help-49

[midnight plank]

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[naive flint]

how is mass kg and amu

the same?

there are equal somehow?

[hallow scroll]

Hello I was doing some practice problems and I keep getting two answers for one so can someone help

“Find the area bounded by y=x^3 and y=x^2-2x over (-1,1)”

[naive flint]

what the

[sharp coral]

because that's the conversion between kg and amu?

[midnight plank]

@naive flint Has your question been resolved?

[naive flint]

like

is there a process

im jsut use to

si prefixes

[sharp coral]

,w 1 amu to kg

[grand pond]

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Requested by cloudedhorizons

[naive flint]

oh

perfect thnx

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[burnt hornet]

hello! im trying to figure out why this could be marked as wrong

[burnt hornet]

heres the other ones for context.

[brazen palm]

would it just be "missing a •"

[soft garnet]

i think ur supposed to keep it as just 7m+10 m^2

[burnt hornet]

hm alright lemme try

[soft garnet]

cuz you factored it when all the other answers are simplified

[lavish venture]

maybe they want it to be m(7+10m)

[soft garnet]

communative property

[lavish venture]

no kidding

[burnt hornet]

???

i think my thing is just messed up 😭

[lavish venture]

but they might have coded the answer to be accepted as ^

[lavish venture]

this isn’t simplified

you just rewrote it

[soft garnet]

thats as far as u can simplify it

[lavish venture]

samm

please leave

[burnt hornet]

yeah i had it written what i thought was correct beforehand but then rewrote it per the advice i got and now the answers locked and its telling me its 999 lol

oh well

[lavish venture]

7m+10m^2 is m(7+10m) this is a gcf factoring assignment

i assume

[soft garnet]

the answer is 999 and ur talking about gcf

telling me to leave

tf

[midnight plank]

@burnt hornet Has your question been resolved?

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[potent wind]

need calc help on hw i understand the concept but just need some guidance

[potent wind]

To solve $ \int^{3}_{1}5xln(x)dx$ ik that $u=ln(x)$ and $dv=5x$, meaning $du=(1/x)dx)$ and $v=(5x^2)/2$

wait i wrote that completely wrong let me fix the notation

[grand pond]

fre

[chilly cobalt]

seems alright

[potent wind]

that part is intuitive

idk why i get tripped up next tho

i got $ln(x)\frac{5x^2}{2}-\int\frac{5x^2}{2}\frac{1}{x}$

im so bad at this notation thing wow

one sec

[grand pond]

fre

[potent wind]

there we go

this part is correct right

cause after this is where i get tripped up

anyone?

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[midnight plank]

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[midnight plank]

Please don't occupy multiple help channels.

[graceful frost]

this is what I did but I feel like I messed up

[midnight plank]

@graceful frost Has your question been resolved?

[fallow scarab]

!1c

[midnight plank]

Please stick to your channel.

[fallow scarab]

#help-25

[midnight plank]

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[quiet river]

the number of vectors in a basis is fixed for a vector space

[quiet river]

it is annoying trying to prove this bc it could be like the linear independence part can be true given two basis w m and n vectors

and the one w less vectors is the subspace

how can i use unique representation

help pls

[midnight plank]

@quiet river Has your question been resolved?

[quiet river]

https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/null-column-space/v/proof-any-subspace-basis-has-same-number-of-elements

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[cosmic river]

can someone help me figure out where I went wrong please

[cosmic river]

My work

Ok I think I see it derivative of 1 would be u not theta

[dawn dagger]

antiderivative, but yes

else everything looked fine

[cosmic river]

@dawn dagger yo

do you understand why it is 1/3

you have 216 * (sec^3/3)

where sec = sqrt(x^2+36)/6

oh baby I see it

Math is king

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[dawn dagger]

my absence helped you

[cosmic river]

Lol

I'm really liking the integration parts its fun math

I hate the disks and washers and cylindrical shells stuff

is that common

[dawn dagger]

imagine you would have to find the volume around the y=x axis

[cosmic river]

In other words yes?

[midnight plank]

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[odd raptor]

hello

[odd raptor]

this question is about transformations of functions

im just wondering where the vertical shift of 1 came from

[fallow scarab]

there is no shift

[odd raptor]

oh

how did they figure out that z is 1

[fallow scarab]

that's just e^(-(0^2 + 0^2))

[odd raptor]

oh

[fallow scarab]

G(0, 0) = 1

[odd raptor]

which is just 1

got it

thank you

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[floral bronze]

I don’t get why the second set’s the way it is

[last slate]

Why?

it asks u for which x values do the y-values of the function is more than or equal to 1

[midnight plank]

@floral bronze Has your question been resolved?

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[last slate]

I need helpp

[carmine sigil]

What is g(3)?

[slim steeple]

Just look at y values

[last slate]

i don’t understand the concept & i can’t find the answer

[carmine sigil]

You don't understand how to read the graph?

ok, I'll give an example. g(1) = 0

because if we look at the graph of g, at x=1, the line of g(x) it at y=0

However, you also need to know that an open circle means that the graph is not defined at that point

[last slate]

oh

[carmine sigil]

whereas a closed circle means it is

So based on this, what is g(3)?

[neon dome]

heello i need to learn absolute value and i barely undeeerstand a thing is anyone down for vc?

[carmine sigil]

!occupied

[midnight plank]

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

[last slate]

umm i think it’s

undefined right?

[carmine sigil]

It is

[last slate]

yay

[carmine sigil]

so we know that undefined/anything = undefined

[last slate]

so put undefined?

[carmine sigil]

but just for practice.

what is f(-5)?

[last slate]

it’s a closed circle soo

defined?

[carmine sigil]

it is defined

[last slate]

yayy ty

[carmine sigil]

but what is it defined as?

what's the y value?

[last slate]

-5..?

[carmine sigil]

yes exactly

f(-5) = -5

[last slate]

how about this

[carmine sigil]

ok what are these two values?

what is f(3) and g(7)?

[last slate]

idk honestly

wait is it the same concept as last time

[carmine sigil]

yes

[last slate]

ohh

both are defined

bc closed circles

[carmine sigil]

is that true?

[last slate]

wait im wrong😭

[carmine sigil]

ok, if the circle was closed, what would f(3) be?

[last slate]

=1?

[carmine sigil]

f(3) is not 1

f(3) would be -1 if the circle were closed.

[last slate]

ohh

[carmine sigil]

(but because it's not it is undefined.)

[last slate]

that makes sense

i have one more

is it undefined

oh nvm it’s not

[carmine sigil]

it's defined yes.

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[wise tusk]

Hi! How do I solve int x^3e^x dx from 0 to \pi? I’ve solved the int x^3e^x dx using Tabular method. So, do I just have to insert the limits now?

[atomic magnet]

so youve found the indefinite integral, now just use the fundamental theorem of calculus. assuming your answer is given by F(x) - to find the answer we use the FTC, which gives us F(pi) - F(0) to find the answer

[grand pond]

icannotdoanymorecauchy

[fallow scarab]

,tex .FTC1

[grand pond]

riemann

[wise tusk]

thank you

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[carmine tide]

no comprendo

[fallow scarab]

which part don't you understand first

[carmine tide]

how Dp/dt = Bo - Do goest to M = BoPo/Do. i dont rlly get population models i tried watching videos and i know the equations but just dont understand

[fallow scarab]

that's the whole point of the problem

solve the differential equation first

"limiting population" is the limit of P(t) as t goes to infinity

[carmine tide]

do i keep it as dP/dt = aP-bP^2 to solve it?

[fallow scarab]

solve the differential equation means to find P(t) in terms of t

[carmine tide]

yes

[robust isle]

tbh you could just look for equilibria instead of solving the DE fully

[midnight plank]

@carmine tide Has your question been resolved?

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[drifting wigeon]

Why is my answer (b) partially incorrect? Am I misinterpreting the notion of interval of convergence?

[sharp coral]

the ratio test should use absolute value

which means the final inequality should be |x - 3| < 1 instead

[jaunty canopy]

so you got that it has a radius of convergence of 1, but then how did u conclude that the interval of convergence is (-infinity, 4]?

[tribal temple]

(and for all x less than or equal to 4? what do you mean here?)

[midnight plank]

@drifting wigeon Has your question been resolved?

[drifting wigeon]

Problem solved, I blindly applied the absolute value to the initial function incorrectly, which led to the problem. Thanks

[midnight plank]

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[cosmic river]

Hey @carmine sigil still available by any chance?

[carmine sigil]

what's up?

just for a little bit

[cosmic river]

ok I see here u put cot^2 = 1 - csc^2

its csc^2 - 1 right

[carmine sigil]

I'd have to double check.

sin^2 + cos^2 = 1

divide by sin^2

1 + cot^2 = csc^2
cot^2 = csc^2 - 1

yes

my bad

[cosmic river]

np, what confuses me is I got

-[ln(sin)]5,2 - [u^2/2]5,2

I put the problem into integral calculator and the whole answer is only from the first term

the second term isn't part of the answer

do you happen to know why

[carmine sigil]

I haven't sat down to crunch the numbers. There is a possibility that two terms cancel somehow.

[carmine sigil]

also only one of these should be negative.

[cosmic river]

if you got integral of cot csc^2

u = cot
du = -csc^2

[carmine sigil]

oh yeah

dang I'm screwing up a lot today

[cosmic river]

do u know the mechanism behind this

if du contains a negative

but its not part of the problem

[carmine sigil]

no, you are correct.

just use 1 = -1 * -1

[cosmic river]

I don't completely understand the logic behind it

so du = -
Is the logic you are multiplying by a negative you don't have in the problem
so you have to divide by a negative in the problem?
and if you divide by -1 its just -integral?

[carmine sigil]

If you have integral cot u csc^2 u du, and let v = cot u, then dv = -csc^2 u du, so we have

integral cot u csc^2 u du
integral cot u (-1)(-1) csc^2 u du
integral cot u (-1) (-csc^2 u du)
- integral v dv

[cosmic river]

Ok so multiply by -1
I guess dividing by -1 is the same

[carmine sigil]

yeah, same thing

[cosmic river]

so its -[(1/sqrt(x^2-1))^2/2]5,2

the second term

I don't think it cancels

itself

guess I'll just revisit later

[carmine sigil]

can you remind me what the original question is again?

[cosmic river]

is the answer, the same thing obtained from the first term

The two terms are -[ln(sin)]5,2 - [u^2/2]5,2

[worthy wing]

Sin of what

[carmine sigil]

sin(u)

the limits are in terms of x.

and this is following a trig substitution

[worthy wing]

Exactly

[carmine sigil]

but he's not replacing the u values with x values I don't think

[cosmic river]

sin theta because we still have to plug back in sin theta

The two terms are -[ln(sin)]5,2 - [u^2/2]5,2

sin theta = sqrt(x^2-1)/x

-[ln(sin)]5,2
becomes
-[x/sqrt(x^2-1)]5,2

and - [u^2/2]5,2
u is cot
cot theta = 1/sqrt(x^2-1)
becomes -[1/2sqrt(x^2-1)]5,2

this answer is just obtained from the first term -[x/sqrt(x^2-1)]5,2, I don't understand whats happening with the -[1/2sqrt(x^2-1)] 5,2

[midnight plank]

@cosmic river Has your question been resolved?

[midnight plank]

@cosmic river Has your question been resolved?

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[dry crescent]

Just to make sure, if I have this for example(see image)

Is the "maximum a posteriori likelihood "just the highest of these probabilities?

[midnight plank]

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[midnight plank]

@dry crescent Has your question been resolved?

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[alpine gyro]

How do I solve $$\lim_{x \to (\pi/2)^-} (x-\frac{\pi}{2}) \tan x$$

[grand pond]

Good

[alpine gyro]

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[boreal needle]

Why do you distribute the exponents here?

[boreal needle]

How do you know when you are supposed to distribute and when to leave like that

[wintry eagle]

wdym?

[boreal needle]

5 being distributed to the nominator and denaminator

[wintry eagle]

It is not a must, that's just simplifying it

[boreal needle]

How do you know when you should do it and when not

Cuz some of the answers don’t have it in simplified form

[wintry eagle]

I mean typically on an exam, you would get correct answers for both

[boreal needle]

And otherwise that’s one of the answers there

Well that’s answer A and that’s the wrong one

[wintry eagle]

Wrong according to answer sheet or?

[boreal needle]

Ye

According to the answer key it’s C

[wintry eagle]

I mean the only thing I can tell you is that it doesn't matter if you distribute the 5 to the nominator or denaminator, both expressions are completely fine

The most important part is that you do the chain rule + product rule correctly, wether you distribute the 5 or not, is just preference of how to write the expression

[boreal needle]

Wait this is the product rule??

Not the quotient??

Then I’m beyond confused now

[wintry eagle]

You can also use quotient yes

[boreal needle]

So I can do it without distributing?

[wintry eagle]

Yes

[boreal needle]

F(x)=Cos(x)
G(x)=1+sin(x)
?

[wintry eagle]

Yes, that's correct, and then you just apply quotient rule

I personally prefer using product rule, but both methods work equally as fine

[boreal needle]

I see thank you lemme try

[wintry eagle]

In math there are multiple ways to write expressions and multiple ways to take derivative

[boreal needle]

It’d confuse me putting it at the numirator haha

[wintry eagle]

No method is more correct than the other, so it's all up to preference really

[boreal needle]

Thank you, I’ll try

[wintry eagle]

haha yea, I know. Some might find it more complicated and confusing

Good luck!

[boreal needle]

Thank you!

[midnight plank]

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[oblique raptor]

help with 30

and 35

<@&286206848099549185>

[midnight plank]

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[sudden harbor]

I think a is correct but I'm stuck on b

[lavish venture]

is this a test

[midnight plank]

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[ancient dragon]

for ii is it that x_4 = -1 and there are no solutions for x1 2 and 3?

[proper anchor]

w

no idea

and

is that basic 8th math

😭

[last slate]

yeah its matrices

[ancient dragon]

if you find it so basic then go for it

[proper anchor]

😭

[proper anchor]

bro baited me

k fine

🙄

[ancient dragon]

I mean whats the point of coming in saying its basic and just going

[proper anchor]

what do u need help with

[ancient dragon]

^

[proper anchor]

ill solve and let u know

hey @ancient dragon

so

[ancient dragon]

hi

[proper anchor]

there are solutions for x1,x2,x3 and x4 but those solutions arent unique cos the system is underdetermined and uh x1,x2,x3 can take many vals by the equation 2x1-x2+3x3 = 9/2, while x4 = -5/2 is fixed

@ancient dragon

dis is why i hate helping

they take so long

[ancient dragon]

?

[proper anchor]

and not say thanks

😤

[ancient dragon]

thank you?

[proper anchor]

well did u understand it

[ancient dragon]

yes

[proper anchor]

see told u it basic math 😄

u want help with iii) since u said thanks

[ancient dragon]

no thank you

Im just saying dont come in say its basic and just leave

if thats your defn of help then I'm good

[proper anchor]

i dindt leave

[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.

[proper anchor]

shut it

[ancient dragon]

its alright I just needed to know that there are sols for the others

thank you

[proper anchor]

k bye now

gl

[ancient dragon]

.close

[midnight plank]

Channel closed

Closed by @ancient dragon

Use .reopen if this was a mistake.

[last slate]

howw to do thi s

[proper anchor]

simplify the numerator

then

use an identity or factorizations

identify the deno//

simplify the whole expression

the answer will be 0

[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.

[proper anchor]

GOD

WHOS THAT

STOP

IM TRYING TO TEACH HIM

OMD

@tacit rose shut it pls

@last slate

[glacial fiber]

.close

[midnight plank]

Available help channel!

Send your question here to claim the channel.

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• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[worthy kestrel]

how blue

[tribal temple]

Size of subgroups divide size of the original group

Oh wait, hang on, did I read the line before that right

[worthy kestrel]

so many reacts 💀

[tribal temple]

Wait hang on, uncross that, H and K are assumed to be (Sylow 2-)subgroups of order 16, and H is a subgroup of N(H intersect K)

[hard umbra]

it literally is just that

[worthy kestrel]

make up your mind 💀

[copper mirage]

Hello umm, is help available here?

[hard umbra]

,, H \subsete N(H \cap K) \subsete G

[worthy kestrel]

not here katheryn

[copper mirage]

Oh okay, sorry

[worthy kestrel]

sorry, the normalizer has to contain H int K. but why would it contain H?

[hard umbra]

H \cap K has index 2 in H

so H \cap K is normal in H

so H is contained in the normaliser

[worthy kestrel]

wait

given subgroups A, B of G. if A in B then A is a subgroup of B?

[hard umbra]

well yeah

thats definition of subgroup

[worthy kestrel]

seems about right

thanks snow bit

.close

[midnight plank]

Channel closed

Closed by @worthy kestrel

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[slate oar]

What's the definition of infinum?
Shouldn't it be 3?

[worthy kestrel]

what is the full question

[slate oar]

Infinum of the subset

3,6,9

[hearty rune]

the results make little sense to me unless theres missing context

[slate oar]

Hmm idk

I don't understand

Greatest lower bound for the subset should be 3 i think

[cedar coral]

infimum should be the greatest of the lower bounds

[slate oar]

Yep, mistake I guess

[midnight plank]

@slate oar Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[viral dagger]

A,B,C,D are points on a circle S, AB is the diameter and C and D is not the diameter and C and D are on opposite sides of AB, let P be the intersection point of the tangent at C and D, let Q be the intersection of AC and BD, let R be the intersection of AD and BC, prove that PQR are on a single line

[viral dagger]

status 1, this question is horrifying

[slim sail]

looks like

[jaunty canopy]

so we know RCDQ is cyclic

and it also kinda looks like P is the centre of that circle

so maybe try doing a phantom point

let P be the centre of that circle

and then prove that it lies on the tangents at C and D

[viral dagger]

whats a phabtom point

[jaunty canopy]

oh basically it's doing it in reverse

so as in if we let P' be the centre of the circle and show that it lies on the tangent at C and at D

then P' must be the intersection of those tangents, so P' = P

[viral dagger]

oo

how do you know RCDP is a cyclic quadilateral?

[jaunty canopy]

RCQ = RDQ

[viral dagger]

huh

how do you know that?

am i stupid lol

[novel lion]

geometry moment

[viral dagger]

ok yea im stupid

looks cyclic quadilateral ish enough

[opaque fjord]

is it not enough to prove that ADQ = 90 degrees?

similar thing with ACR

[viral dagger]

adq is 90 yea

[opaque fjord]

well theres that theorem right

[jaunty canopy]

yeah RCQ = RDQ = 90 because ACB = ADB = 90

[opaque fjord]

if you draw a triangle with two vertices on the diameter, and third vertex on circumference, then that angle is 90 degrees

so if ADQ is 90, then point d will be on the circle with diameter RQ

i dont know if my logic is faulty cuz i havent done geometry proofs in over 3 years, but i think thats abt it?

[viral dagger]

vut how does that prove PQR is in a line

[opaque fjord]

sorry i meant RQ is the diameter

[jaunty canopy]

they haven't

i think they're explaining why RCDQ is cyclic

[opaque fjord]

wait no im stupid that doesnt prove P is on it

[viral dagger]

yea basically ACB=ADB=90=ACR=ADQ

[opaque fjord]

yes you can extend that into cyclic quadrilaterals to prove it

ill try to solve it on paper to get a proper proof, my brain isnt braining rn 😭

[viral dagger]

hmm gow about the tangent to C and D part

wait can you go backwards

[jaunty canopy]

yeah so basically this was what i was saying

[viral dagger]

assume you draw line CP' DP' and extend it, there is only one circle S that fits

[jaunty canopy]

since PD = PC and from the problem statement we know that P lies on RQ, the diameter

that means that P has to be the centre of the circle

it feels like there's more things we can say if we know P is the centre of the circle rather than the intersection of 2 random tangents

so instead let's set P (probably should use P' in ur write-up but i'm lazy)

to be the centre of this circle

and then show that PD and PC are tangent to our circle

(this also means it's hard to "accidentally assume" something is a straight line)

[viral dagger]

wait

isnt the proof done then

cause P'=P veing the center of the circle means RQ is the diameter which means P is on RQ

[jaunty canopy]

we still need to show this

as in we still need to show P'D is tangent to our original circle

also looking at the diagram, it looks like CD is antiparallel to QR

[viral dagger]

antiparallel?

[jaunty canopy]

oh wait yeah nvm it's obvious

[jaunty canopy]

antiparallel means that BRQ = BDC

[jaunty canopy]

which is true by cyclic quads

it feels like we have quite a lot of information actually so proving that P'C is tangent to our original circle shouldn't be too hard hopefully

[viral dagger]

hmm

can you go backwards like

[jaunty canopy]

yeah ok i've done it

[viral dagger]

let P and S be some arbitrary point

[jaunty canopy]

from that point you just angle chase

[viral dagger]

oh

[jaunty canopy]

go on

[viral dagger]

actually nvm

[viral dagger]

i mean thats the title for tbis chapter :(

[jaunty canopy]

wait is this EGMO?

[viral dagger]

no

[jaunty canopy]

nvm

but yeah basically to recap:

• we observed that P must be the centre of RCDQ
• so instead let's set P' to be the centre of RCDQ, we'll show P' = P
• it suffices to show that P'C is tangent to our original circle (cus by symmetry we get the other
  one)
• now angle chase to show that QCP' = ABC

can u finish it off from here?

[viral dagger]

is this lije fir proving P'D is tangent?

[jaunty canopy]

yeah

[viral dagger]

gimme a sec

idk :(

[viral dagger]

i was kinda hoping for some random P you choose C and D and theres only 1 S that is tangent to both, so just find some location for C and D and you get your S

but i dont think this is valid lol

[jaunty canopy]

yeah no

[jaunty canopy]

so we know CRQ = CDB

there's lots of angles equal to CDB

similarly, we know RP = CP

[viral dagger]

wait what

oh nvm

CDB=CRP'=RCP'
RCQ=90=RCP'+QCP'
QCP'=90-CDB

yea idk what after that to angle chase that

CDB=CSB/2
ABC=CBS=180-CSB=180-2CDB

where did i go wrong

[jaunty canopy]

CDB = CAB

similarly, PRC = PCR

[viral dagger]

lemme repost this so i can see

[viral dagger]

ok i went wrong somewhere here

CBA=90-CAB=90-CDB

got it

so P'=P

after that its just the fact that RQ is the diameter so P is on it right?

.close ty

[midnight plank]

Channel closed

Closed by @viral dagger

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• 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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[untold oriole]

I'm trying to follow the following proof. How are they getting to it's false that Pr(e | A & ~T) is many times greater than Pr(e | T)?

[midnight plank]

@untold oriole Has your question been resolved?

[midnight plank]

@untold oriole Has your question been resolved?

[midnight plank]

@untold oriole Has your question been resolved?

[midnight plank]

@untold oriole Has your question been resolved?

[midnight plank]

@untold oriole Has your question been resolved?

[midnight plank]

@untold oriole Has your question been resolved?

[real pivot]

😭

[stuck girder]

Wtf

😭

[midnight plank]

@untold oriole Has your question been resolved?

[candid olive]

hello

IF h=12 and p= 45 and b=66 then what is the value of 4h/p x 78p/h / bb

[rough lance]

Make a new channel

[brazen spire]

your notation is confusing, is the final /bb meant to mean its all multiplied by b^(-2)?

[midnight plank]

@untold oriole Has your question been resolved?

[last slate]

here you don't ask for a solution read #❓how-to-get-help

right now if I understand how you wrote things correctly

you have one fraction with a numerator that are made of two fraction that are multiplied and divided on the value of b multiplied by itself (denominator)

correct?

if so, you will work on the numerator first like the two fractions above

then handle what will remain above with the main denominator which is bb

[fresh sparrow]

it would be helpful if you provided context regarding what each of those variables are (e, B, T)

this looks like some variation of bayes theorem

[untold oriole]

They're arbitrary propositions. Like A, B, C

[fresh sparrow]

$Pr(e | t) = [Pr(B | T)\cdot Pr(e | T \cap B)] + [Pr(B^c | T) \cdot Pr(e | T \cap B^c)]$

[grand pond]

artemetra

[fresh sparrow]

just for myself

[untold oriole]

i believe its an instance of the law of total probability

[fresh sparrow]

it is

[untold oriole]

like per the law of total probaility,

Pr(e) = [Pr(B) x Pr(e | B) + [Pr(~B) x Pr(e | ~B)]

and then i think it works with conditional probability the same

Pr(e | T) = [Pr(B | T) x Pr(e | T & B) + [Pr(~B | T) x Pr(e | T & ~B)]

sorry idk how to write latex

[fresh sparrow]

yeah right

we are given that T happens so we apply it as a condition to every probability

[fresh sparrow]

but i honestly don't get anything written from line 7 and onwards. what is A? why is it restating the exact same thing on line 9 as on line 3?

[untold oriole]

here i got all the text ```Pr(e | T) = [Pr(B | T) x Pr(e | T & B)] + [Pr(~B | T) x Pr(e | T & ~B)]

The idea here is that, P(e | T) will be an average of Pr(e | T & B) and Pr(e | T & ~B). What makes it an average is the fact that Pr(B) + Pr(~B) = 1. But it is not necessarily a straight average because Pr(B | T) and Pr(~B | T) may not each equal 1/2. Thus, for example, if Pr(B | T) = 2 / 3 and Pr(~B | T) = 1 / 3, then Pr(e | T & B) is given twice as much weight as Pr(e | T & ~B) in calculating the average.

Suppose Pr(B | T) is very high, and also that Pr(e | B & T) is much greater than Pr(e | A & ~T). Then it's false that Pr(e | A & ~T) is many times greater than Pr(e | T)

This is because Pr(e | T) = [Pr(B | T) x Pr(~B | T)] + [Pr(~B | T) x Pr(e | T & ~B)]```

[fresh sparrow]

where did you get it

[untold oriole]

Pr(B | T) is very high, i think it means Pr(B | T) > Pr(~B | T)

[fresh sparrow]

nvm it's not restating the same thing

[fresh sparrow]

are you sure there isn't a typo or something on the last line

[untold oriole]

from here

[fresh sparrow]

yikes

[untold oriole]

i replaced HI with A & ~T

[untold oriole]

?

the author is making an argument using technical probability formulas

[fresh sparrow]

this is really difficult to understand

at least for me

[untold oriole]

what added information would you be looking for

(i had the same reaction)

[fresh sparrow]

let ‘O’ stand for a statement reporting what we know about the pain and pleasure in the world.

‘hypothesis of indifference’ or ‘HI’ for short, which states that neither the nature nor the condition of sentient beings on earth is the result of actions performed by benevolent or malevolent non-human persons.

the belief that there is life after death (‘L’ for short).

Suppose that P(L/T) were very high
so we suppose that given theism, there is a high [chance of] belief that there is life after death

not really sure what O is supposed to mean but

and also that P(O/L&T) were much greater than P(O/HI).
and given theism and belief of life after death, we know much more about pain and pleasure in the world than if we were given 'hypothesis of indifference' (???)

okay i think i am starting to sorta get it

[untold oriole]

ideally, it shouldnt matter. the inference should all mathematically follow by virtue of the form of the formulas, irrespect of the semantics of the formula

but i believe O is supposed to mean "there is a mixed distribution of pain and pleasure in the world"

[fresh sparrow]

yeah, ideally. the author is making vague assumptions about certain probabilities and uses them as arguments for inference, mixing probability theory and just sentential logic

[untold oriole]

yeah it is unnusual

[fresh sparrow]

average philosophy text

wut

[untold oriole]

so from my gathering, he's trying to show if Pr(L | T) >> Pr(~L | T), and Pr(O | L & T) >> Pr(O | HI), then Pr(O | T) ≥ Pr(O | HI)

and he left this unstated, by HI entails ~T

so Pr(HI) ≤ Pr(T)

maybe its easier to work with if we focus on a simplified version of the problem

lets say he's instead arguing that if Pr(L | T) >> Pr(~L | T), and Pr(O | L & T) >> Pr(O | ~T), then Pr(O | T) ≥ Pr(O | ~T)

and he's appealing to two theorems for his reasoning:

Pr(O | T) = [Pr(L | T) x Pr(O | T & L)] + [Pr(~L | T) x Pr(O | T & ~L)]
Pr(O | ~T) = [Pr(L | ~T) x Pr(O | ~T & L)] + [Pr(~L | ~T) x Pr(O | ~T & ~L)]

any ideas on how this derivation would go?

[fresh sparrow]

i see how to prove the first one but not the second one

prove
or at least show it's true. draw a venn diagram

[untold oriole]

the first one? which one

[fresh sparrow]

Pr(O | T) = [Pr(L | T) x Pr(O | T & L)] + [Pr(~L | T) x Pr(O | T & ~L)]

[untold oriole]

oh

Pr(O | ~T) = [Pr(L | ~T) x Pr(O | ~T & L)] + [Pr(~L | ~T) x Pr(O | ~T & ~L)] is true by the same reasoning

the law of total probability

oh wait

[fresh sparrow]

yes but idk where that B came from

[untold oriole]

typo!

[fresh sparrow]

yeah all good now

[hallow wraith]

@keen saddle sorry man, i just closed discord cause i wasnt able to proceed but i did try your method and didnt know what to do with the other terms.

[keen saddle]

Hey man

Dm me

Because this is someone else's channel lol

@hallow wraith sorry for responding late

[untold mulch]

how do I use higher order exponents? like, if I have x raised by (x raised by x) = 2 for example, how do I solve it?

[last arch]

Make a new post

#❓how-to-get-help

[untold mulch]

Thanks

I know I didn't follow the procedure because of my ignorance, but do I have to do the close to open the channel or is it still open?

[keen saddle]

Dm me

@untold oriole I think you forgot to close this lol

[midnight plank]

@untold oriole Has your question been resolved?

[untold oriole]

not yet

[remote tangle]

#help-49 message

[remote tangle]

The question.

[west dagger]

Has this equation or math been solved sorry for my bad english hehe

[hushed atlas]

ur english is good

[tepid field]

Tru

[west dagger]

=>Pr(e | T) means the chance of e happening if T is true. To calculate this, we look at two possible situations: When B happens with T. When B does not happen (~B) with T.
We calculate Pr(e | T) by combining both situations. This combination is a "weighted average," where the weights depend on how often B happens when T is true (Pr(B | T)) and how often B does not happen when T is true (Pr(~B | T)).
If B happens a lot when T is true (meaning Pr(B | T) is high), then Pr(e | T & B) will matter more in the overall average. So, if Pr(e | T & B) is big, it makes Pr(e | T) bigger.

=>Now, let’s look at the statement:
"It’s false that Pr(e | A & ~T) is many times greater than Pr(e | T)." This means they’re saying Pr(e | A & ~T) is not way bigger than Pr(e | T).
Why? Because if B happens a lot, and when B and T happen, e also happens a lot, it pushes Pr(e | T) higher. So, it becomes hard for Pr(e | A & ~T) to be much bigger than Pr(e | T) since Pr(e | T) is already high due to the impact of B.

i dont know if it correct translation because i translate it to my language to english universal language hehe

In short:
Pr(e | T) gets bigger when B happens often and makes e happen more.
Because of this, it's not true that Pr(e | A & ~T) is much bigger than Pr(e | T).
This happens BECAUSE Pr(e | T) is already boosted by the likelihood of B happening and making e happen more often when T is true. So, Pr(e | A & ~T) can't be much bigger than Pr(e | T).

why can't they became can't be much bigger than Pr(e | T)???
Pr(e | T) because Pr(e | T) already takes into account two key factors:

Pr(e | T & B) (when B and T both happen).
Pr(e | T & ~B) (when B doesn’t happen, but T does).
if B happens often with T (meaning Pr(B | T) is high), and e happens much more when B and T occur together (Pr(e | T & B) is high), this makes Pr(e | T) bigger.
Since Pr(e | T) is already large due to B happening frequently and pushing up the chance of e, it's unlikely that Pr(e | A & ~T) will be many times larger. This is because Pr(e | T) has already been raised by the influence of B and can’t be easily overshadowed by Pr(e | A & ~T).
Sooooooooooooooooooooooooooooooooooooooooooo Pr(e | T) is higher, making it harder for Pr(e | A & ~T) to be much bigger than that.

In simple terms: Pr(e | T) is boosted by B and can't easily be overshadowed by Pr(e | A & ~T), BECAUSE Pr(e | T) is already high due to how much B affects it. That's why Pr(e | A & ~T) can't be many times larger than Pr(e | T).

are you saying it true that false that Pr(e | A & ~T) is many times greater than Pr(e | T)?

The statement is saying it’s false that Pr(e | A & ~T) is many times greater than Pr(e | T). In other words, it is not true that Pr(e | A & ~T) is way bigger than Pr(e | T).
This happens because Pr(e | T) is already boosted by the likelihood of B happening and making e happen more often when T is true. So, Pr(e | A & ~T) can't be much bigger than Pr(e | T).

[untold oriole]

i get why Pr(e | T) is getting bigger. Pr(e | A & ~T) might be higher than Pr(e | T), no? I mean, what if Pr(e | T) is still really small even after being boosted.

[west dagger]

damn it i am so stupid my bad bro

[west dagger]

I am sorry, but I am a little confused about your what if. Hehe, I don't get it that really, but this is my own understanding of your what if, sir, maam, or any gender:

Even if Pr(e | T) is small, B has already given it a boost. So, while Pr(e | A & ~T) could be higher, it probably won’t be much higher because Pr(e | T) already got some help from B, and that makes it harder for the difference to be huge.

I am sorry, is this cheating? I am using a QuillBot to fix my grammar.

[untold oriole]

"it probably won’t be much higher" but is it guaranteed that it wont be much higher?

[vagrant marlin]

do you still need help

[brave stag]

Fortunately, your thinking is wrong

[brave stag]

Don't post in others' channel

[brave stag]

Don't help this person

[remote tangle]

Why..?

Oh

He's not OP

[brave stag]

Yeah

[glacial valley]

18n + 18×19/2 =144,169,225,289
Check which gives integer value
If 2 or more integer value then select the lowest one

[west dagger]

OKIE OKIE My answer, sir/maam/any pronoun, is no; it's not guaranteed that Pr(e | A & ~T) won't be much higher than Pr(e | T). It’s just less likely if Pr(e | T) already got a boost from B. But there's no absolute rule that says it can't happen. It just depends on the specific numbers for Pr(e | A & ~T) and how much Pr(e | T) got boosted by B. So, Pr(e | A & ~T) could still be higher—it just wouldn’t be many times bigger unless something special happens with A and ~T

[untold oriole]

okay, so i think there is a rule or collection of rules and am trying to figure out how it all works. if you & i arent sure, maybe someone else can aid

[untold oriole]

.close

[midnight plank]

Channel closed

Closed by @untold oriole

Use .reopen if this was a mistake.

[rose rapids]

this has 1 free variable correct? Since either x_4 or x_5 can take on the role

[last slate]

Could you reask this in an unoccupied channel

[rose rapids]

I cannot see any

[last slate]

ah I see

[rose rapids]

And I have enabled them

I don't think the bot is working

[last slate]

as soon as one is available I’d recommend moving bc this channel will automatically close 🙂

[eager geyser]

puros weones

ayudenme con mi caga de tarea

[tidal turret]

mas perdido que el teniente bello, si quieres ayuda necesitas abrir tu propio canal

[last slate]

Holy moly

[wild hinge]

.closed

[twilit jetty]

.reopen

[midnight plank]

✅

[twilit jetty]

.close

[midnight plank]

Channel closed

Closed by @twilit jetty

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[haughty ibex]

hola

[haughty ibex]

A box of spheres has 8 rows with 5 spheres each, how many spheres are there in each box?

.close

[midnight plank]

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[next rover]

.reopen

[midnight plank]

✅

[next rover]

.close

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[haughty ibex]

.close

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[midnight plank]

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[jolly roost]

can someone explain to me

what this question is asking

im confused

[upper crag]

well first can you write out the cubic polynomial they describe?

[jolly roost]

i got the equation

2(x-2)(x-3)(x+5)

do i just set this equal to 120

[upper crag]

yeah

[jolly roost]

and solve for x?

[upper crag]

yup

[jolly roost]

how is this the answer

at the back of the textbook

[upper crag]

looks wrong to me

you should expect 1, 2, or 3 different values of x as the answers depending on how many times it hits 120

you know it has to hit at least once because it's an odd function

[jolly roost]

right

-2, -3, 5

are those the solutions

[upper crag]

yeah that's what I get

[jolly roost]

what is this even supposed to mean then

[upper crag]

books have mistakes, that or maybe you turned to the wrong section of answers

you're in the same chapter for the answers right, idk what else it might be

[jolly roost]

mistake for sure then, cuz im on the right section

yeah its fine

[upper crag]

like they changed some things for a newer edition of the book but forgot to fix the solutions

[jolly roost]

thx for ur help tho ❤️

[upper crag]

probably, idk

yeah you're welcome

[midnight plank]

@jolly roost Has your question been resolved?

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[neat notch]

.reopen

[last slate]

Hamilton-Jacobi-Bellman (HJB) equation

[last slate]

can someone explain this

[carmine sigil]

Did you have a specific question? Otherwise the best thing I can suggest is perhaps watching a lecture on YouTube or something

[midnight plank]

@last slate Has your question been resolved?

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[leaden matrix]

how was the lamba cubed - 8 factored?

[grim vector]

You have (lambda² -1) as common factor for both

They actually factored by (lambda² - 1) instead of factoring by (lambda³ - 8)

[lethal path]

you have $ac - bc$ if $a = \lambda^3, b = 8, c = \lambda^2 - 1$

[leaden matrix]

I thought the lamba 2 -1 was the (lambda-1)(lamba+1) part

[grand pond]

south's secret twin brother

[lethal path]

ahhhh wait I see what you are asking now

difference of two cubes

[grim vector]

Yeah after

[lethal path]

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

[grand pond]

south's secret twin brother

[leaden matrix]

after the first line what was done to get to the second line

[grim vector]

Oooooohhh right that was this asking

They think lambda³ - 8 as lambda³ - 2³ and use the formula that south remind us

[leaden matrix]

I see

and then what was done here?

[lethal path]

completing the square

so note that x^2 + 2x + 1 = (x + 1)^2

[leaden matrix]

ah so was it like you factor x and the 1/2 of the constant B

and then subtract (1/2 b ^2) ?

[lethal path]

oh yeah, so (x + b)^2 = x^2 + 2bx + b^2

so if you have the coefficient of the middle term, 2b

halve and then square it

(2b/2)^2 = b^2, bingo

[grim vector]

Bingo

[leaden matrix]

got it thanks

[midnight plank]

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[thorny delta]

Can i cut ln both sides

[dawn dagger]

wdym by "cut"

you can apply e^(...) on both sides

but be careful with the -1/10

[midnight plank]

@thorny delta Has your question been resolved?

[olive matrix]

essentially no you can't just drop the ln on both sides

what you can do is use the power rule to move the -1/10 inside the ln

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[thorny delta]

cancel

[thorny delta]

oh why

[dawn dagger]

yea i saw already

[thorny delta]

why cant we cut it

[dawn dagger]

you can either drag the power into or not

[grand pond]

bacc (unhelpful)

[dawn dagger]

both lead to the same

[thorny delta]

ohhh ok thank u :- )

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[thorny delta]

which form to apply for (i)?

.reopen

[midnight plank]

✅

[dawn dagger]

do you know the partial fraction decomposition method?

[grand pond]

bacc (unhelpful)

[thorny delta]

yes but i didnt know which form to use , linear form or quadratic form

i was doing like tis but it was wrong in the key

[dawn dagger]

nonono

You only need a linear term in the numerator if the a factor doesn't have real roots

or is irreducible into linear factors

like for example x²+1

[thorny delta]

ohh if that happens only then we use the form i used ?

[dawn dagger]

yes

[thorny delta]

thank you 😊 so much

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[honest bridge]

i’m pretty stumped on how to go by answering this problem

i know i have to use fundamental theorem of calculus but im having a hard time interpreting it from this problem

[dawn dagger]

do you want me to explain it?

[honest bridge]

yes please

[dawn dagger]

You wanna find g'(x)

[honest bridge]

ok

[dawn dagger]

We can apply FTC

[grand pond]

bacc (unhelpful)

[honest bridge]

ok

[dawn dagger]

We now use the theorem

[grand pond]

bacc (unhelpful)

[honest bridge]

okay

[dawn dagger]

f is the antiderivative of f'

like in the theorem

[honest bridge]

yup

[dawn dagger]

Now we differentiate both sides with respect to x

[grand pond]

bacc (unhelpful)

[dawn dagger]

are you familiar with this notation

[honest bridge]

yes

[dawn dagger]

by definition d/dx g(x) = g'(x)

and we can distribute d/dx on the right side

[grand pond]

bacc (unhelpful)

[dawn dagger]

d/dx f(x) is also by definition just f'(x)

[honest bridge]

okay..

[dawn dagger]

do you understand this?

or not

[honest bridge]

yes

[dawn dagger]

you need to tell me

[honest bridge]

i do

[dawn dagger]

ok now your turn

what is the derivative of f(-4)

[honest bridge]

do i plug that in into the equation

[dawn dagger]

no

you think

what does f(-4) resemble

what is it really

[honest bridge]

-4x

[dawn dagger]

ok let me rephrase

f(-4)

represents a constant

[honest bridge]

okay

[dawn dagger]

it represents any number

it has some value

and good thing

you dont have to figure which value

just that it has one

and by that

f(-4) is a constant

and any constant differentiated

what do you get

[honest bridge]

0

[dawn dagger]

exactly

so d/dx (f(-4)) = 0

[honest bridge]

okay

[dawn dagger]

because you literally plug in x=-4 into the antiderivative

the antiderivative would give you any value

and that differentiated would be 0 because it's a constant

[grand pond]

bacc (unhelpful)

[honest bridge]

okay that part makes sense

[grand pond]

bacc (unhelpful)

[honest bridge]

yes

but what about x

would it stay like that

[dawn dagger]

We said this in the beginning

If f'(t) = ... what is f'(x) = ?

[honest bridge]

d/dx f(x)

[dawn dagger]

no...

just plug in for t = x

Again $f'(t) = e^{8t^2-9t}$ we defined it like that in the beginning. What is now $f'(x)$?

[grand pond]

bacc (unhelpful)

[honest bridge]

e^(8x^2 -9x)

[dawn dagger]

ya

[grand pond]

bacc (unhelpful)

[dawn dagger]

And in fact that's not coincidence

there is a theorem about this that the derivative ends up to be the integrand itself

[honest bridge]

is that theorem part of 1 or part 2

[dawn dagger]

part 1

[honest bridge]

taking the antiderivative or derivative of e to the x is still e to the x right ?

[dawn dagger]

the derivative yes and antiderivative it would be e^x + C

[honest bridge]

oh okay

[dawn dagger]

I wish you a good day.

[honest bridge]

let’s say in that case it was instead r (top) and 0 (at the bottom) , would 0 still be a constant and therefore we only have to plug in r?

[grand pond]

bacc (unhelpful)

[honest bridge]

yess

[dawn dagger]

If f(0) is defined, then yes

by FTC1

and f'(t) is continuous

[honest bridge]

what if it was reversed where we have a number at the top and r at the bottom? does that mean anything different

[dawn dagger]

no

[honest bridge]

like -4 at the top and x at the bottom

[dawn dagger]

you end up with a minus

[ \int_r^0 f'(t) : \dd t = -\int_0^r f'(t) : \dd t]

[grand pond]

bacc (unhelpful)

[honest bridge]

why is that?

[dawn dagger]

It's an integration rule

[honest bridge]

i see

so

[honest bridge]

if there is a number we consider it a constant

[dawn dagger]

yes

[honest bridge]

and if it’s in the format of r to 0 it’s positive

but if it’s other way around it’s negative

[dawn dagger]

no

what if r = -1

achtung

think of it like this

when you flip the bounds

you put a minus infront of the integral

that's it

[honest bridge]

okay..

and flipping the bounds

means

that x would be at the bottom instead of the top resulting in a minus in front of integral ?

[dawn dagger]

here visually

instead from a to b you integrate from b to a (backwards) and you make up for it with a minus

[honest bridge]

oh i see.. also can’t see attachment

[dawn dagger]

oh

that explains it

[grand pond]

bacc (unhelpful)

[honest bridge]

ohh okay

thank you

[dawn dagger]

geometrically you are integrating backwards

let me do an example

[honest bridge]

ok

[dawn dagger]

,, - \int_1^0 x : \dd x = \int_0^1 x : \dd x

[grand pond]

bacc (unhelpful)

[dawn dagger]

That would make a triangle

,w plot x from 0 to 1

[grand pond]

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[dawn dagger]

the region under the curve

but instead you are integrating backwards which would make the area under the curve negative

[honest bridge]

okay makes sense

[midnight plank]

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[abstract cedar]

Help me with c part

[midnight plank]

@abstract cedar Has your question been resolved?

[abstract cedar]

<@&286206848099549185>

[midnight plank]

@abstract cedar Has your question been resolved?

[midnight plank]

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[cinder bloom]

im trying to compute this integral to find Cn, to find 0th and first 4 coefficients,

[cinder bloom]

i have the worked solution, and im correct up to this point

on line 2 i cant figure out how the J terms are pulled outfront of each expressoin

<@&286206848099549185>

[nova yoke]

!15min

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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.

[nova yoke]

observe that $e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = i$

[grand pond]

Bungo

[cinder bloom]

so it uses polar form?

Re ^ j * theta?

[nova yoke]

yea, the terms in your numerators on the first line are all in that form

(with R=1)

[cinder bloom]

okay i see that, but R e ^ j * theta = R Cos theta + j * R sin theta, im still confused as to how each term now gets a J

[nova yoke]

did you agree with this:

[cinder bloom]

yes

[nova yoke]

ok, so look at your first term, you have $$e^{i\frac{\pi}{2}(1-n)$$

[grand pond]

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

[nova yoke]

which equals $$e^{i\frac{\pi}{2}}e^{-i\frac{\pi n}{2}}$$

[grand pond]

Bungo

[cinder bloom]

oh fml

[nova yoke]

that first factor $e^{i\frac{\pi}{2}$ is just $i$

[grand pond]

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

[cinder bloom]

i didnt multiply the brackets in

[midnight plank]

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[strong tapir]

Hello

[strong tapir]

If a group G is abelian for two operations, + and ·

can we say that it fulfills the distributive property?

$\forall a,b,c \in G \Rightarrow a \cdot (b+c) = a \cdot b + a \cdot c$