help-38

[jolly saffron]

I think you're gonna have to do the same thing here

[marsh forum]

got it

I can then just draw a path that adds a circle around the origin

and I'm done

[jolly saffron]

basically this .(dx, dy), and evaluate the line integral clockwise on the unit circle around the origin

[marsh forum]

so it's just 2π

thanks

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[vagrant imp]

For this Question I am trying to find the equation of the line in the form a+λb, and I have found (the direction ratio for the line) b=-i+2j-k by using the cross product but Im not sure how to find a

[jolly saffron]

hmm

[dull pilot]

more jee bros

[jolly saffron]

i fucking hated this topic given how boring it was

[chilly bobcat]

shocking how many

[vagrant imp]

I hate 3d too 😭

[jolly saffron]

man didn't you have some dumb formula for this

[vagrant imp]

its in a few days so yeah

[jolly saffron]

with like a determinant

or something

[dull pilot]

not really

[vagrant imp]

idk man i can't remember them formulae anyways, is there like a method or smthn

[dull pilot]

id like

assume a random point on one of the lines

find the normal

do stuff

idk

[jolly saffron]

yeah the procedure was to use a random point

snd then you project it to find the smallest distance

[ember fern]

ngl this is the most important step

[jolly saffron]

which is the perpendicular

if I was near a computer I'd use paint but I'm not

[vagrant imp]

but isn't that for finding the smallest distance, i need the point for my equation

[jolly saffron]

yeah I know

so you assume the point is governed by a parameter t or something

and the random point that you take

the projection of that would coincide with t

[vagrant imp]

oh so basically like (2t+1)-λ=3u+2
and do that for y and z and solve for u or t

[vagrant imp]

yeah I got the answer, thank you very much

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[spark wagon]

People, how do I find general solution to this ODE? I got as far as dy/dx = (-2yx^3)/(x^4+y^4) but this doesn't seem to be separable

[spark wagon]

and I don't know what substitution to use here

my notes mention something called "Integration Factor" but I don't know how to apply it here

if it's applicable to begin with

[wraith hinge]

divide the equation by y^4

substitute x/y=t

[spark wagon]

ok let's see

oh wow, i am not completely stuck anymore

i think i'll figure out the rest myself, thx

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[wraith hinge]

I am happy to help 👴

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[pliant thistle]

In triangle ABC, (b sin C)(b cos C + c cos B) = 42. Compute the area of the triangle.

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[pliant thistle]

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[zealous sigil]

Hi, I was hoping someone could help me double check my answer for this part c of this question. I think it's correct but I'm not 100% sure to be honest. Thanks!

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@zealous sigil Has your question been resolved?

[zealous sigil]

<@&286206848099549185>

[zinc raptor]

it's correct

[zealous sigil]

Thanks so much. I appreciate it.

If it's okay, would you be able to help me with one or two more?

[zinc raptor]

sure

[zealous sigil]

This next one is a bit tricky for me

I have this so far

[zinc raptor]

it can be easy to go down the wrong path here

the key is to think about each of the 50 balls independently

for starters, what is the expected value of the first number

[zealous sigil]

So the expected number of times that the sum will be 1275?

Oh wait sorry do you mean the expected value of the number of times a 1 is observed?

That would be 1/20 I think

That's from the previous question

[zinc raptor]

i meant the expected value of the number on the ball drawn

[zealous sigil]

Oh okay

To be perfectly honest I'm sure what it would be

I'm trying to define some variable but I'm not even sure where to start

[zinc raptor]

maybe it helps to scale it down first

suppose you only have two balls, numbered 1 and 2

and you draw one of them at random

what is the expected value of the number you get?

[zealous sigil]

1.5?

I think

Yeah I think its 1.5

[zinc raptor]

yeah

how did you get that

[zealous sigil]

Well I found the probability of getting a 1 first which was 1/2

and the probability of getting a 2 first which was 1/2

and then (1)(1/2)+(2)(1/2)

=1.5

[zinc raptor]

ok nice

now you can do the same thing if you have 1000 balls from 1 to 1000

[zealous sigil]

But in this one aren't we continuously taking out 50

In the example you just gave we only choose 1 ball

and that's it

[zinc raptor]

yes

[zealous sigil]

So I thought it would be different

[zinc raptor]

but if you look at the first ball

nothing else has been chosen yet

so it's the same scenario

[zealous sigil]

Oh so we are only thinking about the very first one

That we choose

[zinc raptor]

yeah for now

[zealous sigil]

Is it 500.5

I got that by doing this summation in a calculator

[zinc raptor]

yes that's correct

do you know how to compute this sum without a calculator

[zealous sigil]

To be honest I don't

I think there's some formula or something

Oh wait we could take out the 1/1000 constant term and then use that formula

[zinc raptor]

yes

[zealous sigil]

This one

[zinc raptor]

btw, one way to arrive at the formula is to pair up the numbers

1 + 1000 = 1001, 2 + 999 = 1001, 3 + 998 = 1001, etc

but i guess this is not directly relevant to the problem

anyway, we got that the expected value of the first ball is 500.5

[zealous sigil]

Ah that's really clever

[zealous sigil]

Yup

[zinc raptor]

now what about the second ball

[zealous sigil]

Well this time I'm not entirely sure to be honest, because I thought about doing this but there aren't 1000 items to choose from anymore

But we don't know which one was picked on the very first one so I'm kind of confused a bit

[zinc raptor]

ok so, the way we calculated it for the first ball is

we found the probability that the ball is each number

which is just 1/1000 for any number

[zealous sigil]

Yeah

[zinc raptor]

so maybe for starters, what is the probability that the second ball is 1?

[zealous sigil]

1/999 right

But hold on, if the first ball picked was actually 1 then it's 0

[zinc raptor]

yeah

[zealous sigil]

I think

[zinc raptor]

that's correct

and the first ball has a 1/1000 chance of being picked, right

[zealous sigil]

Yeah

[zinc raptor]

so how do we combine these to find the probability the first ball is picked on the second turn

[zealous sigil]

I'm so sorry, this sentence doesn't make too much sense to me

[zinc raptor]

no worries

[zealous sigil]

First ball picked on the second turn

I'm not too sure what that means

[zinc raptor]

by first ball i meant ball 1

ok i see the misunderstanding

[zealous sigil]

Oh so we want ball 1 to be picked on the second turn

[zinc raptor]

we want P(second ball picked = ball #1)

[zealous sigil]

Then (999/1000)*(1/999) right

[zinc raptor]

yep exactly

[zealous sigil]

Oh what the heck it's just 1/1000

[zinc raptor]

yeah

that's very interesting

[zealous sigil]

Yeah honestly that's weird

I didn't expect P(first ball is 1) to be equal to P(second ball is 1)

[zinc raptor]

what about P(second ball picked = ball #2)?

[zealous sigil]

Wouldn't that also be 1/1000

[zinc raptor]

yeah

because there's nothing special about ball #1

the same math works for everything else

[zealous sigil]

Yup, I agree with this

[zinc raptor]

so what would be the expected value of the second ball picked

[zealous sigil]

Is it also 500.5?

By any chance

[zinc raptor]

yeah

because all the probabilities are still 1/1000

[zealous sigil]

But how would that make sense from the E(X) formula though

Let X2=The ball picked on the second turn

P(X2=1)=P(X2=2)=...=P(X2=1000)=1/1000

But there aren't 1000 balls to choose from anymore right

So E(X2)=Sum(X2*P(X2=x2)) or whatever it is

[zinc raptor]

there aren't

however

[zealous sigil]

I get how the probability is the same

[zinc raptor]

the 999 remaining balls can be any 999 out of the 1000

so there are still 1000 possibilities

[zealous sigil]

But it just doesn't make sense how the expected value is the same

[zinc raptor]

maybe, let's go back to the example with only 2 balls

we have ball #1 and ball #2

either we pick ball #1 first and ball #2 second, or ball #2 first and ball #1 second

so the expected value of the second ball picked is 1/2(1+2) = 1.5

which is the same as the first ball picked

[zealous sigil]

Okay yeah that does make sense

But if you had to do it by defining a variable and then applying the E(X) formula how would you do it?

Or is it not possible to do it like that

[zealous sigil]

Something like this by any chance

[zinc raptor]

what are you referring to by the E(X) formula

[zealous sigil]

The top one

[delicate solar]

Oh

[zinc raptor]

i presume f(x_i) refers to the probability of x_i?

this is what we're using

1(1/1000) + 2(1/1000) + ... + 1000(1/1000)

[zealous sigil]

Yup

[zealous sigil]

That's for E(X1) where X1 is the first ball picked right

But what about E(X2)

Where X2 is the second ball picked

[zinc raptor]

same thing

because the probabilities are all 1/1000

[zealous sigil]

Yup I agree

But X2 can't take values {1,2,3,4,...,1000} right

Because one ball has already been picked

So you can't sum from 1 to 1000 right

Which is why I'm confused

[zinc raptor]

one ball has already been picked

but

we don't know which

[zealous sigil]

Yup

I agree with that

[zinc raptor]

i guess

it might be better to think about it

as starting before any ball has been picked

we pick 1 ball, throw it away without looking, then pick a second

[zealous sigil]

Okay, so this means we still sum to 1000 then?

Or maybe I'm missing something here

[zinc raptor]

yes

like, the second ball could be any of the 1000

there are 1000 possibilities

here's another way to think about it

generate a random permutation of balls 1-1000

and look at the second ball

(if this is harder to think about don't worry about it but maybe it helps)

[zealous sigil]

Okay yeah this makes sense, we are just looking at all the different ways we could order 1000 balls

[zinc raptor]

this might also help to understand the symmetry

i.e. why all the probabilities are still 1/1000

[zealous sigil]

I get that part

But what's just really confusing me is the values that X2 can take

Like I get the thing that you said about throwing the ball away as soon as it's picked

But I don't know it just feels weird

[zinc raptor]

i guess another way to think about it

instead of drawing from a bag of 999 balls

actually nvm

what i was going to say was

you are drawing from a bag of 1000 "shrodinger's cat" balls

where each exists with probability 999/1000

this might just be more confusing because their existence events aren't independent

[zealous sigil]

Yeah it's dependent right because one influences the other

Or something like that

[zinc raptor]

i think the permutation phrasing probably makes it most clear

any ball could end up in position 2 with equal probability

it doesn't matter what happens in position 1

[zealous sigil]

OHHH WAIT

Because we could have 1 in the second position

We could have 2

We could have 3

Etc. etc. all the way to 1009

1000*

[zinc raptor]

yeah

[zealous sigil]

That's genius okay that makes a lot of sense

Okay that makes complete sense why E(X2)=500.5

[zinc raptor]

yay

so what about the other 48

X3 through X50

[zealous sigil]

I'm guessing they also have an expected value of 500.5 but I'm just going to quickly double check

Yup they're all 500.5

[zinc raptor]

yep

[zealous sigil]

So E(S)=E(X1+X2+...+X50)=E(X1)+E(X2)+...+E(X50)=50*500.5=25025?

Is that maybe how we would do it then?

[zinc raptor]

yes exactly

[zealous sigil]

Absolutely genius man

[zinc raptor]

glad i could help

[bright quarry]

thanks

🙏🏻

[zealous sigil]

Just one last thing, it says to compare to when we do replacement

I have a feeling that the expected value doesn't change

Because X1 can take {1,2,3,...,1000}

P(X1=whatever number between 1 and 1000)=1/1000

P(X2=same thing)=1/1000

It's the same thing

[zinc raptor]

yep that's right

[zealous sigil]

Dude thank you so much, you spent so much time explaining that to me. Have an amazing day.

[zinc raptor]

no problem, you too 🙂

[zealous sigil]

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[rancid gate]

Anyone who can help me understand why this is true, the topic is RSA cryptographics?

[zinc raptor]

are you familiar with euler's theorem?

[rancid gate]

Yes

[zinc raptor]

so what happens if you take m^(ed)

and then you add or subtract phi(n) from the exponent

[rancid gate]

ed = 1 mod n ?

[zinc raptor]

i guess i was just explaining why the centered line is true

are you given that ed = 1 mod n here?

[rancid gate]

no this is for an assignment, i need to explain how RSA works

but i dont understand how to make proof that RSA can be decrypted and output same input

[trim lichen]

maybe state euler's thm in full

so we are on the same page about it

[rancid gate]

a^phi(n) is congruent to 1 mod n, given that gcd(a, n) = 1

[trim lichen]

ok right

so you do then have that m is coprime to n, yes?

therefore m^(phi(n)) ≡ 1 (mod n)

[rancid gate]

yes

[trim lichen]

therefore you can add or subtract phi(n) to the exponent on m however many times you wish

[rancid gate]

yes

so we can remove it?

so it says:
m ≡ 1 (mod n)

?

[trim lichen]

no, we get m^(ed) ≡ 1 (mod n)

[rancid gate]

how so?

[trim lichen]

er

oh my bad, do we?

no, i think i got ahead of myself

i think maybe im missing a bit of context, but can you provide the theorem this is meant to be part of the proof for

[rancid gate]

i found an assignment that simply stated the text in this picture

[trim lichen]

but you do have that ed = 1 mod phi(n), no?

[rancid gate]

yes

[trim lichen]

well then

do you understand that since $m^{\varphi(n)} \equiv 1 \pmod{n}$ we can also say that $m^{ed} \equiv m^{ed - \varphi(n)} \pmod{n}$?

[solid kiln]

Ann

[rancid gate]

this is cause of inverse elements right?

[trim lichen]

i... wouldn't say so explicitly?

[rancid gate]

i dont understand this part

[trim lichen]

ok maybe it would be easier to illustrate my point with some numerical examples

let's consider some stuff mod 10

phi(10) = 4

since 3^4 ≡ 1 (mod 10), we have that for any natural n, 3^n ≡ 3^(n-4) (mod 10)

in other words the exponent can be shifted by multiples of 4 (aka phi(10)) for free

does that make sense to you

[rancid gate]

AH

yes

[trim lichen]

yeah so

in the same way in your example we can shift the exponent on m by multiples of phi(n) for free

and this can take us from ed all the way down to the remainder of ed mod phi(n), yes?

[rancid gate]

✅

[trim lichen]

yeah

and ed mod phi(n) is just 1

so you get m^1

[rancid gate]

thanks

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[scenic lion]

Intuitively, I understand that the triple scalar product gives you the volume of the the parallelepiped. I know that each of the three scalar triple product check if the line is to the left of each of the edges when going trough them counterclockwise. What I dont understand is how to know that the line is on the left or on the right. Note that it also determines which normal of the triangle (front face vs back face) the ray is alinged with. So if going through the triangle in counter clockwise directions you expect the ray to be on the left of all three triangle edges. I just dont understand how the math lets you know whether the ray is on the left of an edge or not. I have attached an image showing the ray-triangle intersection visual and some context.
One more though, if you take A X B and A X B > 0 you know A is on the "left" of B, but nothing here is directly using any edge. all the vectors in any scalar triple product consist of two vectrs from the ray origin to the respective vertices of the edge, and the third is the ray origin going towards the ray direction. So I dont understand how this triple scalar product says if the ray is to the left of the edge or not when we arent directly using the edge.

[scenic lion]

also to show definition of the triple scalar product here

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@scenic lion Has your question been resolved?

[fleet bear]

Still trying to parse this myself but here are some thoughts I guess

If we consider
v = [PQ PA PC]
for that particular viewing direction

v = PQ • (PA × PC)
The vector (PA × PC) points into the triangle. I suppose if we think of the plane with normal PA × PC placed against edge AC (AC must in lie in the plane), dotting the normal with PQ would tell us which side point R is on

[scenic lion]

youre not suppose to put that here

im using this channel lol

please use one of the open ones

[gray elm]

Ok im sorry

[scenic lion]

its okay!

[scenic lion]

the open ones are under MATH HELP for example like help-15 or help-45 or help-49

[scenic lion]

why does AC lie in the plane? im imagining a fixed triangle parallel to the xz plane (y going up) and as p goes from being above triangle to below the normal made from PA X PC changes

idk if that makes sense or not

[fleet bear]

It's like the triangle normal:
You have points P, A, and C. There necessarily exists a plane they all lie in. PA and PC will lie that plane too

PA × PC is normal to the plane

Reading your message now

[scenic lion]

oh wait i think i see what you mean

[fleet bear]

This makes sense

Oh btw

I made a visualisation in demsms 3d

Desmos

Which might help

[scenic lion]

is it okay if i see it

[fleet bear]

Ofc

https://desmos.com/3d/wpyicnkaqe

[scenic lion]

dang that looks good

[fleet bear]

Thanks

Unfortunately no labels cuz desmos 3d doesn't support them yet

The points are defined at the start

[scenic lion]

im still trying to understand why you can tell if something is on the left or right of an edge but im going to play around and see if it clicks. either way thanks for doing that and the effort you put into it

[fleet bear]

It's no problem

I feel left and right are the wrong way to look it tbh

[scenic lion]

whats a better way to look at it? im open to anything tbh

i just want to understand it better

[fleet bear]

Since you're comfortable with planes, you may take it as which side the line PQ enters/leaves the edge-plane

[fleet bear]

Edge plane being the plane containing the edge as described earlier

Or even as whether the vector PQ has a positive component in the direction of normal (this is for the CCW case)

I think the viz should help anyway

Updated viz: https://www.desmos.com/3d/ch8je1sxt5

[scenic lion]

i understand it from the first visual now thank you, going to look at this one too now

[fleet bear]

You're welcome 💯

[scenic lion]

.close

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[lavish cosmos]

Working through textbook questions and there are no solutions for this question

I found this proof online but the last part doesn’t make sense to me

If someone could explain or propose a different proof that would help

[next vessel]

What app is this?

[half marsh]

goodnotes

[next vessel]

ight

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[lavish cosmos]

<@&286206848099549185>

[fallow bane]

I think it looks okay

Are you meant to write your proof in this list format or is a worded proof sufficient

If the former is true you need to add the axiomatic logic rules at the end of each step to justify what you did

[lavish cosmos]

its meant to be a semiformal proof

mainly words with justifications for each line

I Understand that Y is an element of both but why is it neccessarily true that f(D^E) a subset of f(D) ^ F(E)

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[wraith hinge]

so i have trouble trying to simplify this

[solid kiln]

Suika

[simple panther]

I think this is as far we can go

[wraith hinge]

uhhhhhhhhhh

[simple panther]

Or did u mean something else?

[wraith hinge]

no

im

bad

at

this

[wraith hinge]

im trying to see how to get to that expression

the factorisation is a bit confusing 😭

[solid kiln]

Suika

[wraith hinge]

idek how

[solid kiln]

Suika

[wraith hinge]

im gonna cry

[solid kiln]

Suika

[simple panther]

this is ur question

[wraith hinge]

oh

i am slowly

[simple panther]

multiply the square brackets

[wraith hinge]

starting to envision this

so from 16xx and 2x

uhh

[simple panther]

yes

[wraith hinge]

2x

take out 2x

[simple panther]

ye

[wraith hinge]

idk

oh

[simple panther]

from that, we can take 2x common out

and from the rest, we can take (1-4x)³ common out

[wraith hinge]

2x(8x^(1-4x)^3)+(1-4x)^4

idk

or do i need one extra bracket

[simple panther]

yes

[wraith hinge]

2x((8x^(1-4x)^3)+(1-4x)^4)

like this?

[simple panther]

yes

you're missing the 2 next to x^

it's x^2

[wraith hinge]

wait

[simple panther]

2x((8x^2(1-4x)^3)+(1-4x)^4)

[wraith hinge]

oh

no wonder why it looked confusing

okay

yayyyyyy

thank you

i lvoe you

[simple panther]

and in the remaining part of the terms, there is (1-4x)³ common, that can also be taken out

[wraith hinge]

is that actually the simplest form

[wraith hinge]

oh holy

now what

[simple panther]

take it out

[wraith hinge]

wait

[simple panther]

we get this after that

[wraith hinge]

let me write it first

[wraith hinge]

wait why 8x^2 when we have 2x out of the bracket

[simple panther]

oh, yeah

it's just x

8x

[wraith hinge]

should it look like this

also excuse my funny shape of x

[simple panther]

yes

[wraith hinge]

okay now

how do we

take out

1-4x

[simple panther]

as you can see, there is (1-4x)³ common in the 2 terms

the inside part can be written as -8x * (1-4x)³ + (1-4x) * (1-4x)³

[wraith hinge]

hwat

oh

[simple panther]

the part inside the [ ]

[wraith hinge]

i get what u maent

yes, process

I mEAN

PROCEED

😭

[simple panther]

then, we can take the (1-4x)³ common out and we get

(1-4x)³ * (-8x + (1-4x))

[wraith hinge]

ok this is where im stuck

[simple panther]

so, it will be 2x(1-4x)³(-8x + (1-4x))

Simplifying it gives 2x(1-4x)³(1-12x)

[wraith hinge]

im tryinh to udnerstad

[simple panther]

u get this

??

[wraith hinge]

oh

this is what i did

[simple panther]

Yes

If that heps you understand it, that be it

[wraith hinge]

yY

yayayayay

thanksks yuou so much

hearts 4 u

.close

[trim jolt]

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[wraith hinge]

how do i integrate 2/(2x-a)

[wraith hinge]

dont i take the 2 to the left of the integral

[hybrid vine]

Sure

[wraith hinge]

then divide it by the derivative

so ln (2x-a)?

[hybrid vine]

Derivative of?

[wraith hinge]

2x-a

[hybrid vine]

Yup

[wraith hinge]

so this is correct?

[hybrid vine]

Yes

ln(2x -a) + C

[wraith hinge]

oh

thanks

.close

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[obsidian zealot]

!status

[trim jolt]

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

[jolly saffron]

if the radicals give you trouble sub n = x^60

[obsidian zealot]

perhaps some roots should turn into fractional powers

orrrrrrrrrr that

[jolly saffron]

yeah i think that's right

,w limit as n goes to infinity (5n^(4/5) - 2n^(2/3))/(2n^(3/4) + 5n)

[solid kiln]

Results provided by WolframAlpha

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

Requested by dyxn06

[trim jolt]

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[late rivet]

this is correct or no

granted i can simplify the 54/16

[trim lichen]

looks vaguely correct but i feel like it could be less complicated to begin with.

[late rivet]

how

vaguely

i just put it in like that as i was going by the terms

i got it correct

but how did they simplify it to this

v

[obsidian zealot]

16=4^2

[dapper swift]

yeah the trick is just $\frac{9/4}{1 - x^6 / 4}$ so $\sum_{n = 0}^{\infty} \frac{9}{4} (x^6 / 4)^n$

[solid kiln]

south

[obsidian zealot]

4^(n-1+2)=4^(n+1

[late rivet]

ohh

okay i see

thanks

.close

[trim jolt]

Channel closed

Closed by @late rivet

Use .reopen if this was a mistake.

[late rivet]

.reopen

[trim jolt]

✅

[late rivet]

what did i do wrong

[surreal bone]

can you show your solution , please ?

[late rivet]

my solution is the bottom

with th esunm

sum

[surreal bone]

well you need to solve it step by step

[late rivet]

this

[late rivet]

i did

and it is wrong

idk what i did wrong

[obsidian zealot]

!showwork

[trim jolt]

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

[obsidian zealot]

not “stuck” but like

[late rivet]

100% my work will not help but ill show it

here

[surreal bone]

did you use this ?

[late rivet]

yeah

[surreal bone]

you need to take care when the sum starts from 0 or from 1

[late rivet]

meaning what

it starts at 1

[surreal bone]

in the first line the sum starts from 0

[late rivet]

what do i do

[surreal bone]

you need to take care of the bounds of the sum

[late rivet]

how

[late rivet]

so for every n i have

i do -1?

[surreal bone]

as you see

then put t = x^9 /2

multiply it by 3/2

then find f'(t)

[late rivet]

would this be it then

[late rivet]

i did all of this

[surreal bone]

try it

[late rivet]

😔

okay

no

that wasnt correct

how did they get this

<@&286206848099549185>

[surreal bone]

what is the summing bounds here ?

[late rivet]

n=1 to inf

[obsidian zealot]

they used n=0 lower bounds probably

[late rivet]

no it is it n=1

in their answer

[obsidian zealot]

I’m just saying the answer is correct if you use n=0

[wraith hinge]

If I followed the standard geometric series approach, This is Definitely a bit different

[surreal bone]

are you sure ?

[trim jolt]

@late rivet Has your question been resolved?

[wraith hinge]

My solution

[trim jolt]

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[late rivet]

.repoen

[wraith hinge]

what is happening here

[late rivet]

oh

oopsies

[wraith hinge]

sorry

[late rivet]

nah its all good

[late rivet]

which part

[wraith hinge]

how did we get 1/2 ln (1+x^2)

[dull pilot]

integration

[wraith hinge]

and why didnt we add C to the left side

[wraith hinge]

😭

can i swear?

[late rivet]

cause youre going to set it all equal to y

also

[wraith hinge]

so we only add C to the dx part?

[late rivet]

if you added the C to the left side as well they would just cancel out

[dull pilot]

fuck yeah

[late rivet]

u sub

[wraith hinge]

no fucking shit man

[late rivet]

insta ban

bye bye

[wraith hinge]

anyways

[late rivet]

they did u sub

[wraith hinge]

for the x/1+x^2

[late rivet]

u = 1+x^2, du = 2xdx

[wraith hinge]

ohhh there is a rule for it

[late rivet]

du/2 = xdx

[late rivet]

?

[wraith hinge]

nvm no rules

only if the numerator was 1 i couldve done sometrhing

[late rivet]

wdym

like a trig inte??

[wraith hinge]

this rule

[late rivet]

do you understand how they got 1/2 ln|1+x^2|

[late rivet]

oh yeah

[wraith hinge]

no

[late rivet]

they used u-substitution

[late rivet]

.

[wraith hinge]

hold on

oh let me try on a piece of paper

[late rivet]

okie

[wraith hinge]

kinda lost

u = 1+x^2

[late rivet]

yes

[wraith hinge]

and i got du = 2xdx

now what

oh make dx subject

[late rivet]

yes

[wraith hinge]

dx = du/2x

[late rivet]

no

not all the way

you have xdx

[wraith hinge]

what

ohhh

so du/2 = xdx

[late rivet]

yes

thats where the 1/2 comes from

[wraith hinge]

now i have (du/2) / (u)

[late rivet]

you can pull the 1/2 out

[wraith hinge]

alright

[late rivet]

so 1/2int(1/u)

[wraith hinge]

where did the 1 come from

[late rivet]

where ?

[wraith hinge]

1/u

i have du/u

[late rivet]

sorry

thats just how i do it

i mean
1/2 int(1/u)du

du/u works to

[wraith hinge]

ohhh

[late rivet]

mb

[wraith hinge]

alright thanks

[late rivet]

no problem

[wraith hinge]

.close

[trim jolt]

Channel closed

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[late rivet]

how did you get this

well i see it

.open

didnt know that doesnt work

anyway

[trim jolt]

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[late rivet]

@surreal bone

[late rivet]

^

[dull pilot]

huh

what

[late rivet]

the stuff above

[surreal bone]

?

[late rivet]

i couldnt reopen the channel in time

[late rivet]

.

[dull pilot]

...you used .open lol

it's .reopen

[late rivet]

no someone took the channel before

[dull pilot]

anyway if you have a guy ill leave you to it

[late rivet]

i did .reopen

also the the is n=1

[trim jolt]

@late rivet Has your question been resolved?

[trim jolt]

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[proper ferry]

Where does the pi come from when the y do fourier ttansform

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@proper ferry Has your question been resolved?

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[next vapor]

The function f(x) = kx + m is given for x ∈ (0,1) and 0 otherwise.

a) What conditions must k and m satisfy for this function to be a probability density function?
Let X be a random variable with density f(x) where now m = -1.

b) Find P(< x <).
c) Compute E(X) and D(-2X).
d) Let Y = 3X - 2. Find the density of the random variable Y.

[nova spire]

!status

[trim jolt]

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

[next vapor]

2

i want someone to go throw together

<@&286206848099549185>

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@next vapor Has your question been resolved?

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[thin vine]

can someone please help me with this problem

so im pretty sure youre supposed to use half angle formula for tan

oh wait

[gusty bone]

yeah just sub and solve the cubic or whatever for cos(x)

[trim musk]

Convert cosx to half angle

[trim jolt]

@thin vine Has your question been resolved?

[thin vine]

.close

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[chilly sparrow]

Two cars start simultaneously from two reference points A and B at a distance of 500 m, heading towards each other. Car A maintains a constant acceleration of 2.0 m/s2 until it reaches a speed of 20 m/s, continuing in uniform motion (constant speed). Car B always maintains a constant acceleration of 1.0 m/s2.
A. How long after the start do the cars meet?
B. How far from reference point A does the meeting take place?

[chilly sparrow]

I haven't been able to solve this problem 😦

[supple copper]

what have you tried

[chilly sparrow]

It's a bonus for mechanical physics, but I really didn't understand the topic.

[supple copper]

do you understand what the question is asking?

[chilly sparrow]

I know he asks about mechanical physics and something about the MRU

[supple copper]

do you know calculus?

[chilly sparrow]

so so

[supple copper]

can you write down the position of the 2 cars x(t) and y(t)?

[ornate bobcat]

u dont need to apply calc for this qs tbh

theres basic speed formulae u learn in elementary phy but rest is regular algebra

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[old glen]

can you explain in second example its 31 / 33 but it could have been with 26 / 215

[old glen]

can i still do it like 3/6 as 1/2 and do not do first step like he has done what is better?

[kindred jungle]

"can you explain in second example its 3*1/3*3 but it could have been with 2*6/2*15"

wdym?

[kindred jungle]

you mean skip this step?

[old glen]

oh no my bad

[old glen]

Yes

[kindred jungle]

absolutely

[kindred jungle]

how would you do it with 2?

[old glen]

here its second

can also do with 2

[kindred jungle]

the 9/15, or the 12/30?

[old glen]

i am sorry moonful so my question was incorrect

[kindred jungle]

thats okay

[old glen]

12/30

[kindred jungle]

yes of course, but youll have to simplify it more

try doing it with 2

see what you get

[old glen]

moonful why did they not do it with2

[kindred jungle]

youll see

just try doing it with 2

see what you get

[old glen]

12/30 = 2 * 6 / 2 * 15

cancel out 2s and 6/15 = 3/5

[kindred jungle]

6/15 is not 3/5

you're very close tho

[old glen]

2/5

[kindred jungle]

yes

6/15 is 2/5

[old glen]

moonful you think i can improve my math

[kindred jungle]

oh look you got the same answer!!

you can do it with 2, yes, but itll eventually simplify to 2/5 anyway

[kindred jungle]

anyone can

[old glen]

so like there are many ways?

[kindred jungle]

yes

as long as you get a fully simplified answer in the end

[old glen]

but if i was good at math i could do it like he has done it

[kindred jungle]

in general, though, to do it the fastest, you need to look for the largest number you can cancel

[old glen]

i liked this way of doing

[kindred jungle]

which way?

[old glen]

9/15 = 3/5

so like i dont have to cancel 3s

like directly

[kindred jungle]

yes thats okay

nothign wrong with that

[old glen]

is this what i was taught at school

[kindred jungle]

what he's doing is just showing how it works

if you can do it directly you're doing great

[kindred jungle]

like in my head when i see a fraction like 24/30 i dont think about this i js go straight to 4/5

[old glen]

oh

thanks for a lot of the help @kindred jungle

[kindred jungle]

no probelm

[old glen]

.close

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[cyan forum]

How can i find a y vector that is equal to: y = a + c

[pliant gale]

Hint : Vectors are not static

[cyan forum]

hear me out

[cyan forum]

let’s name that point O right

i can’t prove that BO = DC to be able to do anything

the only hint the exercise gives is that ED // AB, DC//FA and FE//BC

[deep prairie]

Would it be possible if you sent the original question?

[cyan forum]

yeah hold on

[deep prairie]

Thank you!

[pliant gale]

In this case you can prove it (Even if it's more BO = -DC)

[cyan forum]

here

[cyan forum]

hm?

why would BO = -DC

i get that the vectors have different directions

but for now i need to prove that |BO| = |DC|

so i can then say that BO = c

i’m trying to somehow prove that DCBO is a rectangle

it can even work out if it’s a parallelogram

but ive honestly tried everything

[deep prairie]

It’s so sad that it doesn’t say the diagram is drawn accurately

Have you tried using angles to prove it?

[cyan forum]

yeah 😭

[deep prairie]

🥲

[cyan forum]

yeah

i’m trying to utilise that those sides are parallel to play with angles but nothing seems to be working out

[pliant gale]

Are u french ?

[cyan forum]

nope

[deep prairie]

I wish I could be more use but proving stuff with vectors is like a mind game to me

Sorry

[pliant gale]

Normaly, 'till u r not in university, it is not necessary to say to you that the diagramm is accurate

Do they told u how works Vectors' coordonate

Or vectors' distance ?

[cyan forum]

hm?

[cyan forum]

i agree lol

[pliant gale]

"Coordinates"

[cyan forum]

well it’d be too much if i used cords

[pliant gale]

Without it, I don't rly know how to prove it

[cyan forum]

welp

[trim jolt]

@cyan forum Has your question been resolved?

[cyan forum]

<@&286206848099549185>

[trim jolt]

@cyan forum Has your question been resolved?

[trim jolt]

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[gusty bone]

Let $e(x)$ be an even function, and let $o(x)$ be an odd function, such that
[e(x) + o(x) = \frac{6}{x + 2} + x^2 + 2^x]for all real numbers $x \neq -2.$ Find $o(1).$

[solid kiln]

Dork9399

[gusty bone]

All i know is that an even function implies that f(x) = f(-x) and odd implies -f(x) = f(-x)

[jolly saffron]

did you also know that any function can be broken down into odd and even components like that

atleast continuous ones can

[trim lichen]

so you know e(1) = e(?)

[jolly saffron]

I suppose that's simpler

[trim lichen]

(there is a meaningful answer and a cop-out answer to this. i want the former!)

[gusty bone]

e(-1)?

i think i got the cop out answer 😭

[trim lichen]

no

[jolly saffron]

the cop out answer would be 1

[trim lichen]

the cop-out would have been to say e(1)

yes so

e(1) = e(-1)

can you do the same thing for o, but obviously keeping in mind that it's odd

[gusty bone]

o(1) = -o(-1)

[gusty bone]

oh lol

[trim lichen]

yes

ok

now take this: $$e(x) + o(x) = \frac{6}{x + 2} + x^2 + 2^x$$ and put $x=1$.

[solid kiln]

Ann

[trim lichen]

don't simplify e(1) + o(1) yet. keep it as it is.

[gusty bone]

e(1) + o(1) = 2 + 1 + 2 = 5

[trim lichen]

good

now put x=-1 instead.

[gusty bone]

e(-1) + o(-1) = 6 + 1 + 1/2 = 15/2

oh wait

oh lol

what a solution

x + y = 5, x - y = 15/2

[trim lichen]

yes

and your goal is y

[gusty bone]

so o(1) = -5/4

oh i thought it was so much more complex

[trim lichen]

you sure you didnt find e(1) on accident

[gusty bone]

whoop

i thought i had to separate the function into odd and even components 😭

this is so much simpler

.close

[trim jolt]

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[lunar stirrup]

i put a=b=c=0

then got val of determinant as x^3

so it should be divisible by all right

but answer key is showing only 1 and 2

[frozen plover]

you might lose some information with that sub

put everything to be 1 instead

[lunar stirrup]

oh

that kinda makes my life harder

but okay ill try that

[frozen plover]

not like you're meant to be doing it this way anyways lol

[cunning cliff]

why that sub

[frozen plover]

maybe you could let one parameter be 0 actually

[lunar stirrup]

who said!

[cunning cliff]

that sub tells you that you can't rule out x, x^2, x^3

if you got the determinant to be 1 for example, with that sub

then you know the answer is "none of these"

but since you got x^3, it gives you nothing to go on

[frozen plover]

scamming is an art

[lunar stirrup]

sorry i didnt really understand this

[frozen plover]

maybe with a different sub you get x^3 + x

so it's no longer divisible by x^3

with your sub you know at most it's divisible by x^3, x^2, and x

[lunar stirrup]

no it is supposed to be independant of values of a,b,c

[frozen plover]

not exactly y

[lunar stirrup]

otherwise the answer would be subjective

[frozen plover]

yeah sure

but that doesn't mean that it would hold for your case

[cunning cliff]

that's not how it works

let me give an example

[frozen plover]

if the answer is x and x^2, x^3 is still divisible by them

[cunning cliff]

"Is 5x divisible by 3"

when someone asks this

it is implicitly for all integers x

clearly if we choose x = 0 it is divisible by 3

[lunar stirrup]

huh

so what would u suggest i do to solve it

[cunning cliff]

compute the determinant normally

algebraically using whichever method you usually do

[lunar stirrup]

dude? i would get too many terms

[cunning cliff]

stop being lazy

[lunar stirrup]

,w det{{a^2 + x,ab,ac},{ab,b^2+x,bc},{ac,bc,c^2+x}}

[solid kiln]

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[cunning cliff]

sigh

[lunar stirrup]

lazy? if this question comes in my exam im supposed to solve it within 2.5 minutes

[cunning cliff]

it shouldnt take you 2 minutes to compute the determinant with some practice

[lunar stirrup]

fair

[frozen plover]

and yeah this time letting one of the parameters be 0 and the other 2 be 1 works

[lunar stirrup]

oh ok yeah

[cunning cliff]

why though

again, that doesn't give you certainty

[frozen plover]

have faith

[cunning cliff]

You only have certainty after a substitution if it turns out its not divisible by x

[lunar stirrup]

,w det{{a,-1,0},{ax,a,-1},{ax^2,ax,a}}

[solid kiln]

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Requested by yoda_576

[lunar stirrup]

@cunning cliff @frozen plover ty guys i got this one right (didnt put a =0 )

i think this is one specific type of problem where u are not allowed to put = 0

cause it worked for most of the others

,w det{{1,3,1},{3,4,1},{2,3,1}}

[solid kiln]

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Requested by yoda_576

[lunar stirrup]

,w det{{1,9,1},{9,16,1},{4,9,1}}

[solid kiln]

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Requested by yoda_576

[lunar stirrup]

.close

[trim jolt]

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[wraith hinge]

<@&286206848099549185>

I NEED HELP

😭

[umbral hazel]

!15m

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

[wraith hinge]

ohh

[umbral hazel]

What can you tell me about the sequence?

[wraith hinge]

uhhhhhh

for the first one

6n+9

but

it says 3n

so shouldn't it be

18n squared +9

18n²+9

[trim jolt]

@wraith hinge Has your question been resolved?

[wraith hinge]

<@&286206848099549185>

[umbral hazel]

My bad

I’m at an event for my Younger Sister’s school

[umbral hazel]

Plug in 3n in place of n

So 6(3n) + 7

@wraith hinge Still here?

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@wraith hinge Has your question been resolved?

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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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[vocal cobalt]

how do I solve part d? I'm kinda stuck

[burnt mulch]

Consider substituting $u=\frac{x}{2}$

[solid kiln]

Civil Service Pigeon

[vocal cobalt]

i do u sub?

[burnt mulch]

That is what I said 😕

b/c atm, you only have info about the arc length of f, and hence about the integral of sqrt(1+(f'(x))^2)