help-28

[hollow summit]

hi ty but where does it come from

[sudden chasm]

from coulomb's

[umbral dome]

$\epsilon_0$ is the electric permittivity of a vacuum

[glossy valve]

cloud

[umbral dome]

it's related to the coulomb constant by the above formula

[hollow summit]

isnt coulombs this

[sudden chasm]

that k equals this

[hollow summit]

ah okay ty

it wasnt mentioned anywhere so i was confused

[sudden chasm]

this is the expression for field around an infinite plane sheet no?

[umbral dome]

for now you can think of it as being an "electric constant" which relates different quantities in electricity, like the coulomb constant. it turns out that formulas are usually nicer if expressed using this constant rather than k, which is why it's usually used more outside of coulomb's law

[hollow summit]

yes but i didnt learn it the teacher kinda just threw it out like that

[hollow summit]

ah oay ty

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[nimble crane]

Coulomb!

[hollow summit]

single handedly steam rolling me

[nimble crane]

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[nimble crane]

you'll get the hang of it

[nimble crane]

huh, I opened this?

interesting, I don't see my message below the bot message

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[hollow summit]

hahahah tytyy

[viral jasper]

@nimble crane do you still need help?

[nimble crane]

please do my hmw for me

[viral jasper]

It is done

[nimble crane]

I didn't even send the problem yet!

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[neat sentinel]

,rotate

[neat sentinel]

,rotate

[neat sentinel]

,rotate

[neat sentinel]

Q5

I found alpha beta and alpha + beta and alpha squared plus beta squares

[sudden chasm]

do you know the formation of a quadratic from the roots?

[neat sentinel]

Ye isn't it x-alpha x-beta

[sudden chasm]

if u and v are the roots of a quadratic then the quadratic is given by

k[x² - (u+v)x -uv], where k is a constant you can generally set as 1

[neat sentinel]

Ye

[sudden chasm]

so you just gotta treat (apla-beta)² and ( alpha +beta)² as your new u and v

[neat sentinel]

Isn't it difference of two squares

[sudden chasm]

yeah the coefficient " b" in the new expression is the difference of two squares

you'll need to find alpha and beta though by solving the given quadratic

[neat sentinel]

One bracket is 25 correct

[sudden chasm]

lemme check

[sudden chasm]

the discriminant is negative :((
how'd u get 25

[neat sentinel]

-5^2

[sudden chasm]

oh i dont think you need to solve the quadratic
you just need the reaction between roots

relationship **

alpha + beta = -b/a
alpha * beta = c/a

you can find the new quadratic now

[neat sentinel]

Ye equation 1 is -5 and equation 2 is 7

[sudden chasm]

but you need ( alpha - beta)² so expand and try to transform it to a from you know

(a-b)² = (a+b)² - 4ab

[neat sentinel]

Isn't is 2

O nvm

So that equals 11

[sudden chasm]

just simply substitute now

[neat sentinel]

Is it u-v because we are finding alpha-beta

[sudden chasm]

no no
your u now will be (alpha-beta)²
and v will be (aplha+beta)² or vise versa

why its u-v is a different reason
it comes from the comparison of
a standard quadratic a(x-u)(x-v) with its standard form ax² + bx + c

[sudden chasm]

yes this is it

[neat sentinel]

Have we found aloha -beta squared

[sudden chasm]

it can be easily found with this identity

[neat sentinel]

So it's 11

And 25

[sudden chasm]

are you sure 😈

[neat sentinel]

So I did 4+7 instead of times 🙄

🤫

So -3

R they supposed to switch

Bc I get -38

28

Instead of -22 or 22

[sudden chasm]

yess

you've got this

[neat sentinel]

But how was I supposed to know say it wasn't multiple choice

[sudden chasm]

the sign can sometimes change depending on which one you consider u and which one v

[sudden chasm]

both of the answers would've been valid and you'd have to write both

[neat sentinel]

ok thanks

ill write both

[sudden chasm]

lovely

[neat sentinel]

wait 25 --3 is 28

im stuck

[sudden chasm]

yeah

oh shit i made a tiny mistake while telling you the formula

[sudden chasm]

corrected

u+v

and negative sign is there
also -uv

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[neat sentinel]

Q7 I'm trying to find alpha

[neat sentinel]

,rotate

[onyx glen]

@neat sentinel do you know vieta's formulas?

well, i guess this is what the chapter's on

[neat sentinel]

ye

i got alpha* alpha beta * alpha/beta =16

product of roots -d/a

[onyx glen]

your a is 2 tho

[neat sentinel]

i c where i went wrong

how do i find beta

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[stiff blaze]

Sorry really easy question but I graphed this correctly right?

[unique wagon]

looks right

make sure that highlighted point is the max

[stiff blaze]

I looked on Desmos and the x intercepts were ( -pi,0)

[stiff blaze]

Oki tyyy

[unique wagon]

2π/2 = π

[stiff blaze]

omg ima. Baka

Bruh

This is so embarrassing

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[clever sequoia]

from the graphs of 2 polar functions, how do you know which one is the inner and which one is the outer

[umbral dome]

the one that's closer to the origin is the inner function

imagine drawing a line extending straight out from the origin, then the first one you hit is the inner function, and the second one you hit is the outer function

[steel solar]

You can also partition the area

[clever sequoia]

but that line will hit the function depending on the theta

[steel solar]

Yes

[umbral dome]

if that's the case then you have to split the integral

just like in the cartesian case

[steel solar]

^

[clever sequoia]

wdym by that

[steel solar]

It’s typically easier to partition it when you’re given a visual graph

[clever sequoia]

wdym

[steel solar]

Take on function till it intersects the other, then continue the integration with the other function

[umbral dome]

if you get two separate sets of bounds then you have to split into two integrals with those bounds

[steel solar]

Like here

$2\qty(\frac12\int_0^{\frac{\pi}{3}}(1+\cos(\theta))^2d\theta+\frac12\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}(3\cos(\theta))^2d\theta)$

[glossy valve]

;(

[clever sequoia]

ah nvm i gtg thanks for the help anyways

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[cobalt atlas]

Can someone check this i am supposed to find the quadratic equation

[cobalt atlas]

Btw the (-2,3) is the vertex its just that my lines on the graph were off

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[pallid tundra]

how do i solve this without power series? im in an introductory diff eq class

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<@&286206848099549185>

[gritty rose]

try using variation of parameters
https://tutorial.math.lamar.edu/classes/DE/VariationofParameters.aspx

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[woeful flax]

Pls help

[woeful flax]

Why 1/3

[gritty flax]

But here f' is 3

[torn jolt]

^

[gritty flax]

So you must compensate by dividing by 3

[torn jolt]

And the numerator is one

So you must make thr numerator 3 without changing the value of the integral

So make it 1/3 * integral 3/(3x+2)

[woeful flax]

OK, I need help with integrating ln, how do you do this.
I thought we just put ln|something|

How do you integrate ln?

[torn jolt]

You want to integrate ln or 1/smth

[woeful flax]

You cannot

Well, tell me like that then

How to fix it?

Also, why not?

I thought that "S1/something not raise to power dx" you integrate by ln|something not raise to power|

If you understand me!

[torn jolt]

.

[woeful flax]

I feel like I don't know ln anymore

[torn jolt]

The integral is ln you're not integrating ln tho

[woeful flax]

This is actually the original task

but I replaced dx with 1

is that not the same?

hmmm

[torn jolt]

It is but that's not the issue, for the integration to be ln|f(x)| you need to have it in the form f'(x)/f(x)

[torn jolt]

Do you understand this

[woeful flax]

f'(x)/f(x)
We don't have this?

[torn jolt]

We will change the integral so it's in that form

What's the differentiation of 3x + 2

[woeful flax]

derivation?

[torn jolt]

Yes

[woeful flax]

3, right?

[torn jolt]

Correct

So if f(x) = 3x + 2 then f'(x) = 3

[woeful flax]

aha

[torn jolt]

So if we put it in this form

We should make it integral 3/(3x+2)

But

We just multiplied by 3

So we need to compensate for it

By also dividing by 3

So outside of the integral multiply by 1/3

[torn jolt]

No

[woeful flax]

outside multiply by 1/3, okay, got it

[torn jolt]

The integration of 3/(3x+2) is ln|3x+2|

[gritty flax]

Its the same as writing 1 as 3*1/3

[torn jolt]

Yes

[torn jolt]

^

[torn jolt]

Ok now we integrate 3/(3x+2)

[torn jolt]

Where are you getting the 3 from

The integration of 3/(3x+2) is ln|3x+2|

[woeful flax]

3 integrated is 3x?

[torn jolt]

Ok?

[woeful flax]

where did the 3 go

[torn jolt]

We don't integrate the 3

We integrate this as a whole

When you integrate 1/x you don't integrate the 1 correct?

[woeful flax]

one goes in front?

one goes in front, I know

you don't integrate it

[torn jolt]

What's the integration of 1/(x+3)

[woeful flax]

ln|x+3|

[torn jolt]

Ok so what's the integration of 1/(2x+1)

[woeful flax]

1/2 * ln |2x+1|

[torn jolt]

Yes

[woeful flax]

So this is it then?

[torn jolt]

You need to make it in this form then multiply it by half

[torn jolt]

Yes

For your question you will make the numerator 3 and multiply by 1/3

So it's like you changed nothing

Because you did 3*(1/3)

[woeful flax]

ya

You can solve this assignm. through substitution as well?

We I mean

t=(3x+2)

[torn jolt]

Well you technically can but you don't need to

[woeful flax]

okay

Can you check some theory for me?

[torn jolt]

I gtg but send it and someone will check it for you

[woeful flax]

Is this ok?

@torn jolt okay, thanks for help 🙂

Is this okay?

these two last images that show theory answers

P means Area

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[proud girder]

I have watched many videos on this and I still can remember or figure out how to solve this. I need to find the primeter of triangle SOW. I need to know a formula or a away to solve this.

[pure nymph]

h^2 = m * n, where h = 6

m is equal to 8, n = NW

you can find n from here

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[proud girder]

What is m

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[proud girder]

.reopen

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✅

[proud girder]

what does m mean

[pure nymph]

take a look at this picture

you'll figure it out

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[torn jolt]

How are they getting 2/11? for this example? Once again my college uses Pearson and it skips steps

[violet haven]

they wrote $5\frac{1}{2}$ as 11/2

[glossy valve]

aerial.ace

[torn jolt]

That doesn't really explain much? If they wrote it as 11/2 why is it 2/11?

[violet haven]

hours to complete job is 11/2 then what would be part of job done in 1 hour

[torn jolt]

Would an easier way be of thinking of it as multiplying the hours completed by 2 becuase it was done by two workers?

I don't understand your question

[violet haven]

if it takes 11/2 hours to complete a job then in 1 hour 2/11 of the job is done

[torn jolt]

Ahhh okay okay

It's clicking

Thank you once again aerial ace

[violet haven]

np

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[worn lion]

no clue how to find v'

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[worn lion]

<@&286206848099549185>

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[worn lion]

<@&286206848099549185>

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[misty oriole]

I don’t think I can proceed anymore, I’ll need a tip of some sort.

[misty oriole]

I’ve tried using by parts to simplify I_n+1, but to no avail

I’ve checked the validity of the given relation

I’ve checked my differentiation

I’ve tried playing with fraction reduction

Or try using the arcsin form, didn’t work

<@&286206848099549185>

[royal holly]

Hello I can try, but I’m not sure if I’ll arrive at the answer

[misty oriole]

Ya sure, thanks

Damn, I kinda wanna ping helpers again

I still can’t figure it out

Is the question too uninteresting or what

[royal holly]

try x = 2 sintheta

i think its begging for it

[misty oriole]

I don’t think it works

Doesn’t lead to anywhere

[royal holly]

i could solve till here

the 2sintheta^n-2 part is troubling

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[misty oriole]

I saw ChatGPT use Beta function to solve it

But that's out of my scope

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[wicked zodiac]

why doesnt the - cancel out

[hot lily]

Which -s are you talking about

$$\textcolor{red}-\frac 17 \sum_{n = 0}^\infty \qty(\textcolor{orange}-\frac{x \textcolor{yellow}- 8}7)^n$$

[glossy valve]

King Leo

[hot lily]

@wicked zodiac which two -s should cancel out (red/orange/blue)

[wicked zodiac]

So why is there the -1

@hot lily

[hot lily]

Its because the orange - is being raised to the nth power. it is not always negative

On the other hand, -1/7 is merely a constant, which is always negative

Wait i didnt see your new question

[wicked zodiac]

Yea

[hot lily]

$$-\frac 17 \sum_{n = 0}^\infty \qty(-\frac{x - 8}7)^n$$
$$-\frac 17 \sum_{n = 0}^\infty (-1)^n \qty(\frac{x - 8}7)^n$$
$$\frac 17 \sum_{n = 0}^\infty (-1)^{n + 1} \qty(\frac{x - 8}7)^n$$
$$\frac 17 \sum_{n = 0}^\infty (-1)^{n + 1} \frac{(x - 8)^n}{7^n}$$
$$\sum_{n = 0}^\infty (-1)^{n + 1} \frac{(x - 8)^n}{7^{n + 1}}$$

[wicked zodiac]

Ohh

I see

+1 is the othe one

The other- right

I see

Thanks

[glossy valve]

King Leo

[hot lily]

!done

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[honest quest]

Dont know where to go from there? I know it’s ln but them I’m stuck

[spiral viper]

Just integrate both the sides

[honest quest]

Also is there a better way do this ;-:

[spiral viper]

Nah I'd say its fine. You just showed every step of solving ( even the cross multiplication ones ) thats why it looks long

[honest quest]

im following a video as i do it so genetic fear of fucking up... you need make sure that dx or dy is on top correct? not as 1/dx or 1/dy

[spiral viper]

Yes

[honest quest]

Is this correct? I really dont use e or ln a lot

wait, do you move the 1/5 over before hand?

[spiral viper]

Nice

[honest quest]

got to say ... that ugly

[spiral viper]

You can simplify it

$e^{ln(x)} = x$

[glossy valve]

@spiral viper

[spiral viper]

Oh yeah btw you forgot the + C

[honest quest]

yea hw just ping me as wrong

why is it C times the answer

C(x+4)^5

[spiral viper]

C times? Or C +

[spiral viper]

Oh okay see

When integrating you have a + C at the end , which we call an Arbitrary constant

So you'd say

ln|y| = 5ln|4+x| + C

Now I can rewrite 5ln|4+x| = ln|(4+x)^5|

And I can also rewrite C = ln(c)

Cuz you both are constants

Now we have a rule that ln(a) + ln(b) = ln(ab)

So here we get
ln|y| = ln|c(4+x)^5|

Or e ^ ln|y| = e^ln|c(4+x)^5|

Which further simplified gives
y = c(4+x)^5

[honest quest]

oof

ouchy my brain

[spiral viper]

lol

[honest quest]

i appercate it though

🙂

i think im too smooth brain to understand

but ur epic for explaining it B)

[spiral viper]

Thanks

[honest quest]

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[little plover]

im stuck on a

and im not even sure if i did aroc right either 😭

is that correct or is it (r"(10)-r"(7))/10-7

i might be overthinking

[onyx glen]

yes you are

it is (r'(10)-r'(7))/(10-7) that they want you to calculate

[little plover]

...

thats it?

this feels too easy

😭

[lime ether]

it is

[little plover]

do i have to justify to get points as well?

[little plover]

KNIEEFFFFF

long time no see

(tickledpickles)

[lime ether]

hello oh

you’re ticklemypickle?

[little plover]

yeah

i forgot my old user 😭

[lime ether]

ahh

tough

[little plover]

but anyways, r"(8.5) = aroc of r?

r' i meant

[lime ether]

this is AROC

yes

[little plover]

this isnt so bad then i guess im js paranoid 😭

i might need help on the next questions though

hang tight 🙏

do i justify b by saying "there is not a time t between t=0 and t=3 because r' is negative?"

i might be wrong with my answer too

[lime ether]

what

you’re looking at plenty of times where it’s negative

wdym

[little plover]

lemme reread

[lime ether]

is -6 in between -6.1 and -3.5?

[little plover]

BOMBOCLATT I THOUGHT IT MEANT r(t)=6 😭

[little plover]

yes

[lime ether]

what theorem should you state

[little plover]

"there is a time t between t=0 and t=3 where r'(t)=6 because r'(t) is increasing between 0<t<3?"

[lime ether]

no

try again

[little plover]

mean value? (i think)

[lime ether]

no

[little plover]

avg value?

[lime ether]

no

there aren’t many left

you have a few other theorems

[little plover]

https://tenor.com/view/gon-hx-h-gon-thinking-gif-19587492

its somewhere in the back of my head

[lime ether]

list the ones you know

[little plover]

ima go with the last one i can remember

fundamental theorem of calculus

[lime ether]

🤔

no

what theorem is concerned with guaranteeing a function takes on a particular value in an interval

[little plover]

what unit is this

[lime ether]

uhh

idk 3,4?

2?

early

[little plover]

searching my first semester notes

[lime ether]

maybe unit 1 actually

[little plover]

Differentiability Implies Continuous

[lime ether]

that’s true yes and you’ll use it

but what else

[little plover]

instantaneous velocity?

[lime ether]

no

[little plover]

?

[lime ether]

nah mate

[little plover]

thats all i could find 😭

[lime ether]

IVT

[little plover]

ivt?

[lime ether]

intermediate value theorem

[little plover]

i completely skipped it when scrolling

how would i justify this efficiently for the ap reader

"according to intermediate value theorem, f is a continuous..."

[lime ether]

since r is twice differentiable, r’ is differentiable and therefore continuous implying that there exists a time t in the interval [0,3] such that r’(t) is between r’(0) and r’(3) or equivalently, r’(t) is between -6.1 and -5, thus since -6.1 < -6 < -5 there exists a time t such that r’(t) = -6

word it how you’d like

but mention intermediate value theorem

i should’ve said that in there

"by IVT.."

[little plover]

will i still get points by js putting "intermediate value theorem?"...

😭

[lime ether]

yes

that’s probably one point

"student mentions IVT"

[little plover]

so i dont have to type out a full complex sentence right?

[lime ether]

i mean

youll want to state the conditions necessary

r’ is differentiable

and so continuous

and explain that you can apply IVT

[little plover]

that should do

[lime ether]

sure and maybe say -6.1 < -6 < -5

[little plover]

c i can do but maybe not d 😭

i hate related rates

[lime ether]

it’s light

[little plover]

im already lost

im going back to my notes

wait i thought the ap exam would be nice enough to provide the formula for cone volume 😭

[lime ether]

they did

[little plover]

wait

maybe im js sleepy

😭

yeah maybe i js need sleep

[lime ether]

alright

[little plover]

im going to bed and studying tmrw

.close

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[neat sentinel]

i found one value or gamma which is -1

[onyx glen]

original question please?

[neat sentinel]

q9

then i found alpha and beta but the signs are inversed

the server locked me out of my original channel

[onyx glen]

i see your mistake

this plus should have been a minus

[neat sentinel]

i c bc i moved the 20/3 to the lhs

thanks

[onyx glen]

also btw your work is quite messy with the "scribble individual numbers out" thing

if a line is mistaken, you should cross it out entirely and rewrite it correctly

[neat sentinel]

ye i thought i made a different mistake and went to fix it but i was correct originally

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[twin wolf]

$\text{Prove that if } f: \mathbb{R} \to \mathbb{R} \text{ is a continuous function of order 3, then f(x) = x.}$

[glossy valve]

water beam

[twin wolf]

No clue where to start

Part A) was show f(x) = -1/1+x was order 3

[midnight stone]

!help

[full forum]

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

[twin wolf]

Get your own channel

[onyx glen]

you don't need to put a thousand \text commands btw

you can include formulas in the middle of your message $y=x^2$ like this

[glossy valve]

Ann

[onyx glen]

anyway

[twin wolf]

I would do that but I’m outside on a phone so instead I got gpt to latex for me

[onyx glen]

what do you mean by order 3? do you mean that the composition of 3 copies of itself gives the identity?

[twin wolf]

So in the question it defined a function to be order 3 if f(f(f(x))) = x

[onyx glen]

ok so yes

what you want to prove essentially is that every real number is a fixed point for f

so perhaps try to suppose that there is a real number a which isn't a fixed point, and see what you might derive from that

in particular you will know that f(a) and f^2(a) cannot be fixed points either

[midnight stone]

how

[onyx glen]

read #❓how-to-get-help for instructions

also important is that you know f is a bijection!

[twin wolf]

so if there is a real number a where f(a) neq a then can f(a)>a and f(a)<a

Since a isn’t a fixed point it can be anywhere on the graph

[onyx glen]

"not a fixed point" means f(a) != a not anything else

[twin wolf]

O

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[rancid skiff]

It's a theorem that if P(A| B) = P(A), then P(A | ¬B) = P(A). Do you happened to know the name of this theorem?

[full forum]

@rancid skiff Has your question been resolved?

[slate violet]

that's just the definition of independent events
P(A | B) = P(A) implies A, B are independent

[rancid skiff]

thanks :)

[slate violet]

yeah and then there's some sort of derivation afterwards, idk

no worries

P(A and B) = P(A) P(B) , then P(A and B') = P(A) P(B')

now add the two equalities together and you get P(A) = P(A) * (P(B) + P(B')), so a valid proof would start from this and work backwards to find P(A and B')

[rancid skiff]

so, um what about P(C | A¬B)=P(C |¬B), does that imply P(C | AB)=P(C |B) as well?

[full forum]

@rancid skiff Has your question been resolved?

[rancid skiff]

<@&286206848099549185> does anyone happened to do know? :) Thanks in advance !

[crude kayak]

!showwork

[full forum]

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

[rancid skiff]

To be proved: P(C | A¬B)=P(C |¬B) implies P(C | AB)=P(C |B).

I proved that
(1) P(C | A¬B)=P(C |¬B) -> P(C | ¬A¬B)=P(C |¬B)
and
(2) P(C | AB)=P(C | B) -> P(C | ¬AB)=P(C | B)

But I don't know how useful that is for proving that P(C | A¬B)=P(C |¬B) -> P(C | AB)=P(C |B)...

[full forum]

@rancid skiff Has your question been resolved?

[full forum]

@rancid skiff Has your question been resolved?

[sacred rivet]

so first lets recall some basic syntactic laws (and hopefully their semantics are intuitive)

definition of conditional probability: P(X|Y) = P(XY)/P(Y)

joint probability:
P(X) P(Y) = P(XY)

complement:
P(A) - P(AB) = P(A -B)

where the -B is my shorthand for "not B"

you can prove this by bashing

for convenience of writing, im going to ignore the P( ) bits, but just know that there is a P() around each term, and we are allowed to do this joint probability means the P function distributes over multiplication

but even if you dont believe me, you can go back and rewrite it with the P() notation to double check

$$\frac{A\neg B C}{A \neg B} = \frac{\neg B C}{\neg B}$$

[glossy valve]

Cozmogrgdfschkipkhrshtensi

[sacred rivet]

this is the given, the precondition

$$\frac{AC - AB C}{A - A B} = \frac{C - B C}{1-B}$$

[glossy valve]

Cozmogrgdfschkipkhrshtensi

[sacred rivet]

complement

now multiply out the denominators

(AC-ABC)(1-B) = (C-BC)(A-AB)

after some rearranging you should get
P(ABC) P(B) = P(AB) P(BC)

and that can be turned into your conclusion

not sure if i made a mistake somewhere

something feels a bit off

[sacred rivet]

oh ahaha i think i see it

so the statement you are trying to prove

if you can write it in the form "If f, then g"

i think the idea is that im assuming joint probability when it may not necessarily hold in general

so the missing piece i think is to show that f implies joint probability

and then joint probability implies g

mmmmm i think im close

yeah this last step seems tricky, maybe im approaching it wrong

damn i thought i had it

sorry

[full forum]

@rancid skiff Has your question been resolved?

[rancid skiff]

thanks for your thoughts!!!!

very much appreciate it

I ask chatgpt and they told me this

We want to analyze whether:

[
P(C \mid A, \neg B) = P(C \mid \neg B)
]

implies:

[
P(C \mid A, B) = P(C \mid B).
]

---

Step 1: Using the Law of Total Probability

We express ( P(C \mid A) ) using the law of total probability:

[
P(C \mid A) = P(C \mid A, B) P(B \mid A) + P(C \mid A, \neg B) P(\neg B \mid A).
]

By assumption,

[
P(C \mid A, \neg B) = P(C \mid \neg B),
]

so substituting this into the equation,

[
P(C \mid A) = P(C \mid A, B) P(B \mid A) + P(C \mid \neg B) P(\neg B \mid A).
]

Similarly, applying the law of total probability to ( P(C) ):

[
P(C) = P(C \mid B) P(B) + P(C \mid \neg B) P(\neg B).
]

If we assume that ( P(C \mid A) = P(C) ) (which is a necessary assumption for full independence),

[
P(C \mid A, B) P(B \mid A) + P(C \mid \neg B) P(\neg B \mid A) = P(C \mid B) P(B) + P(C \mid \neg B) P(\neg B).
]

Since we are given ( P(C \mid A, \neg B) = P(C \mid \neg B) ), this condition alone does not necessarily force ( P(C \mid A, B) = P(C \mid B) ).

To conclude ( P(C \mid A, B) = P(C \mid B) ), we would need additional assumptions, such as independence between ( A ) and ( C ) given ( B ).

---

Step 2: A Counterexample

Consider a probability space where:

• ( B ) affects ( C ),
• ( A ) only influences ( C ) when ( B ) is true.

For instance, suppose:

[
P(C \mid A, \neg B) = P(C \mid \neg B) = 0.5.
]

But suppose ( A ) significantly alters ( C ) when ( B ) is true:

[
P(C \mid A, B) \neq P(C \mid B).
]

Then the assumption ( P(C \mid A, \neg B) = P(C \mid \neg B) ) holds, but the conclusion ( P(C \mid A, B) = P(C \mid B) ) does not necessarily follow.

---

Conclusion

$P(C \mid A, \neg B) = P(C \mid \neg B) \quad \not\Rightarrow \quad P(C \mid A, B) = P(C \mid B)$

The implication does not necessarily hold in general unless additional assumptions (such as independence or symmetry) are made.

[glossy valve]

! If in Oxford, see my about me
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sacred rivet]

"but suppose A significantly alters C when B is true"

thats not particularly rigorous or formal

otherwise i think its basically saying exactly what i said

but more confusing

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[torn jolt]

Be more specific

Which circle theorem are you talking about?

[hot herald]

not sure what answer you're expecting

[torn jolt]

Me too

[hot herald]

they're proofs of the established circle theorems...

you can look at a concise list of circle theorems,
you usually won't have to proof them every time when doing questions
the video shows you where they come from

learning where they come from usually makes it easier to remember

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[visual mural]

Can I solve this problem using double identities?

[honest hill]

probably

[visual mural]

Like this

[honest hill]

yeah that looks correct

[visual mural]

alright thanks!

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[honest hill]

you might get a different C though

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[carmine linden]

If 4 points A(a,b) , B(x,y) , C (s,j) , D(m,n) are concyclic then what’s the formula for it

[hoary ember]

formula for what

[carmine linden]

Concyclic

for 4 points

[hoary ember]

for what? the radius? the coordinate of the center?

you cant just say "whats the formula for triangle"

[carmine linden]

Coordinate center ig

[hoary ember]

choose 3 points of your liking and find the circumcenter's coordinate

[carmine linden]

Oh alr

Ty

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[torn jolt]

Check please

My solution is kinda messy

btw my answer is 2x^2+72

It says to factor and to simplify

[bright bronze]

arent 2x-12 and x-6 in denominator?

[onyx glen]

(2x-12)(x-6) is not equal to 2x^2 + 72

[bright bronze]

why did you pull them up to the numerator in the answer?

[torn jolt]

Huh?

[bright bronze]

• what Ann just said

[onyx glen]

that also

[torn jolt]

wait how do i do that

[torn jolt]

?

[cursive dock]

can't do that with the (x-4)s - you'd get (x-4)^2 not 1

[torn jolt]

i think idk bro

[torn jolt]

so what do i do now

[onyx glen]

those (x-4)'s ought to stay in the numerator

as (x-4)^2

[torn jolt]

ahhh hold on lemme fix it

[onyx glen]

also you probably are not expected to re-expand stuff afterwards

(but if you do want to do it anyway, then you have to do it correctly. you messed it up)

[cursive dock]

why are you now cancelling the thing you cannot cancel, but not cancelling the thing you can cancel

[torn jolt]

which can i cancel

[onyx glen]

you can cancel out a factor on the top with an identical factor on the bottom

the key thing is that one of them has to be on the top and the other on the bottom

[torn jolt]

In one fraction or

?

[onyx glen]

NEVER two on the same side

[torn jolt]

Okay

[onyx glen]

also your product of fractions might as well be one big fraction anyway

so that part does not matter

[torn jolt]

i dont see any

[gusty blade]

You forgot to take the reciprocal

[torn jolt]

ohh okay

wait so what do i do now

[cursive dock]

oh that makes the answer much neater

[onyx glen]

actually yes this also

your first step

[torn jolt]

do i reciprocate it back?

to division?

[onyx glen]

you just swapped out the $\divisionsymbol$ for a $\times$ between the fractions

[glossy valve]

Ann

[onyx glen]

and then you like

[bright bronze]

Uhh, I just noticed, you swapped division sign by multiplication sign, but did no other changes

[onyx glen]

^

[torn jolt]

yeah

[onyx glen]

keep change flip, yes?

[torn jolt]

wait so what do i do

[onyx glen]

you must do the flip part

[gusty blade]

Flip the second fraction

[onyx glen]

it's part of the package

you cannot just omit it because you don't want to

[torn jolt]

oh

oop

hold on

[onyx glen]

basically your factorizations seem to be correct

but your entire second column needs to be redone

erase it all and redo it from the beginning but do it properly

[onyx glen]

you still have not flipped the 2nd fraction.

again. redo the thing.

from the beginning.

do not try to edit your earlier work, it will only be confusing for you AND for us.

[torn jolt]

ok wait

[onyx glen]

ok see NOW we're cooking.

[gusty blade]

You can factor out 2 from (2x-12)

[onyx glen]

this also, but she's now fixed her mistakes.

[gusty blade]

So you can cancel (x-6)

[onyx glen]

so now you have $\frac{(x-4)(x+2)(x-6)(x-5)}{2(x-6)(x+2)(x-5)(x-4)}$ to follow tines' suggestion

[glossy valve]

Ann

[torn jolt]

yes

[onyx glen]

and a lot of shit cancels out don't it

[torn jolt]

yes

[onyx glen]

so what'll we be left with

careful w/ this

tines no

you spoiled it

i was asking lei

[gusty blade]

Oops

Mb I forgot

[onyx glen]

@torn jolt

[torn jolt]

(x-6)/2(x-6)

?

[gusty blade]

??

You can cancel out the x-6

[torn jolt]

oh

so its 1/2?

[onyx glen]

bingo

[torn jolt]

okayy thankss

[gusty blade]

🎉

[torn jolt]

i have other questions lol

but ill try to answer them first

[gusty blade]

[onyx glen]

.close this channel and open a new one when the need arises, then

[torn jolt]

ok

.close

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[torn jolt]

.reopenI

i forgot but our teacher said to do it in standard form.

[gusty blade]

?

[torn jolt]

is it possible to do it in standard form

or no

just 1/2?

[sudden chasm]

repost your question once more

[onyx glen]

what do you mean by "standard form"?

[gusty blade]

5*10^-1 ?

[onyx glen]

is that the exact wording your teacher used?

1/2 is as simple as things can get

@torn jolt did your teacher write the requirement(s) down anywhere?

[torn jolt]

yes

hold on

[onyx glen]

ok show us

[cursive dock]

I have a feeling the question was y = <that big fraction division> and therefore standard form is y = 1/2

[onyx glen]

ok well

1

and 2

are already in as standard a form as you could possibly think of

[torn jolt]

oh ok nvm

thankss

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[warm grotto]

Hi again um I forgot to ask if my new solution is right im a bit braindead 🤕

[hot herald]

yes if asked to round
otherwise ideally you should simplify the sqrt(4) and leave the value exact

[warm grotto]

Thank you

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[night musk]

For this when we make the assumption of x greater than a how can we jump to the next part where they say n(x-a) greater than y

Because we are saying x greater than a not x greater than a +y/n

[gritty flax]

Limit not working here ?

[cursive dock]

have a look at theorem 1.30

I have a feeling it's going to be the archimedean principle

[devout valley]

(and in particular, x > a means that x - a > 0)

[night musk]

It is but I don't get how we make the jump

[night musk]

But how can we reach n(x-a) greater than y from here

Because 1.14 says x is less than or equal to it

How can we flip the sign

[cursive dock]

so x-a > 0
and by the principle of archimedes, we know that we can find some n such that n(x-a) > y
right?

it's a weird way of saying "we can make numbers as big as we like"

[night musk]

Ye

But how does this make x greater than a not possible

[devout valley]

(the whole aim here is that you're proving this by contradiction: you assume what you have isn't true, so x >= a and x != a, that forces x > a and leads you to something that is clearly crazy, so really x has to be equal to a)

[cursive dock]

and if n(x-a) > y, then x-a > y/n

(n is positive nonzero integer so we gucci)

which means x > a + y/n. for this specific value of n.

but we've assumed for every integer x above 1, that x <= a + y/n.

[night musk]

Wait isn't this only x greater than a

[night musk]

So you are saying that we assume x greater than a. Which means x-a greater than 0. Using nx greater than y where x equals x-a

Then we say this contradicts the second part of the inequality so x cannot be greater than a?

[cursive dock]

So you are saying that we assume x greater than a
yup

Which means x-a greater than 0
yup
Using nx greater than y where x equals x-a
yeah but I guess it's better to say nk > y where k = x-a, or just n(x-a) > y

Then we say this contradicts the second part of the inequality
we have reached a contradiction because we have proven "starting from the assumption x <= a + y/n for all integers >=1 n, and for x > a, we can prove that there is some n' such that x > a + y/n'"

both of these cannot be true

our assumption has led us to deduce nonsense. a contradiction

because our assumption includes "a <= x", we don't even need to write out the case for that. "if x < a, then contradiction immediately" is just unnecessary

[night musk]

Wait im losing you at the end how does we prove that just x is greater than a+y/n

[cursive dock]

so can we start at "we know there exists n' such that n'(x-a) > y"

I want to use n' to make it clear this is a specific value of n'

[night musk]

Ok

But then we would need n' to be greater than 1?

[cursive dock]

we would!

and the archimedean property/I.30 says that n' is a positive (>0) integer

so we only actually need greater or equal to 1

and rearrange the inequality
n'(x-a) > y
x-a > y/n' (this does not need sign switching because n' is positive)
x > a + y/n'

[night musk]

Two questions I'm confused how this connects to our original thought where we started with x greater than a. And also by saying n' greater than 1 aren't we already contradicting our boundaries for the inequality

[cursive dock]

'Two questions I'm confused how this connects to our original thought where we started with x greater than a'

we are breaking down the cases of what x can be relative to a into 3 cases: less than, equal, or greater than

less than doesn't make sense so we can ignore that case

if we can prove that greater than doesn't make sense either, then the only remaining option is x=a

[night musk]

Ok

[cursive dock]

' And also by saying n' greater than 1 aren't we already contradicting our boundaries for the inequality'

this is why I wanted to use n' - it's not the same as the n in the original theorem. the n in the original theorem is any integer n such that n >= 1

which, of course, must include the specific n'

right? 🙂

if the question says for all of something, then finding one case when it doesn't is good enough to prove it's a contradiction

[night musk]

Wait I'm confused on this so your saying if it says for all n less than equal to 1 then finding one case where n is one of those values and it contradicts any part of the inequality of enough to contradict our original statement

[cursive dock]

if we had some other theorem that said "blah blah blah, for all n <= 1, then something is true" and we could prove that when n = -3 it isn't true, then we've got a contradiction

[night musk]

Ok

So going back to our original we said that n(x-a) greater than y correct?

[cursive dock]

yep

[night musk]

So is this our contradiction

[cursive dock]

we can find some n that makes this inequality true

[night musk]

Just making sure this is the n less than or equal to 1

[cursive dock]

it exists. it doesn't matter what it is, but we do know it is >= 1 and an integer

(we know it is >= 1 and an integer, because that's what the archimedean principle tells us)

[night musk]

Wait so for that we use a different n

[cursive dock]

yeah, they are shortcutting a bit

[night musk]

Wait but then it doesn't apply to our inequality?

[cursive dock]

I would write it in full:
...there is a positive integer k satisfying k(x-a) > y. we can rearrange this into there is a positive integer k such that x > a + y/k because k is a positive integer, it must be >= 1.
(and then the bit that may click for you)
by our assumption of I.14, every integer n >= 1 satisfies <condition>. however it is not true for the case n = k

therefore contradiction, and x cannot be greater than a. the only remaining case must be that x = a

[night musk]

Wait could you explain again where we got this condition for k

[cursive dock]

theorem I.30 says, probably "for all real numbers x and y, x > 0, there exists a positive integer k such that kx > y"

aka no matter how big y is, and how small (but positive) x is, you can multiply x by something to make it bigger than y

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[woeful anvil]

Can someone help me with bounds on problem 25

[woeful anvil]

i set r^2 = r^2

which gets me pi/6 and 5pi/6

so are those my bounds?

[brittle steeple]

That should get you every real number as a solution

oh

[woeful anvil]

?

[brittle steeple]

,w solution set to r^2=r^2

[glossy valve]

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Requested by a_mog

[brittle steeple]

I would probably sketch the curves

[woeful anvil]

ok

[brittle steeple]

I don't completely trust the bounds although it's not unreasonable what you did

[woeful anvil]

ok i try

.close

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[chilly talon]

Hello

[chilly talon]

Is there a name for a therorum that says the maximum value of some set of positiove reals contrstained to be a certain sum, can be maximized if they are all equal

[wild sleet]

isn't it the opposite? if you make them equal, you minimize the maximum

[chilly talon]

sum = 4
x1 = 1
2x = 3
product = 3

x1=2
x2=2
product = 4
4>3

[wild sleet]

there was nothing about product

[chilly talon]

oops

i typed it wrong I am sorry

I keep forgetting to say product.

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[void sleet]

Hi I’m new to this server, I’m trying to study for the tsi with the accuplacer practice tests but I’d like to know what type of problem this is so I can practice solving it. Like what steps do I take and etc.

[onyx glen]

this is an arithmetic problem

you might see the acronym PEMDAS thrown around to describe the order in which the operations are to be done

this one in particular involves negative numbers and decimals as well

[void sleet]

I see

Man I forgot all of this. Okay so use pemdas

[onyx glen]

do you know how to add, subtract and multiply?

[void sleet]

Yeah I do

I just don’t know what order to do it

Do I add 0.5 + 0.75 first?

[onyx glen]

no

ok, let's dial it back

do you know what the acronym PE(MD)(AS) stands for?

[void sleet]

Yes

[onyx glen]

can you tell me?

[void sleet]

Parenthesis exponent multiplication division add subtract

[onyx glen]

ok

so, in your expression:

-14 - 8 * 0.5 + 0.75

you've got a subtraction, a multiplication, and an addition.

according to the priority order you just spelled out, which has to be done first?

[void sleet]

Multiply?

[onyx glen]

yes.

so you do the 8 * 0.5 first.

what does that work out to?

[void sleet]

Lemme see

4

[onyx glen]

yes

so you got this down to:

-14 - 4 + 0.75

now one important thing about PEMDAS that trips people up:

addition and subtraction are on the same tier. despite the A coming first in the acronym, it does not hold higher priority than subtraction.

so our thing now has only additions and subtractions, and you do these left to right.

do you understand how to proceed now?

[void sleet]

I think so

Sorry something came up

I understand, thank you

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[hot lily]

<@&268886789983436800>

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[cursive condor]

What happened

Scam?

[hot lily]

Ya

it was a server invite

[lunar veldt]

yes it was a server invite to a nsfw

[cursive condor]

Lol

Loser

[safe trench]

There should be a bot that mutes server invites

[gritty flax]

There is

[hot lily]

Nah there are non-scam invites I’ve sent

Let it stay

[gritty flax]

Thats why it got rekt that fast

[gritty flax]

It makes the difference dw

[hot lily]

?

[gritty flax]

Between scam and not scam

[hot lily]

Oh it auto mutes only if the server is a scam?

[gritty flax]

Maybe ban actually but yeah

Like

https://discord.gg/tUU9dPp

No prob with this one

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[terse dove]

Hey can someone help me with this

[terse dove]

What's the bigger inside angle either in

regular hexagon

or in a octagon

regular one

[cursive condor]

Think about it

If sides get bigger does degree get bigger or no?

Triangle versus quadrilateral

[terse dove]

yes

[cursive condor]

Ok

That answer your questino

[terse dove]

So then the octagon is bigger?

[cursive condor]

Yes

Good boy

[terse dove]

:3

But also I dont understand why is the interior angle on the edge of the octagon

[cursive condor]

Because it is in between two sides

Not really on "edge"

[clever anvil]

Hello

I have a question

[cursive condor]

Ok

Fire away

[gritty flax]

https://discord.com/channels/268882317391429632/488120190538743810

[clever anvil]

f(x)=-2sin(1/3x-pie/6)-1

I need help

[gritty flax]

.

[cursive condor]

With?

[clever anvil]

Thats it

[cursive condor]

Finding zeroes?

[clever anvil]

No

what the formuala would be

after doing a part

Hello?

hello?

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[clever flicker]

What does it mean when it says "f is the principal branch"