help-36

[ivory bear]

yes maybe ill try your ques man

but idk mate

monday i have a school test too

so maybe ill leave in middle

[valid carbon]

lmao how do we have freaking jiya as a mutual

small world

never noticed before

[ivory bear]

her name is jiya?

i too noticed that fact long before

ig you were in r/jnt and left it

[valid carbon]

jeeneetards?

[vague anchor]

me tomorrow😭

[ivory bear]

yes

[valid carbon]

no i know her from ages ago

[ivory bear]

SUNDAY?

[vague anchor]

yes

[valid carbon]

i was wondering if you were a irl friend or smth

[ivory bear]

ask her

[valid carbon]

thatd be very small world

anyway

[ivory bear]

@vague anchor .

[valid carbon]

i hate this question

[ivory bear]

you from thailand or smth right?

[vague anchor]

im on laptop, isnt it zoomable

[valid carbon]

ye afvahein kaun faila raha hai bruh

[ivory bear]

no not visible clearly

the e^stuff isnt visible

[ivory bear]

the fuck man

[valid carbon]

skill issue dude

[ivory bear]

tu bhi india se?

[ivory bear]

my god mera laptop old hogya then

mb.

[valid carbon]

yeah sadly

[vague anchor]

2sin^2 x on RHS at numerator

[valid carbon]

no

[vague anchor]

2

[ivory bear]

teri photu hi shiv ji jaisi dikhri thi so i firsthand suspected that you are an indian

[valid carbon]

,av

[soft zealot]

percykel's Avatar

Click here to view the image.

[ivory bear]

but i asked you too about it but you told thailand or some other country

[valid carbon]

its fan art of thanatos

[ivory bear]

still quite the resemblence of him at a glance right?

[valid carbon]

the greek god of non-violent death

[valid carbon]

less blue for one

[vague anchor]

IG I pingged to early😓 😓

[ivory bear]

||yes for sure but physique dekh ke lagra hai||

delete it

[valid carbon]

dont make him ghost ping

!nosols

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vague anchor]

when did i say it as answer

[ivory bear]

ye cheex bekar lagti fr

saala no soln chep dete

[valid carbon]

take it up with the mods

or stick to jnt

[ivory bear]

jnt pe hu but timeout daalke rakha hai rn

school test spree

[vague anchor]

its minimum better than
no helping and yapping here

i have answer

what i want is method and solution approach
nosols thop detae ho saalo

[ivory bear]

yes sir ill too try

[valid carbon]

what i want is method and solution approach

yeah so

!nosols

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[valid carbon]

nobody is gonna give you the solution

[vague anchor]

that is answer man

[valid carbon]

i mean i dunno these people might

[ivory bear]

dms might work

[valid carbon]

maybe its an ego trip for them

[ivory bear]

chodo usko wo anime fan lagra door se

[vague anchor]

bro first know why was that made and then use

then?

ik

but help me find max value of RHS only

[valid carbon]

ok

[vague anchor]

that is the topic headed with

[valid carbon]

either you stop or im mod pinging

we also have the insane amount of shade thrown towards me for no reason

[vague anchor]

pls go bro dont annoy us rn

pls

no abuse @terse sorrel

[valid carbon]

<@&268886789983436800> see

indeed

also its greek

thanks roketto

@vague anchor id suggest closing this and reposting while telling us what you've already tried

[bold turtle]

Hard agree on this; this actually should be restarted anew

[valid carbon]

only reason im not closing it myself is to not seem like a total dick

[vague anchor]

<@&268886789983436800>
can u also take action on this guy who was yapping here nepou or whatever

.close

[final saddle]

Channel closed

Closed by @vague anchor

Use .reopen if this was a mistake.

[vague anchor]

.close

[opaque ember]

wot

[valid carbon]

[tired walrus]

tf happened over here

that was one trainwreck of a convo

[bold turtle]

It was a train that lost its engine

[valid carbon]

you guys cant even follow half of it cuz its hindi

but yeah

[bold turtle]

oh I can and it's still a trainwreck

[valid carbon]

it def doesnt make it better

[vague anchor]

😭

[tired walrus]

i couldnt follow half of it cause this guy literally asked me to block him so he can't DM me anymore

[vague anchor]

now u might unblock me
i have got control on my senses

[valid carbon]

.reopen because this is hilarious

[final saddle]

✅

[vague anchor]

why in this world was it reopened😵‍💫

[vague anchor]

😑

[old quarry]

u can close it again

[vague anchor]

im not gonna its none of my duty now

[vague anchor]

is it?

[old quarry]

nah u can do what u want

[vague anchor]

still i dont wanna make trash of this server
maths🥹

.close

[final saddle]

Channel closed

Closed by @vague anchor

Use .reopen if this was a mistake.

[bold turtle]

For the record - the only people who can close channels are the person who opened the channel to begin with and anyone with at least a Helpful* role or greater

[dusty quarry]

Helpful*

[bold turtle]

This has been noted, no need to keep this train off its tracks.

[old quarry]

🚋

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[round cedar]

hey hru, i wanted to prove that the red brackets diverges so i did those steps

is that ok?

btw ive shown that the red brackets is >0 for every n which is natural

[tired walrus]

0*e^∞ = e^∞

[round cedar]

e^x growing faster than x

hru btw

[tired walrus]

btw was this limit given to you \textbf{exactly} in the form $\frac{\frac1n - e^{-n^2}}{\frac1n}$ or did you mess with it in any way

[soft zealot]

Ann

[round cedar]

no i wanted to show that the red brackets diverges (cuz ive came to a conclusion it is since 1/n diverges and e^-n^2 converges)

[tired walrus]

ok gonna ask for the original then

[round cedar]

in retrospect i could have shown that e^-n^2 converges and the whole red brackets diverges since its diverges + converges

[round cedar]

ok

[tired walrus]

i really think you're on a road to completely confuse yourself (partly because even i am confused)

[round cedar]

lol

[round cedar]

i was needed to prove this one converges but not aboslutely converges

[tired walrus]

erase the "was"

but yes ok

the most straightforward way is in fact to break it into the sum of two series

[round cedar]

"conditionally convergent series"

[tired walrus]

$\sum_{n=1}^{\infty} \frac{(-1)^n}{n} - \sum_{n=1}^{\infty} (-1)^n e^{-n^2}$

[soft zealot]

Ann

[tired walrus]

which you can analyze with about 97% less headache as: $$\underbrace{\sum_{n=1}^{\infty} \frac{(-1)^n}{n}}{\text{cond. conv.}} - \underbrace{\sum{n=1}^{\infty} (-1)^n e^{-n^2}}_{\text{abs. conv.}}$$

[soft zealot]

Ann

[round cedar]

if i prove each left and right how can i prove the sum of it cond converges

[tired walrus]

proving that it converges (without specifying cond. or abs.) is straightforward

from then you need to show that it fails to converge absolutely.

it will be best to do this abstractly, unless that's considered non-kosher in your class

meaning that you prove the more general result

[round cedar]

that makes sense

[tired walrus]

if $\sum_{n=1}^{\infty} c_n$ converges conditionally and $\sum_{n=1}^{\infty} a_n$ converges absolutely, then $\sum_{n=1}^{\infty} (c_n + a_n)$ converges conditionally.

[soft zealot]

Ann

[tired walrus]

the choice of names is deliberate on my part

this can be done with some funny business involving the triangle inequality

which i think would be $|c_n + a_n| \geq |c_n| - |a_n|$ or something like this

[soft zealot]

Ann

[round cedar]

wish we had a thorem like this which we could use, but as you say i need to prove its converges & not abs converges --> then it cond converges

[round cedar]

if i understood correctly

[tired walrus]

this is a theorem that you would prove on the spot

[final saddle]

@round cedar Has your question been resolved?

[round cedar]

@tired walrus is that ok to prove red series abs converges ? so then it converges

[tired walrus]

i can't read hebrew so i can't follow the logic.

but im p sure the limit of the ratio of adjacent terms is 0 not 1.

even better for you, but you have to not lie in your proof.

[round cedar]

but it goes 0 for me

[tired walrus]

what is this symbol

[round cedar]

lol its some sort of greek "p" idk

if p<1 then it converges

[tired walrus]

... maybe you mean rho?

but if it's a greek letter, you drew it in such a wonky manner that it looks like phi instead.

[round cedar]

yeh it is

i googled it

rn

[round cedar]

ohhh thats the difference ok good to know

@tired walrus used to chatgpt to translate what i have written

not the best translation but does the job

[tired walrus]

ok yeah that will do

[round cedar]

thxxx

.close

[final saddle]

Channel closed

Closed by @round cedar

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bold zenith]

!status

[final saddle]

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

[bold zenith]

Also, do you have a diagram/sketch? If not, draw one and post a photo of it.

[clear rivet]

I do yeah

[robust mulch]

where is this question from?

looks like an imo q

[clear rivet]

egmo 2

[robust mulch]

ohh

cool

[bold zenith]

Top tip: Make sure to make your sketches big

[clear rivet]

anybody know how to solve

[bold zenith]

I don’t actually I suck at geometry but I hope another helper can use what you’ve posted so far

[clear rivet]

alexis man plzzz

[bold zenith]

You can ping helpers role after 15 mins of posting

Not yet

It’s a high traffic time with like a few helpers and I don’t know geometry

[robust mulch]

your T is on the wrong side of BC

[final saddle]

@clear rivet Has your question been resolved?

[clear rivet]

yeah ur right

[final saddle]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chilly raven]

i wanted to ask which calculator is better for 0580 IGCSE
991 ex
991 cw

[scarlet sonnet]

the cw is abysmal to use

if you're used to the old design stick to ex imo

[final saddle]

@chilly raven Has your question been resolved?

[chilly raven]

k

[chilly raven]

if I haven't used any then?

[scarlet sonnet]

not sure

i definitely prefer ex but that's because i've always used that design calculator from casio

[final saddle]

Channel closed

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vague trench]

is this good?

[vague trench]

some guy told me that i should be "solving by inspection"

i dont know what that means at all

[bold zenith]

Yes

[atomic moon]

Yes its good

[bold zenith]

Basically trial and error i think

The answer you got is correct assuming x is real

If you don’t know complex numbers yet you’re good but given it’s x not z we should be good

Ignore what I said, (basically there are weird numbers you might not know about that can give negative answers after exponentiation)

[atomic moon]

Guess the solution from what i read, its not a real method and more about hit and trial of some values

[vague trench]

oh

that sounds confusing lol

[atomic moon]

Probably based on the fact that this function is strict increasing + continious and so only one solution

Btw when you do a change of variable, make sure to write the domain of the new variable

Here u = 2^x is > 0

[vague trench]

what do u mean by domain

[atomic moon]

So writing u = -3/2 is quite a mistake but it depends the rigor of the course

[atomic moon]

The values that the variable can realisticaly take

[vague trench]

oh

[bold zenith]

I haven’t seen it being called a mistake before, to me it is just accepted as a value of u but with no solutions

[atomic moon]

Yeah it depends on the one who grades

Anyway its just that since ive seen this '?' After "no solution" i just make it clear that its normal to not have any solutions to u = -3/2

[final saddle]

@vague trench Has your question been resolved?

Channel closed

Closed by @vague trench

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[versed terrace]

I understand why a^-n equals the last one, but what is the point of having the last one there? Isn't it the same as the second one cause 1 to the power of anything is 1?

[bold zenith]

Yes

[versed terrace]

wait so why is the last one there

[bold zenith]

Because it is, just in case you don’t realise it’s the same

[versed terrace]

is the last one unsimplified then?

[formal trail]

the last one is there in case it isn't immediately obvious to you that the last two are equal

since it is a useful fact sometimes

[versed terrace]

oh

[bold zenith]

What I said with more words

[versed terrace]

alr thanks guys

.close

[final saddle]

Channel closed

Closed by @versed terrace

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[golden cave]

idk where to start

maybe find BCA?

[bold zenith]

First step in many of these questions is to try drawing the edges in the angle you’re looking for

[golden cave]

ye

25 5/7

is BCA

[bold zenith]

!15m

[final saddle]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[golden cave]

oh mb

i still need help

i got 33.43

[robust mulch]

maybe show your work instead of having us guess where you got your numbers

heres a hint: you can tell youre wrong because the diagram is basically drawn to scale and you can see BCD is about 90

[golden cave]

ye

[golden cave]

so

i calculated BCA and got 25 5/7

and i tried calculating ACD

and I messed up

[robust mulch]

you telling me a number is not showing your work

[golden cave]

let me send ic

pic

[bold zenith]

I can’t believe the number isn’t showing the helpees work

It’s just saying I am 25 5/7

[robust mulch]

homie what

[robust mulch]

ok looks good

[golden cave]

i need to find a way to calculate Acd

[robust mulch]

any ideas for finding ACD?

[robust mulch]

im not sure thats necessarily a line

[golden cave]

is DCE 180

yea

[robust mulch]

can you prove thats a line?

[severe verge]

that's most likely not a line

[golden cave]

no

[golden cave]

ye

[robust mulch]

good, you dont need to

and dont assume it

[golden cave]

ok

[robust mulch]

all you need is the triangle ACD

what do you know about ACD?

the triangle, not the angle

[golden cave]

um

iscocelese

AC=AD

[severe verge]

exactly

[robust mulch]

thats good

do you know about any of the angles? if so, we can crack this

[golden cave]

hm

BAD is an exterior angle

and BAC is 900/7

[robust mulch]

youre close

[golden cave]

BAD shorter + BAD longer = 360

longer is 198

and BAC is 900/7

[robust mulch]

ok

keep going

youre on the right track

[golden cave]

DAC 69 3/7

[robust mulch]

personally, i wouldnt use mixed fractions, but thats just personal preference

either way

[golden cave]

oh shoot i understood mistake mb

wait nvm

so each is 55 2/7

55+25+1

so prolly 81

considering i did not calculation errors

ima review

[golden cave]

am i correct? @silent

tysm

tbh i was tryna find similar triangles

[jagged flare]

looks good

[robust mulch]

ohp sorry

yes 81

[jagged flare]

coordinate bash :3

[robust mulch]

coordbash is for cowards

draw line segment from H to the intersection point called it I

GHI has a special property

first off this does involve knowing BAI is isosceles

that one is an angle chase

[final saddle]

@golden cave Has your question been resolved?

[jagged flare]

i was thinking of extending FE and AD to intersect at like X, cause IFX and IBA are simmilar and you can get the ratio of FI and IB

then trig

[robust mulch]

nah my methods cooler

[final saddle]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[robust mulch]

im back

did you try my method?

[heady reef]

Bro

There’s 1 number

What the hell

[robust mulch]

its a regular octagon

so technically 8 numbers

actually 16

[golden cave]

math with no numbers lol

[robust mulch]

8 side lengths 8 angles

[heady reef]

Thankfully ima go into finance so I don’t need to do allat

[barren hound]

the number is symbolic really

[golden cave]

lol

[robust mulch]

yeah theyre basically asking the ratio of x to the side length

[heady reef]

They made all this up btw

Wtv tho I can’t help so ima js leave

[robust mulch]

isnt all math made up

[heady reef]

Yeah I think so

[robust mulch]

i presented a method before the channel closed

how have you got along with that?

[golden cave]

mid

HIG i don't see anything abt it

it has 1 and x

[robust mulch]

have you showing BAI is isosceles?

shown* not showing

[golden cave]

ye

well

not rlly prove

[robust mulch]

instead of looking at the side lengths of BAI, look at the angles

[golden cave]

oh

what abt them?

i dont see much

[golden cave]

i think i understood

[robust mulch]

so that means AI is what length?

[golden cave]

1

[robust mulch]

now draw HI

[golden cave]

ohh

[robust mulch]

what do we know about HAI

[golden cave]

root(2)

its isocelese

is it a right triangle

x is prolly root(3)

ngl

[golden cave]

its right

[robust mulch]

yup

you get two right triangles

[golden cave]

HAI and GHI

if im not wrng

[robust mulch]

yes

[golden cave]

so x is root(3)

[robust mulch]

yup

[golden cave]

tysm

.close

[final saddle]

Channel closed

Closed by @golden cave

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[upper orchid]

So I'm currently trying to calculate the expected value of the Eurasian Nutcracker card in the game, Wingspan.

Translating this into normal dice terms, there is a tray that holds 5 six-sided dice. On a given turn, a player can move a specific number of dice from the tray outside of the tray. With that said, there are 2 special cases to consider:
case 1, there are no more dice in the tray -> in which case, all dice are rerolled and added back into the tray
case 2, all the dice in the tray have the same face turned up -> in which case you may reroll all 5 dice and add them back into the tray.

What my goal is, is to find out how many 5's & 6's I can move out of the tray on a given turn, when I may choose 1, 2, 3, 4, 5 or 6 dice at once; note, picking dice happens one at a time, so if they tray had 44435 and I could pick 3 dice, I could move the 5 and the 3, then reroll to hope I get a 5 or a 6 to pop up for my third pick.

My main question is if there are any techniques I can use to limit the amount of cases I have to consider or if this problem seems like I will need to brute force the many many cases to figure out. I don't need a specific answer, as I do have a math ed degree, but if I can cut down the work I need to do, that would be nice.

[upper orchid]

.close

[final saddle]

Channel closed

Closed by @upper orchid

Use .reopen if this was a mistake.

[upper orchid]

Upon rereading the rules, it's not a homework question

[latent dragon]

I feel like you can still recieve help with this?

[plucky rover]

These channels don't exist purely for homework questions lol

[upper orchid]

oh ...

[latent dragon]

use .reopen

[plucky rover]

.reopen

[upper orchid]

.reopen

[final saddle]

✅

[opal plinth]

Tbf it says:

These channels are for pre-university homework-type questions.
But "homework-type" and "homework" aren't exactly the same thing

[plucky rover]

Pre-university?!

That just feels like a rule that exists in letter but is just ignored entirely

[opal plinth]

Yeah I know, might be a bit outdated

a tray that holds 5 six-sided dice
when I may choose 1, 2, 3, 4, 5 or 6 dice at once
How do you choose 6 dice from a tray that holds 5 maximum?

[formal trail]

it's not a rule but the likelihood of being helped effectively drops dramatically the farther away from university the question gets

[plucky rover]

The farther from pre-university you mean

[upper orchid]

either case 1 or 2 I mentioned happened, where dice will get readded to the tray. For example, after moving 5 dice outside of the tray, the tray is empty so you reroll the 5 dice back in and take one more dice out.

[opal plinth]

Oh ok so no matter what happens you can choose to take 6 total

[upper orchid]

yesn't, really 99% of the time you'll likey only be able to grab 4 dice in a turn, but there are edge cases I am trying to cover

hense why I have the different cases to consider.

[opal plinth]

Ok I'm even more confused

[upper orchid]

ok, where are you getting lost

[opal plinth]

Say it's my turn, the tray can contain 0 to 5 dice, what's my goal? To end in a state where there are as many dice that rolled 5 or 6 out of the tray?

[upper orchid]

So on a given turn you are allowed to "take" X amount of dice, where X is an integer ranging from 1 to 6 (X is decided for you).

when you "take" a dice, you gain a resource and set off to the side to be re-added to the tray when the tray gets rerolled or becomes emptied. 2 faces of the 6 sided dice allows us to gain a resource we want (in the game, it's wheat, but that's not important here).

The reason why "take" is in quotations, is because you don't own the dice or keep it, you just move it outside of the tray and collect what is shown ontop of the dice.

[opal plinth]

Ok so X is not in the player's control, and the gains happen whenever you pick up a dice, not just after you're done manipulating the dice

[upper orchid]

X is not in the player's control

the gains happen when then the dice is move by the player from inside the tray to the outside.

If a dice is added back into the tray from the outside or is rerolled, nothing is gained.

[opal plinth]

Nothing is gained but the previous gains are kept

[upper orchid]

correct

[opal plinth]

And you're only interested in the 5s and 6s

[upper orchid]

yeah

if you prefer, I can put everything in the language of the game, but those are the 2 faces I want to see

[opal plinth]

No it's fine, I just think there are a lot of conditions and the player has very little control

Like you would obviously pick all the 5s and 6s that you can from the tray as a first step

[upper orchid]

yeah

It's the being able to reroll all of the same face that is being the bane of my existance

[opal plinth]

Then if you still have moves, either the tray is empty already or you try to leave it with only one "kind" of dice (same face)

But that can also take moves

6 is not a lot of moves, and if you say it's almost always 4, that's even worse

Like there's really not a lot you can do

What exactly do you want an answer for? Some kind of probability, expected value?

[upper orchid]

expected value

[opal plinth]

And you consider all possible starting configurations (1 to 5 dice in the tray, 6 faces per dice)?

[upper orchid]

30 total cases

[opal plinth]

Are they even equally likely?

[upper orchid]

Ok

So, the dices are fair

The issue is, there can be 1 to 5 dice in the tray at the start of the turn and 6 possible amount of dice you may be able to select, so each turn you have 30 different cases to consider.

Again, I am just trying to see if I can limit the calculations I need to do here

[opal plinth]

But don't the starting configurations depend in part on the other players previous turns? Maybe that's insignificant?

[upper orchid]

The reason why I want to know the expected value, if I know I can do some something better on my turn, I do that instead
so it is insignifigant

[opal plinth]

Ok so you consider all starting configurations to be equally likely

[upper orchid]

it doesn't matter,

my goal is to get 30 expected values, although I expect the math to get to most of them to be roughly the same

That is to say, once I find out how to do a couple, the rest will be easy

[opal plinth]

Fair enough

Let N be the number of dice in the tray and X the number of dice you can "move" (pick)

With N=1, X=1, either the dice is 5 or 6, or you reroll

The expected value E in this case is 5/6 + 6/6 + 4/6(expected value after reroll)

For the after reroll part, you have a 1-(5/6)^5 chance to get at least one 6, and a (5/6)^5-(4/6)^5 chance to get no 6 but at least one 5

Honestly I don't think there is much you can do to "limit the amount of cases"

The reroll parts with other values of N and X are going to be ugly

[upper orchid]

welp, it was worth a shot

at the very least, the different cases are going to show up in each other, which is nice

[opal plinth]

Yeah some parts will

I'm not the greatest at probability stuff, maybe someone else could help at https://discord.com/channels/268882317391429632/326138783181438976

[upper orchid]

ok, I'll check there

[opal plinth]

But you probably want to formalize your problem and cut out as much as possible

[upper orchid]

after I see if my geometric sequence idea goes somewhere

anyways, thanks for your help :)

[opal plinth]

(the more concise your question the better your chances at an answer from other people - that includes making use of the domain's language, things like random variables, discrete uniform distribution, etc)

You're welcome

[upper orchid]

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[thin cloud]

So I'm making a cake in desmos for my friend birthday ( yes you read it right ) And I want to make that red part wrap around the cake (idk what's it called )

[thin cloud]

I tried to use trig function but it doesn't seems 3d

[bold zenith]

What would be ideal is the projection of a sine wave on the surface of a cylinder and then project that onto the “screen”

Doing ellipses and straight lines were easy but this is difficult

You could try faking it with trig like you said

[formal trail]

not perfect but you can add a sine wave to a curve and it sort of gives that effect

where the curve acts sort of like the "midline" of the sine wave

[thin cloud]

How can i do that

[bold zenith]

Copy the blue ellipse section just above the wave onto it by adding it to the sine wave

Like if you just draw a curve where you want the icing and then just add sin x to it what does it look like

[thin cloud]

I'm not sure what you mean by adding it to the sine wave , I tried to re-write the ellipse as y=f(x) and then make a function sin(f(x))

Ig I misunderstood what you said

[bold zenith]

F(x)+sin(x)

[thin cloud]

oh

[bold zenith]

Sin(f(x)) is called composition

I meant adding it in a literal way

[thin cloud]

Yeah i understand what you said now

that's not bad, I'll try to fix it

[bold zenith]

Change frequency and period so the end points have sin(x)=0, we could work it out if you know the start and end x

Then you should shift it down

Like remove the +5

It appears the start and end x are -14 and 14

[thin cloud]

Yup ik

Looking gud

[bold zenith]

!bnuuy

[final saddle]

[thin cloud]

awesome

thanks y'all!!!!

.close

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[ornate tree]

Translate each of the following statements into logical expressions using predicates, quantifiers, and logical connectives.
Let:

• P(x): “x is perfect.”
• F(x): “x is your friend.”
• Domain: All people.

1. Some of your friends are not perfect.
2. If someone is perfect, then they are your friend.
3. Everyone who is not your friend is not perfect.
4. There exists someone who is not your friend and is perfect.
5. Every perfect person is your friend.
6. You have at least two friends who are perfect.
7. If someone is your friend, then either they are perfect or they are not.
8. There exists exactly one perfect friend.
9. Not all perfect people are your friends.
10. If everyone is your friend, then everyone is perfect.

[ornate tree]

1 6 7 8

[plucky rover]

Hullo

[ornate tree]

Okay so, I was told by a senior helpful that question 1, 6, 7, and 8 are incorrect

[ornate tree]

hiiii

[plucky rover]

Well 1 is immediately incorrect cuz it doesn't contain the friend information at all

[ornate tree]

I wanna double check if that's true. Then I'll work on fixing them

[plucky rover]

Ah cool Lemme do that then

[ornate tree]

Tysm 🧡

[plucky rover]

2, 3 are fine

4 is fine

5 is fine

6 doesn't work cuz what does x ≥ 2 even mean

7 is.. interesting

Same problem with 8 as with 6

9 is fine

10 is fine

The reason I say 7 is interesting is cuz... Uh

If you're working in classical logic, that's just a true statement

[ornate tree]

ye, that works on the truth table

[plucky rover]

But yea you can still write it using logic notation

[ornate tree]

Alright

Can 10 be simplified further?

[plucky rover]

It will be a tautology but well whatever

No, 10 is as simple as it can be I think

[ornate tree]

$\forall x (F(x) \to P(x))$

[soft zealot]

This is sad 😢

[plucky rover]

Nope, this just means all your friends are perfect

[ornate tree]

oh bruh

alright, I'll start working on the correction

[plucky rover]

Good luck

[plucky rover]

Tag me if you need anything

[ornate tree]

Alright, appreciate it

1. $\exists x (F(x) \to \neg P(x))$\
2. $\exists x \exists y (F(x) \to P(x)) \wedge (F(y) \to P(y))$\
3. (I'll skip this part as this is classical logic)\
4. $\exists x (P(x) \wedge F(x)) (x = 1)$\

[soft zealot]

This is sad 😢

[plucky rover]

1 is wrong I think

[ornate tree]

oh wait

yeah, incorrect notation

[plucky rover]

6 needs a little more information

Actually no 6 is wrong too

As a little nudge

Abandon implications

For 1 and 6

[ornate tree]

yeah.....that's "and"

[plucky rover]

Yes

[ornate tree]

idk why I'm so obssessed with arrows

[plucky rover]

People starting out with logic always are

[ornate tree]

1. $\exists x (F(x) \wedge \neg P(x))$\
2. $\exists x \exists y (F(x) \wedge P(x)) \wedge (F(y) \wedge P(y))$\
3. (I'll skip this part as this is classical logic)\
4. $\exists x (P(x) \wedge F(x)) (x = 1)$\

[soft zealot]

This is sad 😢

[plucky rover]

6 needs a little bit more

And again, x = 1 doesn't make sense for 8

You can't say people are greater than 2

But yes smth close to that

Don't delete stuff lol, it's fine logic is a lot of book keeping

Oh also 1 is correct

[ornate tree]

Let's begin with 8

[plucky rover]

Sure

[ornate tree]

My idea is to confine the domain at x = 1

[plucky rover]

6 is easier to fix though

Like you're missing one little bit

And the idea with 6 will help with 8

But up to you

[ornate tree]

Alright, let's begin with 6

[plucky rover]

Cool

[ornate tree]

ye....since you mention it lol

[plucky rover]

So currently the only problem 6 can run into is..

x and y can refer to the same person

And then you might only have one friend who's perfect

But you want at least two

[ornate tree]

damn.....I never come up with this problem

[plucky rover]

So what are we to do

[ornate tree]

x != y is the easiest solution I presume

[plucky rover]

Yup there you go

Just an extra and

Cool now that we are done with 6

Does this idea give you a hint for 8

[ornate tree]

1. $\exists x (F(x) \wedge \neg P(x))$\
2. $\exists x \exists y (F(x) \wedge P(x)) \wedge (F(y) \wedge P(y)) \wedge (x \ne y)$\
3. (I'll skip this part as this is classical logic)\
4. $\exists x (P(x) \wedge F(x)) (x = 1)$\

[soft zealot]

This is sad 😢

[ornate tree]

gimme a minute

[plucky rover]

Sure sure

[ornate tree]

There exists exactly one perfect friend.

[plucky rover]

Yes

[ornate tree]

I don't have any idea actually, aside from missing wedge

[plucky rover]

Fair enough

Well we want there to exist an x such that both are true

But we want there to exist exactly one such x

So what do we want for the others

[ornate tree]

the others are false

[plucky rover]

Yes

So how can you write it

[ornate tree]

yeah but what's wrong with this version?

like if I just add wedge in it

$\exists x (P(x) \wedge F(x)) \wedge (x = 1)$\

[soft zealot]

This is sad 😢

[plucky rover]

x = 1 doesn't make sense

What does it mean for a person to equal a number

[ornate tree]

hmm...so if I understand it right. x means a person's name

or simply means a person's identity

[plucky rover]

Let's not get philosophical about what differentiates a person

[ornate tree]

1. $\exists x (F(x) \wedge \neg P(x))$\
2. $\exists x \exists y (F(x) \wedge P(x)) \wedge (F(y) \wedge P(y)) \wedge (x \ne y)$\
3. (I'll skip this part as this is classical logic)\
4. $\exists x (P(x) \wedge F(x)) \wedge (P(x) = 1)$\

[soft zealot]

This is sad 😢

[plucky rover]

Still doesn't work

You're trying to use the idea of cardinality

Unfortunately we do not have that yet

[ornate tree]

hmmmm

[plucky rover]

Like I said

You want there to exist one person such that both properties are true for them

AND for everyone else, you want at least one of them to be false

Try just writing this thing exactly

[ornate tree]

why would it be at least one

[plucky rover]

Well if none are false then both are true

[ornate tree]

but not just all of them

[plucky rover]

And then your perfect friend isn't unique

[plucky rover]

Cuz non-perfect friends and perfect non-friends can still exist

We just want exactly one perfect friend

We eventually define a quantifier that reduces the amount of writing for this, $\exists !$

[soft zealot]

Xavier 🌺

[plucky rover]

But I imagine the goal of this exercise is to make you define that yourself

[ornate tree]

So basically we exclude this special guy, and for others we don't want them to fulfill the characteristc of this special guy

is that correct?

[plucky rover]

Yes

There's multiple ways to do this

And funnily enough the cleanest one uses implication this time

[ornate tree]

$\exists x (P(x) \wedge F(x)) \vee \neg (P(x) \wedge F(x))$

[plucky rover]

Close to the idea

[soft zealot]

This is sad 😢

[plucky rover]

You want to incorporate other people into this as well

[ornate tree]

yes

[plucky rover]

So $\exists x (P(x) \land F(x))$ is fine

[soft zealot]

Xavier 🌺

[plucky rover]

Now you want to add a condition saying that no one else satisfies this

[ornate tree]

no one else satisfies this
Makes me want to add forall in front of the second

[plucky rover]

Good thinking

And it's correct

[ornate tree]

$\exists x (P(x) \wedge F(x)) \vee \forall x \neg (P(x) \wedge F(x))$

[soft zealot]

This is sad 😢

[plucky rover]

Close

[ornate tree]

but what's the difference between having and not having forall?

[plucky rover]

You want to exclude this special person from the second condition though

Also maybe use a different variable for the forall

[plucky rover]

Not having forall doesn't allow you to talk about arbitrary people

[ornate tree]

$\exists x (P(x) \wedge F(x)) \vee \forall y \neg (P(y) \wedge F(y)) \wedge (x\ne y)$

[plucky rover]

Having forall, you can make generalizations about every element of the set

Sky you don't just get to add a bracket there needs to be an operator in between

[soft zealot]

This is sad 😢

[ornate tree]

bad habit fr

[plucky rover]

Uh, no that is wrong

It's close but wrong

Wait

No it's correct I'm blind

[ornate tree]

This is god damn complicated 💀

[plucky rover]

Now needs one bracket around the entire second part

[plucky rover]

There is a much easier solution to this

[ornate tree]

$\exists x (P(x) \wedge F(x)) \vee (\forall y \neg (P(y) \wedge F(y))) \wedge (x\ne y)$

[soft zealot]

This is sad 😢

[plucky rover]

$\exists x (\forall y (P(y) \land F(y)) \to (y = x))$

[ornate tree]

oh....makes sense

[soft zealot]

Xavier 🌺

[ornate tree]

it's a miracle that I can read this

[plucky rover]

The x ≠ y is also in the forall

So the bracket should end at that

[plucky rover]

Lol it comes with experience

Also actually wait yours doesn't work

Cuz if y = x, the forall fails

What you want instead is, forall y, either (not perfect friend) or (equal to x)

So you actually want an or with equality, instead of and with inequality

[craggy plank]

hmm

[craggy plank]

No, we’ve already mentioned that x!=y

sry, I was changing my seat

[plucky rover]

You can't do that at this stage lol

Forall includes everything unfortunately

So you're gonna have to make the thing inside true for everything

It is extremely annoying, I know

[craggy plank]

yeah, but if x = y the entire system becomes false right?

[plucky rover]

Yes, hence it fails

[plucky rover]

Look at this, it works for any y

[craggy plank]

This works for any y instead of x = y

But if x = y the entire system fails

[plucky rover]

Yes that shouldn't be the case

Let me show you what I'm aiming for

∃x (P(x) ∧ F(x)) ∧ (∀y (x = y) ∨ ¬(P(y) ∧ F(y)))

You don't want your system to fail at all

For any given y, either they're equal to x, or they are NOT a perfect friend

You might notice that

[plucky rover]

This is just unfolding this implication

[craggy plank]

Yep, they either fall into the first or the second category

[plucky rover]

This is probably one of the very few cases where the implication is easier to reach

And easier to understand

[craggy plank]

So the only problem in this system is that when x = y it immediately fails

[plucky rover]

Yes

[craggy plank]

Hmm

Is there anyway to patch this drawback on the current basis as far as you know?

[plucky rover]

I mean what I gave here is a patch

[craggy plank]

Oh, bruh. Didn’t notice that

Lemme check

[plucky rover]

Lololol

It just turns your inequality to an equality and uses or instead of and

[craggy plank]

I see, I’ll check it around later.

Appreciate your help

Have a good one 🙂

[plucky rover]

You too! 🌺

The default hibiscus looks so much better than the discord one

It's sad

\🌺

They don't even let you do this anymore

[craggy plank]

I’m too lazy to take out my laptop again

[plucky rover]

Gonna get Nitro just to be able to send default hibiscuses

Lololol you can close it from this account

[craggy plank]

200% not worth it 💀

[plucky rover]

Lmao yeah for sure

[craggy plank]

Sadly I can’t because I’m not helpful

[plucky rover]

Sky

[craggy plank]

:sigh:

[plucky rover]

Look at which account you're on

[craggy plank]

-# in terms of real contribution, I only farm for easy problems

😭😭

[plucky rover]

Smh no

The bot has deemed you helpful, and so you are

Your opinion is irrelevant on this

[plucky rover]

Feel free to tag/dm me whenever you need help, helping you with logic is refreshing compared to the other stuff I usually do lol

[craggy plank]

I thought this is considered an auto-piloting question to you lmao

[plucky rover]

It isn't no

I have not even formally studied predicate logic yet

That isn't a joke

It's just from years of doing related stuff and having conversations with people

[craggy plank]

Sadge

Alright imma close this one

Bye :))

.close

[final saddle]

Channel closed

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[stray notch]

If i have a step function like this. Will the square of that function be what i wrote inside the pic?

[plucky rover]

Yes

Which is just equal to 1 everywhere

[bold zenith]

$f(x)^{2}=1$

[soft zealot]

BBMaths

[bold zenith]

This highlights an important fact that f(x) might not be constant if you started from this ^

[stray notch]

Ohh

Alr thx a lot

.close

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[short wraith]

hints

[twin pivot]

Good morning. @short wraith

Quick note: if this is from a live exam, the rules say we can’t provide help in that case. If it’s just practice material, then it’s fine.

[quaint rampart]

first thing you've gotta do is look at the first summation

as we know f(1), f(2) can be found by taking n=2

[short wraith]

can you leave the channel lol

[short wraith]

yeah the series i got is

[quaint rampart]

after knowing f(2) we can find f(3) from n=3 substitution

[short wraith]

1,2,6,18,54

[ivory bear]

true

[quaint rampart]

try to find a pattern

[quaint rampart]

hey! That's rude

[short wraith]

i see

[short wraith]

a=1

[ivory bear]

no other options....

[quaint rampart]

;-;

he's tryna help dude

that's very rude

[short wraith]

lol is funny no?

[short wraith]

couldn't find the pattern

[valid carbon]

what

[valid carbon]

ye wot

[short wraith]

https://tenor.com/view/waiting-little-rascals-black-and-white-tv-gif-7351141631919267269

[short wraith]

.

[quaint rampart]

hmm

[short wraith]

then?

should I close it?

[quaint rampart]

1, 2 * 3^0, 2 * 3^1, .....

first term doesn't fit

[valid carbon]

escape them

[short wraith]

why>

[twin pivot]

Strange. Usually people who aren’t cheating don’t mind when someone mentions the anti-cheating rule.

[short wraith]

i seeeeeeeeeeeee

[quaint rampart]

sus indeed.

[short wraith]

will you focus on my work?

[quaint rampart]

;-;

[short wraith]

but options are odd

[twin pivot]

If this is an exam question, I can’t provide help. That’s against the rules.

[short wraith]

keep troll !!

[valid carbon]

no theyre not

[craggy plank]

I’m not sure why you came up and accused him of cheating in the first place

Is there any reason behind this?

[quaint rampart]

he's asking.

no accusations were made.

[short wraith]

no he is accusing

[quaint rampart]

just tell him it ain't an exam question then

[valid carbon]

If it’s just practice material, then it’s fine.

[craggy plank]

You don’t simply ask this to every helpee lol

[short wraith]

like he is sitting behind me in the exam hall....lool

[valid carbon]

insane accusation

[short wraith]

please hint

pattern is invisible to me

1,2,6,18

[valid carbon]

drop the 1.

[short wraith]

yeah i did

[twin pivot]

Call me observant, but a blurry photo of a multiple-choice question, paired with a rushed request for ‘hints,’ statistically screams ‘exam situation.’ And according to the rules, that’s academic dishonesty.

[valid carbon]

its a gp

[bold zenith]

We don’t have to debate over if this is an exam question, just take a more zoomed out screenshot

[short wraith]

2(3^0,3^1,3^2)...

[short wraith]

alfred alt

[valid carbon]

anyway no its not an exam its a worksheet you can see the watermark

[bold zenith]

lol no

[valid carbon]

OP is still obnoxious tho

[valid carbon]

yeah so whats the issue then

[short wraith]

i got the paatenr

it is C

[bold zenith]

We don’t even talk the same

[short wraith]

3^m-1

[bold zenith]

And Alfred says so much stuff I wouldn’t have time to type my own stuff

[short wraith]

it is GP simple

[bold zenith]

(No disrespect) I think more detail is good as long as it doesn’t confuse the helpee

[twin pivot]

If this is practice material, please confirm the source (book/worksheet). Once that’s clear, I’ll be happy to give hints.

[short wraith]

I don't need irrelevant advices...troll stop or I complaint in modmail

[short wraith]

why do you not leave the channel

i said 10 minutes ago

.close

[final saddle]

Channel closed

Closed by @short wraith

Use .reopen if this was a mistake.

[craggy plank]

Just modping if you find it necessary 👍

[short wraith]

thanks

[twin pivot]

<@&268886789983436800>

Could you please clarify? I just quoted the rules about exam material, but the OP disagreed. I’d rather get an official confirmation.

[short wraith]

just wait for the conclusion

[bold zenith]

Why go to all this trouble just do this

[craggy plank]

.reopen

[final saddle]

✅

[short wraith]

.reopen

[quaint rampart]

I vouch for alfred.

[brisk charm]

The rules about not being able to help with exams are correct, but I don't see any indication that this is a live exam

[twin pivot]

.

[bold zenith]

Could you just send a more zoomed out photo?

[short wraith]

your self observation doesn't prove anything

[bold zenith]

Why are we arguing

[short wraith]

he accused me

[valid carbon]

id just take their word for it

not worth t

[craggy plank]

There are thousands of questions asked and presented in similar way

[short wraith]

the way he was giving the time in the channel for 15 minutes is tottaly wasting

[brisk charm]

Okay please stop arguing about this

Let him get math help

[valid carbon]

hes done

[craggy plank]

Is he?

[short wraith]

yesh

already done

[valid carbon]

.close

[final saddle]

Channel closed

Closed by @valid carbon

Use .reopen if this was a mistake.

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[foggy wren]

I'm a little confused on what to do here

[magic walrus]

😰

[foggy wren]

I think you can split it into two cases but I don't know if that encompasses all of s

So far what I've got is for small values:

$I(s) = \frac{s^3}{3} - \frac{7s^5}{30} + \frac{141s^7}{840}+...$

[soft zealot]

Katrro

[foggy wren]

and for large:

$I(s) = \frac{Cos(s)}{s} + \frac{Sin(s)}{s^2} - \frac{2Cos(s)}{s^3} +...$

[soft zealot]

Katrro

[valid carbon]

$\cos$

[soft zealot]

Περσυ

[twin pivot]

Good morning. @foggy wren

Source?

[foggy wren]

source of what?

[twin pivot]

Source of the problem.

[foggy wren]

from one of my classes

[twin pivot]

Strange.

[foggy wren]

how is that strange

[twin pivot]

What course are you doing?

[foggy wren]

it's an applied math class

[twin pivot]

Can i see the exercise 3?

[proper dagger]

what does that show here though? are you suspecting exams again?

if you are, then I'd recommend not immediately assuming that unless it's super obvious or something? or you can always modmail/modping, I suppose.

[twin pivot]

This exercise is very complicated

https://math.stackexchange.com/q/2176306/1259698

You should start by reading this.

[foggy wren]

I don't need to solve the integral for an exact answer

It's just an expansion

Also this is question 3

Which is how I got the answer for large x since if you assume x>>1 then the integral turns into this one

[twin pivot]

for $s\ll1$ (Maclaurin development)

[soft zealot]

Alfred Eisenberg

[twin pivot]

$$
\frac{x\sin x}{1+x^{2}}
=(x-x^{3}/6+x^{5}/120-\cdots),(1-x^{2}+x^{4}-x^{6}+\cdots)
= x^{2}-\frac{7}{6}x^{4}+\frac{47}{40}x^{6}-\frac{5923}{5040}x^{8}+\cdots
$$

[soft zealot]

Alfred Eisenberg

[twin pivot]

Okay so far ?

[foggy wren]

you forgot the first x before the sin expansion but that's what I've got

[twin pivot]

Integrating deadline

$$
\boxed{;
I(s)\sim \frac{s^{3}}{3}-\frac{7}{30}s^{5}+\frac{47}{280}s^{7}-\frac{5923}{45360}s^{9}
+\cdots\qquad (s\to0)
;}
$$

[soft zealot]

Alfred Eisenberg

[twin pivot]

Okay so far ?

[foggy wren]

That's what I got yea

[twin pivot]

With the first 4 terms the relative error is $<1%$ for $0<s\lesssim0.65$.
(Adding more terms from the series product extends the range: 12 terms $\Rightarrow$ $<1%$ up to $s\approx0.8$; 20 terms $\Rightarrow$ $<1%$ up to $s\approx0.95$.)

[soft zealot]

Alfred Eisenberg

[twin pivot]

Yes?

[foggy wren]

I don't know how to compute it with these coefficients because they seem random to me

[twin pivot]

Let me explain.

[undone matrix]

as s -> infinity the integral goes to pi/2e so you might want to check this

[twin pivot]

$$
\frac{x\sin x}{1+x^2}=\underbrace{\bigl(x\sin x\bigr)}{\sum{m\ge0}\frac{(-1)^m x^{2m+2}}{(2m+1)!}};
\underbrace{\frac{1}{1+x^2}}{\sum{k\ge0}(-1)^k x^{2k}}
$$

[soft zealot]

Alfred Eisenberg

[twin pivot]

Okay so far?

[foggy wren]

yes

[twin pivot]

It is just Cauchy product.

Shall we skip this?

[foggy wren]

yea you can skip that

[final saddle]

@foggy wren Has your question been resolved?

[undone matrix]

Step 1: By contour integration you can prove that
[
\int_{0}^{\infty}\frac{x\sin x}{1+x^{2}},dx = \frac{\pi}{2e}.
]

Step 2: Writing $u = x/(1+x^{2})$, $dv = \sin x,dx$ and integrating by parts repeatedly for this integral
[
\int_{s}^{\infty}\frac{x\sin x}{1+x^{2}},dx
]
produces an asymptotic series valid for large $s$.

[soft zealot]

Warg

[foggy wren]

I had that idea initially for the asymptotic series but I feel like the way I've done it would be valid still

this way seems like it takes a lot more effort computing everything

[undone matrix]

hmm, then you may want to check your work and verify that it converges to pi/2e for s->infinity

[foggy wren]

thank you for the help I should be good now

.close

[final saddle]

Channel closed

Closed by @foggy wren

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[soft heart]

.close

[final saddle]

Channel closed

Closed by @soft heart

Use .reopen if this was a mistake.

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

[bold zenith]

Nvm it didn’t scroll down sorry, just saw image

[craggy plank]

Do you still need help?

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[silk fjord]

My idea was to use the midpoint of y-values for A, and the midpoint of x-values for B

so I had A(0,4) and B(7.5,0)

[pliant shore]

have you heard about *the law of reflection?

[silk fjord]

but that resulted in ~17.23 which is greater than the actual answer, 17

[silk fjord]

no

have not

what is snell's law?

[pliant shore]

what I mean is that you must have a setup like so

[old quarry]

isnt that the refraction of light thing

[pliant shore]

where you have the angle of incidence = angle of reflection

basically, it's symmetrical

[silk fjord]

I'm not sure I understand how that would work here though

are you saying PA and BQ need to intersect?

[pliant shore]

let me draw it

[silk fjord]

Oh I see

So it's like perfect bounces

[pliant shore]

yes

[silk fjord]

How do I determine the perfect coordinates or lengths though?

Is it just midpoint of Py and 0?

And then midpoint of Qx and 0

[pliant shore]

no, so given that A = (0, a)

can you find the equation of the line AP?

then that gives you the equation of AB

[silk fjord]

Using distance formula?

[pliant shore]

no

point-slope form or just y = mx + c

[silk fjord]

Oh yeah that makes more sense

It would be y=((5-a)/(4-0))x+a?

$y=\frac{5-a}{4-0} x + a$

[soft zealot]

UCYT5040

[silk fjord]

because a is the y-intercept

[pliant shore]

yep

and then, how are the slopes of AP and AB related?

[silk fjord]

they are perpendicular

[pliant shore]

no

[silk fjord]

oh wait its just

opposite

[pliant shore]

wdym by opposite?

[silk fjord]

negative

[pliant shore]

correct

[silk fjord]

isnt that def of opposite or did i say that wrong

?

[thin cloud]

Isn't the least distance is just the distance of reflection of P through y-axis and reflection of Q through x-axis?

[pliant shore]

opposite is not well defined

[silk fjord]

ah ok

anyway, then the slope of BQ should be the same as PA then right?