resolved?

[sturdy forge]

<@&286206848099549185>

[pearl patio]

there is couple ways to do it

one way is to transform the integeral over D to one over dx and dy

1: what does D bound y by?
2: if we fix y, what are the possible values of x?

[sturdy forge]

Are u saying that x will become √(4-y²)? I was doing it like integration of small circle through the radius as it is a disc

[pearl patio]

x will be between -√(4-y²) and √(4-y²)

fixing a y value, we are integrating the horizontal strip between -√(4-y²) and √(4-y²)

[sturdy forge]

Yeah and y will be in between -2 to 2?

Because if I put zero in the place of y then I got -2 and 2

[pearl patio]

yeah

think of it this way:

the integral over D = integral_{-2}^2 something dy

what is this "something"?

it is the red lines that are formed when we fix y

[sturdy forge]

Oh ..yes and then we are ranging x as -2 to 2 which is also radius

[pearl patio]

the way we are ranging x depends on y

as you see the length of the red line sort of depends on the y coordinate

so $$\int_D =\int_{-2}^2 \text{length of red line} \ dy$$

[thorny flame]

qwertytrewq

[pearl patio]

and the length of the red line, weighted by some function $f$ is our normal one variable integral!

[thorny flame]

qwertytrewq

[sturdy forge]

Yeah. ..I think if I do first integral w.r.t x then I will get the equation of a line

[pearl patio]

yeah then can you figure out what the integral is?

$$

$$\int\int_D x+y+10\ dx\ dy= \int_{-2}^2\int_?^? x+y+10\ dx\ dy$$

[thorny flame]

qwertytrewq

[sturdy forge]

X limits are from -√(4-y²) to √(4-y²)

[pearl patio]

OK!

so now you can compute the integral this way

$$\int_{-2}^2\int_{-\sqrt {4-y^2}}^{\sqrt{4-y^2}} x+y+10\ dx\ dy$$

[thorny flame]

qwertytrewq

[sturdy forge]

Yess... thankyou!

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[sullen herald]

How to show without calculator

[sudden temple]

assume both the arcsine terms to be a,b or whatever you like

[sullen herald]

yeh

oh

[sudden temple]

then say those a+b=y

[sullen herald]

yeh

@sudden temple ?

[sudden temple]

yeah

[sudden temple]

now you may proceed by taking sine on both sides

[sullen herald]

sine (a+b) = siny

[sudden temple]

yeah

[sullen herald]

oh ok

yeh now i get it

ill close chat after i do the question

close

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[sullen herald]

like b4 i aint got a clue on what to do 🤣

[pearl patio]

sin is the opposite over disgonal, so i would draw a triangle with oppsite as 2 and diagonal 3

[sullen herald]

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[sullen herald]

i found without drawing i think

[pearl patio]

oh 🔥 if u solved it

[sullen herald]

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✅

[sullen herald]

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[eternal plume]

What's the difference between system of equations and simultaneous equations?

[cursive patrol]

in almost all contexts, these phrases are interchangeable

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[neon iron]

Question

[neon iron]

i was able to notice that if we get together cot 28 , cotcot32 ,cot 88 together it helps

it would be cot84 all together

anf will look like cot84 * cot12 * cot24 *cot48

cot84 * cot12 * cot24 *cot48 (cot 72) tan72

cot84 * cot12 * {cot24 * cot48 * cot72 }* tan72

cot84 * cot12 * {cot72}* tan72

cot84 * cot12

now what

[surreal gulch]

cot is cotangens here? remember what cotangens is supposed to express

[neon iron]

cos/sin

[surreal gulch]

no simpler

maybe use that $cot(90-a)*cot(a)=1$

[thorny flame]

Jigglyproff

[neon iron]

i dont see any such complementary angles

[surreal gulch]

ah whoops

nvm

90=100 sorry about that

[neon iron]

mm

[neon iron]

what next ?

<@&286206848099549185>

[supple gulch]

look

Bottom 2 lines

Cot is cotg

[neon iron]

oh wait i made a mistake LMAO

the answer should be 1

[supple gulch]

yea

[neon iron]

thanks

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[fallow torrent]

Hi everyone.

I'm stuck with this: $\lim_{x\to +\infty} \sqrt[3]{2+x^3} - \sqrt[3]{1+2x^2+x^3}$

[thorny flame]

Shadow91518

[plush tulip]

(a-b)(a^2+ab+b^2) = a^3 - b^3

[coarse tusk]

rationalisation!

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[half field]

Can someone guide me part (c) ?

[mint crescent]

Hint: Write the area of the shaded region as a difference of areas

[half field]

Ok

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[empty gyro]

This is not really a math problem but a mod told me asking was ok
It's a physics circuit related question. And more of a conceptual question than a specific problem. I did a capacitor circuit problem with the following diagram today. Where "L" is closed/connected. C2 and C3 and evidently in parallel.
But I argued with a classmate since I thought (C2 and C3) were in series with (C1 and C4), while they thought they were in parallel
My confusion is: What exactly defines that 2 branches are in parallel? Just that the two of them start in the same node, and end in the same node?

[empty gyro]

This is what the diagram looks like with L closed

and, if I "replace" C2 and C3 with an equivalent capacitor C23 according to them being in parallel, it would seem that now there are 3 capacitors in series

[cursive patrol]

hm

[empty gyro]

but the truth is that originally there were 3 branches in parallel right?

[cursive patrol]

which branches?

[empty gyro]

a C2 branch, a C3 branch, and a C1 -> C4 branch
All three of which share their starting node and their ending node

I tried to color the nodes in green and red here, and in purple what I'm considering the 3 "branches"

like, it would be just as valid to first combine C1 and C4 into an equivalent C41 capacitor (As they are in series), and then the diagram looks like this right?

Can I just combine any 2 of these "in parallel" , and then that combination is left "in series" with the remaining branch?

is it wrong to combine these 3 at once "in parallel"?

I'm probably just deeply misguided at a fundamental level on something here

[cursive patrol]

it depends on how the battery is placed/connected

[empty gyro]

there was no battery in this circuit

just a capacitor with a starting charge, and then the circuit was closed and the question was about the resulting charge configuration

[cursive patrol]

ah good enough then

[empty gyro]

"like a loop" here meaning, a closed circuit that is not connected anywhere else?

and does "go with the series interpretation" mean choosing any 2 of the 3 branches to first "combine" in parallel, and then "combine" the resulting circuit "in series" ?

I had a followup question about what would change by adding a battery like this, if that changes what things are "in parallel" or "in series". But you kinda already answered that maybe? that it does
Are things still technically the same, just that there is no way to "replace" a battery and a capacitor in series with a single capacitor, so we solve/simplify things in the only way/s that we can? (Being forced to leave the battery as-is, thus having to "merge" C2 and C3)

[cursive patrol]

the issue is something being in parallel or series implies there’s some concept of current

[cursive patrol]

if there was a battery like this then yes the equivalent circuit would have three capacitors in series

however say you took another wire and connected the two nodes in that circuit to a battery. then your equivalent circuit would have three capacitors in parallel

[empty gyro]

Mhm
that makes sense

I'm not sure how I'd solve circuits with multiple batteries then but

[empty gyro]

before going there, I tried to "solve" the equivalent circuit in different "ways" for the no-battery one, and it gives different results. So the reference point taken matters

I suppose maybe I should explain the original exercise that made me question this better

C1, C2 and C3 hold no charge
C4 holds a charge Q4
then L is closed
And it asks what the charge is at C3 after the system stabilizes

would C4 here act as a battery?

in terms of what I should consider "in series" or "in parallel" I mean

[loud oasis]

once everything is stabilized, all capacitors act as capacitors. the one thing that holds over is that the total charge on all capacitors must be the same as the original charge on C4

[empty gyro]

well, the charge at C1 is the same as the original one in C4, right?

the question is how the charge in C2 and C3 is distributed

but their sum is the same as Q4

with this polarity, I think?
Where Q1=Q4=Q2+Q3

[loud oasis]

the polarities on the top capacitors should be reversed. note that C4 is in parallel with the equivalent C(1,2,3) capacitor

[empty gyro]

doesnt the charge on each plate of C4 generate an opposite charge in the other end of each node?

[loud oasis]

when L is released, + charge from C4 must flow out and accumulate on the left plates of C2 and C3, causing + charge to leave the right plates (hence a - charge build up) and so on throughout the circuit

we should also note that on C4, the left side has a higher potential than the right side, so that should be reflected on the top capacitors

[empty gyro]

ah, you're right. I'm being dumb

so here, "parallel" or "in series" is defined by the polarities?

like, if they share charge

Thanks a lot to both of you, I think I get it now

[loud oasis]

here i would say that they are in parallel if both plates are connected, and in series if one plate is connected

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[lucid junco]

how do i do part B?

[cedar wagon]

What is the shaded area to the left of point P ?

[modern rock]

just set up 2 integrals that are equal to each other

the integral from 0 to P should be equal to the integral from P to k

[lucid junco]

Oh yea

tru

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[iron tapir]

whats the formula for counting
if i have 5 = 100%
then i need 6 = .. ?

[iron tapir]

oh nvm im dumb
6/5 * 100% ?

[wise ivy]

Yeah, it's proportion

[chilly walrus]

yes

[iron tapir]

the result is 120% but something is odd

well, im playing a game and i have a pistol with 5 ammo
with bonus stats 14,8%
somehow i got extra bullet

[wise ivy]

You are lucky

[iron tapir]

lucky for having a broken game

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[raw zenith]

What is the difference between an implicit and an explicit function?

[raw zenith]

If it is neatly written in terms of x, then it is explicit and if not it is implicit, but what if I have
x + y = 1
y = 1-x

These two are basically denoting the same relation of x and y

But is the first one explicit and second one implicit?

also what are the times I will not be able to convert an implicit function to an explicit function

One example I can think of is when y has a square on it, like y^2 + x^2 = 25

This will result into two functions if we try to separate it

[prisma mesa]

Opposite, first one is implicit and second one is implicit

but yeah, they denote the same relation

"explicit" and "implicit" are tied to the way of writing the relation, rather than the relation itself

[prisma mesa]

Usually it will be when it's either not even a function (such as the circle example you provided), or when it's hard / impossible to write it explicitly in closed form

$y^{2}+y\cdot2^{y}-\frac{x^2}{y}=0$

[thorny flame]

MæthIsAlwaysRight

[prisma mesa]

something like this would be probably really hard to write explicitly

[raw zenith]

closed form meaning?

[prisma mesa]

it means that it's writable in terms of standard functions and operations

[raw zenith]

oh ok

[prisma mesa]

e.g. you cant convert this to an explicit closed form

at least im pretty sure you cant

[raw zenith]

I have studied about functions a lot but relations I haven't studied about much, they just popped up in differentiation lol

I was studying implicit diffrentiation

[prisma mesa]

they are quite similar, function is just a special type of relation

i see

[raw zenith]

yes

[raw zenith]

got it

k thank you that clears it up

I'll ask more doubts see ya again

👋

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[supple gulch]

Hello

[supple gulch]

Given a trapezoid

ABCD

Ad is 6 , BC is 8, and the angle between the diagonals alpha is 90

CD is x, AB is 14-x

Find x

[wise ivy]

AD and BC is top and down sides?

[supple gulch]

No

[wise ivy]

Aha

[supple gulch]

AB is lower CD is top

I rly need help

<@&286206848099549185> @amber locust

[mild stirrup]

u have to find x right?

[prisma mesa]

Be more patient, helpers are supposed to be pinged after at least 15 mins of waiting

[flint stump]

wait, whats that 4sqrt2 side?

[supple gulch]

Yes

[supple gulch]

I tried sketching something

I tried constructing CQ perpendicular to the diagonla BR