help-28

[devout valley]

Aha, that’s for the guided question

I forgot that it was part of the overall thing but we are gonna use that of course

[supple plover]

ohhh

so the n -> infinity is part of the final epxression?

and theta -> 0

[devout valley]

We’re gonna use this one

[supple plover]

oh okok

[devout valley]

What we’re gonna do is to take more and more of the triangles, with smaller and smaller thetas, until the polygon turns into the circle for us

The “taking (…) smaller thetas” is the theta -> 0 part

And we of course worked out the area of the polygon in terms of r and theta already, didn’t we

[supple plover]

mhmhm

the 2pi/theta * (1/2) r^2 sin theta

[devout valley]

Yep, all of that, we basically are doing $\lim_{\theta \to 0} \text{that}$

[glossy valve]

@devout valley

[devout valley]

(I’m not writing out that expression because I’m on mobile and laying in bed )

[supple plover]

im doing the math on my bed, and ong im feeling the eepy but i gotta lock in

so we doing that limit expression

[devout valley]

Awwww, let’s keep pushing so that we can both rest as soon as we can

[supple plover]

mhmhm

[devout valley]

That’s the last part writing the expression is our part 8, so we wanna have that written down

Then for part 9, we actually work it out

[supple plover]

ah so just for clarification the lim theta approaching 0 = this expression ^

[devout valley]

I’m gonna do a bit of simplification for us, we wanna find $\lim_{\theta \to 0} \frac{\pi r^2 \sin(\theta)}{\theta}$

[glossy valve]

@devout valley

[supple plover]

ohhhh

[devout valley]

Yep, turns out I lied to myself and I would write it, look how friendly I am

[supple plover]

an angel even

too nice

[devout valley]

Awwww

Anyways, do you know how to work out that limit?

(Assumedly they’ve taught you it )

[devout valley]

Well, they say they have, so if they haven’t

[supple plover]

uhhh the lim theta approaching 0 sin theta/0 =1 ?

[devout valley]

We wanna use that, yep

(The denom being theta of course )

You’re also happy that constants just “scale” the limits, right?

(That you can basically find the limit, then multiply it by the constants)

[supple plover]

multiply it by the constant 2?

[devout valley]

Well, not quite 2 here-

[devout valley]

For a start, I cancelled out the 2 and 1/2 to make our lives easier-

[devout valley]

So we only have to consider this

But $\pi r^2$ is a constant here, so we can equally work out $\pi r^2 {\color{green} \lim_{\theta \to 0} \frac{\sin(\theta)}\theta}$

For which, you told me you know how to do

So, what’s the area of a circle?

[supple plover]

pi r^2

[glossy valve]

@devout valley

[devout valley]

Sounds very familiar to me

As it should be, as the green is 1

And that’s part 9 done for us!

[supple plover]

OMG

and then also answers part 10 i think

because we get the formula for the area of the circle

OMG

[devout valley]

Yep

[supple plover]

WE DID IT CHART

WE'RE FREE

[devout valley]

I told you we would

[supple plover]

lemme write all this down real quick, just in case i have any questions before we depart

[devout valley]

Yep sure get it all down so we can rest easy

[supple plover]

okok i believe I got it all, thank you so much can I buy you a coffee or somehting

[devout valley]

Awwww, happy you got it all down and you’re very sweet, don’t worry about getting me anything, working with you has been pleasure enough

[supple plover]

AH MY HEART that so sweet, well have a good nights rest chart, it was a pleasure working with you <3

[devout valley]

Awwww, I hope you have a wonderful night too! I’m gonna get myself some sleep
hopefully see you around ❤️❤️❤️❤️

[supple plover]

yes yes, see you around!

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[candid tinsel]

what does this mean

what am i supposed to pick of the fomulas

[gritty rose]

find f(4x) using f(x) = 16-x^2

[candid tinsel]

what

[gritty rose]

whereever you see an x, put 4x instead

using parentheses will be helpful

i recommend going through this free khan academy lesson 1 unit 8 https://en.khanacademy.org/math/algebra/x2f8bb11595b61c86:functions/x2f8bb11595b61c86:evaluating-functions/v/what-is-a-function

[candid tinsel]

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[full vector]

I need help with this question.
I derived the formula to solve this question.

[full vector]

Here is my work

I Want to ask if this formula is correct to solve the question. If not which formula I have to use

[full forum]

@full vector Has your question been resolved?

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@full vector Has your question been resolved?

[full vector]

Anyone. I just want to know if the formula is correct or not

[sand escarp]

It's correct

[full forum]

@full vector Has your question been resolved?

[full vector]

Ok thanks

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[formal spoke]

lol its alr bro go ahead

[neat terrace]

ty

why am i not getting the right answer here?

I took moments about c btw

ping me

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[neat terrace]

Im also so lost on this question

this is about all i've done

idk what to do after

[rain merlin]

do you know how to resolve a vector

the tension isnt vertically upwards, it's in the string

in this direction

[neat terrace]

ah

[neat terrace]

not sure what that means exactly

[rain merlin]

youre given the angle for a way to DETERMINE the vertical component of this vector

[rain merlin]

where we can express Fy for example as
sin theta = Fy/F
so Fy = Fsintheta

[neat terrace]

oh yh

[rain merlin]

so here can you find the vertical component

given the angle of the tension

[neat terrace]

but u dont have hypotenuse?

[rain merlin]

it's given in the question

[neat terrace]

oh is the tension

[rain merlin]

"the tension in the cable is 80N"

[neat terrace]

the hypotenuse

ok cool i was thinking

[rain merlin]

yep

[neat terrace]

that

nice

[rain merlin]

do you think you can do it now?

[neat terrace]

ahh i rly dont understand what they're asking

isnt the centre of mass from the wall

just going to be half of 4?

i mean thats my understanding

[rain merlin]

the question specifies it's non uniform

[neat terrace]

ohhh

[rain merlin]

recall the definiton of centre of mass

[neat terrace]

ok

[rain merlin]

where all the forces APPEAR to be acting

it doesnt necessarily have to be in the middle

the question gave you an equilibrium case to determine it with moments

[neat terrace]

wait ik i labelled it wrong

but would there still be these reaction forces

and u take moments with that?

[rain merlin]

does this help

[neat terrace]

i have the side

is 40

[rain merlin]

the 80N tension is in the string

yep

40 is right

[neat terrace]

Like i got the sides of the triangle

but idk what to do from there

[rain merlin]

does this help visualise?

@neat terrace

[neat terrace]

oh wait

i basically got the answer

thanks

[rain merlin]

which was?

[neat terrace]

well i put x = 32 cause i used 40 instead of 40sin30

40sin30 is just 20 tho

[rain merlin]

😭

[neat terrace]

so it was actually 16

😭 got the idea tho thank you

[rain merlin]

fire

np

[neat terrace]

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[rain merlin]

1.6 btw

not 16

[neat terrace]

oh shoot wait

ah yeah

[rain merlin]

,w 40 * 4=100 * x

[glossy valve]

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Requested by technobladenever_dies

[rain merlin]

just to make sure IM not tripping

[neat terrace]

ahhh

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[neat terrace]

yeah no no ur right

[neat terrace]

im trupping

[rain merlin]

🔥

just close it again

[neat terrace]

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[sharp osprey]

hi

[sharp osprey]

i heard or read somewhere that non-real roots come in conjugate pairs

is the complex number -1 -i

considered non-real

[solar aurora]

yes

[sharp osprey]

and if that complex number shows up as my root

then i would expect the other root to be

-1 + i?

[solar aurora]

yes

[sharp osprey]

okay then im confused because i just worked through a problem and looked at the solutions

and the other root turned out to be 1 + i

[solar aurora]

can you show the solution

[sharp osprey]

yes

[solar aurora]

/problem too

[sharp osprey]

original problem, in swedish unfortunately

it says "solve the equation z = 2i, the answer shall be given in the form z = a + bi where a and b are real numbers."

and then the proposed solution

and i got the same answer, but i thought that the rule was they come in conjugate pairs

[solar aurora]

my bad , the conjugate complex roots applies to quadratic equation with real coefficients

like ax^2 + bx + c =0
a,b,c should be real

[sharp osprey]

ah and we have something like

z^2 - 2i + 0 = 0

[solar aurora]

yup

[sharp osprey]

okay thanks gunsh

that clears up all my confusion

[solar aurora]

also the method you used is nice
i tried it by taking z=x+iy

lol

[sharp osprey]

sry that isn't my solution

[solar aurora]

ohh

[sharp osprey]

that was what the examinator did

my solution is messier

[solar aurora]

did you try by eulers or x+iy
i think x+iy is easier for this question atleast

[sharp osprey]

i did (z^2)^(1/2) = (|z|^2)^(1/2)(cos((theta + 2pik)/2) + (sin((theta + 2pik)/2)i)

and checked k = 1 and k = 2

but i could have checked k = 0 instead of 2

anyway thanks again

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[stuck fiber]

Part (c)

Sorry if im opening a lot of help channels back to back i just did a mock exam and im self correcting all the ones I didnt understand 😅

So we need to turn that parametric form into a parabola

My only idea is to do

y(t) = a(x(t))^2 + b(x(t)) + c

But that wouldnt get us very much I think...

[devout valley]

Alright, let's look at the equations we have

We have that both
[
x(t) = {\color{pink} \cos(t) } + 1
]
and
[
y(t) = 4{\color{pink} \cos(t)} - 2{\color{orange} \sin^2(t)}
]

[stuck fiber]

Ooo colours

[devout valley]

Any more ideas now?

[stuck fiber]

Hmmmm

I feel like this is a massive hint that im not getting 🤡

[devout valley]

Funny enough, I don't really like the colour pink

[stuck fiber]

Its my favourite 😝

[devout valley]

(and yet my favourite colour is purple )

[stuck fiber]

Also W colour, doesnt get enough love

I think guys look really good in purple, not enough of them wear it

But OK, maths 😂

[devout valley]

Well then, in that case, what's your least favourite colour you can think of?

[stuck fiber]

Orange

[devout valley]

Cool, one moment please

[glossy valve]

@devout valley

[covert heath]

${\color{orange} \text{yeah fair tbh}}$

OH NO

i give up

[devout valley]

color

[covert heath]

oh

[stuck fiber]

American spelling hehe

[devout valley]

Grrrr

[covert heath]

britihs

[glossy valve]

Percy

[devout valley]

Anyways, look at that nasty orange

[stuck fiber]

I get that we can convert the sin to cos

But I dont really know what were working towards

[devout valley]

Just do it for now

[stuck fiber]

y(t) = 4cos(t) - 2sin^2(t)

y(t) = 4cos(t) - 2(1 - cos^2(t))

y(t) = 4cos(t) - 2 + 2cos^2(t)

OH

Its already a quadratic

I think I can solve

2cos^2(t) + 4cos(t) - 2

[devout valley]

(It is, but not the one we want yet!)

[stuck fiber]

No?

[devout valley]

Cool, so now, what we have is
[
{\color{pink} x(t)} = {\color{orange} \cos(t) } + 1
]
and
[
{\color{green} y(t)} = 2 {\color{orange} \cos^2(t)} + 4{\color{orange} \cos(t)} - 1
]
and we want ${\color{green} y(t)}$ in terms of ${\color{pink} x(t)}$

Do you notice anything yet?

[stuck fiber]

What do you mean by y(t) in terms of x(t) sry?

[devout valley]

You want to write y = [something that only has x in it]

Also one second

[glossy valve]

@devout valley

[devout valley]

Hehe

[stuck fiber]

Ok I thiiink I see what I have to do

x(t) = cos(t) + 1

x(t) - 1 = cos(t)

y(t) = 2cos^2(t) + 4cos(t) - 1

y(t) = 2(x(t) - 1 )^2 + 4(x(t) - 1 ) - 1

And I think I can just solve the rest myself haha

x(t) can just become x, right?

In this case

[devout valley]

Yep, you can just write x there

[stuck fiber]

Awesome!!

I get it all now

Thank you again! (again 😝 )

WRONG EMOJI 😭

Accidentally really freaky

[devout valley]

I didn't even catch it

[stuck fiber]

Oh... then it was nothing hehehe

❤️

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[gloomy briar]

How do I do this?

[gloomy briar]

I have no idea what to do, I missed the day when this lesson was taught

[pure basin]

I think you should have a formula for area

[covert heath]

could you do it if the angle was 180 degrees

[gloomy briar]

Nope I have no clue how to find these sort of areas

The unit just started and I wasn’t really paying attention in class

[pure basin]

Percy is it theta/360 * r^2?

[covert heath]

sigh

[pure basin]

It's sigh

Do I always have to tell everyone that

[onyx glen]

what about an entire circle

[gloomy briar]

Yeah it’s like pi r squared

[onyx glen]

not "like"

it's exactly that

pi * r^2

now

if you wanted a 180° sector

that would be what fraction of an entire circle?

[gloomy briar]

Half?

[onyx glen]

yes

[covert heath]

hey thats what i asked like 10 seconds ago 😭

i envy you ann /lh

[onyx glen]

so then how could you find its area

[onyx glen]

you didn't prime her with the "entire circle" question

[covert heath]

fair

[gloomy briar]

1/2pi(r^2) right

[onyx glen]

anyway

[onyx glen]

yes

[covert heath]

horrible formatting

but yes

[onyx glen]

but the idea is that a sector should be thought of as a fraction of the circle

[covert heath]

please at least seperate the 1/2 and the rest

[onyx glen]

so now 130°

[covert heath]

1/2 pi r^2 yes

[onyx glen]

this 130° sector is what fraction of its circle?

[gloomy briar]

130/360?

[void tusk]

how are you very much active on discord..what do you do for a living @onyx glen ?

[pure basin]

wizard

I don't think you should ask that

Ann gets mad if someone asks that

[onyx glen]

honestly none of your business mate

[void tusk]

oops

[onyx glen]

also you are obstructing

[covert heath]

there we go!

[onyx glen]

@gloomy briar yeah so the area of the sector is that fraction times the full circle.

[gloomy briar]

130/360 pi 18 squared?

[onyx glen]

the same logic you just went through with a half circle goes equally well for any other fantom

[onyx glen]

130/360 * pi * 18^2

[covert heath]

$\frac{130}{360} \pi r^2$ is indeed what you're looking for!

goddamnit

[glossy valve]

Percy

[covert heath]

you can simplify the fraction

but that's about it

[gloomy briar]

Tysm guys I get it now!!!

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[sharp osprey]

hi

[sharp osprey]

i cant tell where i went wrong

i wanted that positive sign to be negative at the end

then it would be perfect

if that + was a -

when i take out the last term out of the p+1 sum

should it be negative?

i dont think it should be it doesnt make sense why it would

this looks right

this looks wrong

and so i assume the error is inbetween

but idk

it looks right too

i dont see what went wrong

<@&286206848099549185> This is an induction proof gone wrong, and I can't find where I went wrong.

i was trying to be cute by doing the base case last, but i just checked and the base case is true

so that's not the reason

there is a proposed solution

for this problem

this is what they do

that's crazy

well ok at least i was wrong about being wrong, i just didnt know how to continue, paradox

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[faint drift]

Greetings! I am stuck on how to u-substitute ^^'

The whole problem: Find the area of the polar coordinates

r = e^-(theta/4), from pi/2 <= theta <= pi

So, I have gotten to this point:

A = (1/2) integral (from pi/2 to pi) [ e^-(theta/4) ] * 2 d * theta

The -(theta/4) * 2 becomes -(theta / 2), but now I need to u-sub this. I feel really lost on how to approach this.

[rough tundra]

you have (A=\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}e^{-2\frac{\theta}{4}}d\theta)?

[glossy valve]

PajamaMamaLlama

[faint drift]

Wait-

To be a little clearer with the e part, it is

(e^-theta/4)^2

[rough tundra]

yeah ur right then

[faint drift]

Where do I go from here? How do I U-sub this part? o>o

[rough tundra]

well really ur only problem is that -theta/2 is the exponent, yes?

[faint drift]

Yeap

[rough tundra]

sooo what's the logical thing to substitute here?

[faint drift]

-theta/2

But how to u-sub it is where I am stuck on

[rough tundra]

yes so we say u=-theta/2

so we int e^u*dtheta

but of course that dtheta make no sense when our integral is in u, right?

so if we have u=-theta/2

[faint drift]

We need to derive -theta/2

[rough tundra]

so what's the derivative of -theta/2 w.r.t to theta?

[faint drift]

This is where I'm stuck ;u;

I have no clue how to derive this part.l

[rough tundra]

what's the derivative of (-\frac{x}{2})?

[glossy valve]

PajamaMamaLlama

[faint drift]

I have no clue what to do with the / 2 part, but x^1-1 will become 1

So far, I have -1/2

[rough tundra]

and that is the derivative

[faint drift]

Just that?

Nothing special to do with the denominator?

I am especially scarred and tired of forgetting so many special rules ;u;

[rough tundra]

nope, because it's just a constant and recall that (\frac{d}{dx}[cf(x)]=c\frac{d}{dx}[f(x)])

[glossy valve]

PajamaMamaLlama

[faint drift]

Right!

But just for my own sake, we have 1/2 [-x]'

[rough tundra]

we actually have -1/2 [x]' :)

[faint drift]

Ok!

So, treat Theta just like an X

[rough tundra]

alright so then derivative of -theta/2 is just -1/2, yes?

[faint drift]

Looks about right. Nothing to drop down from the exponent

So now we have du = -1/2 d*theta

We need to shift -1/2 to the other side

Oop

[rough tundra]

though remember, we don't just shift, we have to multiply both sides by -2

[faint drift]

Warning, what's the warning about?

[rough tundra]

to cancel that -2 guy in the denominator :)

but if this is what you meant then absolutely right

[faint drift]

Was about to say that; just forgot to mention it.

So -2 * du = d * theta

-2 cancels the -1/2 on dtheta

[rough tundra]

now you're nearly there, there's one more thing you gotta do with definite integrals and u-sub, and do you know what that is?

[faint drift]

That is to change the integral range (I dunno what its proper name is) by inserting the u-sub

[rough tundra]

your bounds, yes exactly!

[faint drift]

So -pi/2 for the highest point, - (pi/2) / 2 for the lowest bounds.

Does the lowest bound just change to -pi/4? o>o

Ogie

Time to progress

[rough tundra]

alrigiht now putting it all together

oh wait

I realized something

you did make a mistake here

[faint drift]

Ah dangit

Ah broke the gahd damn wheel

[rough tundra]

in the theta world the top bound is pi and bottom bound is pi/2

you had these as negative

[faint drift]

Negative due to the u-sub

u = - theta/2

[rough tundra]

wait minute

good catch

[faint drift]

Place in the pi and pi/2 there, and we get the negative bounds

[rough tundra]

I was just testing you 😅 /j

Alright I agree, that was my bad

[\frac{1}{2}\int_{\frac{\pi}{2}}^{\pi}e^{-2\frac{\theta}{4}}d\theta\to\frac{1}{2}\int_{-\frac{\pi}{4}}^{-\frac{\pi}{2}}-2e^udu]

[faint drift]

I'm way better at catching smaller details when someone is working than I am working on the bigger picture.

[glossy valve]

PajamaMamaLlama

[rough tundra]

I am the complete opposite, so good thing

[faint drift]

So 2 is a constant.

-2 * 1/2 turns into -1

Whoop. slow down.

Integrate e^u

[rough tundra]

which iss... 👀

[faint drift]

Should come out as e^u ->-

Riiiiiiight...? ->-

Yay

So -1 * [e^u] bounds from -pi/2 to -pi/4

[rough tundra]

Nota bene: you were right to say we are indeed allowed to plug in the bounds right away only because just so happened to be (e^u), but not true in general :)

[faint drift]

Ogie

Plug in the bounds into U

(e ^ -pi/2) - (e ^ -pi/4)

[rough tundra]

what about that minus sign guy in front of the integral 👀

cannot forget him

[faint drift]

We cannot! But I was saving him for last.

But from here, what do I do...? o>o

Do we just have the formula of e - e?

Just ultra-generalizing for now

[rough tundra]

well pretty much nothing lol that guy is pretty much as simplified as it can get

[faint drift]

Ok!

Then if there is nothing else we can do, we plug in our -1

[rough tundra]

you can technically simplify by factoring out a power of e but it will jsut cause confusion I see no reason why ur prof or teach wouldn't accept that answer

[faint drift]

So we get -(e ^ -pi/2) + (e ^ -pi/4)

AAAAAAAAAAAAAAAAA

GODS I HATE THE MANY HIDDEN RULES OF MATH ;-;

Thank you Pajama!

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[sharp osprey]

Hello, I have a question about notation

[sharp osprey]

f:C -> C

i dont really know what f:C means

C is the complex number set

[weak iron]

yes

[sharp osprey]

and -> means we are outputting

[weak iron]

depends how the C is drawn

[sharp osprey]

this one is drawn

the way the set would be drawn

[sharp osprey]

yes

it is drawn like that

[weak iron]

yup complex number set

[sharp osprey]

yes yes

but the f:

[gritty flax]

Thats a function from C to C

[weak iron]

it means we are mapping

[sharp osprey]

idk what mapping means

i thought like if i said

x -> f(x)

[weak iron]

it undergoes something to yield something

[sharp osprey]

that means i have an input x, and i get an output f(x)

[kindred grove]

"f is a function whose inputs are complex numbers and outputs are complex numbers"

that's the english translation

[sharp osprey]

okay f being a function i have seen before

so the : means like

[kindred grove]

it says essentially nothing about what the mapping actually does to the input

[weak iron]

defines or defined by

[kindred grove]

what are the inputs, what can be the outputs, that's it

it's a lot like a type signature in programming

[sharp osprey]

im not really programming yet

unfortunately

im trying to use the : in "a mathematical sentence" that isn't too similar to the example i gave

to see if i understand

If S is a statement, and we can have a lot of different statements, can I say we have S:n statements?

oops

[restive geyser]

Er...

Not really?

[kindred grove]

well set-builder notation is kinda like that

[restive geyser]

You'd have n statements, which you might label S_1,..., S_n

(where "_" is meant to mean "write what follows as a subscript")

[sharp osprey]

can you give a different example of when : is used

than the original

[restive geyser]

Ratios

[sharp osprey]

oh yeah

[kindred grove]

https://en.wikipedia.org/wiki/Set-builder_notation from here

[restive geyser]

Simple numerical e. g. in Germany I have seen are more commonly written as what in UK/US schools would be more familiar as ratios

[sharp osprey]

i feel like i dont get much understanding from reading : as "such that"

like f:C

[restive geyser]

This is not "such that"

"f:C->C" is read "f is a function that maps the set of complex numbers onto the set of complex numbers"

Or, if I were to read/say this at speed, "f maps from C to C"

[sharp osprey]

and sorry what does map mean

like isn't C already mapped to C

idk what mapped means <-<

[restive geyser]

"maps... onto..." = "takes inputs from... and whose outputs are in..."

A mapping is any such object that "maps", like above

A mapping is a function in the proper sense if, when you give it an input, it returns you at most 1 output

[sharp osprey]

i thought that was like included in the definition of a function, not whatever mapping is

[restive geyser]

A function is a type of mapping

All functions are mapping; not all mappings are functions

Technically this should be taught when you learn about functions as an object, though since this never crops up again in HS (iirc) I can see why this might be confusing

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[steep dragon]

I'm a little confused here

[steep dragon]

I want to use uniqueness, but that only tells me that f is identical to tanx on 0 leq x leq 1. Isn't it possible for f to take on other values outside of this range?

Maybe I want to use f's analyticity but idk how that helps me

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[split oak]

posted elsewhere a while ago

[split oak]

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✅

[knotty grail]

is there any syllabus limitations?

like can I just start using archimedes principle here?

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[chrome ore]

Let ABCD be a trapezium (trapezoid), such that (AB) is parallel to (CD).

Let O be the intersection point of the lines (AD) and (BC).

Let O' be the intersection point of the diagonals (AC) and (BD).

Let I and J be the midpoints of the segments [AB] and [CD], respectively.

Show that the points O, I, J, and O' are aligned

[chrome ore]

I think i need to use homothety but im lost

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[chrome ore]

<@&286206848099549185>

[simple yarrow]

Did you draw it?

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[chrome ore]

Yes

But i figured it out so tnx

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[proper estuary]

For this question, would I put in my calculator cos^-1(.5045 or cos^-1(-.5045

[narrow path]

oh yeah cos^-1(-0.5045)

[proper estuary]

ty

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[torn jolt]

Hello world

I got 12 boxes with marbles inside, blue, purple and yellow, the amount is infinite but the base probability to draw a marble when grabbed at random is 94.3%, 5.1% and 0.6% respectively. The probability of blue marbles only decreases where purple increases and yellow dosent change. All blue marbles are the same, but there are 60 different purple marbles and 8 different yellow marbles. Purple marbles have 60 different animals printed onto the marble's surface where yellow marbles have 8 different tools printed on the surface. What animals or tools isnt important, but 3 animals out of the 60, 1 tool out of the 8 has a increased probability when drawing that type of marble. The probability to draw a purple or yellow marble dosent change but the probability for that certain type is increased. This will be called a rate up for simplicity. I have a 50% chance to get any rate up marbles and a 50% chance to get anything else, example the purple marble rate up is dog, lion and iguana, so i have a 50% chance to get either of those when drawing a purple ball and a 50% chance to draw some other animal out of the 57 non rate ups.

All 12 boxes:

1. base probability for blue, purple and yellow, rate up
   Purple = shark, elephant, killdeer
   Yellow = screwdriver

2)probability of blue decreases, purple becomes 56.1%, yellow stays at 0.6% rate up same as 1

3)probability of blue becomes 0, purple becomes 99.4%, yelloe does not change, rate up same as 1

4)probability same as 1 but all purple marbles are the rate up ones only

5)probability same as 2, all purple are rate ups

6)probability same as 3, all purple are rate ups

7)base probability, rate up
Purple = penguin, kingfisher, camel
Yellow = knife

8. same probability as 2, rate up same as 7

9. same probability as 3, rate up same as 7

10)same probability as 7, but all purple is rated up only

11)same probability as 8, but the purple is rated up only

12)same probability as 9, but purple is rated up only

[torn jolt]

Box 1-6 are boxes used for event A where box 7-12 are for event B

Event rules: usually drawing marbles are from the first boxes (1 for A and 7 for B) which has the base probability, after drawing for 8 consecutive times without getting a single purple marble, i will draw from the next box (2 for A and 8 for B), if no purple marble is still seen, i will draw the 10th time in the next box (3 for A and 9 for B) which only contains purple and yellow marbles. When drawing a non rate up purple marble, the same rules apply for the next few draws except box 4,5 and 6 are used for A and box 10,11 and 12 are used for B instead where all purple marbles in those cases are the rated up marbles.

Question: how many draws on average/minimum is required to get a single elephant (one of the 60 animals printed on the purple marbles) in both events?

Situation: i simulated it, event A took 24 trials whereas even B took 15603 trials, the difference is too large so i want to calculate the real rates

<@&286206848099549185>

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[torn jolt]

<@&286206848099549185>

Is my question that tough? TAT

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

[mortal sigil]

its just that nobody likes statistics and probabilities

(me included)

[fiery hemlock]

false, I love them.

[fiery hemlock]

question is not clear to me.
Here's what I understood:
We first draw from boxes 1 and 7, if the first 8 are not purple we move on to the next box but I don't understand how many draws we do there. And nothing even after that
One more thing I don't understand is that you've not told about the total number of colors or the number of marbles of any color or the distribution of the marbles of any color. But I could deduce that there are 60 purple marbles in both events' boxes.

[torn jolt]

I stated that the number of marbles are infinate for simplicity

Its just probability with no amount

[torn jolt]

How many draws is what im trying to find out

[torn jolt]

There are 60 types of purple marbles but the number is not 60

[fiery hemlock]

do you do only 1 draw in the next 5 boxes in each event?

and after that the cycle continues to start from the first box again?

[torn jolt]

Yes

[torn jolt]

But i just keep drawing

Until i get purple elephant marble

How many draws on average/minimum is required?

[fiery hemlock]

isn't minimum 1, lol

are there 2 colors?

[torn jolt]

Oh true

[torn jolt]

3 colours

Blue purple and yellow

[torn jolt]

So then the average then

[fiery hemlock]

but the last 4 boxes in each event doesn't contain blue, right?

[torn jolt]

The 10th box in eavh event dosent contain any blue

Its like

The first 8 times have the qame probability

But the 9th time has increased for purple

And the 10th has no blue

[torn jolt]

10th draw*

[fiery hemlock]

how is it increased in 9th?

[torn jolt]

I change the box

I got 12 different boxes

I draw in them at spesific cases

[fiery hemlock]

but that box still have all 3 colors

[torn jolt]

Yes

Only if i dont draw a purple in the 9th where i draw from the 10th there will be no more blue balls

[fiery hemlock]

I'm currently solving for one event but I just read that you want the probability of getting at least one in any event. I don't have a simple solution and due to large values(60, 6) I'm actually writing a code to solve the problem but if we consider both events then I think it would get more complicated.

[fiery hemlock]

Here's the answer of my code:
89.11081541092710138485

@torn jolt

[torn jolt]

Oh damn

This is for

Which event?

[torn jolt]

It's ok just one is enough

[fiery hemlock]

one event, aren't both events the same?

[torn jolt]

Oh the first event has the elephant in the rate up

The second dosent

[fiery hemlock]

Wait, the first assumption of my sol was that the probability of elephant is 1/60.

[torn jolt]

Oh then it's the 2nd event

[fiery hemlock]

So we want to calculate the expected #draws to get 60 purple marbles.

And that I did with code.

if you want I can share the code.

[torn jolt]

Ooo getting all 60 types works as well

[torn jolt]

Suree thanks

[fiery hemlock]

no, no, as the probability of getting elephant is 1/60. So the expected number of purple marbles such that we get an elephant is 60. And so I calculate the expected number of draws to get 60 purple marbles.

[torn jolt]

Ohhh

Yess it works as well

Thankss btw

[fiery hemlock]

np

#include <bits/stdc++.h>
using namespace std;

#define ld long double

void solve()
{
    ld dp[6][60],p[6];
    for (int i=0;i<6;i++)
        if (i==2) p[i]=(ld)1/(ld)2;
        else p[i]=(ld)1/(ld)3;
    int lim[6];lim[0]=8;
    for (int i=1;i<6;i++) lim[i]=1;
    for (int i=0;i<60;i++)
    { // everything is cyclic, so first we need to fix one variable
        ld cnt=1,pr=1,val=0;
        for (int box=0;box<6;box++)
        {
            for (int draw=0;draw<lim[box];draw++)
            {
                if (i)
                    val+=pr*p[box]*(cnt+dp[box][i-1]);
                else
                    val+=pr*p[box]*cnt;
                cnt++,pr*=(1-p[box]);
            }
        }
        // dp[0][i] = pr*(dp[0][i] + cnt)+val
        // dp[0][i]*(1-pr) = pr*cnt+val
        dp[0][i] = (pr*cnt+val)/(1-pr);
        for (int box=5;box>0;box--)
        {
            // dp[box][i] = p[box]*1 + (1-p[box])*dp[next box][i]
            dp[box][i]=(1-p[box])*dp[(box+1)%6][i]+1;
        }
    }
    cout<<dp[0][59]<<endl;
}

signed main()
{
    cout<<fixed<<setprecision(20);
    solve();
    
    return 0;
}

Here's the code, you may ask an AI to understand it and if you do, please share the response with me too.

[torn jolt]

Suree tqtq

[torn jolt]

What language tho

[fiery hemlock]

C++

[torn jolt]

I see

[fiery hemlock]

np, actually I took it as a challenge when that guy said "nobody likes probabilities".

[torn jolt]

XD HAHAHGA

Thank you anywayyy

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[manic cape]

which inequality should I use?

[hallow walrus]

try to expand the thing first

note that $\abs{v}^2=v\cdot v$

[glossy valve]

kheer257

[remote kite]

RMS-AM Inequality?

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[teal basalt]

rounding in mathematics.

how important is rounding?

assume there is 3 choices:

1. No rounding
2. Rounding to the nearest, where half will be rounded up.
3. Rounding to the nearest, where half will be rounded to nearest even number (Banker's Round).

the normal rounding that I usually see mathematicians use is the number 2 rounding method. why didn't we use the Banker's rounding, especially since it's better?

Actually is rounding really that important?

[unique wagon]

fundamentally you have to round numbers to some decimal precision for any physical machinery to work

banker's rounding is better in general so i really dont know why they dont teach that instead

maybe because it is less intuitive relative to normal rounding

[teal basalt]

hmm...

[rocky vale]

In mathematics there basically is no rounding, you only care about precise values. In the physical sciences, you'll take the best approximation you can get. Rounding a half up or down is just a convention, and doesn't really matter in practice because a physical measurement will never be exactly half of a whole number, or exactly halfway between two known values.

[teal basalt]

i see... thx everyone

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[urban plume]

what happened here? How did the sqrt(x) get on the bottom?

[hot herald]

x = sqrt(x) * sqrt(x)

[urban plume]

1/x isn't sqrt(x) tho? D:

[hot herald]

and then cancel the factor of sqrt(x)

but you have 1/x being multiplied to sqrt(x)

[urban plume]

do I multiply the top and bottom by the sqrtx?

[hot herald]

$\frac 1x \cdot \sqrt{x} = \frac{\cancelto{1}{\sqrt{x}}}{\sqrt{x} \cancel{\sqrt{x}}} = \frac{1}{\sqrt{x}}$

[glossy valve]

ℝαμOmeganato5

[hot herald]

x = sqrt(x) * sqrt(x)

[urban plume]

oH

[hot herald]

you can use that

do I multiply the top and bottom by the sqrtx?
also works

[urban plume]

ok I see how that was done

oki ty :)

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[torn plinth]

Find the volume bounded by z=sqrt(4-x^2-y^2), z = sqrt(x^2+y^2), and z = x^2+y^2, in the first octant

[torn plinth]

the first integral is x^2+y^2 to sqrt(x^2+y^2), the second integral is sqrt(x^2+y^2) to x^2 + y^2, and the third integral is sqrt(x^2+y^2) to sqrt(4-x^2-y^2)

is this correct

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[woeful pasture]

can u put it in proper integral notation

[torn plinth]

is this correct or no

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@torn plinth Has your question been resolved?

[crude kayak]

you can see that the integration region is invariant under rotation
so you may wanna go polar

or cylindrical

a good choice of coord sys can make your life easier

[torn plinth]

are my limits for z correct?

[crude kayak]

first sketch a good graph (on paper or online, just the desmos 3d tool that i've shown you), which will answer your question

[torn plinth]

I basically graphed the zy trace, and got this:

blue curve is basically the cone z=sqrt(x^2+y^2), red curve is the paraboloid z= x^2+y^2, and the black curve is the sphere sqrt(4-x^2-y^2)

[crude kayak]

then imagine the 3D version, by rotating the figure around the z-axis

it reduces to a high school volume of solid of revolution problem

[torn plinth]

why are my bounds wrong tho? I can't use the 2D version to get the bounds for z?

cuz in many other problems it worked

if i assume that that specific curve is part of a larger surface

[crude kayak]

write things in a complete sentence
what actually is this?
integral for what?

[torn plinth]

i trying to find the bounds for z in the integral

that's not the entire integral

[crude kayak]

ofc i know
just to say that it's meaningless to write the first two terms, which cancel each other

so by looking merely at the last term, you'll immediately see that it makes no sense

[torn plinth]

but what if they have different dx and dy

would they still cancel

[crude kayak]

where are the "dx" and "dy" in your picture?
write things out clearly

[torn plinth]

ok i will write it one moment

[crude kayak]

in fact you see the rotational symmetry of the region of integration around the z-axis
that's why using cylindrical coordinates is a smarter choice

[crude kayak]

how come you have '9' inside the root sign?

how do you get those 1.3, 1.6, 2.1?

those look sus

[torn plinth]

oh it should be 4

i will show u wait

[crude kayak]

i suggest that you have a clearer mental image, by colouring the desired solid's (vertical) cross section passing through the origin

you'll then see that your desired quantity should be a sum of two iterated integrals

instead of three

[torn plinth]

This is what I got. “R” is the 2D region for each case.

[crude kayak]

follow what i said in my prev msg 1st

after you've finished, if you have time to kill, retry the problem using volume of solid of revolution

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[undone gulch]

hello! i don't understand algebra at all

[gentle lantern]

like what

[gritty rose]

https://en.khanacademy.org/math/class-7-od/x48ac5f37f7da7ee1:algebra-odia-class-7

[undone gulch]

like basiks

[undone gulch]

like bsics

[gentle lantern]

what have you learned?

[covert heath]

just stay calm and $(x+y)^2=x^2+y^2+2xy$

[glossy valve]

Percy

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[torn plinth]

Is my answer correct? Find the volume bounded by z=sqrt(4-x^2-y^2), z = sqrt(x^2+y^2), and z = x^2+y^2

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[paper frost]

I think substitution is best here?
get like 1/u and then ln(u) is that possible?

[gritty rose]

yes

[clear pumice]

e is just a constant so yes

[paper frost]

ah yeah true

[paper frost]

We're not allowed to use calculators during this exam so how would you go about solving the definite part of the integral yk the e^2-e

[gritty rose]

use log and exponent properties

,tex .exp rules

[glossy valve]

riemann

[gritty rose]

,tex .log rules

[glossy valve]

riemann

[paper frost]

thanks, can I ask another question without opening a new channel?

cuz if so I have no idea how I would derive this one

[gritty rose]

use differentiation rules

,tex .diff rules/chain rule

[glossy valve]

riemann

[paper frost]

yeah but I dont quite understand how I would do e^(e^ex)

[gritty rose]

write f as a composition of multiple functions

[paper frost]

so like e^g(x) where g(x) = e^ex?

[gritty rose]

that works

[paper frost]

okay thanks!

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[solar glade]

Are these right?

[lime ether]

,rotate

[lime ether]

what’s this for

[solar glade]

calc 1

im mainly wondering about b (2nd one) and c because its dne if the left and the right dont match up, but if theres a solid point doesnt that mean that its the answer?

[rustic sequoia]

you're pretty much set :D, except for the last two items

do you want help breaking it down

[mortal sigil]

you love breaking stuff

[solar glade]

so would it be dne then for x->3?

since from the left its -2 and right is 1

[rustic sequoia]

yep, since two sides of a limit must converge at a single point for a limit to exist

[solar glade]

alright

[rustic sequoia]

same for the last item

[solar glade]

so that one would be 3 then

[rustic sequoia]

yep! limits are independent of the function's actual value at x=a

[solar glade]

okay thanks

.close

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[steel jetty]

Were are supposed to use inverse trig functions

[steel jetty]

I know the trig formula im supposed to use but am confused how to change the function inside the sqrt to match that formula

[gritty rose]

Which formula

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[bright thicket]

i need help with this

[bright thicket]

So i got

sinx^1/2=cosx

then in sin^2x + cos^x=1 i subed in

got

sinx^2 + sinx^1/3=1

then i let sinx by y

got

y^2+y1/3=1

then i did

y^6+y^2=1

let y be a

so now i have

a^3+a=1

but i dont know what do next?

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[gritty rose]

Should be 2/3 exponent here

[gritty rose]

Because the power of cosine here should be 2

[bright thicket]

yes

ok

i meant to say that i think it doesnt change anythign

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[granite torrent]

(sinx)^(1 / 3) = (cosx)

let y = sin(x) or smth, you get y^2 = (1 - y^2)^3, which you solve as a cubic

the answer is not pretty since you have to solve an irreducible cubic

$\sin(x) = \sqrt{1 - \sqrt[3]{\frac{2}{3\sqrt{93} - 27}} + \sqrt[3]{\frac{\sqrt{93} - 9}{18}}}$

[glossy valve]

Mqnic_

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[void tusk]

I am getting wrong answers...please help

[austere cove]

,w solve (2y-3)^2 + y^2 = 233

[glossy valve]

Results provided by WolframAlpha

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Requested by omnipotententity

[austere cove]

@void tusk I agree with your work. So does WA

Why do you think the answers are wrong?

[rain merlin]

there's 2 sets of results no?

[austere cove]

Oh right

Fair call

[bright bronze]

The second pair need you to use the y=-28/5 value

[austere cove]

Those are the results for y

To get x you need to double and subtract 3.

[void tusk]

i didn't solve the x and y

[bright bronze]

y=-28/5 gives you x as -71/5 which is the second answer

[austere cove]

For each solution

[void tusk]

how 28/5

[bright bronze]

Both the + and - are for different values of y

its not one is x and other is y

[void tusk]

yes right...

but i am getting both wrong

[bright bronze]

there are two possible solutions to this, which means (x1,y1) and (x2,y2)

what you got is y1 and y2

you need to get corresponding x1 and x2

then the given answer would match the pairs of (x,y)

[rain merlin]

says who

-5.6 = -28/5

[void tusk]

oh

how do i solve for x

[void tusk]

putting value in first step??

[void tusk]

.

[rain merlin]

id reccomend for future problems you write the equations in terms of two variables

x^2 + y^2 = 233

and then choose one of them, either x or y, does not matter if not specified to make the second eq

here you basically did x = 2y - 3

and then you substituted

but you did all this directly

by showing these steps it'll be more clear

[void tusk]

oh 👍

. close

[rain merlin]

you

mistyped it 😭

it's without the space @void tusk

!done

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[void tusk]

.close

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[tame elbow]

Pls help

[onyx glen]

,rccw

[tame elbow]

1001a + 101b + 11c +2d is 2019

I’m stuck after that

[onyx glen]

well clearly a cannot be 3 or greater, yes?

[tame elbow]

Ummm yes

[onyx glen]

so a is either 1 or 2 (bc it can't be 0 either)

consider a=2 first

[tame elbow]

Mhm

[onyx glen]

then you get 2002 + 101b + 11c + 2d = 2019

[tame elbow]

Mhm

[onyx glen]

so 101b + 11c + 2d = 17

[tame elbow]

Mhm

Wait what

[onyx glen]

subtract 2002 from both sides lol

[tame elbow]

But their values can’t be decimal

[onyx glen]

whose

[tame elbow]

b c d

[onyx glen]

and at what point did i say anything about decimals

[tame elbow]

Ok ok

[onyx glen]

do you understand what i did

[tame elbow]

Yes

[onyx glen]

ok do you see how to continue

[tame elbow]

Nope

[onyx glen]

what can b be

[tame elbow]

Um

9?

[onyx glen]

can b be 9?

[tame elbow]

Yes

[onyx glen]

how so?

[tame elbow]

909

Wait nvm

[onyx glen]

909 + 11c + 2d = 17 gives negative values for c and d

b cannot be 9 for sure

[tame elbow]

Wait wait how can this be true if they are not decimal or negative

[onyx glen]

if b is 9 the left hand side is too big

hell even when b is 1 the left hand side would be too big

[tame elbow]

Ye

[onyx glen]

so what's the only thing that b can be

[tame elbow]

0

Do think we should make a 1

Wait 9+9+9+9 is 36. So 2019-N is less than or equal to 36. Which is less than or equal to 1983

[onyx glen]

that will come later

we still aren't done with the a=2 case

[onyx glen]

yes b=0. so now 11c+2d=17.

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[sweet monolith]

can someone help me qués 22

[sweet monolith]

<@&286206848099549185>

.close

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[dusty barn]

"using the table above, list all possible unique results from adding Dice 1 and Dice 2?" what does it mean by 'unique results'

[dusty barn]

can someone help im losing my shit

@full forum

<@&286206848099549185>

[rocky delta]

whats is the problem

[dusty barn]

question 5.

[rocky delta]

list the number u can get when u add two dices

[dusty barn]

so it woupd be 2,3,4,5,6,7,8,9,10,11,12

[rocky delta]

u cant get 1 in that list because there is no number add up to 1 in dice 1 and 2 the minimum u can get is 2 because 1+1 is two and maximum is 12 6+6

[rocky delta]

yeah

[dusty barn]

i never learnt that in yr7 jeez what is my sisters class learning

thanks though

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[orchid lake]

Could someone help me with this question? The one on the bottom

[pure basin]

,rccw

[pure basin]

First of all I suggest you redraw it to get a better understanding

Because the obvious step is not obvious when you draw it like that

[orchid lake]

Oh

[pure basin]

Draw it like a proper right angled triangle

[orchid lake]

Like this?

[pure basin]

Better

Now can you understand what to do?

,rccw

[pure basin]

What is the area of a triangle

[orchid lake]

Do I just square 2x+8 and divide by 2

🤔