help-36

[versed crater]

It’s also proven that for any higher polynomials there are no formulas

[marsh glacier]

also ig if u take the second derivative u can get the inflection points

[versed crater]

Well great idea but we did already do that

[marsh glacier]

and then plug values near those points

[old quarry]

alr ill close this now

thanks frosty

.close

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[untold kiln]

In a book, there is the following passage for an equation for a line.

[untold kiln]

For my own notes, I want to update the definition a tiny bit, in order to make it more explicit for myself. I'd like to confirm that my change didn't mess something up. Specifically, that the (x,y,z) to the left of | is correct.

Is my updated version consistent with the one given in the book?

[cursive bough]

looks good to me

[final saddle]

@untold kiln Has your question been resolved?

[untold kiln]

.close

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[fickle tendon]

Can someone help me please?
In a mixing vat there are initially 200 litres of volatile liquid. 3 litres of fresh liquid are added per hour, but 15% of the current content of the vat is lost each hour due to evaporation. How many litres of liquid are there after 4 hours?

[severe verge]

how much liquid is left after 1 hour

[fickle tendon]

I did 200 + 3 - (200*0.85) and that gave me 173

And i repeated that for the next hours but it was wrong

[severe verge]

ok show your work

one weird question here is whether did they add 3 liters in the 4th hour

[tidal vale]

this feels ambiguous, does the evaporation apply before or after adding the 3 liters?

[severe verge]

my guts tell me evaporation applies before

[tidal vale]

yeah, but perhaps that's the problem

[severe verge]

^

[fickle tendon]

hour 2 - 173 + 3 - (173*0.85) = 150.05, hour 3 - 150.05 + 3 - (150.05x0.85) = 130.5425, hour 4 - 130.5425 + 3 - (130.5425x0.85) = 113.96

That was wrong

[severe verge]

maybe they didnt add the 3 liters in the final hour

[fickle tendon]

What i sent as the question is all that is given to me

[severe verge]

try 110.96

if this is wrong then the additional 3 liters must be accounted for first

[fickle tendon]

Thank you

.close

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[kind mason]

Proof that (cosx)^cosx >= (sinx)^sinx within the domain of [0,pi/4].

[fiery bluff]

!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

[kind mason]

1

[fiery bluff]

Hint: consider the relative sizes of cos and sin in [0, pi/4]

[kind mason]

I'm sorry, I don't know what to do with that yet.

From I remember, I'm supposed to start by solving for (cosx)^cosx - (sinx)^sinx >=0, and then create function h(x) of that, then possibly differentiating it.

I'm lost from there.

[fiery bluff]

I think you dont need calculus for this

[kind mason]

Chatgpt tests values, which i don't think is right either

[fiery bluff]

Oh wait yeah 0^0 -> 1 and 1^1 -> 1 my bad

Guess we need calculus then

[kind mason]

maybe. I'm still lost.

[fiery bluff]

Can you differentiate the function?

cos^cos - sin^sin

[kind mason]

yeah.

If you manage to do the question, could you share your working out

[fiery bluff]

Typically we tend to encourage people to try to reach the solution themselves with help

[kind mason]

right, sorry.

[fiery bluff]

Another way to do is proving
(cos x) ln(cos x) >= (sin x) ln(sin x)

Which might be easier to differentiate

[kind mason]

hmm. I'll try again later. Thank you for helping.

.close

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[bold blade]

qbinom(.75, size=65, prob=p)

[bold blade]

I am trying to understand the qbinom function

https://www.statology.org/dbinom-pbinom-qbinom-rbinom-in-r/ describes it as such

But I do not understand it

Quantile could be a division of any size depending on how many pieces you want to consider

Basically

1. How do we know how many quantiles there are? ex: 4 = quartiles, 100 = percentiles
2. What value are we actually finding for the target quantile? The area to the left of it?
3. Once we've identified how many quantiles there are, which one is being targeted? The first? The last?

I might be way overthinking this

[floral nova]

It is just the inverse CDF just like it says in the document.

Compare the first example to pbinom(2,10,0.4) and pbinom(1,10,0.4) for example

For a random variable X~bin(size = n, prob = p), qbinom calculates the smallest x in the support such that P(X<=x) >= q

[final saddle]

@bold blade Has your question been resolved?

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[undone tiger]

Given answer is A but when I solved out I got B

2× 0 to π/2 integration |sinx|dx

Which is =2

[tired walrus]

given answer is guaranteed wrong

without calculation

you're integrating a positive function over an interval of nonzero length, the answer will have to be positive. and yours is correct.

in this case however, i can see what wrong logic they used to get to their wrong answer! do you want to know? @undone tiger

[undone tiger]

I wanted to know do you agree with my answer?

[tired walrus]

yes i do.

[undone tiger]

Thanks ma'am

Ma'am i have another doubt from group theory

If you allow me here or I open new channel

[fiery bluff]

is it the same as the other channel you had

[drowsy copper]

Can you tell me that?

[undone tiger]

Yeah that same question

[fiery bluff]

we can reopen that channel ig

[undone tiger]

They substraction

positive and negative🤣

=0

[tired walrus]

they simplified the stuff inside the root to sin^2(x) [correct] and then wrote sqrt(sin^2(x)) as sin(x) [wrong]

and then integrated that from -pi/2 to pi/2 to get 0 [GIGO]

[drowsy copper]

Ohhhh yes I get it now

[drowsy copper]

Yes yes

[fiery bluff]

i assume this channel is done, yeah?

[undone tiger]

Yeah

.close

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[undone tiger]

Thank you everyone for collaboration

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[short surge]

i understand everything

except for 2/3

where on earth did 2/3 come from

[signal adder]

(8-6)*(1/3)

Where 8-6 is the height of the unshaded cone

@short surge

[short surge]

OH THEYRE LOOKING FOR THE UNSHADED CONE

GOD WAIT THATS DUMB

😭

my god

[signal adder]

I mean it's the normal approach

[short surge]

no what i meant was i was trynna figure out the shaded portion?

like the height should be 6/3

cause the height of the shaded portion is 6

holy im dumb

ok so then why is the radius of the unshaded cone 3/4

shouldnt it be 1/4

wait why is the radius 3/4 then

cause 6/8?

[signal adder]

It's cuz the cones are similar

[signal adder]

(8-6)/8 is the scale factor

Which is just 1/4

Then 1/4*3 is the radius of the unshaded cone

Where 3 is the radius of the bigger cone

[short surge]

gotcha ok now i understand

[short surge]

ok so the first 8 is the height, the 6 is from the diameter, but where did you get the other 8

[signal adder]

The other 8 is the height again

8-6 is the height of the smaller cone

[short surge]

they said the height of the none shaded cone is 2

so why is it 3/4 here

[hexed sage]

i thought it said the height of the non-shade cone is 2 (sry if im dumb)

[short surge]

youre right its 2

[thin cloud]

The base radius of non-shade cone is 3/4

[short surge]

ok i figured it out

3/2 is the diameter

and you divide it by 2 to get radius

thanks everyone

.close

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[untold sail]

I am getting 0 now the thing is how do I determine

If the answer is pi,2pi or -pi

And shouldn't the answer be in between

-pi/2 , pi/2

[eager shore]

why don't you just put some value in for x

[untold sail]

What do I put tho?

[eager shore]

1 + ep as ep --> 0^+

might work

[untold sail]

Ep?

[eager shore]

epsilon

[untold sail]

Ah l

K

[eager shore]

so alpha becomes -pi and beta becomes pi/2

[untold sail]

But like either way

It will become 0

So the value of x doesn't matter

[eager shore]

this is for JEE right

[untold sail]

Ye

[eager shore]

you've probably got a massive formula sheet

[untold sail]

Yes

Not much for itf tho

[eager shore]

the domains and stuff?

Oh you know what could work

sub x = tan t

[untold sail]

Alr

[eager shore]

t would lie between pi/4 and pi/2

[untold sail]

Right

But we won't get the answer tho idk how the answer is -pi

It's outside of the itfs range

[soft zealot]

maison

[eager shore]

@untold sail is this what you got?

[untold sail]

After I simplified it I got $tan^-1 0$

[eager shore]

hmm

[untold sail]

Tan ^-1 0

[eager shore]

\tan^{-1}

alright

t lies between pi/4 to pi/2

so that's causing us some problems

you can't just cancel

[untold sail]

I didn't use substitution at all in my approach

[eager shore]

yeah but still

there's some domains that you didn't take note of

[untold sail]

Ye must be that

I am checking the domains the only thing I figured out

Is than 1-x^2 / 1+ x ^2 will be -ve

<@&286206848099549185>

[golden flax]

yes?

[untold sail]

Ye i can't figure out where I went wronf

Wrong

I am getting tan ^ -1 0 as my answer

But none of the options match

The answer is -\pi which lies outside of the domain

Of tan ^-1 x

[golden flax]

hmmm

what was the original question?

[untold sail]

Here

Ok i got it

For the 2 tan ^-1 x what I missed was including -pi

Thanks guys

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[cold wedge]

hello

[cold wedge]

.close

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[magic coyote]

[dull grotto]

hi

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[sour kindle]

learnt a new differential method and i don’t know how to integrate the last bit, chat gpt said it wasn’t integrable but then got a solution not in terms of an integral with some other method

is there anything else i can do

[final saddle]

@sour kindle Has your question been resolved?

[tired walrus]

!nogpt, first and foremost!

[final saddle]

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

[sour kindle]

Yup thanks just looking for a solution wont use chat gpt going forward, do you know how to solve it

[final saddle]

@sour kindle Has your question been resolved?

[dense badge]

the last integral for u can be integrated

[final saddle]

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[wet horizon]

((x+5)(x+2))

[soft zealot]

Tikhomirov

[terse crypt]

what ya wanna do w/ it

[wet horizon]

can somebody help?

[sour zealot]

you wanna expand it?

[wet horizon]

ye

[wet horizon]

hmmm...

(\xpmhighlightbox{bgcolor=#3A3F50, underlinecolor=#3A3F50}{x\cdot x+2x+5x+5\cdot 2})

[vital crag]

yea that looks right

just simplify

[wet horizon]

(x\cdot x+2x+5x+5\cdot 2)

[soft zealot]

Tikhomirov

[wet horizon]

?

[sour zealot]

yes

that's correct

now, can you combine like terms?

[wet horizon]

hmmm

[sour zealot]

(and simplify)

[wet horizon]

(x^{2}+2x+5x+\xpmhighlightbox{bgcolor=#3A3F50, underlinecolor=#3A3F50}{10})

?

(x^{2}+2x+5x+10)

[soft zealot]

Tikhomirov

[wet horizon]

?

and then...

[sour zealot]

hmm

can you combine the middle two terms somehow?

[wet horizon]

(x^{2}+7x+10)

[soft zealot]

Tikhomirov

[sour zealot]

yep!

thats right

[wet horizon]

yay

thanks

.close

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[brazen thunder]

hey can i ask a physics q here, its medium level...

[vital crag]

there's a physics server in #old-network

[final saddle]

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

[brazen thunder]

That server is lil slow, if you could answer, ans otherwise ignore
q. A body A is dropped from a height H, and at the same moment, another body B is projected vertically upward from the ground with a speed u. They cross each other at height H/3. Find u
My approach-
let Obj A be the one being dropped and B be the one being thrown:
(Taking downward direction as +ve) = g =10m/s_2
y(a)-H= 5t^2
y(a) = H+5t^2
 
now for B,
y(b) = -ut+5t^2 (since i have taken dwnwrd as +ve)
y(a) = y(b)
getting t = -H/u
now substuting this in either of the eq , i'm getting t = sqrt(-15H/2)... this -ve idk why...

idk where and how i am going going!

[runic tulip]

5t^2 - H

becuase H is towards up and you took downward as +ve

[brazen thunder]

ahh so is it actually y(a) -(-H) ???

so to give y(a) = 5t^2 -H?

[runic tulip]

i recommend taking up as +ve to avoid such confusions

it becomes way more intiutive too then

H - 5t^2

[brazen thunder]

i generally take down as +ve so that g remains +ve always SO that i don't have to care abt g, but looks like in such cases up +ve is good approach👍, thanks

[ripe jewel]

it shouldn't matter in the end

as long as you write something like

v = 34 km/s down

[brazen thunder]

yEah i know , until i am implementing the correct directions...

will take care abt that, in a hurry i forget to think practically a lil bit...

Thanks y'all

[runic tulip]

!close

[brazen thunder]

!close

!close

[runic tulip]

nah thats supposed to trigger the bot to send you a msg to closse it

[sour zealot]

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[brazen thunder]

close()

[runic tulip]

.close

[final saddle]

Channel closed

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[brazen thunder]

Lol

[runic tulip]

its this

[brazen thunder]

XD

.close

[runic tulip]

its already closed

[brazen thunder]

i opened it again...

[runic tulip]

nope its closed

.reopen

[final saddle]

✅

[brazen thunder]

alr GN

[runic tulip]

this is opened

now try closing

[brazen thunder]

.close

[final saddle]

Channel closed

Closed by @brazen thunder

Use .reopen if this was a mistake.

[runic tulip]

!close

bot must be down

[serene panther]

you can ask physics questions here as well, but keep in mind you won't get answers as concise or quick as math questions

idk about the policy here on phy questions, but ive seen tons of ppl ask them

and nobody seems to object them

[brazen thunder]

It'd be better to expand this Chanell beyond maths, may be more dc bots would take the job with ease...

[serene panther]

no, that would be completely against server policy
This server was meant to be only for maths discussion and help. If you want, you can join a physics server but i dont think this server will ever do that

[runic tulip]

this server is mainly for maths

[sour zealot]

see #old-network

theres a physics server there

[magic coyote]

It's !done

[sour zealot]

(and a cs server, and a chem server, and a bio server, and an ai server...)

[runic tulip]

for easier phy questions like motion and stuff might get answered here but for more completed go to the dedicated servers

[runic tulip]

ohh

!done

[final saddle]

If you are done with this channel, please mark your problem as solved by typing .close

[brazen thunder]

I was saying, btw what's the problem with that, amalgamation would be better , wouldn't that?

[runic tulip]

right

it would be chaos

and harder to moderated

[sour zealot]

i mean, math + phys would be already 300k+ ppl

[brazen thunder]

Man i have seen 700K+ people channels working smoothly

[sour zealot]

(granted, they are also the two most popular servers, but some other ones have a good amount of people as well)

[sour zealot]

i think we have enough trolls/spammers/rude people to deal with already 😅

[runic tulip]

you could also send your suggestions to modmail

[sour zealot]

mods are really hardworking

[brazen thunder]

Can't the bots handle them, like i have seen in many servers irrelevant messages are instantly blocked

anyways it is what it is, GN

and btw thanks to the mods

[serene panther]

you're south indian?

[serene panther]

although that server is very much dead

also no channel for help i believe

[final saddle]

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[magic coyote]

Ye

[magic coyote]

Oh

Oopsie

.close

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[sour zealot]

gasp executor you need help? /jkjk

[serene panther]

🫔

[magic coyote]

Kannada

[serene panther]

you are an undergrad now

no longer pre uni

unless you arent pursing a degree

[magic coyote]

Well yeah but

I'm not undergrad matu

Math*

The role says undergrad math

So im gonna keep my pre uni

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[spring haven]

#1398993464338808872 message

[final saddle]

@spring haven Has your question been resolved?

[final saddle]

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[nocturne cedar]

Is this right?

[nocturne cedar]

Can someone use the bot command to turn the pic around

[nocturne cedar]

Idk if I can do this when adding

Fractions

[edgy mauve]

U cant

[edgy mauve]

,rccw

[nocturne cedar]

Both denominators had a sqrt 3 one was positive one was negative

[nocturne cedar]

I’m not simplifying denominator and numerator

[edgy mauve]

Yes but they r in addition and subtraction form

[edgy mauve]

Then?

[nocturne cedar]

I’m simplyfing the respective colors

[edgy mauve]

U defo cant

[night raft]

you can't cancel them like that

[edgy mauve]

They need to be in multiplied form

To be cancelled out

[nocturne cedar]

Even to simplify numerator with numerator?

[edgy mauve]

If it said 2sqrt3/2sqrt3 then u could

[edgy mauve]

Everywhere

[nocturne cedar]

Dang

[tired walrus]

you can cancel out common FACTORS in a fraction

[jagged siren]

You cant cross cancel over addition or subtraction signs

[nocturne cedar]

Wdym

[jagged siren]

So like 2x/ 3x

[edgy mauve]

Like 4/2=(2×2)/2

[jagged siren]

The x can cancel

[tired walrus]

it means that if you wanna cancel something in a fraction then it needs to be MULTIPLIED to the rest of the shit

$\frac{a \cdot b}{a \cdot c} = \frac{a}{c}$

[soft zealot]

Ann

[nocturne cedar]

I get that

[tired walrus]

but importantly there needs to be nothing else added or subtracted

[nocturne cedar]

So this isn’t ok canceling

[tired walrus]

it is incredibly un-OK

[nocturne cedar]

Ok

[tired walrus]

you should not even try to invent stupid cancellation tricks like these.

[jagged siren]

This is breaking two laws

[edgy mauve]

It isnt

[tired walrus]

you should simply add the fractions the honest way.

[nocturne cedar]

I wasn’t trying to

[tired walrus]

ie common denominator and all that shit

[edgy mauve]

Or multiply conjugate

Of denominator

[jagged siren]

First to a LCD, then go from there and show us your work.

[nocturne cedar]

👍

Something’s wrong

The answer to my question is 4

[jagged siren]

Send a photo

[nocturne cedar]

Just find the answer basically

When I cut the denominator and stuff I got 4

Probably a fluke but

Now I got 16

[jagged siren]

Send your work too

[jagged siren]

Where did the top go?

[nocturne cedar]

What

[jagged siren]

On the second image, the second line with the brackets

What's going on there

[nocturne cedar]

I did it one at a time

That was the denominator

I did cross multiplication

For adding fractions

[jagged siren]

Your surd rules are wrong and so are your fraction

Ok let's start from basics

[nocturne cedar]

Uh ok

[jagged siren]

So fractions,

Like we explained ealier it was the canceling out

[nocturne cedar]

Ok

[jagged siren]

However in THIS example you have

There's + and - signs

[nocturne cedar]

Ya

[jagged siren]

In and fraction AND jn the middle

So therefore you need to find an LCD

Do yk what an Lcd is

[nocturne cedar]

But cross multiplication works

[nocturne cedar]

Sorta

[jagged siren]

No it does not

[nocturne cedar]

How

I can do any fraction

[jagged siren]

There's a + sign why are you multiplying

[nocturne cedar]

Lemme show u

It’s called butterfly method

I think

U can try it for urself

[night raft]

,rotate

[jagged siren]

Thanks

[jagged siren]

This is very wrong

[nocturne cedar]

How

[nocturne cedar]

Is this wrong?

The calculator looks ok

[jagged siren]

No the answer foe that shoukd be 32/15

[night raft]

yeah

I'm not sure what 3.1s means though in the denominator

[nocturne cedar]

Bruh that’s just simplifying

[jagged siren]

Multiplying (butterfly method) ONLY works with × signs

And when there's one term on the top

[night raft]

actually 96/45 does simplify to 32/15

[jagged siren]

No I know

[nocturne cedar]

https://youtube.com/shorts/NpWiXh_oUfE?feature=shared

[jagged siren]

But it doesn't apply to the example he's asking

[nocturne cedar]

Watch this pls

[nocturne cedar]

How thou Th

I don’t get it

[jagged siren]

Bcs there's 2 terms at the top.

Ok so

[nocturne cedar]

Ok explain

[jagged siren]

When you work with algebra

There's smt called terms

[nocturne cedar]

Mhm

[night raft]

oh nvm i'm blind, it reads 3 * 15

[jagged siren]

Terms are separated by + and - signs

So 5×b is one term

But 5+z is two

[nocturne cedar]

Yes

So

[jagged siren]

When the butterfly method comes into play

That's for one term

[nocturne cedar]

That’s why u use brackets

[jagged siren]

But when you have more than one

You need an LCD

[nocturne cedar]

No

U use brackets

[jagged siren]

And multiply the tops by thier respective brackets

[nocturne cedar]

Yes

[jagged siren]

But that's what I'm saying

[nocturne cedar]

Check the response to

[jagged siren]

Where'd the top go

[nocturne cedar]

The thing u pointed out was just denominator

Denominator =1

[jagged siren]

Can I ask what grade you are in

[nocturne cedar]

9

I’m 15

March 8th 2010

I’m in Brazil so school is a little different

[jagged siren]

Ok so do you guys know that squared brackets give you three terms

Or multiple brackets

[nocturne cedar]

Ya

[jagged siren]

Have u learnt that

Ok cool

Can I write it out for you

[nocturne cedar]

Sure but idk why u think I added the fractions wrong

U can legit check it

It checks out

[nocturne cedar]

,rotate

[nocturne cedar]

This is = 14

[jagged siren]

This is with your lcd

Then you multiply out

Surds cancel

[nocturne cedar]

So how much did h get

I did the same thing

[jagged siren]

Lemma do it rq

[nocturne cedar]

I just wanna know how to get the answer 4

[jagged siren]

OH I SEE UR PROBLEM

Sorry it took me a while to figure out where u were going wrong

Ima screenshot ur image rq and draw over it

[nocturne cedar]

Ok

[jagged siren]

The top already has each one. You need to multiple it by the one it DOESNT have

So for the 2+sq3 would multiple it by the 2-sq3

And visa versa for the other one

And when you multiple surds, it falls away so sq3 × sq3 =3

[nocturne cedar]

But that’s right

[jagged siren]

No you said the brackets multiple together

You just doubled the same bracket

[nocturne cedar]

If its addition

Then u add them

If it’s subtraction u subtract

[jagged siren]

Ur not understanding

Let me write it out

[nocturne cedar]

He saying it should multiply the fraction numerator

Ur*

[nocturne cedar]

What are u showing me

[jagged siren]

The fraction together

[nocturne cedar]

This is the formula

[jagged siren]

All together

Then simply

Simplify

[nocturne cedar]

That’s not it

It’s squared

Cause u gotta cross multiply it

[jagged siren]

It's not.... ur working with the lcd...

U can't cross multiply it's two terms

[nocturne cedar]

Bro

U just bracket it

And it’s the same terms so just square it

[jagged siren]

Ok what exactly is the right answer so I can do the whole thing and send it

[nocturne cedar]

4

This is it

I got it

In the beginning i put everything power of 2

Then I forgot to undo it later

So then the fraction we were talking about was 14

So then the fraction +2 =16

[jagged siren]

Ok I did it two ways

[nocturne cedar]

Then I undid the square and it’s 4

[jagged siren]

The line divides the two options

[nocturne cedar]

Ok

[nocturne cedar]

Bro what

Is the 1st one

[jagged siren]

Read the top bit

[nocturne cedar]

Ok but why did u say either a or b is 4

[jagged siren]

Bcs when you put it to the power of 2 it doesn't matter

Lemma show u rq

[nocturne cedar]

Ok but like what if u didn’t know the answer is 4

U can’t just come to that conclusion

[jagged siren]

You can

If u don't understand that method go with the second one or the one you did.

[nocturne cedar]

I see

Ty

[jagged siren]

Yeahhhh!!!

Ofc no worries hope u understand better now❤️

[nocturne cedar]

.close

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[drowsy epoch]

So if you had 6 rolls you are guaranteed to always get a sixes?

[bold turtle]

!15m

[final saddle]

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[bold turtle]

For one

For two, this then means your logic is flawed, then, doesn't it?

Assuming two events don't influence another, call these A and B, how do you find P(A and B)?

AND being the operative word

Work through what I'm saying first, so that you can see why this is wrong

You don't ADD probabilities for both A and B to happen

I mean I guess yes

We could work with this

Suppose P(A) = x and P(B) = y; then what does this equal?

(wait no this is far too confusing)

If A and B are independent, P(A and B) = P(A) TIMES P(B)

If A OR B can happen, and they're mutually exclusive (i.e. they can't BOTH happen), then P(A OR B) = P(A) PLUS P(B)

Do you see then why this is not right?

Can you correct it?

[runic lichen]

blud just multiply the probabilites since they are independent and its one after another

[bold turtle]

Does the second roll depend on the first?

[bold turtle]

Blud just don't

If they knew that from the start, this question wouldn't have been asked

Again no

[bold turtle]

@vital kelp

Right

And does the first depend on the second?

No this isn't mutual exclusivity; this is independence

Mutual exclusivity is when one or the other but not both can happen

(i.e. P(A and B) = 0)

Which is certainly not the case here; you can roll 2 out of 2 sixes

Well they're independent

Reply to the message that therefore applies

yeee

[bold turtle]

@vital kelp

Well what two events are we talking about if we're rolling two sixes?

You're looking at the first six and the second six

[runic lichen]

lol

[bold turtle]

!xy then

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[runic lichen]

yea then use complemtnary progability 1- P(0 sixes)

[bold turtle]

(xy is so-called because you're asking Y when you mean X; statement Y might come from X, but it's not clear from context that you're talking about X)
https://xyproblem.info/

[runic lichen]

so the probability of u rolling a six is 1/6 right every roll

so whats the probability of u not rolling it every roll

oh

u can just casework then

like make 2 different cases

one where u roll one six

one where u roll two sixes

then add em up

[bold turtle]

(okay but at this point they haven't got that yet)

[runic lichen]

ok sorry idk where they at in terms of understanding rn

[bold turtle]

Dear lord we got there at last

[bold turtle]

But this is exactly 1 six, this is a little finnicky

!original

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[bold turtle]

I'd still like to see this (question) to better understand how to approach this (explanation)

The "probability of rolling 1 six" is not clear-cut

[bold turtle]

Spoilers, this is 5/18

I'm assuming 1 out of 2 dices are a roll of six

[runic lichen]

yea i think like add up the probabilties of getting EXACTLY one six like waes said and then two sixes

so like 1/6 * 5/6 * 2 + 1/6 * 1/6

[bold turtle]

(again this is 1) not what I said, and 2) not how I'd approach this)

[bold turtle]

...

Is this a problem to show the original question/

[runic lichen]

nah i think we answered his question idk

it was just like u roll twice whats the probability u roll at least one six

in a dice

[bold turtle]

Then this is computationally heavy as an approach

[bold turtle]

It's this approach

[runic lichen]

bro we alr went over them

[bold turtle]

Why?

[runic lichen]

either complementary counting or casework

[dense badge]

Another approach, you can consider all pair (x,y) of dice outcomes, there are 36 of them with the same 1/36 chance of happening then you can count the number of them where x or y is 6. There is this formula

[runic lichen]

yea but when u get problems like "at least once" or something like that using the complement is always the way to go just remember that

[tepid narwhal]

At least or at most have to use 1- P' (A)

[runic lichen]

well another one is casework but then can get very bashy and annoying for certain problems

but sometimes u have to so

mutually exclusive means they are completely separate so they have no intersection so you can just add their probabilities and nothing will be overcounted

like P(A or B) = P(A) + P(B)

[tepid narwhal]

Addition rule: Counting Total numbers of ways that doesn't affect each other

[runic lichen]

this is an AND problem not an OR problem

[tepid narwhal]

Multiplication Rule: (Including number of Ways in order)

Doesn't affect the order

You add up since it counts as ways

You multiply when they have an arrangement order.

Like numbers of ways to dress from blue, red, green shirts pants and underwear

So you multiply it

It's just the way you differentiate when you should add probability and multiply probability based on the question

[tepid narwhal]

^^

[dense badge]

yes, if you add the probability of having one 6 in the first roll event A and one 6 in the second event B you are counting 2 times the case when you have two 6 in a row so you subtract that P(A n B)

[tepid narwhal]

One picture here can solve your question idk if this helps...

👍

Probably you are one of the visual learners

[tepid narwhal]

It's best to draw it and give examples like this

[tepid narwhal]

Or this

[dense badge]

the case (6,6) two six in a row contributes 1/36 to P(A) AND also contributes 1/36 to P(B)

you have one 1/36 extra if you just sum P(A) and P(B)

[tepid narwhal]

Give him a visual picture

Where 1 roll show like this result and that 2nd roll shows that

[dense badge]

each (x,y) pair of possible dice values have 1/36 probability, each square in my picture is the event of observing a certain (x,y).

Event A and B have one square of overlap

writing each (x,y) was kinda a pain

those are all the outcomes possibles for two consecutive dice rolls. for instance the probability of first roll is a 1 and second is a 1 i.e. the (1,1) in the picture is 1/36. it’s 1/36 for any (x,y). then having a 6 in the first roll is event A and a 6 in the second event B. They overlap only at (6,6)

gtg

throw it now

if you can

yeah the 1/36 from two 6 in a row is in event A and event B so it’s counted one time extra

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[coral wedge]

guys when you do questions like this, and it’s <5 or <0, would that mean it’s invertible?

[coral wedge]

I’m assuming not because it’s not actually = to 5?

[dense coral]

hm, what do you mean by that?

a function is invertible on an interval when it passes the horizontal line test

if a horizontal line passes through the graph of f twice on an interval, then f is not invertible on said interval

[versed crater]

(And if the horizontal line doesn’t pass through the graph at all the “inverse” won’t even be a function!)

[coral wedge]

like for example A, 0<=x<5, would not pass because u get infinitely close to 5, but never 5?

[glossy bone]

Can you find a horizontal line in that interval that passes through two points on the graph?

If not, the graph is invertible

[coral wedge]

only for 5<=x<8 I believe right?

[glossy bone]

Why do you say that?

[coral wedge]

Right?

because these two aren’t <=

[glossy bone]

but that doesn’t make it fail the horizontal line test

if it were a <= sign it WOULD fail the test, but it’s not

[versed crater]

The domain doesn’t need to include 5

You can have a function on the domain [0, 5)

[coral wedge]

ohhh im dumb i just messed up the signs lol

[coral wedge]

meaning that each of the 3 are invertible

got itt

I’ve been doing this for wayyy too long today

[versed crater]

Are you sure

[coral wedge]

wait

lemme check again

[coral wedge]

C is invertible for sure

then B also should be invertible because domain doesn’t include 0, and the same logic for A?

[versed crater]

Yeah you’re right, what if there was a fourth option that said 4 <= x < 8?

[coral wedge]

Non invertible

Or wait

Invertible actually

😅

[versed crater]

Yeah anything that’s less than 5 will be invertible provided you restrict to codomain to the image

The inverse of f when you restrict to 4 <= x < 8 has domain of [0, 3.5] ∪ [4, 4.5)

[coral wedge]

Okay I see

[versed crater]

Cos in the middle the horizontal test doesn’t cross any part of the graph

If you just blindly use im(f) as the domain for f inverse it will sometimes not be a function

It needs to be im(f restricted to 4 <= x < 8)

[coral wedge]

alright I think I got it now

thanks a lot for the help

.close

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[round cedar]

hey

[tired walrus]

f is literally differentiable...

any differentiable function is continuous

[round cedar]

thats ok, but how u come up with the fact that f is diff? can i say it since F'(x)=x^2 is cont for every x?

[tired walrus]

yes...

its derivative exists

you said it

[round cedar]

thx

.close

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[terse crypt]

open new channel

this one's killing itself soon

[narrow lichen]

oh

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[tranquil pine]

uh anyone?

i dont even know where to begin, its a trigonometry question

[thin cloud]

Is $\sum \sin^2(A)= \sin^2(A) \sin^2(B) + \sin^2(C)$ ?

[soft zealot]

Alexis_Fx

[thin cloud]

I'm confused by how they use sigma notation

[tranquil pine]

ok im back

[tranquil pine]

ya im confused too

[glossy nacelle]

looks weird tbh

[tranquil pine]

oh

well then ill leave that q for now

i had another one

from same assignment

[glossy nacelle]

what

the

[tranquil pine]

?

uh im kinda stuck in that q

[whole halo]

I dont have a nice way but its a simple way

this question wants you to prove that
[
\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B\sin C=1
]
given $A+B+C=\pi/2$

[tranquil pine]

pls ensure that an 11th grader from india will understand it

[soft zealot]

mtt

[whole halo]

you do that by solving for C
C = pi/2 - (A + B)

then put that in the equation

[tranquil pine]

i didnt understand the pi and sigma notations there

like what are they exactly meaning

in that specific question

[thin cloud]

He mean it looks disgusting and I'm agree with it

[whole halo]

they should be teaching you that in school

[tranquil pine]

well they did but in that question i didnt understand

[whole halo]

well from the context you should be able to infer that they want a cyclic sum

they gave you conditions on A, B, and C

where none of A, B, or C can really be told apart from each other

a cyclic sum then just adds across A, B, C

so $\sum_{\text{cyc}}A=A+B+C$

[soft zealot]

mtt

[whole halo]

and $\sum_{\text{cyc}}\sin^2A=\sin^2A+\sin^2B+\sin^2C$

[soft zealot]

mtt

[whole halo]

similarly $\prod_{\text{cyc}}\sin A=\sin A\sin B\sin C$

[soft zealot]

mtt

[whole halo]

this is really common in competitive math, so they are using sums and products to shorten a pattern like the above

in this case, the "cyc" is left out since it is clear from context

though your teacher shouldve told you about it first

that context comes from being familiar with these kinds of questions already

[whole halo]

so with this, solve for C to get C = pi/2 - (A + B)
then put that in the equation
then keep simplifying, eventually youll see that its 1

[tranquil pine]

alright

ill try 1s

ill need like 5 mins or smth for that

disgusting question ngl

[tranquil pine]

yoo tysm i got the first one, do you know anything for the 2nd one?

this one

uhh@whole halo?

[final saddle]

@tranquil pine Has your question been resolved?

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[tranquil pine]

uh

[tranquil pine]

no

it hasnt 💀

[ivory bear]

how was it done?

such a brilliant one

[ivory bear]

mate whats the answer?

92?

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

yah

mb for late reply

i thought no one was answering so i went on to do the next one

howd you do it btw

[ivory bear]

yes here my friend responded to the same question-

thats how he did it

better thank him

[tranquil pine]

oh tysm

to both you and your friend XD

[ivory bear]

[tranquil pine]

tysm once again!

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[burnt maple]

If p(p>0) is a root of
x¹¹-x¹⁰+x⁸-x⁷+x⁵-x⁴+x²-x-20=0
Then p¹² is greater than
(A)57
(B)59
(C)60
(D)61

[thin cloud]

I'm thinking of $p^11-p^10+p^8-p^7+p^5-p^4+p^2-p-20=0$

[soft zealot]

Alexis_Fx

[thin cloud]

Then $p^12-p^11+p^9-p^8+p^6-p^5+p^3-p^2-20p=0$

[soft zealot]

Alexis_Fx

[rustic light]

(put the exponent in {})

[thin cloud]

$p^{12}=p^{11}-p^9+p^8-p^6+p^5-p^3+p^2+20p$

[soft zealot]

Alexis_Fx

[thin cloud]

And find the minimum or smth

Maybe try complete the square on the rhs

[burnt maple]

Square?

[thin cloud]

Oh wait

It's odd degree,

Hmm

Oh yeah

Still

Since p>0

We use complete the square here might still be possible

[burnt maple]

What square to complete

I don't see any anything here

[thin cloud]

Try to make a bunch of square might work

I don't have any pen or paper atm so

[lusty seal]

why not use any inequalities

since p is +ve

[thin cloud]

Hmm, yeah that should work too, i didn't notice p>0 until I said using square

[burnt maple]

Which inequalities?

[lusty seal]

Thats where we will focus

First, group minus and plus seperately

[burnt maple]

Ok

After that?

[lusty seal]

let me think

AM-GM is a firsthand

[final saddle]

@burnt maple Has your question been resolved?

[final saddle]

@burnt maple Has your question been resolved?

[final saddle]

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[sinful knoll]

@nocturne elbow

[sinful knoll]

i didnt get the question

<@&286206848099549185>

[terse crypt]

what are u

confused

u have a set of six letters

and u are asked to create a four-letter long string

[opal plinth]

??

[terse crypt]

i think??

[opal plinth]

Does that say "boxes"?

[terse crypt]

now this is confusing

[sinful knoll]

i think its 6 letters that be arranged in 4 letter boxes

but is it ??

[terse crypt]

ok if that is that then

[sinful knoll]

letters boxes

[opal plinth]

If the answer is supposed to be 6^4 then I imagine it's just like putting 4 distinct objects in 6 distinct boxes

[sinful knoll]

it is

[opal plinth]

Right, so what are you confused about?

[sinful knoll]

that the number is not reduced and repeated

[opal plinth]

6 choices for the first object, 6 for the second, ...

[terse crypt]

shouldnt it be the other way around?

[sinful knoll]

but it says 4 words is it not ?

im confused on the question

itself

[opal plinth]

It says four letters not four words

[opal plinth]

No?

[sinful knoll]

4 letters be posted in 6 letter boxes i was thinking 4^6

[terse crypt]

putting 4 objects in 6 boxes would give 4 * 4 * 4 * 4 * 4 * 4 = 4^6???

[sinful knoll]

right?!

[opal plinth]

What

How many ways to put 1 object in 6 boxes?

[sinful knoll]

maybe ill type the question

just for clarity

how many different ways can 4 letters be posted in 6 letters boxes

is the question

so is it not 4 objects or letters

??

[terse crypt]

ok i get it now. @opal plinth

• what im thinking: we have infinite no. of objects of 4 types
• what u're thinking: we have only 4 objects but 6 boxes which allow force boxes to be vacant when fully filled

[sinful knoll]

wtf

wait why infinity where did this come

[opal plinth]

There's no mention of infinity anywhere

[sinful knoll]

oohhh bro ure that guy lol

k

[terse crypt]

in the sense that i can repeat the one letter multiple times

[opal plinth]

You might be confused by the wording, if I'm understanding correctly they're talking about paper letters, not letters as in ABC

[terse crypt]

ye 😭

[opal plinth]

Hence why they are "posting" them

[terse crypt]

thats my confusion

6^4 makes sense now

[sinful knoll]

how

[terse crypt]

at least 2 of the 6 boxes will be left vacant

[sinful knoll]

yea

no

why

[opal plinth]

Why would you care about boxes left vacant

[sinful knoll]

right

[opal plinth]

Imagine you're playing a weird game where you have to deal 4 card to 6 players

[sinful knoll]

ok

[terse crypt]

becuz this doesnt happen in the 'letter case'

[opal plinth]

You can deal the first one to any of the 6 players

Same for the second card

Same for the third and the fourth

6*6*6*6

[sinful knoll]

okay

wait

i didnt understand

[nocturne elbow]

you need to understand the restrictions being communicated by the question

there is no restriction stating that each box can only have one letter

there is no restriction stating the letters are different

you have four identical objects

[opal plinth]

Idk what you're talking about

[nocturne elbow]

and 6 places to put them

[sinful knoll]

4 objects to put in 6 different places

thats what im getting

[nocturne elbow]

yes

[terse crypt]

yes

[nocturne elbow]

so how many options are there for the first object?

[sinful knoll]

1st place

[terse crypt]

i misinterpreted the question initially

[sinful knoll]

??

[nocturne elbow]

options

how many different places can the first letter be placed?

[sinful knoll]

6

[nocturne elbow]

okay

so the first letter has 6 options

how many options for the second letter?

[sinful knoll]

6 still

[nocturne elbow]

okay

so 36 arrangements

[sinful knoll]

umm

makes sense

[nocturne elbow]

lets number the letter boxes 1 to 6

if i say 1 that means a letter has been placed into box 1

lets look at two letters - what are the options?

11, 12, 13, 14, 15, 16
21, 22, 23, 24, 25, 26 ... and so on

11 = two letters in box 1

12 = letter in box 1, letter in box 2

in total youahve 36 different options

then each letter you add you just multiply by 6

---

here is a question that might feel more "normal"

How many options do you have for a four digit numeric pin?

[sinful knoll]

this kind is kinda principle of counting

prob

i guess

in permutation

right??

[nocturne elbow]

im really not confident in the names sorry mate

how many options do you have for a four digit pin?