help-49

[midnight plank]

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[orchid grove]

|x|<1 in this question

It seems insane to me

[rough birch]

um whats the problem?

oh

wait so are you asking for help on what?

what do you not understand?

why it is option (3)?

[midnight plank]

@orchid grove Has your question been resolved?

[tidal turret]

|x|<1 has possible values in between [0,1) and (-1,0]

when x approaches 0, the piecewise takes the second case and 0/(1-0)=0 for left and right limit of x approaches 0

so it's continuous at x = 0

[orchid grove]

Ohh thank you very much @tidal turret

[midnight plank]

@orchid grove Has your question been resolved?

[midnight plank]

@orchid grove Has your question been resolved?

[midnight plank]

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[obtuse totem]

are these two not the same thing

yes I think so

[sharp coral]

the critical difference is in using the population vs sample standard deviation

[obtuse totem]

oh i see ok

thanks

[obtuse totem]

so if a question gives observed SD then you would use the 2nd one

t-distribution

[sharp coral]

yes

[midnight plank]

@obtuse totem Has your question been resolved?

[untold rune]

What is the problem here

[midnight plank]

@obtuse totem Has your question been resolved?

[obtuse totem]

.close

[midnight plank]

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[midnight plank]

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[brazen shuttle]

!original

[midnight plank]

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.

[edgy knot]

Determine an equation for the tangent plane to the surface on the top left at the point (1,-1,1)

[brazen shuttle]

ohh

ok so do you know what you need to do to get to the tangent plane

[edgy knot]

do tell.

[brazen shuttle]

huh

[edgy knot]

pretty much means tell me

lol

[brazen shuttle]

ok

soo

or how about we start with the tangent plane

what do you need to make a plane

2 things

[edgy knot]

gradient and the point?

[brazen shuttle]

i mean yes but im saying in general

you need a normal vector and point

[edgy knot]

aha

yeah

[brazen shuttle]

we get that normal vector from the gradient

yessss

ok so find the gradient of your function

[edgy knot]

(3,3,2)

with the point then

(3x^2,3y^2,1+1/z) before point

[brazen shuttle]

i mean im not going to check your work just walk you through it

but thats right

ok so now you take that vector and the point and make a plane

[edgy knot]

(3,3,2)*((x,y,z)-(1,-1,1))?

[lavish venture]

wheezy outta here

[edgy knot]

brrrrrrrrrrbrrrrrrrrrrrrr

[brazen shuttle]

no

[edgy knot]

its the same thing

[lavish venture]

n dot X

lol

[edgy knot]

this proof is what i was looking for though, thanks.

[brazen shuttle]

youre welcome ❤️

[brazen shuttle]

he got the wrong thing..

[edgy knot]

lol what

whats wrong with n dot x

[brazen shuttle]

uh thats not a plane equation

[lavish venture]

well it wasn’t an equation at all

[brazen shuttle]

frs

[edgy knot]

lmao

[brazen shuttle]

but its ok you got it now 🙂

[edgy knot]

it legit works tho !

[brazen shuttle]

uhhh

thats concerning

[edgy knot]

think i get a + for creativity

[brazen shuttle]

sure

[edgy knot]

but ill use the proof this time

[brazen shuttle]

we can go with that

[lavish venture]

bobby

[brazen shuttle]

might as well say ai did it for you 😭

[edgy knot]

nah i can show you gimme a sec

answer sheet

its in swedish but u can just look calculations

[lavish venture]

yea but you didn’t write = 0

[edgy knot]

wdym its there look

[lavish venture]

you didn’t

[lavish venture]

^

[edgy knot]

yeah ur right

anyways, thanks now i know what the formula is.

.close

[midnight plank]

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[midnight plank]

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[proud mist]

This is ez?

7 y + 10x = 8

2 y - 5x = 18

[quiet parcel]

you're not supposed to do it for them

you're supposed to assist

[gloomy snow]

Easy?

Idk if your helping rn

[dawn dagger]

do you know the elimination method?

[gloomy snow]

Nope

[dawn dagger]

what methods are you familiar with?

[gloomy snow]

Trying to use random numbers to replace x and y with until I get the answer

[dawn dagger]

that could take up lot of time potentially

[gloomy snow]

What would you do

[dawn dagger]

I would eliminate one variable from one equation

[gloomy snow]

Oh

[buoyant yoke]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dawn dagger]

(1) 7y + 10x = 8
(2) 2y - 5x = 18

that's your system

[gloomy snow]

Ok

[dawn dagger]

You are allowed to add one equation on to the other

[quiet parcel]

Dip, let the person that started finish their job. (@dawn dagger)

I think it's fair.

[gloomy snow]

So for number two can I do 2(1)-5(4) to get 18

[dawn dagger]

For example getting rid of 10x seems reasonable because the other equation contains 5x which is a factor of 10x

So you could multiply the 2nd equation by 2 and add it to the first

[gloomy snow]

Oh ok

[quiet parcel]

bruh

mods?

you're not supposed to solve it

for them

[gloomy snow]

Uh

[quiet parcel]

i said 5 times

[dawn dagger]

This is difficult because you could find infinite many variations that still doesn't satisfy the other one

[dawn dagger]

you can ping mods

[gloomy snow]

Which is the answer

[dawn dagger]

Let's multiply the equation (2) by 2

[gloomy snow]

Ok

[dawn dagger]

what do you get then

from
(1) 7y + 10x = 8
(2) 2y - 5x = 18

what is the new system

[gloomy snow]

Uh

I don’t understand math as good

[dawn dagger]

<@&268886789983436800>

[quiet parcel]

2y - 5x - 18 = 0 and then you said 2y = 18 - 5x

no sense

it was incorrect

you cant give solutions

and @dawn daggeralready started

so let him

finish

[buoyant yoke]

you can do it on your own but not give the solution here

[gloomy snow]

You don’t even have the helpful role

[paper inlet]

erm so can return to problem now (It was wrong though fyi)

[dawn dagger]

So again we were at
(1) 7y + 10x = 8
(2) 2y - 5x = 18

[gloomy snow]

Yes

[dawn dagger]

The next step was to multiply the 2nd equation by 2

[quiet parcel]

I already told you.

[gloomy snow]

Ok

[dawn dagger]

what would you get?

[quiet parcel]

No it's 18 + 5x

Stop misleading him.

And stop giving answers.

[paper inlet]

<@&268886789983436800>

[gloomy snow]

[quiet parcel]

<@&268886789983436800>

[paper inlet]

You literally gave the answer and it wasnt even correct

ok im gonna break it down for you rq

[gloomy snow]

K

[hollow sail]

just put one stranger on one side and remplace this stranger with the other stranger in the other equation

a lot of stranger

put all x on a side and all y on a side

w'ere extinguished

[paper inlet]

wrong person

[hollow sail]

ohh

@gloomy snow still need help?

[hollow sail]

what did you tell him btw

[paper inlet]

I was going to tell them to put it in the form
$ax_1+by_1=c_1
ax_2+by_2=c_2$

[grand pond]

Banana Steeler

[paper inlet]

but they probably left after the spam

[midnight plank]

@gloomy snow Has your question been resolved?

[midnight plank]

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[midnight plank]

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[alpine oracle]

please help me with this problem

[alpine oracle]

i tried to bound lcm

what should i do after this

<@&286206848099549185>

[midnight plank]

@alpine oracle Has your question been resolved?

[midnight plank]

@alpine oracle Has your question been resolved?

[midnight plank]

@alpine oracle Has your question been resolved?

[alpine oracle]

<@&286206848099549185>

pls help

[spring wave]

hm

[alpine oracle]

pls help me

[spring wave]

so wlog, a<=b<=c

[spring wave]

lemme think of a soln

[alpine oracle]

yeah!

[spring wave]

so
ab<=ac<=bc

[onyx dawn]

use lcmab * gcdab = ab

[spring wave]

and we know
bc+ca<=3ab

[alpine oracle]

yes

[spring wave]

that doesnt always work for more than 2 numbers

hm

lcm(a,b,c)=lcm(lcm(a,b),c)=lcm(ab/gcd(a,b),c)=abc/gcd(ab,bc,ca)

that works tho

4*gcd(ab,bc,ca)=ab+bc+ca

let us say

gcd(ab,bc,ca) = k

[alpine oracle]

ok

[alpine oracle]

can u tell how u reached to 4 equality from 3rd one

[spring wave]

basically

lcm(ab/gcd(a,b),c)

=abc/gcd(a,b)gcd(ab/gcd(a,b),c)

=

abc/(gcd(ab,c*gcd(a,b)))

=abc/gcd(ab,gcd(ac,bc))

=abc/gcd(ab,ac,bc)

[alpine oracle]

ohk

tks

[spring wave]

so

k >= (3/4)ab

[alpine oracle]

ok

[spring wave]

sry for delay

but

from here

we know

actually this is wrong

[spring wave]

this

[spring wave]

this

[alpine oracle]

i dont understood actually how it came

[spring wave]

i mean

[spring wave]

did you get this?

[alpine oracle]

yeah

[spring wave]

so

4abc/(ab+bc+ca) = k

k | ab, k|bc, k|ca

so

k^3|a^2b^2c^2

[alpine oracle]

ok

[spring wave]

do you get this?

[alpine oracle]

ya

i ll lyk when i will dont understand anything at that point

[spring wave]

so

k|abc

let ab = ki ac = kj, bc = kn

[alpine oracle]

ok

[spring wave]

kin/j = b^2

[alpine oracle]

ok

[spring wave]

kij/n = a^2

kjn/i = c^2

[alpine oracle]

ok

[spring wave]

4(k^3ijn)^1/2=k^2i+k^2j+k^2n

[alpine oracle]

ok

[spring wave]

4(ijn)^1/2 = (i+j+n)k^1/2

i<=j<=n

[alpine oracle]

ya

[spring wave]

16(ijn)/(i+j+n)^2 = k

(16/9)n>=k

since

[alpine oracle]

ok

[spring wave]

16(ijn)/(i+j+n)^2 is maximum at i = j = n

[alpine oracle]

ya

[spring wave]

hm

given ab and ac, what can we say about bc ?

only

that

(ab*ac)/bc is a perfect squae

or

their product divided by a perfect square

is bc

hm

ki, kij, kn, can be any values such that k^3ijn is a perfect square

or

kijn is a perfect square

[alpine oracle]

yes

[spring wave]

we also know

that

is p|i p|j p does not divide n

[spring wave]

what ideas so you have?

[alpine oracle]

i was thinking of some inequality ideas like trying to bound them

so that i could limit the value and then make some equation in ab, bc and ac

[spring wave]

oh this is kinda useless, we know k <= n

[spring wave]

what did you get yet?

[alpine oracle]

i am trying different things like gcd >= a and trying to manipulate all that

[wicked tide]

I actually think i could provide some help, can i do that? Even tho i am not a helper

[next rover]

it's totally fine

as long as you don't copypaste from a chat bot

[midnight plank]

@alpine oracle Has your question been resolved?

[buoyant yoke]

you get helper from helping, so if you never help then you can never become a helper

[hollow sail]

https://tenor.com/view/gojo-absolute-cinema-gif-8932131737123141805

[midnight plank]

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[maiden forge]

Hello, Im studying for a trigonometry test using some old tests to practice, and I cant figure this one question out:
From a ship located at a point P in the sea, two lighthouses situated on land at points A and B respectively are observed. The ship sails 4 km toward lighthouse A, reaching a point D in the sea (on the segment that connects P and A). The angle ∠APB is π/4​[rad] and the angle ∠ADB is π/3​[rad].

a) Draw a diagram of the described situation.
b) Calculate the distance between points
c) Calculate the distance between points

[maiden forge]

Im gonna edit the question cause the text got all messed up

This is what I think the problem looks like

[round parcel]

That looks fine. Parts b and c look the same, though, so you may have left something out of them.

[maiden forge]

Well yes, Its only to illustrate, those are not the real proportions

[round parcel]

No, I mean you say "b) Calculate the distance between points" and c is the same words.

[maiden forge]

oooh

1 sec

it was
b) Calculate the distance between points D and B.
c) Calculate the distance between points P and B.

[round parcel]

OK, your part a looks good. Have you made any progress on part b?

[maiden forge]

I tried using the cos theorem ending up with d = sqrt(a^2-4sqrt(2)a + 16)

then I got a mental block

[round parcel]

OK, but that has the problem that you don't know a yet.

[maiden forge]

yep

[round parcel]

What you can do is use the BDA angle to get the BDP angle.

[maiden forge]

The problem with that is that, since its just a diagram, the triangle I made might not be actually representing the actual physical situation

[round parcel]

Right, the lengths and angles are probably not drawn exactly.

[maiden forge]

Im lowkey thinking the problem might be wrong in some way

but I dont know, it was a test question

[round parcel]

But the ship was heading directly toward A, right?

[maiden forge]

I mean yes

[round parcel]

So, the angle PDA will be 180 degrees or pi radians.

Because it's a straight line and those are 180 degrees.

So, BDA + BDP = pi.

[maiden forge]

You might be right

My head was trying to imagine the boat just heading towards A

cuz since its water

its not perfect

[round parcel]

Right, but the idea here is that we're pretending it's perfect.

[maiden forge]

True true

Imma try with that

[round parcel]

In the real world, you might have to make some adjustments.

[maiden forge]

I hate when problems like these dont specify stuff like that

for me, it was hard to even get to the diagram I sent

I firstly thought about making it from a direct view

like from the coast

instead of looking down at the situation

[round parcel]

Well, in math classes with trigonometry or geometry, the idea is that everything is perfect. Then, in physics classes, they start to add in the imperfections of the real world, like maybe the water is flowing in a certain direction at a certain speed or something.

Oh, I see.

You want to get it into the form of a triangle to use trigonometry.

So, you want it to be looking straight down at it.

Or straight up.

[maiden forge]

That is really helpful

Imma add it to my thought process

[round parcel]

OK.

[fading ore]

i mean the main thing to remember about math is that its an abstraction of a more general question

thats why its so "perfect"

[maiden forge]

I hate radians, Imma change everything to degrees

[round parcel]

Radians will be important later, like if you get into calculus, they use radians instead of degrees.

[maiden forge]

TY @round parcel, that was it.

[midnight plank]

@maiden forge Has your question been resolved?

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[storm spindle]

.close

[midnight plank]

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[cursive granite]

can someone help with this question

[cursive granite]

i mean these two problems

[kindred dagger]

What are you trying to do here

-# also did you take a screenshot of a photo 🤣

[kindred dagger]

Your theta should be in the bottom right (because thats where john starts). this screws up your opposite and adjacent sides, which screws up your trigonometry

[midnight plank]

@cursive granite Has your question been resolved?

[cursive granite]

yes i did it was mine because i couldn't retake it sorry 🤣

[cursive granite]

ohh

wait

right

got it

and you place the bearing at theta to find the displacement

[midnight plank]

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[cursive granite]

.reopen

[midnight plank]

✅

[cursive granite]

.close

[midnight plank]

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[tidal turret]

Let $n$ be a positive integer. If $a\equiv (3^{2n}+4)^{-1}\pmod{9}$, what is the remainder when $a$ is divided by $9$?

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[heady plume]

!show

[midnight plank]

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

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

[viral dagger]

cant you just plug in like n=1

[tidal turret]

1

cant you just plug in like n=1
wlog?

[fading ore]

No, there's more to it

[viral dagger]

ok so ig it would be interms of n

[west iron]

but it always has the same value mod 9 regardless of n

[tidal turret]

I don't know how modular inverse works

[west iron]

it's the thing that you multiply to get 1

[tidal turret]

is it bad if I say that

[west iron]

so like the inverse of 2 is 5

[heady plume]

9^n is 0 mod 9

[west iron]

since 2*5 is 1 mod 9

[west iron]

?

[tidal turret]

because n is positive integer

[heady plume]

What you're asked is to find an a such that a(9^n + 4) is 1 (mod 9)

[fading ore]

Well this means we need to find a*9^n + 4a is congruent to 1 mod 9

Obvious that first term is 0 mod 9

So ultimately 4a is congruent to 1 mod 9

So a is congruent to 4 inverse mod 9

What number times 4 gives you a number that is 1 mod 9

4 * 7 = 28 = 27 + 1

So the answer is 7

[west iron]

oh right lol i'm being dumb

[fading ore]

Nah it's ok this is some like modern algebra type stuff

Not really but kinda

[west iron]

nah my brain just didn't work

[fading ore]

Fair enough lol

[west iron]

also i'm not the original helpee lol

[fading ore]

I gotta be honest tho I wasn't aware that the -1 exponent here just meant that we need a number that multiples to 1 mod n

[viral dagger]

ok but n=1 means a×13=4a=1 mod 9 => a=7

a is probably constant so done :3

[tidal turret]

a = 4^-1 mod 9

[fading ore]

I suppose but how do we know that holds for all n

[fading ore]

4^-1 mod 9 is the same as 7 mod 9

[viral dagger]

you guess (try other values)

[tidal turret]

yeah, is just hard to grasp but I think I get what the modular inverse is now

[fading ore]

Alright good stuff

[midnight plank]

@tidal turret Has your question been resolved?

[midnight plank]

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[midnight plank]

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Remember:
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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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• Be polite and have a nice day!

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[twilit field]

Prove that every convergent sequence is a Cauchy sequence.
\
Proof: Let $(a_n)$ be the seqeunce. As it converges, it follows that $\forall n >N, N \in \N$, $\abs{a_n-a}< \varepsilon; \forall \varepsilon >0$. Let $m>N$. It then follows thats $\abs{a_n-a_m+ a_m - a} < \varepsilon \implies \abs{a_n-a+a -a_m} + \abs{a_m-a} < \varepsilon \implies \abs{a_n-a}+ \abs{a-a_m}+\abs{a_m-a}<\varepsilon$.

I'm lost beyond this

[grand pond]

math rocks(wai)

[nova yoke]

right general idea but try starting with $|a_n - a_m| = |a_n - a + a - a_m|$

[grand pond]

Bungo

[twilit field]

Hmm, yeah, that makes more sense

$\abs{a_n-a} +\abs{a_m-a} < \varepsilon/2 + \varepsilon /2 = \varepsilon$

[grand pond]

math rocks(wai)

[twilit field]

and we're done

[nova yoke]

yep

[twilit field]

Cool

thanks

.close

[midnight plank]

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[lethal path]

the idea is that $a_n$ must be within $\epsilon/2$ of $a$ for all $n > N$

so the supremum of the distance between $a_m, a_n$ for any $m, n > N$ must be $(a + \epsilon/2) - (a - \epsilon/2) = \epsilon$

[grand pond]

south

[lethal path]

I like thinking about it this way

[twilit field]

Got it

thanks

[midnight plank]

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[gusty birch]

Hi! I need help understanding the axiom of regularity in ZFC.

"Every non-empty set x contains a member y such that x and y are disjoint sets."

How is y and x disjoint if y is contained within x?

[gusty birch]

∀x[∃a(a ∈ x) ⇒ ∃y(y ∈ x∧¬∃z(z ∈ y∧z ∈ x))]
This is the formal definition but to be fair it just makes it more complicated...

[sharp coral]

for example let's say we have A = {1, 2, {3,4}}. A is disjoint from {3,4} because it does not contain 3 or 4

[gusty birch]

How about A = {1}

1 isnt even a set

how can i say that A is disjoint to 1

[sharp coral]

strictly speaking in ZFC we define the integers as sets

[gusty birch]

Or do we define 1 as a set

[gusty birch]

oh

this is from the construction of N?

so 1 would be {o,{o}}?

o being null

[sharp coral]

that's 2, but the same idea applies

[gusty birch]

oh yea my bad

is everything just a set then?

like what if i have A = {apple}

or is thay just not a set under ZFC

[sharp coral]

for the purposes of ZFC everything being discussed is a set

[dull girder]

@gusty birch trivially this is to circumvent Russell's paradox.

[gusty birch]

oh so like in this case apple would be a set containing some amount of elements which for sure isnt contained in A since A contains only apple

[gusty birch]

Yes its written in the notes

[dull girder]

Which part of it's complicated?

[gusty birch]

i didnt really understand the disjoint sets part

but now i think i get it

[sharp coral]

and we're guaranteed that apple cannot contain apple by the axiom

[gusty birch]

ohh right

[ember bane]

by regularity?

[sharp coral]

yes, we're guaranteed that apple is disjoint from {apple} (whatever those things are) by regularity

[ember bane]

but if apple = {apple, pear}, then apple is regular and it contains apple

(given that pear is disjoint from apple)

[gusty birch]

Axiom schema of specification with the axiom of regularity toghether eliminate russels paradox iirc

i'll close now, thanks for the help :)

.close

[sharp coral]

with the axiom of pairing we know that {apple} is a set (given apple is) then it must be that apple is disjoint with {apple} (so it cannot contain itself)

[midnight plank]

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[midnight plank]

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[twilit coyote]

Consider in $\mathbb{R}$ the equation (E): x^3 + 2 = 3x

1. Verify that 1 is a solution of (E).

2. Hence, solve (E).

[grand pond]

prodigydude
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit coyote]

For question 2, is my proof valid:

.close

[midnight plank]

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[twilit coyote]

found my mistakes

[midnight plank]

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[oak granite]

$\int_0^{\frac{\pi}{2}} \frac{x}{tan(x)} dx$

[grand pond]

! AzizKitten

[oak granite]

help pls

[buoyant yoke]

!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

[oak granite]

I have no idea how to solve it

[lone hound]

im not sure either but i think i can give ideas

try putting the tan up

make it a cot

and then use uh

integration by parts

differentiate x and integrate cot

u will ln|sinx| as the integral of cot

and then try using substitution, u = sinx

i gtg

[oak granite]

ok I'll try

[delicate sage]

Kings rule

[oak granite]

what is that?

[fallow scarab]

https://www.magnamathematica.com/2024/04/23/kings-rule/

[delicate sage]

int 0 to pi/2 x/tanx = int 0 to pi/2 , pi/2-x/cotx

[oak granite]

oh

[delicate sage]

Then just add the two integrals

Wait

It's 1-tan²x !

One moment

[lone hound]

i dont think u can use kings rule

u'll get tan(pi/2 - x²) in the denominator

tan(a-b) = (tana - tanb)/1 + tanatanb

[fallow scarab]

Memes go in #chill

[lone hound]

and tan pi /2 is literally undefined or infinity

whichever u prefer

@delicate sage

wtf

[delicate sage]

x/tanx = xcosx/sinx = x(cos²z-sin²z)/2sinz cosz, z = x/2
= z(cotz-tanz)
Now apply kings rule.

This is from 0 to pi/4

Ig this should solve the integral, if not then ping me, shall try again

[lone hound]

what is z

[delicate sage]

I haven't perfectly computed, I meant to apply the substitution x/2 = z

so dx = 2dz

[oak granite]

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(cot\left(\frac{\pi}{4}-z\right) - tan\left(\frac{\pi}{4}-z\right)\right) dz$

[grand pond]

! AzizKitten

[oak granite]

like that?

[delicate sage]

tan(pi/4 -z) = 1-tanz/(1+tanz

[lone hound]

XOR

wouldn't you have to convert the limits to the u world?

[delicate sage]

0 to pi/4 he has done it

[lone hound]

wonderful

did my method work?

I gotta study english so could u pls check if mine works

[oak granite]

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(cot\left(\frac{\pi}{4}-z\right) - \frac{1-tan(z)}{1+tan(z)}\right) dz$

[delicate sage]

Hey i want you to write cot in the same way

[grand pond]

! AzizKitten

[lone hound]

x/tanx = xcotx,
apply integration by parts, dif x and inte cotx and then perform u sub with u = sinx

[delicate sage]

Integration of ln(sinx) seems rather difficult

[oak granite]

then?

[lone hound]

oof

didn't think about then

lnu / sqrt(1-u²) du

right

what about adding a parameter?

[oak granite]

cot(pi/4 - z) = (1+tan(z))/(1-tan(z))?

[lone hound]

I'(t) = u^t lnu / sqrt(1-u²) du

integrate with respect to t

and then perform trig sub

[oak granite]

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(\frac{1+tan(z)}{1-tan(z)} - \frac{1-tan(z)}{1+tan(z)}\right) dz$

[grand pond]

! AzizKitten

[lone hound]

after that differentiate with respect to I

it should work right

[oak granite]

excusme wait wait

how?

u = sin(x) ?

or you canceled that?

@lone hound

[lone hound]

yes

i never canceled

[oak granite]

ok

u = sin(x)

[lone hound]

im kinda studying english rn cuz i got exam tmrw and its 11 PM for me

[oak granite]

du = cos(x) dx

[oak granite]

ok sorry if I dirturb you

[lone hound]

no no

please do continue

I just wont always respond quick

[oak granite]

ok

np

[oak granite]

after that?

[lone hound]

write the function

if im correct

[oak granite]

ln(u) cos(x) du ?

[lone hound]

it would be lnu/ sqrt(1-u²) du

bruh

[oak granite]

ohh

[lone hound]

ur dividing by cosx

not multiplying

[oak granite]

I forgot /

[delicate sage]

integration of ln(sinx) over 0 to pi/2 gives - pi/2 ln2

If I remember correctly

[lone hound]

also u gotta covert the x to u

[oak granite]

our integral will gives pi/2 ln(2)

the original one

x/tanx

[delicate sage]

Maybe !

[oak granite]

yes

I saw it on wolfram alpha

[delicate sage]

if you do IBP we get lnsinx and ik to integrate it, but I wanna try this method

[lone hound]

but i wanna try ln|sinx| method

[oak granite]

ln(u)/sqrt(1-u²) du

[lone hound]

yes

now add a parameter

[delicate sage]

Do you mind if I come back after dinner ?

[lone hound]

u know how that works right

yes please do have dinner

[delicate sage]

Bagel let me tell you, you can't indefinitely integrate that function

You will have to use the properties of definite integrals

[oak granite]

$\int_0^1 \frac{ln(u)}{\sqrt{1-u^2}} du$

[lone hound]

its a definite integral and a parameter should do the trick right

[grand pond]

! AzizKitten

[oak granite]

sure

[lone hound]

ok u know how parameter works right

[oak granite]

what do you mean? sorry I don't udnerstand english too much

[lone hound]

I'(t) = int(0,1) u^t lnu / sqrt(1-u²) du

this way I'(0) gives us the answer we want

[oak granite]

hmm idk that

[lone hound]

idk if it will work but try it

then integrate with respect to t

to get I(t)

then use x = sin(theta) and integrate with respect to x

then differentitate with respect to t

im not sure if it will get u to the answer cuz u gotta find 'c' and i forgot how to do that but its worth a shot

we'll just see where it gets us I guess

[oak granite]

what is theta?

I have a question
does this is correct? and why?
$\int_0^{\frac{\pi}{2}} ln(sin(x)) dx = \int_0^{\frac{\pi}{2}} ln(cos(x)) dx$

[grand pond]

! AzizKitten

[delicate sage]

Yeah !

This is the way to solve that integral

[delicate sage]

We were on this step right ?

Can you simplify it*

4tanz/(1-tan²z) = 2tan2z

(tanz + tanz)/(1-tan²z)

@oak granite there ?

[oak granite]

I am here

[delicate sage]

Can you understand?

This step ?

[oak granite]

let me rewrite it

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(\frac{1+tan(z)}{1-tan(z)} - \frac{1-tan(z)}{1+tan(z)}\right) dz$

[grand pond]

! AzizKitten

[oak granite]

what did you do?
make both fractions in one?

[delicate sage]

(1+tanz)² - (1-tanz)² = 4×1×tanz

[delicate sage]

Yeah

Simple algebra

[oak granite]

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(\frac{4tan(z)}{1-tan^2(z)\right) dz$

$\int_0^{\frac{\pi}{4}}\left(\frac{\pi}{4}-z\right) \left(\frac{4tan(z)}{1-tan^2(z)\right) dz$

[delicate sage]

Hey wait wait

This doesn't yield any results

[oak granite]

how?

[delicate sage]

Leave it

[oak granite]

ok

[delicate sage]

We shall proceed to integrate

ln(sinx) the traditional way

[oak granite]

ok

[delicate sage]

x/tanx

IBP

[oak granite]

look what I got now

[oak granite]

what is that?

[delicate sage]

Gives int 0 to pi/2 lnsinx

[oak granite]

yes

[delicate sage]

Integration by parts

[oak granite]

u = sinx

[oak granite]

oh ok

[delicate sage]

No

[oak granite]

why?

ok

[delicate sage]

ln sinx = ln cosx

[oak granite]

yes

I was asking for that

[delicate sage]

I shall not write int okk

[oak granite]

why is it correct

[oak granite]

.

[delicate sage]

Now I+I = lnsinx + lncosx

[oak granite]

yes i did

[delicate sage]

Kings rule

[oak granite]

oh ok

[delicate sage]

2I = lnsin2x - ln2

Ok so int 0 to pi/2 ln2 gives us pi/2 ln2

That's our answer

We're left to prove int 0 to pi/2 lnsin2x = i

To do this

[oak granite]

let me write on the paper to follow and I'll be back

[delicate sage]

We consider a substitution 2x = z

or dx = dz/2

[oak granite]

ok continue

yes

[delicate sage]

Apply the substitution

What do you get

[lone hound]

OMG

i forgot

always use

int by parts for natural log

[delicate sage]

Int 0 to pi ln(sinz) dz/2

[lone hound]

ahh

im so stupid

its so easy

[oak granite]

no to pi !

[delicate sage]

Now if you know that

[oak granite]

?

z = 2x

[delicate sage]

Yes yes

Sry

[oak granite]

ok

[delicate sage]

Now you know that property?

[oak granite]

int 0 to pi ln(sin z) dz

[delicate sage]

Int 0 to 2a fx dx if

f(a-x) = f(x)

Then int becomes

2 int 0 to a f(x).dx

[oak granite]

nope

[delicate sage]

You recall this property

One moment

[lone hound]

f(a+b-x) = f(x)

[oak granite]

then int 0 to 2a f(a-x) dx

[oak granite]

u = a - x ?

[delicate sage]

No not this

Look at property 6

And 7

[oak granite]

why is it correct

idk it

demonstration

its name?

[delicate sage]

Property 6 demonstrates

[delicate sage]

Idk most probably Jacks rule

Or queens rule, i really mess up

[delicate sage]

Whatever it maybe

The int again becomes

[oak granite]

ok nvm

continue

I'll see it later

[delicate sage]

0 to pi/2 lnsinz dz

You agree ?

[oak granite]

wait

[delicate sage]

Now look at property 1

[oak granite]

we have int 0 to pi ln(sin z) dz right?

[delicate sage]

lnsinzdz/2

[oak granite]

oh yes /2

[delicate sage]

Recall 2x = z so dx = dz/2

[oak granite]

oh I got it

[delicate sage]

So yes we again have int 0 to pi/2 lnsinz dz = lnsizdx = I

[oak granite]

= int 0 to pi/2 ln( sin z ) dz

which is I

[delicate sage]

So 2i-i = -pi/2ln2

[oak granite]

then I = -pi/2 ln(2)

and our answer is -I

[delicate sage]

Now put this in our integral

[oak granite]

so our integral will be equal to pi/2 ln(2)

yeyyyyyy

[delicate sage]

Ours was -I only ?

[oak granite]

yes

[delicate sage]

Ours was xlnsinx - I

[oak granite]

the first part = 0

[delicate sage]

ln 0 is not defined indeed

So we take this consideration

[oak granite]

with limit

[delicate sage]

But always know there's other reason

[delicate sage]

Int 0 to pi/2

When you put x = 0

[oak granite]

yes

it is not defined

[delicate sage]

But no worries it will be zero in this case

[oak granite]

but prolongeable par continuité

thanks anyways

[delicate sage]

Sry i wasted some time

Trying my own shit ! Thanks for following!

[oak granite]

no problem

I have a lot of free time

[delicate sage]

Won't you like to

See how we define this in the case ?

I thought you would probably ask

🥲

[oak granite]

wdym?

[delicate sage]

ln 0 case !

[oak granite]

.

[delicate sage]

I don't understand the language

[oak granite]

it is going to be
pi/2 * ln(sin(pi/2)) - ...
this first equal to 0
then the case of ln0
lim x->0+ -x ln(sin(x))
apply l'Hopital's rule twice and we get the result 0

[oak granite]

I don't know how to say it in english

[delicate sage]

To apply L hopitals

[oak granite]

it is not defined but it has a limit finit

[delicate sage]

What is the necessary condition

[delicate sage]

We need to apply L hopitals only once too

You're partially correct, L hopitals can only be applied

When we have 0/0 or ∞/∞ form

[oak granite]

ln(sin(x))/(1/x)

[delicate sage]

Yeah now you got it !

[oak granite]

I know

I was just lazy to type it

[delicate sage]

Lmao, anyways ! Have a good day !

[oak granite]

but it is clear 0 times infinity can be converted to 0/0 or inf/inf

have a good day!

thank you

[midnight plank]

@oak granite Has your question been resolved?

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[green cipher]

ok so here someone on yt solved it this way

probability of the 4 samples having defective bikes=5/30 X 4/29 X 3/28 X 2/27=1/5481

probability of having at most 3 defective bikes
1-1/5481=5480/5481
and i really dont understand how does he solve it that way like it doesnt make sense to me

[green cipher]

i think i should use combinations however idk how exactly am i going to use them

[drifting birch]

The restrictions is that the max amount of defects is 3, ie 4 doesn't work

1 - not P= P

ie the probability of winning and losing, when added should be equal to 1

w+l=1
w=1-l

he found l which is the "loss", the one we don't want by calculating the probability of gaining 4 defects

so if we want to find the probability of the ones that do satisfy the condition we take 1-1/5481

[green cipher]

ohh i get it now idk why but i thought that when we do 1- not P ill get the probabiliy of having 3 defects only and not the probability of having 0 defects and 1 and 2 defects

but i understand now

thank you 🙏

[drifting birch]

anytime♥️

[green cipher]

.close

[midnight plank]

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[midnight plank]

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[green cipher]

here question 21 why isnt the answer D (T4)

bc
if we expand (2x+1/x^2) ^3 the forth term will be

(2x)^0 + (1/x^2) ^3

[livid python]

Because it’s not the independent term

[grim vector]

You will have 1 + 1/x⁶

[midnight plank]

@green cipher Has your question been resolved?

[green cipher]

Oh okay thanks guys 🙏

[midnight plank]

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[opaque wyvern]

Hello I've been asking for help for like an hour now can you help?

[midnight plank]

@opaque wyvern Has your question been resolved?

[midnight plank]

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[vocal briar]

can i js say the probability of a boy and a boy is 1/4 and then since the probability of boy girl is 1/4 + boy boy is 1/4 + girl boy is 1/4, its [1/4]/[3/4]

[vocal briar]

so 1/3

im not sure if that reasoning would be correct

like cna you add branch probabaility like that

basically boy boy / all combos without girl girl ofc

[royal sable]

i am nearly 100% sure that since those are two independent things the probability should just be 50%

if i throw a coin once and it lands heads, there is still a 50 percent chance it lands heads the next time

[viral dagger]

it has chances of GG, BG, GB, BB, GG isnt included so BG GB BB, BB is only one out of 3 so 1/3? or am i dumb

[tawdry laurel]

Think so

[royal sable]

potentially my reasoning is incorrect

[subtle kiln]

yas

its 1/2 * 1/2

[viral dagger]

thats only true if you know the first one is heads no? for this question wouldnt it be more like 2 coins are thrown and one of them is heads, not the first one is heads

[midnight plank]

@vocal briar Has your question been resolved?

[vocal briar]

are u talking to me?

theres no coins in the question

[midnight plank]

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[royal sable]

pretty sure i'm wrong, but not for that reason

[vocal briar]

oh wait

im stupid

lol

.reopen

[midnight plank]

✅

[vocal briar]

i read your msesage wrong

.close

[midnight plank]

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[midnight plank]

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[twilit field]

If $(a_n)$ is a cauchy seqeunce, determine if $\floor{a_n}$ froms a cauchy seqeunce

[twilit field]

oops

[grand pond]

What a wonderful world!

[twilit field]

Kind of unsure as to how I'd progress with a proof in this case

[west iron]

Makes sense, considering there isn't one

[twilit field]

hmm?

so a counterexample

[carmine sigil]

(-1)^n / n

[twilit field]

oh yeah

that works

thanks

.close

[midnight plank]

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[visual elk]

think what happens after folding

the point c goes to the midpoint of AD, which is E

then you get a right triangle

[tidal turret]

how so?

[visual elk]

take a square papaer in your hand

and fold it the way it says

you'll understand well in that way

[tidal turret]

I don't have access to a literal piece of paper atm

hmm

[visual elk]

okay

ill draw a picture for ya

[tidal turret]

ok, ty

[visual elk]

look, after folding, CF = EF

[tidal turret]

how did that happened

[visual elk]

by folding

such that the corner C goes to the midpoint E

[tidal turret]

yeah i get that

[visual elk]

we knew that CF + FD = 6

[tidal turret]

how did G happened

hmm is aproximate

[visual elk]

G is basically the point B

after folding

[visual elk]

So, CF = EF
CF + FD = 6 (previously)
so EF + FD = 6

now EF = 6 - FD

now use the pythagorean theorem and solve for FD

[tidal turret]

(6-FD)^2 = FD^2 + ED^2

[visual elk]

yeah pretty much

you know ED is just 6/2 = 3

[tidal turret]

right

(6-FD)^2 = FD^2 + 9

[visual elk]

yeah

[tidal turret]

36 -12FD +FD^2 = FD^2 + 9

[heady plume]

Or you can use this where ∆CXF ~ ∆CBE

So ||CX/CF = CB/CE and you know values for CX, CB, CE||

[tidal turret]

CX we don't know

[heady plume]

CX = 1/2 CE

Because GF is perpendicular bisector of CE

Due to the folding condition

[tidal turret]

CX/CF=CE/CB or other way around

[heady plume]

Other way

Because CXF is the right angle here

[tidal turret]

ohh

[visual elk]

but isn't this approach a bit more complicated than just doing pythagoran theorem?

[tidal turret]

we dont know CE

[heady plume]

Nay, it's the same whether you write (6 - FD)² = FD² + 9 or CX/CF = CB/CE.
It's just that the OP has been posting Qs relevant to the Similarity topic so I wanted to greet them with this other idea

[heady plume]

CE² = 3² + 6² , notice E is a midpoint

[visual elk]

your point is valid

[tidal turret]

,w 36 -12x= + 9

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

I appreciate it guys

[tidal turret]

I appreciate the idea but i wanted to ask, how do you know CX = 1/2 CE

I like the basic proportionality theorem but i am bad at using similar triangles

how come I never notice the similar triangles 🤭

.solved

[midnight plank]

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[midnight plank]

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[twilit field]

A sequence is said to be pseudo -Cauchy if $\forall \varepsilon >0$ , there exists an $N$ such that if $n \geq N$. Then $\abs{s_{n+1}-s_n} < \varepsilon$.
\
Determine if pseudo-Cauchy sequences are bounded.

[grand pond]

What a wonderful world!

[twilit field]

KInd of lost here

[round parcel]

Try to ruin it.

[twilit field]

I'm trying to

[blissful totem]

blast it with a water gun

[round parcel]

Let's say we have n = 1, 2, 3, 4, 5, 6, 7, 8, etc.

The distance from a_1 to a_2 is 1, for example.

The distance from a_2 to a_3 is 1/2. The distance from a_3 to a_4 is 1/2.

The distance from a_4 to a_5, a_5 to a_6, a_6 to a_7 are all 1/3.

And so forth.

a_1 to a_2 moves us up by 1.

a_2 to a_4 moves us up by 1.

a_4 to a_7 moves us up by 1.

[twilit field]

ooh

1,,2,2,3,3,... works

right

thanks

.close

[midnight plank]

Channel closed

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[midnight plank]

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[tidal turret]

what

the guy deleted his message

[last arch]

Me

I thought you didn't know what a triangle inequality was

[tidal turret]

8 + x > 10
10 + x > 8
18 > x

12 + x > 16
28 > x
16 + x >12

4 < x < 18

x in (4,18)

17-5+1

,calc 17-5+1

[grand pond]

Result:

13

[tidal turret]

I appreciate the help

.solved

[midnight plank]

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Use .reopen if this was a mistake.

[midnight plank]

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[novel flax]

Can someone tell the answer

im getting 22

f(0) = 10
f'(0) = 12

is what i got

[visual tiger]

How did you compute it? It might help showing your work

[visual tiger]

did you compute f(x) for all x (tiresome) or did you go for something else?

[novel flax]

i put the value of x as 0

for f(0)

for f'(0), i differentiated all of them and then put the value of x as 0

[visual tiger]

uhhhh

what do you mean by that

just a quick tip

if f(x) = g(x)h(x)

is the derivative f'(x) = g'(x)h'(x)?

[novel flax]

as in i differentiated all the terms in f(x), then put the value of x as 0 and found the value of the determinant

[novel flax]

no

[visual tiger]

well

[novel flax]

its g'(x)h(x) + g(x)h'(x)

ryt?

[visual tiger]

yep

the problem is

[visual tiger]

when you tell me that

[visual tiger]

you kinda did this

[novel flax]

ahhhhhhh

[visual tiger]

depending on if you look at it in terms of rows or columns or whatever

but you have to differentiate row by row (or column by column)

[novel flax]

icic

[visual tiger]

so if det(A) is det(Row1,Row2,Row3)