help-49

[grim sierra]

wait

its revolving about the y axis

[clear inlet]

well it will fill all space

[grim sierra]

its centered at the z axis

and x i guessdepending on perspective

[grim sierra]

yes

hmm so

im not sure what spherical thing describes it?

i know y = rsintheta = 0

rsintheta = 0

is that the spherical counterpart?

im visualizing it as like r = infinity

[clear inlet]

not really. It's the xz plane in 2d, when you resolved it around the x or z axis it will be all R^3

[grim sierra]

yes so

r = infinity but that's not like a thing

[clear inlet]

yes

[grim sierra]

how do i describe all of R^3 in spherical coords

[clear inlet]

not a geometric shape

[grim sierra]

the answer key says sin(theta) = 0

i dont see how that is correct

oh wait it said

spherical

i misread as cylindrical my bad

[clear inlet]

$r \in (0,\inf), \
\theta \in (0,\pi), \
\phi \in (0,2\pi)$

[grand pond]

leonardomoura

[grim sierra]

alright so what are like the general steps to finding the cylindrical counterpart of a rectangular equation revolved about an axis?

also, i'm not sure if this is necessary, but im going thorugh a textbook and have not convered vector functions yet

like functions of form f(x, y, z)

am i meant to do that before this?

[clear inlet]

first of all drawning the curve resolved, then you think about what is the distance from the axis of rotation, that will be your r

[grim sierra]

alright

could u give another question which doesnt jjust have x = r @clear inlet

just to ensure i understand

[clear inlet]

there are many in vector calculus textbooks

look for questions about cylindrical and spherical coordinates system

[grim sierra]

alright

well thank you a lot

imma close it now

.close

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[clear inlet]

you're welcome

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[ember prairie]

is there anyway i can fastly get the prime factors of a number?

like 9450
which has 2 * 3^3 * 5^2 * 7

[vapid trench]

yeah I usually look online

[grand latch]

divisibility tests may help

@ember prairie

[ember prairie]

we divide it with every prime u mean?

[grand latch]

no

like if you are given a number, you first check where it is divsible by a certain prime number or not

like divisibility test for 3 is meaning a number to be divisible by 3 is that the digit sum of the number must be a multiple of 3

[ember prairie]

ohhh

sum of the digits of a number

[grand latch]

there is test for 2,3,4,5,6,7,8,9,11 and many more

[ember prairie]

got it, thanks

[midnight plank]

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[molten bay]

7^m+7^n is divisible by 5 and mn are between 1 to 100

[molten bay]

So how can I get values?

[midnight plank]

@molten bay Has your question been resolved?

[grand latch]

7^1=7
7^2=49
7^3=343
7^4=2401

there are two pairs

7^1+7^3 and 7^2+7^4

you can generalize this

[midnight plank]

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[molten bay]

@grand latch

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[prime garden]

Find the value of the summation:
1/1^2 + 1/2^2 + 1/3^2 + ........... + 1/inf^2

[verbal pumice]

hmm

[prime garden]

Assuming that it is known that this summation has a fixed value

[verbal pumice]

well

there is a quick way

[prime garden]

!status

[midnight plank]

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

[verbal pumice]

do you know the product formula for sin x

[prime garden]

No

No trigonometry is allowed

[verbal pumice]

I think this is the original way Euler solved it

[prime garden]

😄

Euler?

maybe

[verbal pumice]

did you just come to us with the basel problem and are applying your own restrictions on it

[prime garden]

I want to find the value of this summation.

[verbal pumice]

,w sum from k = 1 to infinity 1/k^2

[grand pond]

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[prime garden]

I know

[verbal pumice]

$\sin x = x\prod_{k = 1}^{\infty} \left ( 1 - \frac{x^2}{k^2\pi^2} \right )$

[grand pond]

dyxn

[prime garden]

Well sin(x) is undefined in my dictionary

so

...

[verbal pumice]

good luck!

[prime garden]

ok

[vagrant willow]

lol what is happening here

[prime garden]

Read the above conversation to know.

[vagrant willow]

all the best! i'll recommend you reading some mathematical literature to find the answer to your problem

[prime garden]

Like reading a non-trignometric proof of this summation?

[vagrant willow]

but you will need to know quite a bit of advanced mathematics, i believe

[verbal pumice]

and all the math literature will end up pointing back to euler's product for sin

[vagrant willow]

😭

i'm pretty sure you're right but it's worth trying

[prime garden]

This is a simple summation, there are no terms that can't be defined not using trigonometry

[verbal pumice]

seems like you know how it's done then

[prime garden]

of course, that's the purpose of mine, the purpose of a baby mathematician

I shouldn't have come here, because this server is for fast maths help regarding small questions and problems

Nothing is known at first without any thinking

Did Euler know his solution from the start?

[chrome vessel]

a side note, we don't write 1/1^2 + 1/2^2 + 1/3^2 + ........... + 1/inf^2

the definition of infinite sums are as follows:

we define $\sum_{n=1}^\infty \frac{1}{n^2}$ to be $\lim_{N\to\infty}\sum_{n=1}^N \frac{1}{n^2}$

[grand pond]

qwertytrewq

[prime garden]

yes

[chrome vessel]

we are not summing to 1/infty^2

[prime garden]

That's how I wrote it one way just for fun

oh

okay

I get it

[chrome vessel]

but rather taking the limit of the finite sum

[prime garden]

thanks y'all

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[twilit field]

So here I'm not supposed to use line integrals , I suppose

Using green's theorem, this is basically $\int_{0}^{1} \int_{0}^{1} y^2 -x dy dx$

[verbal pumice]

uh

[grand pond]

What a wonderful world !

[verbal pumice]

are your limits right

[twilit field]

Yea

[verbal pumice]

you're integrating over a right angled triangle right

so it should go from 0 to 1 - x and then from 0 to 1

[twilit field]

I thought it's over a square

oh

okay

I misread the question

[verbal pumice]

(0,0) --> (1,0) --> (0, 1) --> (0,0)

[twilit field]

yea

so $\int_{0}^{1} \int_{0}^{1-x} (y^2-x )dy dx$

[verbal pumice]

yeah there we go

also like

parentheses please :P

[grand pond]

What a wonderful world !

[verbal pumice]

perfect

now suffer

[twilit field]

Thanks!

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[hasty flume]

can someone help me sketch b

[midnight plank]

@hasty flume Has your question been resolved?

[analog laurel]

the line A to center of mass should be parallel to the vertical

otherwise there will be an unbalanced torque about A

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[normal cloud]

The sum of x values satisfying equality?

[normal cloud]

Should i factor it?

[verbal pumice]

have you already written cos 8x in terms of sin 4x?

[upbeat spire]

Break it down to a single trigonometry ratio (sin/cos)

[normal cloud]

Will it helpful?

[verbal pumice]

yes

[normal cloud]

👍

Bismillah im starting

[upbeat spire]

Use the identity of sin2x and cos2x and reduce it to only sin or cos

[normal cloud]

🤔🤔okey

1-2sin^2.4x ?

[normal cloud]

İ think im did

[verbal pumice]

yeah that's right

$\sqrt{2}\sin 4x - (1 - 2\sin^2 4x) = 1$

[grand pond]

dyxn

[verbal pumice]

now this is a quadratic equation in sin (4x)

can you solve for sin 4x using the quadratic formula?

[normal cloud]

💀ill try

[normal cloud]

Can u help

[verbal pumice]

for now call sin 4x u

can you rewrite it

in the form of a quadratic equation

[normal cloud]

Ah

İ get it

1min

This?

[verbal pumice]

,w solve sqrt(2)x - (1 - 2x^2) = 1

[grand pond]

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[verbal pumice]

so sin 4x = 1/sqrt(2)

$4x = \frac{\pi}{4} + (-1)^{n}n\pi$

[grand pond]

dyxn

[verbal pumice]

I think you're right

but again I can't be bothered to plug in values manually here

[normal cloud]

Thx❤️ @verbal pumice

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[normal cloud]

For the function f graphed above in the perpendicular coordinate plane.

Equalities are provided.

According to this

What is the value of the integral?

[normal cloud]

Guys can we say the value of sum of areas is 20?

[hasty haven]

have u done anything with change of variable yet?

[normal cloud]

No idk

[hasty haven]

wdym by sum of areas

[normal cloud]

1 min

İm stuck here

[hasty haven]

ya

[midnight plank]

@normal cloud Has your question been resolved?

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[normal cloud]

@dreamy lichen sry for ping but i realy need help in this

[dreamy lichen]

Yeah, so firstly, what happens to graph of f(x) when its changed to f(2x)

what kind of transformation occurs

[normal cloud]

Ummmm

🤔

😨😵‍💫

[hasty haven]

damn bro died

[normal cloud]

💀

İm bad ad this

To left?

[dreamy lichen]

https://www.desmos.com/calculator/jziujxfhn7

slowly change k from 1 to 2 and tell me what happens

it shows graph of sin(kx) compared to plain sin(x) btw, so at k = 2 it's gonna reach the f(2x) state

[normal cloud]

Hmmmmmmm🤔🤔🤔🤔

[dreamy lichen]

so what does the graph do?

Describe it in words, no math is needed

[normal cloud]

Norrowing

[dreamy lichen]

Yeah, it's basically squishing

[normal cloud]

im sory🤧

[dreamy lichen]

np at all

since our function is f(2x), it's gonna get squished by a factor of 2

what do you think happens to the area when we squish the function

[normal cloud]

Hmmmmmmmmmm

Smaller?

[dreamy lichen]

Yep, by what factor?

by a third? A half? A quarter?

[normal cloud]

Half cuz its 2x

[dreamy lichen]

exactly

here is a sketch

sorry for my awful handwritting

it's a/2, b/2 and f(2x) respectively

so from this, we can conclude that the integral in question is just half of the integral from a to b of f(x)

because the blue area is just half of the green one

so now the question is reduced to finding the green area and then halving it

is that clear?

[normal cloud]

Bro i should be the one sory im wasting ue time🤧

Np

[normal cloud]

Yes👍

[dreamy lichen]

Aight, perfect

so now let's focus on this graph

i just colored the areas so that we can refer to them

[normal cloud]

Alr

[dreamy lichen]

can you express these 2 facts using the green and orange areas?

In 2 equations (the variables being green and orange)

[wet pollen]

None of the helpers ever feel like they are wasting their time. Helping others helps to reinforce our understanding of the same material. 😉

[dreamy lichen]

Yeah, and we are here voluntarily. If we didnt want to be here, we wouldnt

[normal cloud]

İ think i can

İll try

14 and 6?

[wet pollen]

I can show a different approach after you are done here.

[normal cloud]

İ would be hapy sir

[dreamy lichen]

Yeah, 14 is orange and 6 is green

[normal cloud]

😁😁

[dreamy lichen]

so now that you figured out the green area, you can just halve it

and get the answer

[normal cloud]

Btw can we say 2x=u ?

Right

[dreamy lichen]

oh

yeah sure

i thought you said you havent done change of variables

probably just a different name then

u-sub is what you call it?

[normal cloud]

🫠bro idiot turkish system sory..

[dreamy lichen]

if you did this, then you'd also get $\int_{\frac{a}{2}}^{\frac{b}{2}}f\left(2x\right)dx=\frac{\int_{a}^{b}f\left(x\right)dx}{2}$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

is that the approach btw?

through u-substitution?

[normal cloud]

I guess I was focusing on the boundaries in a ridiculous way

Mb

İll fix it

[wet pollen]

That's the Fundamental Theorem of Calculus. Not sure what it would be called in your educational system though.

Which is what I was going to show you.

[dreamy lichen]

I see

[hasty haven]

ya

[normal cloud]

İ messed up again

[normal cloud]

Bro im stuck

Nedd help again...zzz

[hasty haven]

give the question

[normal cloud]

Wdym

[wet pollen]

Math wrote what I wrote.

[normal cloud]

Hmmmm🤔🤔🤔🤔

[wet pollen]

The division by 2 is because of the chain rule.

[normal cloud]

Oh k

👍👍👍👍👍👍

1/2.6= 3 ?

İs this answer?

[wet pollen]

Yes.

[normal cloud]

Hmmmmmm thx

Wait its -3

💀

İ messed again

😡

İt should be - cuz below x asis

[wet pollen]

Oops, didn't see the original problem.

[normal cloud]

I'm very distracted today......

[wet pollen]

You should go for a walk then. Clear your mind. 🙂

[normal cloud]

i played minecraft instead of walking which was a mistake

Now its dark not safe for walk xd

bro what would you rate this question

[wet pollen]

Fairly low. As long as you know how to apply the Fundamental Theorem of Calculus to this problem, it should be a quick solve.

And remember that the answer should be negative.

Which I missed.

[normal cloud]

Bro can i ask you 1 more question actualy i solved but i want to see your way

[wet pollen]

That was it.

Applying FTC was probably the correct method to solve this problem. The translation method works as well but probably not the preferred solution method.

For all of these math problems, come back to them a day and a week later and see if you still remember how to solve them.

[normal cloud]

👍👍👍

[wet pollen]

And not just necessarily remember the steps, you should understand why those steps work.

[normal cloud]

Yes i aggre they asked this for ftc ig

[normal cloud]

😁

[normal cloud]

Yes i always solving ösym problems again and again

[normal cloud]

🥲

[wet pollen]

?

[normal cloud]

can i ask you 1 more problem which i solved but i want to see your way and comment

[wet pollen]

Sure.

[normal cloud]

Accordingly, what is the number of elements in the union of set A and set B?

[normal cloud]

Yes AUB

How many members it have

[midnight plank]

@normal cloud Has your question been resolved?

[wet pollen]

Note in this example, that A \cap B = 2. There are two common elements between both sets. This translates to four common pairs in A x B and B x A.

[normal cloud]

Hmmmm🤔🤔🤔

[wet pollen]

This extends for any number of common elements between A and B. For A \times B \cup B \times A, there will be s(A \cap B)^2 common pairs.

For s(A \cap B) = 2, there are 4 common pairs between A \times B \cup B \times A.

[normal cloud]

Hmmmm🤔🤔🤔

İ think thats above my level🤯

[wet pollen]

Did you learn any set theory?

[normal cloud]

idk what is set theory

And dont think i remembered

Wait wait

İ know bacis ofc

[wet pollen]

"küme teorisi" is what Google says is the correct translation.

[normal cloud]

Yes küme teorisi i know

İts 9th grade

[wet pollen]

Not in the US.

[normal cloud]

lower ornhigher?

[wet pollen]

We learn about the notation(symbols) of Set Theory in the US, but we don't do a deep dive into it.

But that was back in my day, things may be different now.

[normal cloud]

hmmmmm🤔🤔🤔

[wet pollen]

So I don't think any student in the US would ever encounter this question.

[normal cloud]

Your days are best ig

[normal cloud]

Soo this is a hard question?

[wet pollen]

For a typical US high school student, it would be difficult.

But let me explain some of the concepts and hopefully that should help you understand how to solve the problem.

[hasty haven]

we learned abt this as part of our theory of functions

[normal cloud]

Yes keep goin pls

You explain it along with its logic and that is very good

[wet pollen]

Do you understand what the symbols mean?

[normal cloud]

Yes i know symbols meaning

[wet pollen]

In this example with sets A and B, there are two common elements.

This results in four common elements between both A x B and B x A.

Does that make sense?

[normal cloud]

Yes👍👍👍

[normal cloud]

👏👏

But how can we connect this to AUB

[wet pollen]

There are 46 unique pairs between A x B and B x A.

[normal cloud]

Oh i didnt knewed this formula

this goes into my notebook

[normal cloud]

👍

[wet pollen]

Does that make sense from the given matrices?

[normal cloud]

matrices?

İdk what is it

Sry

[normal cloud]

But the entire thing you did was makes sense

[wet pollen]

matris

[normal cloud]

Sry bro idk matris is what

U mean matrix?

[wet pollen]

Yes.

Matrices is the plural of matrix in English.

[normal cloud]

İsnt it a movie😥

idk man sry

[wet pollen]

Matrix (singular)
Matrices (plural)

[normal cloud]

yes bro but what is the meaning matrices in math

İs it a symbol?

İ dont think i learned something like that before

[verbal pumice]

it's a thing

that does many things

but it's basically a grid of numbers

at its core it's used to solve systems of equations

[grand pond]

dyxn

[wet pollen]

In this case, A x B is creating a set of ordered pairs.

[normal cloud]

Like permutation

[normal cloud]

👍👍👍👍👍👍

[wet pollen]

Cartesian Ordered Pairs if I remember correctly.

Cartesian Pairs

[normal cloud]

Yes

ik this

(x,y)

[normal cloud]

What to do next

[wet pollen]

You need to find

s(A x B) + s(B x A).

You can do that from the information given along with the equation I showed.

[normal cloud]

234?

[wet pollen]

AxB + BxA - 5^2 = 209

Yes.

[normal cloud]

Hmmmmm🤔🤔🤔🤔🤔

[wet pollen]

The thing to notice here is that s(AxB) = s(BxA).

[normal cloud]

117?

[wet pollen]

Yes.

Now that is not a perfect square.

[normal cloud]

😨

[wet pollen]

Which means that the size of the matrix is r x c.

[normal cloud]

Can we say like 9.13

[wet pollen]

Yes, but you do need to examine all of the possible options.

1, 117 and 9, 13

[normal cloud]

But we know AnB is 5

İt should be 9,13

[wet pollen]

Correct, so you can rule out 1,117.

[normal cloud]

Yes!!!!!!!

Wait

But thats all possible options?

[wet pollen]

Yes.

[normal cloud]

Soo 9and 13

Then its 19

[wet pollen]

So s(A) = 9 and s(B) = 13 or vice versa.

[normal cloud]

Gg

[wet pollen]

And s(A) + s(B) = 22.

But ... there are five common elements between the two sets.

[normal cloud]

💀💀💀💀💀💀

Lol injust said 22-5=19

Xd......

[wet pollen]

So you have 9 + 13 -5 = 17

[normal cloud]

Ty soo much

[wet pollen]

🙂

Did that all make sense?

[normal cloud]

That was really instructive

Yes!!!

Makes sense

But what woule you rate this bro?

[wet pollen]

Maybe a 5 or 6 for a US high school student. Some students in the US take Statistics and Probability so they should know how to solve that problem.

But if they didn't take a Statistics and Probability class, it would be an 8.

[normal cloud]

Do you think this problem deserve to be in a test for entering university

[wet pollen]

I would say yes. It shouldn't be the one question that determines if you are declined entry though.

It wouldn't be on a college entrance exam in the US.

[normal cloud]

We hav 40 question in math

[wet pollen]

We use tests called the SAT and ACT to determine a student's academic knowledge.

[normal cloud]

Do everyone enter same ?

[wet pollen]

What do you mean?

The ACT test leans more towards STEM degrees while the SAT focuses more on Arts and Science.

[normal cloud]

İ mean we have 2 test
Tyt and Ayt
Tyt 120 question
Ayt 80
Question writer is ÖSYM
everyone enter same exam and select university depends on his score and rank

[normal cloud]

But every uni have his own test right?

[wet pollen]

Some colleges do require an essay to be written for admission explaining why you would be a good student at that school, but I am not aware of any college that has a unique test for admission.

The only two tests used in the US are the SAT and ACT and that is only for high school graduates. Non-traditional students don't usually take it as far as I know.

[normal cloud]

Do sat and yks same level you think in math?

[wet pollen]

The SAT is not very strong on math and science in the way the ACT is.

The ACT would be taken if a student wanted to go into an engineering college.

The SAT would be for other colleges such as business, music, art, arts and science, etc.

[normal cloud]

Ah i get it

ACT is verh hard then

[wet pollen]

I wouldn't say it was. The difficulty lies in where one went to school. My high school was not very challenging but I still scored well enough to get into engineering. I scored a 29 out of 36.

[normal cloud]

Sir 29 out of 36 is awesom😱

[wet pollen]

Schools with a more rigorous curriculum would teach students well enough to get a perfect score of 36 out of 36.

[normal cloud]

Sir maybe in your time but we have internet!!! nobody can stop me xd

[wet pollen]

Well, I would not have passed the JEE in India that's for sure. 😄

But their school curriculum is different than in the US.

[wet pollen]

The internet is great for learning. There are so many resources and you can get real-time help as well. 🙂

[normal cloud]

Sir you still ver wise and good at math

[wet pollen]

ty

And you are a good student. 🙂

[normal cloud]

Ty!

Thx for your helps!🙏🙏🙏

[wet pollen]

yw

[normal cloud]

have a good day

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[grim sierra]

Find the equation of the line tangent to ⟨cos t, sin t, cos(6t)⟩ when t = π/4.

[grim sierra]

$$\textbf{r} = \langle \cos(t), \sin(t), \cos(6t) \rangle|{t = \frac{\pi}{4}} = \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0 \rangle$$
$$\textbf{r}' = \langle -\sin(t), \cos(t), -6\sin(6t) \rangle|{t = \frac{\pi}{4}} = \langle \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 6 \rangle$$
$$l = \langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 0 \rangle + t\langle \frac{-\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, 6 \rangle \quad t \in \mathbb{R}$$

[grand pond]

.RODATA

[grim sierra]

this is the answer the textbook is saying, but im confident im correct. did i make a mistake, or did thetextbook

[night egret]

ur answer is the same as theirs

[grim sierra]

oh wait

their direction vector is scaled

[night egret]

yea

same result tho

[grim sierra]

oopsies!

yeah

alr thx

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[molten bay]

How can i start this one?

[midnight plank]

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[warm island]

is this just illistratig the isomorphism between the multiplication of complex numbers and the summation of their angles?

[vagrant willow]

Not even that, it's just saying that the map $\phi : \mathbb(Z)_n \to U_n $ defined by $\phi(N) = \zeta^N $ is an isomorphism

Where's the latex when you need it

[warm island]

its saying that the product of the n roots is an isomorophism to the summation of them right?

[vagrant willow]

One second

[vagrant willow]

You do have to use the fact that $e^{w + z} = e^w e^z$ of course

[grand pond]

Transience

[vagrant willow]

Let me retype: it's just saying that the map $\phi : Z_n \to U_n$ defined by $\phi(N) = \zeta^N = e^{ \frac{2\pi N i}{n} } $ is an isomorphism

[warm island]

LOL why wont that render

[vagrant willow]

So strange

Okay, that is good enough

The only problem here is that the map here needs to be well-defined

Which means that if $N \equiv M$ modulo $n$ then $\zeta^N = \zeta^M$. This is definitely true

[grand pond]

Transience

Transience

[vagrant willow]

The fact that it is a homomorphism is easily checked by this property, right?

To be crystal clear, the map $\phi$ needs to be defined like $\phi([N]) = \zeta^N$ where $[N] = {kN : k \in Z } \in Z_n$ is the equivalence class corresponding to N.

[grand pond]

Transience

[vagrant willow]

Actually, it depends on how you define $Z_n$, so take that with a pinch of salt

[grand pond]

Transience

[vagrant willow]

The crucial point is that $\zeta^{N + M + n} = \zeta^N \zeta^M$, for all $N, M \in \mathbb{Z}$ this is all that we need to know to show that it is an isomorphism

[grand pond]

Transience

[warm island]

okey this is hard for me to wrap my head around. ill think about this 🙂 thank you!

[warm island]

so this is both a homomorphism and isomorphism?

okay okay i get it

thank you

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[leaden seal]

question: do i solve for A

or have i completely written this wrong

thanks

t is supposed to say 0

[buoyant yoke]

its incorrect

k is the growth factor

not the amount of bacteria

[verbal pumice]

You know N(0) = 1500 and N(5) = 9000

[leaden seal]

hmm

[verbal pumice]

just plug in and solve for A and k

[leaden seal]

is A correct?

[buoyant yoke]

yes

A is your starting amount

usually this formula is written as $N=N_0e^{kt}$ with $N_0$ the starting value

[grand pond]

Bonk

[last slate]

does anyone speak german here and can help me in a high school 11th year math problem

[midnight plank]

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/

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[buoyant yoke]

double whammy, sheeeesh

[leaden seal]

no we speak french

[buoyant yoke]

yup

,w ln(6)/5

[grand pond]

Results provided by WolframAlpha

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Requested by .bonk__

[buoyant yoke]

,w 9000=1500e^(5k)

[grand pond]

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Requested by .bonk__

[leaden seal]

so 0.35835

[buoyant yoke]

yup

[leaden seal]

it says 0.155630 on my calculator

[buoyant yoke]

then you made a mistake

did you fill in ln(6)/5?

[leaden seal]

yes

in degrees

[buoyant yoke]

degrees?

lol

what

[leaden seal]

i meant radians

[buoyant yoke]

radians?

lol

what

[buoyant yoke]

!occupied

[midnight plank]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[leaden seal]

omg

ill show u

There

[buoyant yoke]

thats log...

[pale vapor]

log5 and ln5

[buoyant yoke]

use ln

your calculator uses log base 10

ln uses log base e

very different

[pale vapor]

or use 2.303logx = lnx

[leaden seal]

oh yeah i just saw

im using log not ln

my bad

[buoyant yoke]

wtf...

[pale vapor]

what

[buoyant yoke]

!nogpt

[midnight plank]

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).

[pale vapor]

no but thats true

[buoyant yoke]

its not though

its an approximation, sure, but not exactly the same

[leaden seal]

lmfao

i

[pale vapor]

yeah not exactly

[leaden seal]

i gtg ill finish the rest after 1 hr

[buoyant yoke]

gl

[leaden seal]

Thanks ❤️

[buoyant yoke]

yeah so dont say that theyre equal

[pale vapor]

ok

mb

we dont use calculators here so yeah

mb

[buoyant yoke]

!done

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[buoyant yoke]

.close

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[plucky marsh]

!open

[lyric charm]

this is not a command. to open a channel, you just need to send your question here directly.

(but DO NOT DELETE THAT!)

[plucky marsh]

This mess of an equation (courtesy of Wikipedia) is the formula for the sphericity of a disdyakis triacontahedron. Is there any sort of simpler formula involving cube or sixth roots?

[lyric charm]

i would imagine no

or if there is, it's a bitch to find

what do you need an exact formula for

[plucky marsh]

It's not that I really need it, I just want the equation to look nicer. And there probably is some kind of simplification, because with how the formula for sphericity works, it's not about the numbers for volume and surface area themselves, but rather the ratio between them.

If the volume is 1, the surface area is around 4.90604054942533

I'm not sure if that number is any kind of known constant.

[lost sphinx]

why would this constant help when volume and area dont grow on the same scale

[midnight plank]

@plucky marsh Has your question been resolved?

[plucky marsh]

Because sphericity is a measure of surface area vs. a sphere of the same volume. Since the surface area of a sphere with unit volume is a known quantity (cube root of 36 pi), knowing that constant would give me the simplified equation that I'm looking for.

e.g. the sphericity of a cube is (36π^(1/3))/6 because a cube with unit volume has a surface area of 6.

[midnight plank]

@plucky marsh Has your question been resolved?

[lost sphinx]

you want this number?

[plucky marsh]

No, I already know that. What I'm trying to calculate is what that 4.90604054942533 constant is.

[lost sphinx]

in terms of other famous known values so its exact?

wait 4.91 refers to your polyhedron not a sphere right

[carmine sigil]

The disdyakis triacontahetron, i.e. the d120

@plucky marsh I don't know what the area of one of the triangles is, but if you can find a nice form for it, then your problem is pretty much solved.

Wikipedia has lists of vertices. And you can probably adapt hero's formula to find the area of one. But that will certainly give you something gross.

[midnight plank]

@plucky marsh Has your question been resolved?

[midnight plank]

@plucky marsh Has your question been resolved?

[midnight plank]

@plucky marsh Has your question been resolved?

[bleak ginkgo]

BRUHHH

[plucky marsh]

.close

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[silent torrent]

how to solve this one

[lyric charm]

let X be the number of customers arriving between 11:00 and 11:30 (half an hour)

then X ~ Po(3) and you want P(X ≥ 3)

[silent torrent]

X - po(3) ?

do we need lambda

when i learn about the poisson i only remerber e^-lambda * lamda^n/n!

can u help me to clear this one

[lyric charm]

poisson distribution with λ=3.

[lyric charm]

that is its PMF.

[silent torrent]

why lamda=3

[silent torrent]

im sorry let me check again

[lyric charm]

6 customers/hour * 0.5 hour

[silent torrent]

so e^-3 (lam^3/3!+lam^4/4!+lam^5/5!+lam^6/6!) still right ?

[lyric charm]

no

that gets you the probability that the number of customers is between 3 and 6 inclusive.

i.e. P(3 ≤ X ≤ 6)

not P(X ≥ 3) as you need

[silent torrent]

so integral 3 to 6 right

[lyric charm]

????

[silent torrent]

im sorry =((

[lyric charm]

no, you are way off the mark.

[silent torrent]

u mean im wrong ?

[lyric charm]

yes, very.

if you wanted to do it the dumb way you would have an infinite sum: $$P(X \geq 3) = \sum_{n=3}^{\infty} e^{-\lambda} \frac{\lambda^n}{n!}$$ but if you want to be a reasonable person who doesn't like suffering for 17 years to compute that, you can do it via the complement: $$P(X \geq 3) = 1 - P(X \leq 2) = 1 - [P(X=0)+P(X=1)+P(X=2)]$$

[grand pond]

Ann

[silent torrent]

so this one is true right

[lyric charm]

no

both of the expressions you attempted to give are wrong

except the second one is even wronger

[silent torrent]

=))

so 1 -( E^-3 ( 1+lam^1/1+lam^2/2)) right ?

[silent torrent]

why from 3 to 6 is wrong no difference with 1 -(0 to 2)

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[lyric charm]

no, there IS a difference.

[lyric charm]

a poisson random variable can take any natural-number value.

it is entirely possible that for example 48 customers all show up within the half hour.

@silent torrent do you understand this yes/no

[silent torrent]

i think i get the point its not 6 costomers so i need to cal by 1- p0 p1 p2

[lyric charm]

ok sure i guess that's good enough

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[craggy yew]

Hello, how do I calculate the limit on the bottom?

[thin tree]

,rccw

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[onyx plover]

Hi, I have trouble with understanding the active set A denoted in equation 3.20

And the subsequent update for dz_A denoted in 3.21

my issue in understanding is that to me it look slike that when z is at its lower bound a and F(z) > 0, the update step dz will be fixed to 0

however this to me looks counterintuitive, as F(z)>0 if the gradient J'(z) >0

this would mean z should INCREASE no?

then why would we fix it because it is fixed at the lower bound a? to me it would make more sense if it was fixed when z is at its lower bound a and F(z) < 0, indicating it wants to decrease, and our active set making sure that it cant decrease

however right now it looks like were stopping z from INCREASING when it is at its LOWER bound?

[midnight plank]

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[mellow sand]

smh

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[autumn folio]

hi, i'm supposed to get two lines for the answer here, but i think i'm missing one somehow

[normal prawn]

ooh ik exactlly what's the issue loll

It happened to me too few days before

[normal prawn]

actually when you cancel the '9m²' from both the sides you are ignoring that m can be infinite

[autumn folio]

but if i switch sides then its 9m² - 9m²

[normal prawn]

ik but...
Lemme give u an example
t²+4= t²+2t+1

What do you think the solutions are

[autumn folio]

3\2

[normal prawn]

Yes you r right

This equation has only one real solution

But... Let's say that t is approaching infinity

The equation still holds true

[autumn folio]

lets say its given that m isnt approaching infinity

[normal prawn]

then there is only one solution

btw m is the slope right?

[autumn folio]

thats not possible tho

[autumn folio]

yes i think ill post the full question

[normal prawn]

what does a line with infinite slope looks like?

[autumn folio]

x = ...

[normal prawn]

yes

[autumn folio]

ohh

[normal prawn]

So i think the second solution is smtg like x=k where k is some constant

R yOu GeTtInG mY PoInT!?

[autumn folio]

but shouldnt the equesion of the distance between a line and a point allways give to options - when the line is above and beneth a point?

[normal prawn]

idk the question:(

[autumn folio]

find the equesion of the two lines that pass through p(-2,-2) and that their distance from b(-5,2) is 3.

[normal prawn]

so u think that the answer to this question should be two parallel lines, one of em above the given point and one below?

[autumn folio]

thats the intuative answer for me, ya

[normal prawn]

your intuition is wrong sir :)

[normal prawn]

jk

How is that even possible

how can two distinct parallel lines pass through a common point?

[autumn folio]

how would u explain that when m approaches infinity y = mx+2m-2 = x = -2

[normal prawn]

Wait a sec

I would divide both the sides by m

[autumn folio]

math isnt mathing after 23:00 4 me

[normal prawn]

it happens :(

[autumn folio]

🤔

[normal prawn]

So it become y/m= x+2-2/m

Right?

[autumn folio]

yes

[normal prawn]

And what would happen when m approaches infinity? ♾️

[autumn folio]

its 0

x = -2

nice

[normal prawn]

:))

Hell na

[autumn folio]

then if the +- in the line equesion cancels itself, i should consider the variable when its approaching infinity?

[normal prawn]

wdym

[autumn folio]

when the two possibilities here cancel eachother

i should see whats up when m approaches infinity?

[normal prawn]

yos yos

[autumn folio]

cuz i never did that before

in this types of questions

[normal prawn]

same

I learned it just a few days b4

:)

[autumn folio]

nice

thanks!

my whole class didnt solve it

[normal prawn]

:)

Byee

[autumn folio]

.close

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[molten bay]

Hints for this one?

[last slate]

Guys WTF am I doing wrong??????

[buoyant yoke]

!occupied

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[last slate]

mb

[cyan inlet]

check the sides of the triangle

and what you wrote

[molten bay]

Huhhhh??

[cyan inlet]

sorry I was answering the other guy

what is cot ?

I only know cos sin and tan

1/tan ?

[buoyant yoke]

yes

[molten bay]

<@&268886789983436800>

[cyan inlet]

use :
tan(pi/2 - x) = 1/tan(x)

you'll be left with cot(2pi/11)

[midnight plank]

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[ember fractal]

4^(2x+1 )= 3^(2-3x)

[viral dagger]

take the log of both sides

[ember fractal]

2x+1 ln (4) = (2-3x) ln (3)

[viral dagger]

mhm

left one should prolly be in be in brackets tbo

[ember fractal]

ok

[wind jay]

I would take the log of one of their bases

Not natural log even though you can

Then substitute a variable in for one of them

Solve

Then resubstitute back

[ember fractal]

i got 2 ln (3) - ln (4)/ 2 ln (4) + 3 ln (3)

final

[midnight plank]

@ember fractal Has your question been resolved?

[delicate kraken]

well

so

[delicate kraken]

u got this

well done

now multiply ln(4) by (2x+1) on lhs and ln(3) with (2-3x) on rhs

[ember fractal]

idk

[delicate kraken]

do that and wat u get

@ember fractal

shouldnt take this long lah

[ember fractal]

bruhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[delicate kraken]

wot

[delicate kraken]

oh wait

u got the answer

mb

thats cool

[ember fractal]

am i right

or whaaaaattttttttttt

[delicate kraken]

yah

ur right

i mean technically u could like

further solve it

[ember fractal]

how

[delicate kraken]

by js finding out what the values are

but its gon be messy

thats fine too

[ember fractal]

but i need the actual answer cuz im doing a hw curcuit

[delicate kraken]

thats allg

[ember fractal]

circuit

[delicate kraken]

i mean

js put the values in ur calc

and find it?

[ember fractal]

when did u firts learn logs

freshman year

?

[delicate kraken]

year 11

[ember fractal]

bruh

[delicate kraken]

idk what freshman yea ris

[ember fractal]

me freshman

[delicate kraken]

what year is that

[ember fractal]

wdym

[thorny cradle]

It’s the first year of high school

[delicate kraken]

year 7?

[ember fractal]

year 9

[delicate kraken]

ok

that happens

now close the channel

[ember fractal]

how daddy

[delicate kraken]

???????

.close

@marsh obsidian girlriend this man called me daddy

Kill him

[ember fractal]

.close

[midnight plank]

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[midnight plank]

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[mild jolt]

GUYSBHELPP

[mild jolt]

NUMBER 2 PLEASEE

[noble eagle]

show your work please

[round parcel]

!status

[midnight plank]

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

[mild jolt]

I DONT UNDERSTAND

TRIED USING DIFFERENT RULESS

im actually in the middle of a quiz rn

HUHUHUHU

[round parcel]

OK, so the thing to see is that it's mainly a product when it starts out, so you use the product rule.

[mild jolt]

Wait favor, can u write ur solution in a paper and send it, I’ll study it once I pass my quiz paper 🥹🫰🫰🫰🫰🫰

[round parcel]

So, you have: [(6x^3 + 1)(-2x^2 - 3)^4] You'll get this: [(6x^3 + 1)((-2x^3 - 3)^4)' + (6x^3 + 1)'(-2x^3 - 3)^4]

[grand pond]

Chai T. Rex

[round parcel]

Do you see how I just copied what you started with twice and added them together?

[mild jolt]

Yes

[round parcel]

And then I put the ' differentiation mark on one of them before the plus and the other after the plus.

Does that make sense so far?

[mild jolt]

Yessss

[round parcel]

OK, now we can do the ' marked parts.

Let's do (((-2x^3 - 3)^4)').

[grand pond]

Chai T. Rex

[round parcel]

So, that's a function inside a function.

Like -2x^3 - 3 is a function.

And taking something to the fourth power is a function.

So, we want to use the chain rule.

So, there's two parts to that.

One is that we get the inside part and mark it with '.

((-2x^3 - 3)').

[grand pond]

Chai T. Rex

[round parcel]

Then we pretend that the inside part is just like a variable.

[mild jolt]

Okeiss

[round parcel]

Like if we have (x^4), we would differentiate that to get us (4x^3), right?

[grand pond]

Chai T. Rex

[round parcel]

So, we have ((-2x^3 - 3)^4), and we pretend that the ((-2x^3 - 3)) part is like a variable and do it that way.

[grand pond]

Chai T. Rex

[round parcel]

We get (4(-2x^3 - 3)^3).

[grand pond]

Chai T. Rex

[round parcel]

That's a lot like (x^4) to (4x^3), right?

[grand pond]

Chai T. Rex

[round parcel]

Does it make sense up to here?

[mild jolt]

yesssss

[round parcel]

OK, so [((-2x^3 - 3)^4)'][(-2x^3 - 3)' \cdot 4(-2x^3 - 3)^3].

[grand pond]

Chai T. Rex

[round parcel]

So, what's ((-2x^3 - 3)')?

[grand pond]

Chai T. Rex

[mild jolt]

-6x^2

[round parcel]

OK, good.

OK, so we have (-6x^3 \cdot 4(-2x^3 - 3)^3 = -24x^3(-2x^3 - 3)^3).

[grand pond]

Chai T. Rex

[round parcel]

Filling that in, we have [(6x^3 + 1)((-2x^3 - 3)^4)' + (6x^3 + 1)'(-2x^3 - 3)^4][(6x^3 + 1)(-24x^3(-2x^3 - 3)^3) + (6x^3 + 1)'(-2x^3 - 3)^4]

[grand pond]

Chai T. Rex

[round parcel]

What's ((6x^3 + 1)')?

[grand pond]

Chai T. Rex

[midnight plank]

@mild jolt Has your question been resolved?

[mild jolt]

18x^2

SORRY IN CHEM CLASS RN HEHEHE GO ON

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wanton glen]

I need a challenging question with the answer being 101

[lavish venture]

what’s the pick up line?

[wanton glen]

nah, this thing is just for my other math addicted friends

[main owl]

what kind of math do they know

like anything?

[wanton glen]

gr. 10 math

[main owl]

that lokey says nothing

[wanton glen]

at most

[main owl]

what specific class

[wanton glen]

there isn't specific math classes in my region for gr.10, but algebra

[main owl]

u could just have an AI make a problem where the answer is 101

without knowing specifically what ur friend knows its p much impossible to make a problem like that

[wanton glen]

the AI is so bad 😢

even my dumbest friends know the answer

[main owl]

i mean do you want like an unsolveable problem

[wanton glen]

no, just some hard problem using weird formulas

[main owl]

if its just algebra its not gonna be hard no matter what if they actually know algebra

just tedious at most

[wanton glen]

just something that looks weird

🙏

[main owl]

there you go

theres a very real algebra problem

[wanton glen]

is the answer 101?

[main owl]

they wont be able to solve it either way

so it doesnt really matter

[wanton glen]

alright

@main owl wats the answer though?

[main owl]

shi dont ask me

[wanton glen]

bruh

[main owl]

its also not actually algebra if that wasnt obvious

[wanton glen]

yh, quantum mechanics

[pearl hull]

lmao

[wanton glen]

can u gimme the answer 🥹
I'll call u the better Sky if u do

[main owl]

https://www.reddit.com/r/PhysicsHelp/comments/16o4unk/need_help_with_some_quantum_mechanics_here_i/

original post

if u actually care about the answer

might be somewhere in there

[wanton glen]

alright

[pearl hull]

I’m pretty bad at math yk

[wanton glen]

bruh

[midnight plank]

@wanton glen Has your question been resolved?

[gleaming latch]

pick a crazy indefinite integral and make the particular solution c = 101

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[queen geode]

How to prove Proposition 1.3 of Barwise (Admissible Sets and Structures) using Definitions 1.1 and 1.2 ?

[queen geode]

My offhand attempt:

Suppose not, i.e. that A is E-1 compact and self-definable, then since A is self-definable, there is a theory T that defines A (in the sense of Def 1.2). Since T defines A, T has a model. It follows that every subset of T has a model. Since A is compact, it follows that there is a model M of T and this M is inside A. However from Def 1.2, we also have that any model of T is isomorphic to A, so that M is inside A and yet M is isomorphic to A - a contradiction

.... the part am not sure with the above attempt is that if A is compact and T has a model M, then M is also inside A, but I guess that since A is admissible (i.e. satisfies KPU), then Def 1.1 is framed under the KPU metatheory ..........

[solid iris]

@queen geode try #foundations

[queen geode]

okay can I post this question there ?

.clsoe

.close

[midnight plank]

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[lavish venture]

Let $(X, d)$ be a metric space and let $A, B \subset X$ be disjoint closed subsets. Show that there exist open subsets $U, V \subset X$ such that $A \subset U$, $B \subset V$, and $\overline{U} \cap \overline{V} = \emptyset$.

[grand pond]

knief

[lavish venture]

am i bugging? i had the idea of taking for each point a in A an open ball around a with radius equal to 1/3 of the infimum of d(a, b) for b in B then since A and B are disjoint i know that this radius is greater than zero