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[lilac drift]

Can someone take a look at this?

[lilac drift]

This is how far I got

[charred summit]

You can search on the internet on a formula for the area of intersection between 2 circles depending on their radii and the distance between them

[lilac drift]

Well yes

But this is a calculus problem and i'm not exactly sure the professor will accept it

[charred summit]

I remember doing a research paper on it and yeah I used integerations to get it but if you're interested then here is the formula that I camd with

[lilac drift]

w h a t

i can't take your work dawg 😭

[charred summit]

Ok so I will work it out with you

[lilac drift]

that would be fantastic

by work it out, do you mean work through to your formula?

or just solve the problem?

[charred summit]

No I mean just solve the problem

[lilac drift]

ok

yes, that would be great

thank you so much

[charred summit]

First in my opinion turning the graph to be on the x-axis would help us

[lilac drift]

coordinate plane?

[charred summit]

Yeah

[lilac drift]

Okay, go for it

[charred summit]

Leaving it on the y-axis would work also but I think It is more comfortable doing it on the x-axis

[lilac drift]

Right

So basically rotate 90 degrees? is that what you're saying?

[lilac drift]

Ye

[charred summit]

Then we get the
Formula for the 2 circles

[lilac drift]

Right

piR^2 and pir^2 is the area

what are we supposed to do with them?

[charred summit]

I didn't mean the area
I meant the equation of both the circles

[lilac drift]

Um

Sure, how do we do that

[charred summit]

Hmm I think I am taking it the long way by that approach

[charred summit]

What was you trying here

[lilac drift]

I was trying to find a formula to find A

then I could multiply it by 2

but also I'm not exactly sure whether I'm doing it right

[charred summit]

I am confident of my way but what I am thinking is that it is long

[lilac drift]

Long is fine as long as it is in line with Calc II techniques

[charred summit]

So let's get the equation of the first circle

Which is Easley
x^2+y^2=R^2

[lilac drift]

Ok

[charred summit]

Then we need to get the distance between the 2 circles so that we can get the equation of the second circle do you know how to get it

[lilac drift]

uhhh

No

💀

[charred summit]

Think Pythagorean

[lilac drift]

well, the issue is the triangle doesn't reach the center of the big circle

unless you want to make an entire different triangle?

[charred summit]

I think this will make things clear

[lilac drift]

Right, okay

[charred summit]

So it is obvious now right

[lilac drift]

Yes, keep going

[charred summit]

So what's the distance

[lilac drift]

square root of R^2-r^2

[charred summit]

Yeah

Now that distance is going to the right meaning that the co-ordinates of the center of the second circle is (√(R^2-r^2), 0)

So we can use this to make the equation of the second circle which is

$(x - \sqrt{R^2-r^2})^2 + y^2 = r^2$

Let's put the 2 of them together

[lilac drift]

Okay

[ocean seal]

Sherif Player

[lilac drift]

Ya

That looks right

[charred summit]

$$x^2 + y^2 = R^2$$
$$(x - \sqrt{R^2-r^2})^2 + y^2 = r^2$$

[ocean seal]

Sherif Player

[charred summit]

Here it is
Now we want to know the point of intersections between the 2 circles

[lilac drift]

Ohhhhhhh

Okay, hang on rq

Okay

[charred summit]

You can get it by getting the y^2 value of both the equations and equating them to each other

Then solving for x

[lilac drift]

Holy jesus, okay

give me a second

So

is it (R^2-r^2)/sqrt(R^2-r^2)?

[charred summit]

$$ y^2 = R^2 - x^2 $$
$$ y^2 = r^2 - (x - \sqrt{R^2-r^2})^2 $$
$$ R^2 - x^2 = r^2 - x^2 + 2x\sqrt{R^2-r^2} - R^2 + r^2$$
$$x\sqrt{R^2-r^2} = R^2-r^2 $$
$$ x = \frac{R^2-r^2}{\sqrt{R^2-r^2}}$$

[lilac drift]

damn

holy crap

uhm gimme un momento to see where i went wrong

[ocean seal]

Sherif Player

[lilac drift]

Wait

when you're squaring the sqrt(R^2 - r^2)

Doesn't the negative get cancelled because we're multiplying two negatives?

In the (R^2+r^2) part

[charred summit]

Yeah but there is a negative out

That's why the x^2 is also negative

[lilac drift]

Right, what I dod was move stuff around

Let me check what I did wrong

[charred summit]

You are right

[lilac drift]

Done

[charred summit]

I am wrong there

[lilac drift]

Oh

Now I'm confused

but if you say so

[charred summit]

$$ y^2 = R^2 - x^2 $$
$$ y^2 = r^2 - (x - \sqrt{R^2-r^2})^2 $$
$$ R^2 - x^2 = r^2 - x^2 + 2x\sqrt{R^2-r^2} - R^2 + r^2$$
$$x\sqrt{R^2-r^2} = R^2-r^2 $$
$$ x = \frac{R^2-r^2}{\sqrt{R^2-r^2}}$$

[lilac drift]

So where now?

[ocean seal]

Sherif Player

[charred summit]

I cancelled R^2 together in the third line LOL

By the this is the same as

[lilac drift]

Lol okay

[charred summit]

$$ x = \frac{R^2-r^2}{\sqrt{R^2-r^2}}$$
$$x = \sqrt{R^2 - r^2}$$

[ocean seal]

Sherif Player

[lilac drift]

Ofc ofc

[charred summit]

Which is the same as the distance

I didn't expect that

Now let's get into the integration thing

[lilac drift]

lol

[charred summit]

To be able to do the integration we are gonna need to shave the bottom

[lilac drift]

Now that's a weird turn of phrase

But go off

[charred summit]

Like the bottom here

[lilac drift]

The bottom being?

[charred summit]

What's under the x-axis or the yellow line

[lilac drift]

But like why

[charred summit]

Because if we did the integration like this the position part would cancel the negative part giving us 0 area

[lilac drift]

I'll just trust you, show me what to do

[charred summit]

We can shave the bottom, calculate the area using integration then multiply it by 2

[lilac drift]

Right

[charred summit]

Shaving the bottom is not hard we gonna take the equations put y^2 on one side and the rest on the other side, take the square root and we are done

[lilac drift]

Uhhhhhhh

Yeah, math doesn't translate well to words

Do you think you could do the thing with TeXit again?

[charred summit]

$$ y^2 = R^2 - x^2 $$
$$ y^2 = r^2 - (x - \sqrt{R^2-r^2})^2 $$
$$y= \sqrt{R^2 - x^2}$$
$$y = \sqrt{r^2 - (x - \sqrt{R^2-r^2})^2}$$

[ocean seal]

Sherif Player

[lilac drift]

Yeah, that looks right

Sorry I made you do that lol

[charred summit]

For the bottom one I feel like simplifying it

[lilac drift]

That's fair

[charred summit]

No problem I enjoy doing that it is not much

Nevermind expanding the power won't do anything

[lilac drift]

lol

alright then

[charred summit]

Now we start Integration

[lilac drift]

oh boy oh boy

All in terms of x, R and r are constants, correct?

That requires trig sub...

[charred summit]

X is not

[lilac drift]

Yeah, it's in terms of x

R and r are constants

Ugh, this is gonna be a lot of work

[charred summit]

$$\int_0^{\sqrt{R^2-r^2}} \sqrt{r^2 - (x - \sqrt{R^2-r^2})^2}$$

[lilac drift]

oh wow

[charred summit]

Wait

That would not work best

[ocean seal]

Sherif Player

[charred summit]

Then add this to it

[lilac drift]

no, add the integral you're typing rn

ok

[charred summit]

I think using king therom would help here

[lilac drift]

What's a king theorem 😭

[lilac drift]

add this integral to the first?

[charred summit]

No

I was gonna write another integeral

[lilac drift]

alright

[charred summit]

$$\int_{\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}+r} \sqrt{R^2-x^2}$$

[ocean seal]

Sherif Player

[lilac drift]

I don't quite understand King's Theorem

[charred summit]

It has a proof for it if you understand where it came from but if you don't understand it itself
Then it is just a property that helps us with definite integrals by plugins a and b in the f(x) with x

[lilac drift]

Okay

I'll just cite King's Theorem ig

What next?

[charred summit]

Let's calculate this

[lilac drift]

this problem is a nightmare

my god

[charred summit]

It is called king's property by the way

[lilac drift]

Right

I checked the stack exchange

[charred summit]

Use King's property to make it easier

[lilac drift]

Alright

[charred summit]

Substitute x with (b+a-x)

[lilac drift]

I'll try, but i'm a damn slow worker

[charred summit]

By the way do you know what is the area this integeral gives us

[lilac drift]

The area this integral gives is the distance between the center of the big circle to the center of the second circle'

right?

well

the area under that line

[charred summit]

Yes

It gives us this red part

So after using king's property it is a lot easier right?

[lilac drift]

Uhhhhhh

[charred summit]

$$\int_0^{\sqrt{R^2-r^2}} \sqrt{r^2 - (x - \sqrt{R^2-r^2})^2}
= \int_0^{\sqrt{R^2-r^2}} \sqrt{r^2 - x^2}$$

[ocean seal]

Sherif Player

[charred summit]

I see it is a lot easier

[lilac drift]

Man, i just want to get this done

I've been at it for hours now

Let's do this\

[charred summit]

Ok

So do you know what u substitution we are gonna use

$$\int_0^{\sqrt{R^2-r^2}} \sqrt{r^2 - x^2} dx$$

[lilac drift]

x=sintheta

[charred summit]

Yes

Ok not really

You forgot the r

[lilac drift]

damn

[charred summit]

X = r sin(theta)

[lilac drift]

i'm so lost

not really

but still, this problem is a mess

[charred summit]

Lol I forgot about the dx in the integerals

[lilac drift]

i got the memo dw

[ocean seal]

Sherif Player

[charred summit]

And dx = r cos(theta) d(theta)

[lilac drift]

ofc ofc

let's move to integration

trig sub is my biggest weakness so i'm worried abt it

[charred summit]

Okay

$$\int_0^{\arcsin(\sqrt{R^2-r^2})} \sqrt{r^2 - r^2 \sin^2(\theta)} r \cos(\theta) d\theta$$

[ocean seal]

Sherif Player

[lilac drift]

oh holy crap

alr, alr, here we go

[charred summit]

$$\int_0^{\arcsin(\sqrt{R^2-r^2})} r^2 \cos^2(\theta) d\theta$$

[ocean seal]

Sherif Player

[charred summit]

r^2 can get out

[lilac drift]

ikik

[charred summit]

$$r^2 \int_0^{\arcsin(\sqrt{R^2-r^2})} \cos^2(\theta) d\theta$$

[lilac drift]

power reducing?

[ocean seal]

Sherif Player

[lilac drift]

we doin power reducing formula?

pythag identity

i mean

[charred summit]

Yeah

[lilac drift]

alr, gimme a sec, i'll send a pic

[charred summit]

Or wait

[lilac drift]

you don't need to TeXit, just confirm

[charred summit]

Just use trig identities

[lilac drift]

(1+cos2theta)/2 right

[charred summit]

Yeah

That is what we will get

[lilac drift]

aight gimme a sec to write it down and then we'll go

I mean, plugging in bounds

[charred summit]

Turn sin(2x) into 2 sin(x) cos(x)

[lilac drift]

smart

So now the other integral?

[charred summit]

$$\frac12(\arcsin(\sqrt{R^2-r^2}) +\sqrt{R^2-r^2} U)$$

[ocean seal]

Sherif Player

[charred summit]

You can simplify cos(arcsin

[lilac drift]

uh

what ahppen to r^2?

[charred summit]

,wolf \cos{\arcsin{R^2-r^2}}

[ocean seal]

Results provided by WolframAlpha

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[lilac drift]

uhhhhhhhhhh

[charred summit]

Yeah

,wolf \sqrt{R^2-r^2} \cos{\arcsin{R^2-r^2}}

[lilac drift]

How do I prove that dawg

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by sherif_player

[lilac drift]

MY GOD

I'M NOT DOING ALL THAT

[charred summit]

None of this is useless

[lilac drift]

Bro i have better things to do than try at this dumb problem 💀

[charred summit]

It can be done by using the right angle triangle

[lilac drift]

Okay

It's not really necessary tho...

I'll just write the extra stuff

can we move to the second integral?

[charred summit]

Like this

Ok let's move on to the next one

It is the same by the way

[lilac drift]

NO

YOU'RE LYING

[charred summit]

$$\int_{\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}+r} \sqrt{R^2-x^2}$$

[ocean seal]

Sherif Player

[lilac drift]

ughhhhhhhhhhhhhh

[charred summit]

This is the same thing but with different boundaries

[lilac drift]

Okay

So can I just copy the answer from the first?

[charred summit]

Yeah

But don't forget that when you change x to Rsin(theta)
You are changing the boundaries with it to

[lilac drift]

alright

i'm not doing this algebra

i'm just gonna have the two add up, multiply by two and call it a day

[charred summit]

As you are doing that I remembered that there were an error in the boundaries of the previous integeral

[lilac drift]

i give up

[charred summit]

You still need to subtract that from the area of the small circle

[lilac drift]

arcsin of 0 is 0, no????

[charred summit]

Yeah that's correct

[lilac drift]

i am gonna check if my toaster is waterproof rq

[charred summit]

I am pretty sure it is not

[lilac drift]

let's keep going

a damn shame

[charred summit]

$$r^2 \int_0^{\arcsin(\frac{\sqrt{R^2-r^2}}{r})} \cos^2(\theta) d\theta$$

[ocean seal]

Sherif Player

[charred summit]

That was what I forgot

In the boundaries

[lilac drift]

you're lying

[charred summit]

Let's take the help of wolf here

,wolf $$r^2 \int_0^{\arcsin(\frac{\sqrt{R^2-r^2}}{r})} \cos^2(\theta) d\theta$$

[lilac drift]

ughhhhh

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sherif_player

[lilac drift]

o

[charred summit]

Where did that 2 came from

[ocean seal]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by sherif_player

[lilac drift]

i really feel like giving up right now

[charred summit]

We are about to finish

[lilac drift]

I'm gonna finish writing this, whether it's right or not

I don't need it simplified, i already just replaced all of the bounds and stuff

can we just finish the last calculations

[charred summit]

Ok

So did you calculate the other integeral

[lilac drift]

give me a second to do the bounds

but assume i did so

[charred summit]

Add the 2 results then multiply by 2
Then after that subtract it from
πr^2

[lilac drift]

that's it?

problem over?

[charred summit]

Yeah except if you want to simplify somethings

[lilac drift]

god no

thank christ

okay, god bless you for helping me man

i'm gonna take a nap

.close

[lone heart]

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[charred summit]

You're welcome

[lone heart]

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[astral horizon]

can someone please help me with my homework 😛

[lone heart]

@astral horizon Has your question been resolved?

[lone heart]

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[astral horizon]

.reopen

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✅

[pseudo ice]

!occupied

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

@astral horizon Has your question been resolved?

[astral horizon]

can someone please help me

🙂

[pseudo ice]

A lot of questions you have there

[astral horizon]

i am currenrly in need of help on number one badly

[pseudo ice]

Cool, let's take that one then

You're aware of what average velocity is, right? (as opposed to e.g. instantaneous velocity)

[astral horizon]

yea so basically i already have the velocity calculated for each of the times like the velocity at 1.0 is .3 and 4 is 4.8 but like i am just confused how i got it cus someone helpedme in class

[pseudo ice]

You don't find the [instantaneous] velocities at those times

[astral horizon]

ohhhh

[pseudo ice]

Have you heard of, say, average speed cameras and how they work?

[astral horizon]

erm no...

[pseudo ice]

Anyways, you might know how to do this question, but anyway

[pseudo ice]

the idea of those is that you put two cameras at some distance apart, and to catch people who are speeding, you time how long it takes for vehicles to pass from the first camera to the second

[astral horizon]

ohhhhh yea ive heard of those before, i just drew a blank

[pseudo ice]

anyways, to cut a long story short, average velocity is basically (s(4) - s(1))/(4 - 1)

The numerator corresponding to the "distance" (well, signed) and the denominator corresponding to the time

[astral horizon]

so do i just find the average velocity between 1 and 4?

[pseudo ice]

Yep, literally that (it's what they ask you for, after all )

[astral horizon]

LOL okay so when t=1 its 3/10 or .3 feet per second and then when t=4 its 24/5 feet per second so then id just find the average between the two?

so 2.1 feet per second i think

[pseudo ice]

Well, more that when t = 1, you're at position 3/10ft, when t = 4, you're at position 24/5ft

[astral horizon]

ohhhh oka that makes sense

[pseudo ice]

Remember the "average speed camera" thing: what's the "distance"*, and what's the time it takes to travel that "distance"?
[*displacement here, as we're doing velocity rather than speed]

[astral horizon]

ohhh okay that honeslyt makes it easier to understand the teacher didnt explain it well for me

[pseudo ice]

Awww

[astral horizon]

do you think you could possibly help me with #6 please :3

[pseudo ice]

You know the epsilon delta definiton, right

[astral horizon]

like the formal way to evalulate the limit?

kinda but not really

[pseudo ice]

Well, if I state it out, if you want to show that $\lim_{x\to a} f(x) = L$, what you want to show is that
\begin{quote}
for any $\epsilon > 0$, you can find a corresponding $\delta > 0$ such that whenever you have $0 < \abs{x - a} < \delta$, it follows that $\abs{f(x) - L} < \epsilon$.
\end{quote}

[ocean seal]

@pseudo ice

[astral horizon]

hold on ill be back in like 15 minutes im sorry my moms calling me to help dinner

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@astral horizon Has your question been resolved?

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@astral horizon Has your question been resolved?

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[tulip oasis]

plz

just to be sure i am lost

if i want to take atom number to grams

i should do 7.772x10^22 atoms times 6.022x10^23mol/1 atom times

then it would be

go periodic table and get the g/mol number

so times 293.20g/1mol

[silent iron]

pause

[silent iron]

atom > grams

1. of atoms / avogadros number -> moles

2. moles * mass -> grams

[tulip oasis]

oh

[lone heart]

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[vapid shuttle]

I'm stuck on doing this sentence in red
if |G|=p^(k)m, how can I show that n_(p,l) mod p is always the same regardless of choice of G?

@mint robin Any ideas?

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@vapid shuttle Has your question been resolved?

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@vapid shuttle Has your question been resolved?

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@vapid shuttle Has your question been resolved?

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@vapid shuttle Has your question been resolved?

[tiny sky]

when it says G acts on the set of its subsets of size p^l by left translation, does it just mean left multiplication

[tiny sky]

if this is the same as the normal proof of sylow's

||let our subgroup of order p^l act on the set of subgroups of order p^l by conj.||

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[pale shell]

In a game, there is a box containing 3 red balls, 4 blue balls and 5 green balls. Someone will take 4 balls at random from the box without looking. If one more ball is drawn after the first four balls are drawn, what is the probability that the fifth ball drawn is a blue ball, provided that the first four balls drawn are not all blue?

[pale shell]

Help please

[tiny fable]

!status

[lone heart]

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

@pale shell Has your question been resolved?

[pale shell]

I'm trying using Bayes Theorem
P(B|K')=P(BnK')/P(K')
B is event blue ball being taken
K' is event all 4 balls are not all blue
P(K')=494/495

<@&286206848099549185> help pleasd

[alpine sable]

Ill take a look

[prime badge]

(5th is blue & first 4 are not all blue)

is the same as 5th is blue

and 5th is blue is the same as first is blue, so 1/3

so it should be

[tiny fable]

you're drawing without replacement, so it's not that simple I don't think

[prime badge]

,calc (4/12) / (1 - (4/12)(3/11)(2/10)(1/9))

[ocean seal]

Result:

0.33400809716599

[prime badge]

165/494

[lone heart]

@pale shell Has your question been resolved?

[pale shell]

I tried chat gpt it gives me 1/4

[tacit arch]

Don't use chat gpt for math

[pale shell]

Just trying to get some idea (inspiration)

[tacit arch]

Then just listen to it and stop telling helpers to explain their response to you

[pale shell]

Is this fraction?

Not combinatoric

[prime badge]

which one do you want

this isn't combinatorical, i can do combinatorical

[lone heart]

@pale shell Has your question been resolved?

[pale shell]

This is counting problems thats why i thought you using combinatorial

[lone heart]

@pale shell Has your question been resolved?

[lone heart]

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[tulip oasis]

);

[tulip oasis]

like

id imagine its 18.7g oxygen in the 56.75 grams

idk how to go about 440.6

[hushed locust]

the proportion of oxygen stays the same in both size samples (one of dalton's laws)

[tulip oasis]

uh

this law cant click for me

i feel like if i try to apply it maybe

so for the word problem

i should do like

18.7g oxygen

wait

mass ratio would be the 440.6g?

[hushed locust]

ratio always means a fraction

so here we are interested in the ratio of oxygen to the total

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[formal laurel]

can someone double check this

[buoyant saddle]

submit it and find out

[formal laurel]

this isnt a submission

its just questions from our book for exercise

[buoyant saddle]

yea that’s just not believable lol

[formal laurel]

but i cant see if they're right or wrong since i cant submit

people like you are unbearable tbh

like its not even a quiz or anything its litrly from y book

i cannot submit it and im just trying to check my answers to see if im doing it right

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[viral falcon]

I ended up with $$u(x,t)=\sum_{n=1}^{\infty} B_n \cos(n\pi t)\sin(n\pi x)$$ where $B_n=2\int_0^1f(x)\sin(n\pi x) dx$

[ocean seal]

Michael

[viral falcon]

But can't figure out how to calculate B_n. Can someone help?
If I plug in f(x) = u(x,0), I just get that B_n = 0, and that can't be right

[tacit arch]

! show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[viral falcon]

and yeah, didn't finish the last line since I knew it was 0

so yeah, from the third line and on I'm trying to calculate B_n

According to wolframalpha, the integral is 0, so there must be a mistake somewhere else, or something I'm missing. Not quite sure.
I followed the exact steps of the book, and I got the solution on the same form as them. It's just that I'm setting L=1 and c^2 = 1, while they have it general

[lone heart]

@viral falcon Has your question been resolved?

[tacit arch]

You have to deal with the n=1 and 2 case separately

You're dividing by n-1 and n-2

Those shouldn't be zero

[lone heart]

@viral falcon Has your question been resolved?

[viral falcon]

ah

right

I'll look more into it tomorrow

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@viral falcon Has your question been resolved?

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[strange fractal]

can I pls have some help with this question

5

[alpine sable]

You want to recover 3 variables (a,b,c). To solve this, you need to have at least 3 independent linear equations. To start it off, what does "passes through (0,5)" entail to you

[strange fractal]

C=5

[alpine sable]

yeah

[strange fractal]

Ya

[alpine sable]

now, what about "stationary point at (2,7)"

[strange fractal]

at 2,7 dy/dx is 0

[alpine sable]

uh huh

[strange fractal]

but idk what next

[alpine sable]

so, plug in dy/dx = 0, x = 2 in your derivative equation that you wrote

[strange fractal]

o what ab y

[alpine sable]

there is no y in the derivative equation

[strange fractal]

O ok now wat

[alpine sable]

.

what do you get

[alpine sable]

ok cool. thats your second equation

you need one more

anyways, when we say your curve has a stationary point at (2,7) it obviously means that the curve passes through (2,7) as well, right?

[strange fractal]

Yea

O so I just sub 2,7

[alpine sable]

into y, yes

and then just do the algebra

[strange fractal]

Ohh ok thx :P

.close

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[young light]

Need help trying to plug into a function. I feel like I did something wrong early on

[buoyant saddle]

your first step is wrong

how did you get 3(x+h)^2 - x - h

[young light]

I thought you just plug in whats inside the f() in the problem and subsitute it for x in the function

[buoyant saddle]

f(x+h) = 3(x+h)^2 + 2(x+h)

[young light]

Oh right, im blind

.close

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[wet remnant]

what part do you need help with?

[tardy flume]

Finding the area of the shaded region for the semicircles

[wet remnant]

ok, do you know the formula for finding the area of any given circle?

[tardy flume]

uhh is it 1/2 x pi x r x r?

[wet remnant]

pretty close, its just pir^2 tho

[tardy flume]

o

but its half a circle

[wet remnant]

yea, but if you treat it like a whole circle, then calculate the area you can just halve the area afterwards

[tardy flume]

OH

[wet remnant]

yeaaaaa

[tardy flume]

so it will be inaccurate if i multiply 1/2 tgt?

[wet remnant]

not for a half circle, but when you sent this i thought you meant for a whole circle

but yes, that would work for a half circle

[tardy flume]

ohhh

then what is the formula to find the area of a parallelogram

[wet remnant]

base times height

[tardy flume]

ohhh

alright i think i can solve it myself now

tysmm!

[wet remnant]

np, gl

[tardy flume]

.close

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[nova berry]

how do I make a function?? specifically logarithm and piecewise

[lone heart]

@nova berry Has your question been resolved?

[rose sigil]

wdym make?

[nova berry]

like make a function

f(x) =

blabla

but in logarithmic form?

[rose sigil]

[nova berry]

does that make sense

😭

[rose sigil]

you write it down?

!original

[lone heart]

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

[nova berry]

Formulated Real-Life Problem:
To quantify the impact of rising fuel prices, consider the following scenario: In Lipa City, the price of gasoline was PHP 60 per liter in September 2024. If the price increases by 5% each month due to persistent global market trends, what will be the projected price per liter of gasoline in December 2024?

This is the problem

but I need to graph it

and the function that I need to use

is logarithmic function

[rose sigil]

logarithmic?

[sturdy stream]

May be you get these kind problems related to gaussian and possion distrbution forms that are generally in exponential form then doing some operation you will get log function (this is general aproach)

[rose sigil]

what

[nova berry]

oooooo

okok

tytyyy

!close

(?)

how

[rose sigil]

.close

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[nova berry]

tysm

[sturdy stream]

You got the genral approach?

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[dim pebble]

probably a silly question but if i were to find a domain of a function , do i gotta make it in its simplest form?

[dim pebble]

or no? like
f(x) = (x^2-4)/(x-2) , it can be simplified

[cedar kelp]

generally no

[midnight wharf]

i think that depends on how it is graded

[rose sigil]

what would be "unsimplfied form"?

[dim pebble]

if its simplfied , domain is all real numbers
but rn its all real numbers other than 2

[alpine sable]

layla

[rose sigil]

yajat

[midnight wharf]

yajat

[alpine sable]

ginger

[midnight wharf]

layla

[rose sigil]

you, in fact, cannot "simplify" it that way

[dim pebble]

the form given is unsimplified , i can simplify it more tho

[alpine sable]

a person named ginger joined my clan yesterday

was it u

[midnight wharf]

it was not but cool, i haven't seen another ginger

[rose sigil]

the function doesn't parse if you try to evaluate it ay 2

[dim pebble]

why not (x-2)(x+2)/(x-2)

x+2

[quasi vector]

you can only "simplify it" under the assumption that x is not equal to two

[rose sigil]

that doesn't parse when you evaluate it at 2

[rose sigil]

whereas this does

[alpine sable]

damn ok

u can join tho if u wanna

[midnight wharf]

the simplification x/x = 1 holds unless x = 0

what kinda clan is it

[dim pebble]

ok so i find domain of the given function only? dont try to simplify it further?

[alpine sable]

wym

[midnight wharf]

like clan where, on discord?

[rose sigil]

however... the function $f:\bR\setminus{2}\to\bR$ defined by $f(x) = x+2$ is the same as your original one

[alpine sable]

clash of clans?

[ocean seal]

layla is not harper

[dim pebble]

?

[midnight wharf]

wait is this true @rose sigil that a function R -> R and R{2} -> R can be the "same function"? just curious

like they are effectively the same but

[dim pebble]

nvm

i understand

[midnight wharf]

idk even how equality between functions is defined

[rose sigil]

well if you write f(x) = (x^2-4)/(x-2), it's already implicit that the domain is R\{2}

[quasi vector]

no

[midnight wharf]

ok so it's simply not an R -> R function?

[midnight wharf]

ok

[rose sigil]

if you write f(x) = x + 2, then the domain is implicitly R

[rose sigil]

yea

[midnight wharf]

ok ty

[dim pebble]

ok i understand , so the domain of the given function should be all real except 2

and not all real numbers

[rose sigil]

yea, that's implicit

i would prefer if everyone writes the domain of a function when they declare it but

doesn't seem to be a very popular opinion in <= calculus math education

[midnight wharf]

what if the domain is complicated or unknown

[rose sigil]

when you declare a function you need to declare a domain

[dim pebble]

well the question is to find it , so if domain was given the question would be very easy :d

[midnight wharf]

what about 1/(sin(1/x))

would you define the domain implicitly?

sry if i'm shifting the topic here

[rose sigil]

yea and that's kinda backwards but whatever

finding the "largest domain that would make sense" is a reasonable problem

and that's what "find the domain" problems are

or should be

even though they aren't portrayed that way because functions are taught poorly

[dim pebble]

hmm

nothing i can do tbh xd

ty tho

.close

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[rose sigil]

you can think of find the domain problems like:

if we were to make a function f with domain A defined by f(x) = ...
what's the "largest" set A that would make sense?

[midnight wharf]

.open

[dim pebble]

isnt real numbers set the largest tho

and it would make sense if i were to be able to simplify into x+2

[rose sigil]

not for e.g. f(x) = (x-2)(x+2)/(x-2)

[midnight wharf]

but you can't simplify it

[dim pebble]

yeah ok i understand

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[cinder sundial]

I wonder where I did wrong

<@&286206848099549185>

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@cinder sundial Has your question been resolved?

[cinder sundial]

.close

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[native escarp]

How do i do these?

[native escarp]

i assume we just plug the values into the function or something?

[echo ruin]

[unkempt pollen]

yes depending on the value of x we will need to use different parts of f(x0

[native escarp]

i kind of get it ig

[unkempt pollen]

[native escarp]

actually nah i get this now, f(x) is just y, i keep forgetting

[unkempt pollen]

lol it do be like that at times

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[alpine sable]

How do I find A^k for this matrix, my friend said to use diagnolization but idk how ;-;

[tight pier]

Have you found the Eigenvalues?

[vale crag]

||it's not diagonalizable in the first place||

[alpine sable]

oh ;-; but I still have to find A^k of it

convert to Jordan canonical form first?

[vale crag]

it's already a jordan block lol

[tight pier]

How could tell so fast btw?

[alpine sable]

What that? I just started linear algebra

[vale crag]

was just responding to qylo, yeah jordan stuff is advanced

[alpine sable]

o ok

I wanted to try brute force but idk how to figure it out ;-;

[vale crag]

well char poly is (X-1)^3 the only way to have an eigenspace for 1 of dim 3 is if the matrix is the identity

[alpine sable]

the off-diagonal elements follow a binomial-like structure

[vale crag]

nice spoil

[alpine sable]

maybe not too bad

what is a binominal structure? Im not really fluent in english

[vale crag]

well the 1-off diagonal elements are kinda obvious, it's just 1 2 3 4 5 6 ...
as for the top right corner, you might wanna try computing some products by hand

for example A * A^3, A * A^4, A * A^5

to see what terms make that top right corner

machine is nice, but the computation is hidden from you

[alpine sable]

Im a bit stuck, I did like you said and can clearly see it something + n to get the top right corner

[vale crag]

so like (corner for n+1) = (corner for n) + n

something like that ?

[alpine sable]

ye

but idk how do I put the corner for n into a function

[vale crag]

yeah exactly that's what's going on

well think about it

for n=1, you have 1 in the corner

[alpine sable]

for n =1, I have 0 though-

[vale crag]

yeah nvm I forgot what the matrix looked like

[vale crag]

so yeah n=1, you have 0

n=2, you have 0 (the previous corner) + 1

n=3, you have 1 (previous) + 2

n=4, you have (1+2) (previous) + 3

n=5, you have (1+2+3) (previous) + 4

etc...

quite a cute pattern we have here

[alpine sable]

I see it so do I put it into a function like the sum function? I forgot the name for it symbol ;-;

[vale crag]

you could write it as a sum sure, but as qylo mentioned there's actually something binomial going on

but yeah let's start with the sum anyway and go from there

what sum would that be ?

[alpine sable]

smt like this? Im not sure if I did it correctly

oh wait nvm I just remembered there a formula for sum of a number sequence

[vale crag]

well almost

you have to be careful with the top index there

for n=5, we stop the sum at 4 for ex

[alpine sable]

I dont get what you mean here

[vale crag]

^

for n=5, it's not the sum from 1 to 5

it's the sum from 1 to 4

[alpine sable]

but i start at x at zero so i dont really need to think about it, right?
like for n=2 it 0+1, for n=3 it 0+1+2, and so on
So I just go for a function that calculate sum from 0 to n

[vale crag]

it's not the starting at 0 that's the problem

it's the ending at n that's incorrect

[alpine sable]

ah wait it n-1

[vale crag]

yea

[alpine sable]

ty

[vale crag]

so yeah now you can use the fancy formula for the sum of integers

[alpine sable]

generally speaking, for matrix that I need to find A^k is it better that I go and write out calculation by hand?

[vale crag]

well in general diagonalization is nicer

but this question was specially chosen so that you can't do that

[alpine sable]

I havent learn about diagonalization yet so ;-;

[vale crag]

so you kinda have to make your own way there

[alpine sable]

can I send one more? same question, find A^k

[vale crag]

let's finish this one first tho

[alpine sable]

wait it not finished?

[vale crag]

what did you get as a formula for the sum ?

I thought you wanted a closed form lol

[alpine sable]

this is what I got

[vale crag]

yeah exactly gg

and turns out k(k-1)/2 is just (2 choose k)

that's why qylo was talking about binomial stuff

[alpine sable]

what is "2 choose k" ?

[vale crag]

$\binom{k}{2}$ $C^2_k$ combinatorics stuff

[ocean seal]

aPlatypus

[alpine sable]

I know the C^2_k thingy but the matrix on it left is new for me

[vale crag]

yeah it's just diff notations for the same thing

[alpine sable]

oh alright

[vale crag]

https://math.stackexchange.com/questions/326688/why-does-the-n-th-power-of-a-jordan-matrix-involve-the-binomial-coefficient but yeah in more generality this is what you get

binomial coefficients (the C^k_n) everywhere

[alpine sable]

[vale crag]

when they say J_k(lambda), they mean the matrix of size k x k with lambda on the diagonal, and 1s just above it

what you had was J_3(1)

[alpine sable]

ooo

[vale crag]

ok time for the second one now ig

[alpine sable]

ye I have no idea what happened here

magically they all turned to 0

[vale crag]

✨ magic ✨

well at least it's not hard to see the pattern now

it's an example of a nilpotent matrix

a matrix that ends up being 0 when you take its powers

[alpine sable]

I dont see the pattern ;-;

[vale crag]

well what do you think happens when you multiply A * A^4 now

[alpine sable]

the 0 from A^3 gonna make everything to 0?

[vale crag]

well the 0 from A^4 but yeah

so A^5 = all zeros

[alpine sable]

ah?

[vale crag]

what about A^6 now

[alpine sable]

the same thing, the 0 from A^5 make everything 0

[vale crag]

yeah

we don't have to continue infinitely many times I suppose

I don't have that much time on my hands

[alpine sable]

I get that for k>2 everything equal 0 but for k=2 idk how to get those value out

[vale crag]

just computing A * A isn't really difficult

[alpine sable]

sorry ;-; im just really confused, I dont see how I can write a function with k to get the matrix for k=2 and k=1

[vale crag]

don't

just treat the k=1 and k=2 cases seperately

[alpine sable]

I dont get what you mean, so like for k =1 A = .., k = 2 then A = .., and for k>2 it a 3x3 matrix of 0?

[vale crag]

yeah

[alpine sable]

but the problem tell me to find A^k ;-;

[vale crag]

piecewise functions exist you know

if you're that concerned about writing a function

[alpine sable]

oh- so basically you saying for k=1 and k=2 there is no pattern?

[vale crag]

yes

[alpine sable]

oh oke, tysm

.close

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[rotund quartz]

can anyone help me or give me some ideas to a scenario of real life mathematical modeling with some solutions aswell if can

[opal jolt]

you're gonna need to be a lot more specific

[rotund quartz]

uhm, so i need ideas or examples about making a real life scenario of mathematical modeling(piecewise function) cuz i’m going to make my own problem solving

[lone heart]

@rotund quartz Has your question been resolved?

[rotund quartz]

Or maybe problem solving question examples of piecewise functions

with solutions…?

[lone heart]

@rotund quartz Has your question been resolved?

[hoary nimbus]

Modeling motion

Example the velocity function of a car is 3t^2 for t<5

And it’s 4t^3 for t> or =0

I hope this is real life enough cuz irl these instantaneous changes in velocity aren’t possible

[lone heart]

@rotund quartz Has your question been resolved?

[lone heart]

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[left frigate]

Given a semicircle with center O, diameter BC and a point A on the semicircle (A is different from B and C). Lower AH perpendicular to BC (H belongs to BC). On the half-plane with edge BC containing A, construct two semicircles with diameters HB and HC, they intersect AB and AC at E and F respectively. Let I and K be two points symmetric to H through AB and AC respectively. The line IK intersects the tangent drawn from B of the semicircle (O) at M. Prove that MC, AH and EF are concurrent.
Proved: AE x AB = AF x AC; EF is a common tangent to the two semicircles with diameters HB and HC; 3 points I, A, K are collinear.
Stuck: Don't know how to prove that MC, AH, EF are concurrent.

[dense sleet]

Do you have an image

[left frigate]

I'm drawing it. Please wait.

[dense sleet]

Ok i see

What have you tried

[left frigate]

I've already proved that EF and AH intersects

at O

I'm having a problem about how can I prove that MC crosses O and intersects with the rest

[dense sleet]

O being the center of AFHE?

[left frigate]

Yes

AFHE is a rectangle

[dense sleet]

Is IK tangent to the circle?

[left frigate]

Yes. It is tangent to (O)

(Forget about the O at the center of the rectangle. We can call that O1)

I apologize

[dense sleet]

What is M

Better question

What is CM

[left frigate]

M is the intersection between IK and O's tangent at B

[dense sleet]

Intersection of 2 tangents

Do you know anything about that

[left frigate]

It would be equal to each other?

[dense sleet]

I mean its ok if you dont know symmedians

[left frigate]

I was taught about this but Idk if it's correct

[dense sleet]

Theres prob other solution

Do you know what a symmedian is?

[left frigate]

I know about a point which symmetrics to another point through a line or a point

But other than that, I don't think I know that much

Btw, this is a sample given to us

About the intersection of 2 tangents

Don't know if it's usable in this situation

[dense sleet]

This is a symmedian btw

I mean if i was solving this id probably use something i know about that

[left frigate]

Can you give me further explanations about the lines?

[dense sleet]

AD is A-symmedian of triangle ABC

[left frigate]

What is A-symmedian? (I apologize if I sound dumb)

[dense sleet]

There are some cool stuff with it then but i dont think you need it

[dense sleet]

Symmedian from point A

[left frigate]

Ok.

[dense sleet]

Like B-symmedian would be this

[left frigate]

Ahhh

That's Interesting

[left frigate]

Can you give me a hint of what I can use to prove the statement here with the knowledge provided?

Just a hint

[dense sleet]

Do you know homothety

[left frigate]

Yeah I know that

[dense sleet]

This is prob best hint i can give

Do you see it?

[left frigate]

The only thing I can think of is AEH and ACH

Or something relating to O1CH

[dense sleet]

No in the white circle

[dense sleet]

Reminds you of something?

[left frigate]

CEH is a right triangle

[dense sleet]

And what is O

[left frigate]

OE is a tangent of (P)?

[dense sleet]

And

[left frigate]

OEP = 90 degrees? (Guessing)

[dense sleet]

Thats not important here

[left frigate]

Oh alright

[dense sleet]

Do you see how this

[dense sleet]

Looks like this

[left frigate]

It looks pretty similar

[dense sleet]

How similar

Can you solve the problem from that observation

[left frigate]

I apologize but I still haven't got an idea from this observation.

[dense sleet]

Consider a homothety centered at C

[left frigate]

So I need to prove CEH is homothetic to AHC?

I can't think of anything right now =((

[dense sleet]

Proving there is a homothety that takes O -> M would finish the problem

Find one that does such

[left frigate]

I have 2 options:
CEH and CBM
CH(O1) and CBM

But I think the first one is much more acceptable since you've mentioned that CEH is quite important

[dense sleet]

Why CBM

[left frigate]

It contains (O1)M and it's relatable to the (P) circle

[dense sleet]

How about CBA

[left frigate]

I haven't thought about that

But doesn't it covers only part of (O1)M?

[dense sleet]

Consider a homothety that takes H -> B and E -> A. Prove it takes O -> M

[left frigate]

So it's like CEH is homothetic to CAB

[dense sleet]

Yes

[left frigate]

But I don't know what to do with CAB now

[dense sleet]

Ok its really very simple

If O is on AH on which line does it go when we take the homothety

You know homothety takes lines to lines

Circles to cirlces etc...

Where does line AH go

[valid bramble]

a

[left frigate]

I apologize but I'm not understanding this part =((

[dense sleet]

You know homothety takes lines to lines right?

[left frigate]

Yes, (line-line-line) and (line-angle-line) (That's how we call them overhere)

[dense sleet]

Ok basically AH goes to some line

Which line is it

Important part here is that H -> B

[left frigate]

Is it IK?

I think that is the most reasonable thing I can think of

[dense sleet]

So you are saying AH goes to IK by homothethy centered at C that takes H -> B

[left frigate]

Probably not, I guess

But I think I've got the general idea now. I can solve the rest by myself.

Thanks.

.close

[lone heart]

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[dense sleet]

This is how homothety looks

[left frigate]

Oh yes

[dense sleet]

So do you now see where AH would go?

[left frigate]

Yes. Thank you so much ^_^

[dense sleet]

Where?

[lone heart]

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[tall garden]

how do you integrate this

[tall garden]

I got this as the answer

[tacit arch]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[tall garden]

I just showed you

like every step?

I did u=L/2-x

then got this

then this

then used some integral table to get what I got 2 posts above these 2

[tacit arch]

What's Ke alpha?

[tall garden]

just some constants ignore them

I forgot to add it back in, but yeah just ignore it

[tacit arch]

,w int 1/sqrt(x^2+b^2)

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by rie.mann

[tacit arch]

Is this what your table showed?

[tall garden]

no

well maybe

but heres a more concise version

@tacit arch

[tacit arch]

The second term 2 * sqrt(...) probably shouldn't be there. the integrand u/sqrt (u^2+b^2) is an odd function so when you integrate it over a symmetric domain, you should get zero

So you probably have a sign error somewhere you're not showing

[tall garden]

oh your're right

[lone heart]

@tall garden Has your question been resolved?

[tall garden]

@tacit arch I have a question in general about this problem im solving if you dont mine

this is what im doing

wouldn't the potential integral from a 'dq' charge on the rod to point B be

[lone heart]

@tall garden Has your question been resolved?

[tall garden]

.close

[lone heart]

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[lone heart]

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[quaint viper]

Yes

Hi

Oh no it's very very hard......... I'm so sorry

Im sorry

[lone heart]

@sharp copper Has your question been resolved?

[lone heart]

@sharp copper Has your question been resolved?

[nimble breach]

why did i request the helpers role

what the fuck is any of this

[summer dirge]

@nimble breach if you want the Helpers role removed, go to id:customize and uncheck it

the "I really want to help others with math", that is

[nimble breach]

thanks

[lone heart]

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[lost tapir]

Can someone help me with this problem?

[lost tapir]

this is my work

<@&286206848099549185>

please help me with this problem

.close

[lone heart]

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[wheat saffron]

Provide a trigonometric equation. Considering only the space between 𝑥 = 0 𝑎𝑛𝑑 2𝜋, the
equation must only have solutions at 𝑥 = 1 and 𝑥 = 2. Explain your thought process and
the work you did to create the equation. You may round decimal values to 3 places.