help-28

[tough nest]

How?

im guessing logˇ10 16?

[ancient folio]

Pretty useless if u use log base 10

16 = 4^2

Use log base 4

[tough nest]

1/4^x^2

no?

[ancient folio]

Do u know the basis of log?

[tough nest]

Yeah

kidnda

[ancient folio]

Rewrite 16 = 4^2 then

Whats the whole thing looks like?

[tough nest]

Logˇ4 16 = 2

[tough nest]

dunno how this helps tho

[ancient folio]

Whats 0.25 = ?

[ancient folio]

I didnt ask for this

I want u to rewrite the whole thing down but convert 16 to 4^2

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[maiden knot]

I’m confused about how we factor our 1/2 x but then we also have x^1/2 in the denominator?

[steel cove]

we factor out a 1/2

we keep the x^-1/2

[rancid acorn]

additinal note: x^-a = 1/x^a

[maiden knot]

Oh my god wait lol there’s already an x in front there

I completely missed it

[maiden knot]

Yeah I know this rule but I had just completely missed there was already an x in parentheses in front. Thank you haha

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[pearl trout]

https://proofwiki.org/wiki/Correspondence_Theorem_(Group_Theory)

[pearl trout]

So let's say we have 3Z <= Z

i.e. integers that are 0 mod 3 form a (normal) subgroup of Z

So by the above theorem we find that 3Z/3Z <= Z/3Z

i.e. 3Z/3Z forms a subgroup of Z/3Z

ah, nevermind.

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[void jackal]

hii, is my answer for 17b correct? thanks

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[void jackal]

<@&286206848099549185> sorry for tagging

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@void jackal Has your question been resolved?

[void jackal]

<@&286206848099549185> thanks :)

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[marble roost]

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[torn gust]

one thing off, since it's qj, that adds to the pj, so they're together in the pythagorean formula

since it's sqrt(x^2+y^2+z^2) and there's two y vectors to add

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[merry olive]

hi

[merry olive]

idk where to start

[stiff musk]

which question?

[merry olive]

i guess the first one to start but both

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[merry olive]

<@&286206848099549185>

[torn gust]

integrals don't care about coefficients and adding things, so integral of 13f(x)+20 is 13*(integral of x) + (integral of 20), you can separate things

so you just multiply what they give you by three and add the integral of -5 from 1 to 3

which is just a rectangle

the second question is kind of a trap, with exponentials anything added up top is just a number you can pull in front
like e^(x+6)=e^6 * e^x

[merry olive]

where did you get the 13 from

[torn gust]

just an example

I used random numbers

[merry olive]

oh ok

what do you mean by multiply what they give me?

[merry olive]

got it. thank you. could u help me with one more question or should i go into another help room

[torn gust]

you can ask here

[merry olive]

actually ill ask tomorrow im tired lol

thanks though

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[blazing dome]

I know what I have to do

just curious:

can i integrate the (6-y^2)^2 as a whole quantity and not the stuff inside the parentheses?

meaning: 6-y^2 as a quantity

[boreal stump]

what do you mean

[blazing dome]

like as if i just treated the integrand as u^2 - 3^2

[boreal stump]

no unless you're integrating wrt 6-y^2

[blazing dome]

ok ok

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[calm ridge]

How to prove forward direction?

[calm ridge]

Like how do u prove that N(T) = {0v} in general

do u let some arbitrary x,y in the vector space V

and then let T(x) = T(y)

?

Cuz like backwards, u let T(x) = T(y)

then T(x) - T(y) = 0w

and then due to linearity T(x-y) = 0w

Since N(T) = 0v

this would imply that x-y = 0v

so x = y

[boreal stump]

if v is some non zero element in N(T) then
T(v+w)=T(w) for any w but then T is not injective

[calm ridge]

oh

so to prove it

I have to disprove that N(T) can't have any non zero vectors?

[boreal stump]

you can disprove that N(T) can have any non zero vectors

[calm ridge]

ok

thank u

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[lavish terrace]

i dont understand how this converges

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[calm ridge]

By bct u know that (2^2n + 5^n)/(6^n+2^n) < (2^2n + 5^n)/6^n

nvm

[lavish terrace]

i did a ratio test and got this but the answer is suppose to be converge

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[midnight field]

hi

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Please don't occupy multiple help channels.

[undone vector]

Stay in #help-6

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[visual mural]

Im solving 2 and

This is my graph

How do i know where it intersects with the 2root2 + 1?

[craggy tapir]

sir i repeat please read your gmails

[sinful tartan]

Why

[visual mural]

Im saving it man

[craggy tapir]

equate the function to the value

[visual mural]

gotta reach 5000 before graduation

[visual mural]

Ohhhh

Thays genious

Thanks

I get multiple x though

How do I know which one is the point that intersects?

[craggy tapir]

of course

[craggy tapir]

all of them

[visual mural]

but in the answers

7pi/2

9pi/2

pi/2

are the ones that intersects

like what about 3pi/2

where did that go

oh is it because

from 3pi/2

[craggy tapir]

seems like your graph is wrong

[visual mural]

it becomes negative?

[visual mural]

why?

ohhh i figured it our why 7pi/2

thanks

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[agile ermine]

help please

[agile ermine]

no clue what to do

[lofty vine]

write the general form of a quadratic

[agile ermine]

i did 7 = k(x-2)(x-4)

[lofty vine]

$f(x) = ax^2+bx+c$

[glossy valve]

Obotron

[agile ermine]

oh

[lofty vine]

equate f(2) = f(4)

[nimble current]

note that the maximal value of a quadratic occurs at the midpoint of its zeroes

[lofty vine]

you can find max value by completing the square or sub in the middle point of the parabola

[river stream]

0=(x-4)(x-2) 7 is max point

[agile ermine]

why tho

i don't understand it

[river stream]

f(x) is basically y

[agile ermine]

yeah

[river stream]

and x is x

so when y=0 x=2

and y=0 x=4

[agile ermine]

ohh

[river stream]

-> y=(x-4)(x-2)

and when max point

m = 0

y=x^2 -6x + 8

[agile ermine]

yeah

[river stream]

dy/dx

[nimble current]

💀

???

you do not need calc

[river stream]

i aint using calc

ik is 2x-6

[nimble current]

@agile ermine if a quadratic has zeroes at two values, then its max will occur at their midpoint

so you can use this fact to solve it pretty quickly

[agile ermine]

wdym has 0s

[nimble current]

why are you taking the derivative

[river stream]

doing the long way to practice

[nimble current]

a "zero" of the function in this context is a value of x such that the function outputs 0

in this case, the zeros would be 2 and 4

[agile ermine]

ohh

yeah

the question states that

[river stream]

you need to work out the function first

[nimble current]

i dont think that is very helpful

[river stream]

same time is not bad to do it either

[agile ermine]

pls the quickest way

or easiest way

[nimble current]

yep, that's the way im showing you

[river stream]

just do x=3

[nimble current]

so we've already shown that f(x) = k(x-2)(x-4)

[agile ermine]

yes

[nimble current]

so when we plug in 3, we must get 7

which means you can solve for k

[agile ermine]

where did we get the 3 from?

[nimble current]

3 is the midpoint of 2 and 4

[agile ermine]

oh

we plug 3 into x?

[nimble current]

yes

we subsitute 3 for x and 7 for f(x)

[agile ermine]

into this?

[nimble current]

because we are told that the maximum value is 7 and we've deduced that the maximum value must occur at 3

[nimble current]

yes

[agile ermine]

soo

then that means

7 = (3 - 2) ( 3 - 4)

[agile ermine]

?

[river stream]

you missing k but

[agile ermine]

7 = k (3 - 2) (3 - 4)

and do i expand?

[nimble current]

first simplify the 3-2 and 3-4

and then solve for k as usual

[river stream]

k=-7

so what is k

[agile ermine]

k = -7

[river stream]

then what is the equation

[agile ermine]

-7 (x-2)(x-4)

do you always use the mid point to sub in x?

[river stream]

so that is the function?

i dont think so

[agile ermine]

f(x) = -7(x-2)(x-4)

then how do you figure out x?

[river stream]

you dont

but you put the x back into it to check if the answer is correct

normally a quadratic function is ax^2+bx+c

if that his way of doing it and is correct i guess i learned something new today

for now i do my way

[agile ermine]

i'm still confused on how the x was found or where it came from

[river stream]

i have no idea because i didnt do it this way

[agile ermine]

how did you do it?

[river stream]

i have no idea where k(x-2)(x-4) came from

idk what is k

[agile ermine]

that is the method given in my book

[river stream]

is my eye wrong or is that a different question

[agile ermine]

this is an example given in the text

book

[river stream]

the thing is this is a different thing

[agile ermine]

so this textbook is cooked

[river stream]

the other one was given another one and this is given a max point

in the k one you were given

C is 16

this one you were given the max point + C is 7

[agile ermine]

ohh

so what is the method of solving it?

[river stream]

so you have to workout C this time

y=(x-2)(x-4)

[agile ermine]

yeah

[river stream]

but C is the number that causes the extra chage

so Y = (x-2)(x-4)+c

7=3^2-6(3)+8+c

[agile ermine]

how do you know when the method to solve it requires a c value?

[river stream]

7=9-18+8+c

[agile ermine]

what part of the question clued you on this

[river stream]

you were given a max point

you werent given a gradient

[agile ermine]

the max point is 7 right

what was the gradient

oh

[river stream]

you dont need the gradient on this

because is a quadratic which means every single point on the curve have a different gradient

[agile ermine]

oh yeah

[river stream]

you can workout from the direct f(x)

and differenciate it

but for max point

[agile ermine]

i might continue this tmr, i have to sleep

its alr 1am

thank you for your help tho

[river stream]

so you know it is y=mx+c where x=3 and y = 7

c=8

[agile ermine]

this was the answer

[river stream]

i did this the wrogn way

-7 is the gradient

change in y change in x

ok i am dumb

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[indigo orchid]

how many five-digit natural numbers can be formed from the numbers 0-9
How many are divisible by five?
I? Numbers are not repeated

[indigo orchid]

I need help with this math problem

[steel cove]

First one is easy enough

[indigo orchid]

I can calc the Number of all, but i dont know about the one divided by 5

I know

I got that

Already

[steel cove]

A fifth of all numbers are divisible by 5

[indigo orchid]

You mean divide by 5?

[steel cove]

?

[indigo orchid]

I dont understand rn

[steel cove]

Actually wait

Ok so

First question

What did u get

[indigo orchid]

27 216

[steel cove]

How did u get that

[indigo orchid]

10!/5! - 9!/5!

[steel cove]

Oh wait

Oh i mistyped into my calc

Alr yeah thats right

So the numbers that are divisible by 5 are those that end in 5 or 0

[indigo orchid]

Yy

I know

[steel cove]

Ok

So you just calculate that

How many end with 5 or 0

[indigo orchid]

Yeah

How tho

[steel cove]

Calculate how many 5 digit numbers are of the form abcd5 and of the form abcd0

[indigo orchid]

But its 9!/4!

No?

Times 2

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[odd rose]

.

How is this (c)

[low echo]

Hiii

[odd rose]

hi, any idea why this is c?

[low echo]

I get it probability of choosing cider is 5/8 and beer is 3/8

Now of 16 students it says x = number of students who get exactly two ciders

so the prabability of getting exactly two siders for one student is 5/8 * 5/8 * 3/8 * 3c2

Note it is mentioned all students get exactly 3 drinks

So now this above value is the success = p of distribution

Calculate q by using q = 1 - p

Then find the distribution and expected value of students getting exactly 2 ciders

There could be a easy way but this is what all i see

Got it ?

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[torn jolt]

**I am searching for videos related to a particular topic. However, I would prefer more didactic learner sheets instead. **"The concurrency of arbitrary triangles BKB KBK BBB and right-angled triangles."

[atomic hare]

are you talking about triangle similarity theorems and special right triangles

[atomic hare]

uhhuh

bok kat bok => side angle side or SAS similarity

i think youll have more luck looking up the english names instead of polish

[torn jolt]

Form a playlist of videos from you tube related to these topics.

[atomic hare]

you want me to... make a playlist for you?

[torn jolt]

Yes, by gathering yt videos related to these topics.

[atomic hare]

why cant you do it

i gave you the english names

[torn jolt]

Because, Idk how to search this? lol?

[torn jolt]

Its more than one topic..?

What do you mean?

🤔

[atomic hare]

i... told you what the english name is

look those up instead of the polish names

[torn jolt]

For the whole topics?

Check again.

[atomic hare]

yes.

google those

[torn jolt]

Thats 1 / 13

[torn jolt]

.

[atomic hare]

💀 cant you google the rest

[torn jolt]

.

[torn jolt]

THE NAMESSS?

Omg..

[atomic hare]

that thing is written in english already

[torn jolt]

Its Ai generated.

[atomic hare]

lMaO

[torn jolt]

Are you helping or twerking?

[atomic hare]

anyways

you should be able to google the other topics

i dont see any problems with the terms used there

let me help you get started https://google.com/search?=Properties+of+isoceles+triangles

[torn jolt]

Help me or let someone else help. Don't waste my time. BRUH.

No rude, no help. Good.

[lucid flower]

no one else is gonna help you with this sorry, if you want a youtube playlist you gotta make one yourself

you’re asking someone else to make a google search for you, someone told you how to search it and then you got mad

[torn jolt]

I don't want a YT playlist. I want a group of videos. BRUH.

[torn jolt]

Bro you think I havent tired?

Can just the helper talk?

You have to tell me you can't do it? Just don't answer.

[atomic hare]

no offense but yeah i dont think you tried because literally copy pasting the ai generated curriculum yields promising material

[torn jolt]

Either I am getting old, or I am in the wrong place.

fking .close

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[torn jolt]

Cna anyone check if my answer is correct

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

[river stream]

@torn jolt i can not really see the question

[torn jolt]

Sorry. It was about finding solve of the differential equation using constant coefficient

[river stream]

i dont think there can be a constant coefficient @torn jolt

y=ln(x^2) / x^2y^3+1+x

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[wild fox]

Hi, I have a question about differential equations, when dealing with something like dampened pendulum motion
we have to solve for the differential of $l\frac{d^2\theta}{dt^2}+\frac{d\theta}{dt}b+g(\theta)=0$
which gives $Ae^\frac{-b}{2l}e^{\pm\sqrt{\frac{b^2}{4l^2}-\frac{g}{l}}}$
I have a question about the square root part of the e exponent $e^{\pm\sqrt{\frac{b^2}{4l^2}-\frac{g}{l}}}$
how does this get turned into
$cos(\omega't)$ exacly? ($\omega'$ = $\sqrt\frac{b^2}{4l^2}-\frac{g}{l}$) ?
why does the $\pm$ dissapear? and how is this turned into a complex exponent?

[wild fox]

sorry still learning LaTeX lol

[glossy valve]

Mephisto

[torn gust]

hmm the only way you go from e^ to cos() is with these identities

[wild fox]

oh so the $\pm$ just indicate the two solutions, and then we just take the linear combination by adding them up and deviding by 2?

[glossy valve]

Mephisto

[torn gust]

oh yea e^(positive)+e^(negative) means cosine, that's good

[wild fox]

and I assume that the $\frac{A}{2}$ part of the equation just gets absorbed into the constant?

[glossy valve]

Mephisto

[wild fox]

but where do the i's come from though? I have e to a real exponent

[torn gust]

yea the Ae^(real number) part just stays in front

according to this the omega w's are supposed to have an i in front https://math.libretexts.org/Bookshelves/Scientific_Computing_Simulations_and_Modeling/Scientific_Computing_(Chasnov)/II%3A_Dynamical_Systems_and_Chaos/11%3A_The_Damped%2C_Driven_Pendulum

maybe b^2/4l^2-g/l is a negative number so it becomes imaginary

[wild fox]

ohh so we factor -1 out?

$e^{\pm\sqrt{(-1)*(\frac{-b^2}{4l^2}+\frac{g}{l})}}$

[torn gust]

yea with the b and g terms flipped

[glossy valve]

Mephisto

[torn gust]

assuming that makes physical sense

then you do the 1/2 part and replace it with cosine of the inside

[wild fox]

yeah b is generally smaller then gravity

alrighttt this makes alot of sense now

thank you!

[torn gust]

np 👍

[wild fox]

one more question, when solving for the amplitude after a time t, why does the cos part dissapear? let's say we want to know what the damping factor b is, if we reach 50% of our intial angle, after 5min/ 300sec?

so if we would plug everything in, the cos part of the equation would just give $cos(\omega'300)$

[glossy valve]

Mephisto

[wild fox]

but it become 1 for some reason, why?

[torn gust]

amplitude is like the top of the wave

and cos() is just a thing waving between -1 and 1

so if you plug in cos=1 you get the curve above the wave

[wild fox]

oh so we assume that cos(omegat) is 1, because 50% of the amplitude is reached after 5minutes?

so it has to be at its 'highest' amplitude possible at that time

[torn gust]

yea here's the picture

you just care when the red part hits half A0

[wild fox]

ohhh ok

so when amplitude is ever mentioned, or calculating when amplitude becomes x, the cos part is assumed to be 1 or -1

[torn gust]

yes

it's like ignore the wiggle

[wild fox]

yeah, it's like a discrete function of amplitudes

alright thanks alot again

[torn gust]

np

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[fast marlin]

Wouldn’t you add p(b)*p(a|b)?

[tribal viper]

Y

Don’t u use multiplicative principle thing

[fast marlin]

Because that’s a different event that could occur if b happens first then a could happen given b

Then that’s still p(a and b)

[quartz gale]

Well that’s actually how you define P(a|b)

I’m sure there is some intuitive understanding of this equation

But this equality is how you define p(a|b) so you can’t change this definition by adding p(b)*p(b|a)

[neat aspen]

you don't add but p(b)p(a|b) = p(a)p(b|a) = p(a and b)

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[wintry birch]

I shall show that this mighty thing is a well defined, open, not empty interval in R. I get the task somewhat, but i always struggle to get behind the "well defined" stuff. What do i have to show to proof that its "well defined"? i dont want a solution, rather just the thing that i need to show.

[neat aspen]

!original

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

[wintry birch]

"Let A be open in R with standard metric" was missing

The rest is exactly what i said 🙂

[neat aspen]

basically you need to consider when there's more than one representitive (u,v) which a belongs to

[wintry birch]

Mhhm, but what couldnt be well defined then? i mean then just both (u, v)'s would be in Ja.

[neat aspen]

well that's your job to consider, in this particular example you're asked to show it's well defined so these Ja's will have only one interpretation

i can give you an example of something not well defined if you'd like

[wintry birch]

Yeah, maybe that helps, cuz i dont seem to understand the problem.

[neat aspen]

ok so let X, Y be sets

and define the function $f(c) = 0, c \in X$ and $f(c) = 1, c\in Y$

[glossy valve]

chebyshev's infinite pee norm

[neat aspen]

what happens when X and Y are not disjoint?

then they're going to contain a common element, say d

[wintry birch]

okay, i do see the problem here

[neat aspen]

what should f map d to?

[wintry birch]

yeah, thats understandable.

🙂 Buut....

[neat aspen]

your (u,v)'s are not disjoint

for example

$J_0 = \bigcup {(u,v)\mid 0\in(u,v)\subset A}$

[glossy valve]

chebyshev's infinite pee norm

[neat aspen]

well 0 can be chosen from the representative $(-1, 1)\cap A$ as well as the representative $(-2, -2)\cap A$

[glossy valve]

chebyshev's infinite pee norm

[neat aspen]

it's your job to show that $J_0$ can be uniquely defined regardless of the representatives 0 is coming from

[glossy valve]

chebyshev's infinite pee norm

[neat aspen]

and in particular all $a\in A$

[glossy valve]

chebyshev's infinite pee norm

[wintry birch]

Im sorry, im feeling very dumb, but what does the representative has to do with it? I thought that Ja is just the combination of all of those representatives. Why even should one be choosen that zero comes from?

[neat aspen]

i was just giving a particular example $J_0$ but you need to show it holds for all of $a\in A$

[glossy valve]

chebyshev's infinite pee norm

[wintry birch]

Yeah, but still. Ja includes every representative. Why should one be prioritised if i throw everything into one pot?

[neat aspen]

but it's a generalised union

you need to show that this generalised union is a unique representation

it can't be the case that J_a = one thing = some other thing that's different

[wintry birch]

Okay, but what aside from "The union of all the intervals of A that include a" should it be? I hate this stuff, proofing something is well defined was always the hardest part for me. Even some rather simple exampels like the modulo addition or smth like that. Bc here i have one, imo clear definition. Why should i get something different when i use other representatives if i just throw in every interval?

[neat aspen]

well it's based off the metric/topology right

[wintry birch]

i guess

[neat aspen]

Why should i get something different when i use other representatives if i just throw in every interval?
well you shouldn't in this particular question because it is well defined, you're just tasked in arguing it

i think you may be over thinking it

its likely exactly what your intuition is telling you

if a is a member of two representitives, then a should also be in their union

[wintry birch]

Mhhhm...

[neat aspen]

so if we take a in J_a it shouldn't matter which representative (u,v) a is coming from

when we consider the open sets J_a

[wintry birch]

Ahhh, is my task to show that: if a in (r1, r2) and (r3, r4), then J_a ist the same whether we first do (r1,r2) u (r3,r4) and then put it in J_a or just put them both in J_a?

[neat aspen]

pretty much yeah

i assume you're trying to show J_a is a topology?

er

satisfies the topology axiom

[wintry birch]

The task wants to show that A is always a union of disjoint intervals

Can i ask you a quick follow up question on that task? its on my proof that its really only one interval for every a.

[neat aspen]

do you mean J_a is just one interval for every a?

[wintry birch]

yeah

im not sure if my argument is perfectly fine

I did the following:

If A=R, its trivial
If A!=R, and a in A, w/out loss of g, there ex. an x in R\A s.t. x>a -> M={x in R\A | x>a} has a lower bound and therefore must have an infimum. -> (a-E, inf(M)) is completly in A and open. But for every b<a, x>inf(M), there must exist a point in X\A s.t. its an element of (b,x), therefore (b,x) is definitely not in A.
-> J_a has the supremum inf(M).
same argument for lower bound. Every for every point p in between J_a's supremum and infimum, its either p=a, which is in a, or its p<a, then we take (inf(J_a),a) or its p>a (a, sup(J_a)). therfore J_a = (inf(J_a), sup(J_a))

[neat aspen]

A is just a connected open set no?

[wintry birch]

huh?

[neat aspen]

oh not connected sorry

what is your set X

(a-epsilon, infM) ?

[wintry birch]

Oh, its just R lol

[neat aspen]

o

[wintry birch]

i always switch them up, cause in the uni we always talk about X, not R specifically

[neat aspen]

ok i think i am brain lagging

But for every b<a, x>inf(M), there must exist a point in R\A s.t. its an element of (b,x), therefore (b,x) is definitely not in A.
-> J_a has the supremum inf(M).

why does this imply J_a has the supremum inf(M)

[wintry birch]

Wait, i wrote it a little unclear.
Consider b>inf(M). For b to be in Ja, there must exist an interval (l, u) s.t. l<a, b<u completely in A. But bc b>inf(M), there must exists at least one x in R\A in (inf(M), b), otherwise b would be a greater lower bound for M. --> x in R\A c (inf(M), b) c (l, u) -> (l, u) !c A. Therefore b cannot be in Ja.
-> inf(M) is an upper bound for Ja.

and bc every a<x<inf(M) is in A (otherwise inf(M) wouldnt be a lower bound for R\A), this means that inf(M) is the smallest upper bound for Ja, thus the supremum

[neat aspen]

ok yea

Every for every point p in between J_a's supremum and infimum, its either p=a, which is in a, or its p<a, then we take (inf(J_a),a) or its p>a (a, sup(J_a)). therfore J_a = (inf(J_a), sup(J_a))
what are you doing here exactly?

[wintry birch]

Im sleepy, its again wrong, i would have to use an Epsilon with the a somehow. But its just there to show once again that for every x in between those two bounds, we really find and intervall, which is rather dumb, cause i could just give the interval (inf(Ja), sup(Ja)), i was freestyling and hadnt exactly thought it through

[neat aspen]

lmao ok

i haven't thought too hard but maybe try showing (sup -M, inf M) and J_a are subsets of each other

and once you have that

to show well defined, take a = b, and show that J_a = J_b

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[wild fox]

Hi, I don't really understand divergence of a vector field and its implications
So by definition a divergence of 0 implies no sinks or sources
does that mean that the vector field has no attractor? or does it mean that the vector field never converges?
Should I interpret it as vector field density?

[wild fox]

can it be interpreted as instantenious change of direction?

as in if you would take the corss product of a vector at F(x,y,z) and F(x+dx,y+dy,z+dz), and took the divergence of it, it should give zero?

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[viral jasper]

Divergence, in short, is the instantaneous measure of how much a field is going into, or out of a specific point

Zero divergence means that the field is flowing into a point at the exact rate as it is flowing out

[wild fox]

is this a right way to see it?

[wild fox]

so it's like the flux of a single point?

[viral jasper]

Yes exactly

Check out the definition at Wikipedia

https://en.m.wikipedia.org/wiki/Divergence

Divergence is defined as the limit of the flux as the enclosing volume approaches zero

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[scarlet iris]

how did he get the W to be [1,0]?

[scarlet iris]

isnt the w supposed to be a transpose of one of the rows of the (A-lambda*identity) vector?

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[dreamy torrent]

I need help with part b. I know that there is only 1 way when the first number is 4, but idk how to model the number of ways for numbers 3,2,1

[torn gust]

yea building it up one by one won't work, another way is looking at completed numbers, so how would you make each of these increasing?
759231
126598
etc.

[dreamy torrent]

divide by number of ways they can be arranged

[torn gust]

yup, so just do that and remember you can't use 0

[dreamy torrent]

is this q just combinations then?

[torn gust]

yea it's choosing

same reason why sets are written increasing order {1,2,3}

[dreamy torrent]

ok

thanks

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[slender heart]

What formulas would I use for each of these

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[celest pewter]

should be $P' = P(1 + \frac{r}{n})^{nt}$ for all of the ones where you compound every period of time

if there's no period you can just do $P' = P \cdot r^t$

[glossy valve]

b

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[slender heart]

Thanks

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[simple mica]

just need help with c

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[simple mica]

help c?

the answer key shows 24

but idk how to get that number

<@&286206848099549185>

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[tropic torrent]

This is more compsci than stricly math but I need to make a parser for propositional logic using bnfc so if anyone's familiar with that tool, could you take a look at where my mistake is.

These are the contents of my bnfc file for prop logic.

AtS. S ::= Ident ;
NegS. S ::= N ;
ConjS. S ::= C ;
DisjS. S ::= D ;
ImpS. S ::= I ;
EqS. S ::= E ;

Neg. N ::= "(" "-" S ")" ;
Conj. C ::= "(" S "&" S ")" ;
Disj. D ::= "(" S "|" S ")" ;
Imp. I ::= "(" S "=>" S ")" ;
Eq. E ::= "(" S "<=>" S ")" ;

From this definition:

Definition 
The sentences of propositional logic language L based on the non-empty set of atomic sentences P are formed as follows: 
1. Every atomic sentence p∈P is a sentence of the language L. 
2. If α and β are sentences, then (¬α), (α∨β), (α∧β), (α→β), (α↔β) are also sentences of language L. 
3. Only strings formed on the basis of the two previous rules are sentences of the language L. 

Example 
If P={p,q} , then for example p,q,(¬p),((¬p)∨q) and (((¬p)∨q)→p) are sentences of language L, but not (¬()) nor (p∨r).

Automatic grader gives me 0 points but no feedback as to why lol. I have a little over 100 tests of my own that the resulting parser passes.

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[distant silo]

I had this exercise on an exam today, so I decided to check it with symbolab, but I can't seem to get why it gets a different answer from mine. Could somebody explain? it somehow gets a second -1 in the final solution and ends up with -2/9, I solved it several times and can't seem to find out how. It does solve it in a different way(with substitution), so it's possible it got it wrong but it's more likely that I got something wrong somewhere and can't find out what...

[distant silo]

It's a fairly simple integral too 🤔

[stiff musk]

cos(0) is 1, not 0

[distant silo]

.

oops

[stiff musk]

mystery solved ⚔️

[analog raptor]

blud was doing meth and math at the same time

[distant silo]

Im so stoopid 🤦‍♂️ well thx

[stiff musk]

it happens to all of us haha

[analog raptor]

relatable

[stiff musk]

cheers

[distant silo]

!solved

.solved

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[stiff musk]

username checks out

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[quick sage]

https://cdn.discordapp.com/attachments/909417980532756581/1230508401713217536/image.png?ex=66339340&is=66211e40&hm=67348da145ae69de1b95246ac2225ad73ee77e49e9f4367d6ab201cc4d3c36c6&

how does this work?

i get putting it in terms of b

but not the step after

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[dreamy torrent]

how do u guys do part a without the integration (it's 1 mark)?

[modern oracle]

1/2a is const

[dreamy torrent]

ok so what do I do next?

oh is the graph just a rectangle then?

[modern oracle]

yes

[dreamy torrent]

base times height = $1/2a \cdot 2a$

doesn't this equal 1

but the answer is 2a

[modern oracle]

if you have same probability to choose x from a and 3a

[dreamy torrent]

yes

[dreamy torrent]

it's just the average of 3a and a?

[modern oracle]

yes

[dreamy torrent]

ok thanks

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[ashen laurel]

.reopen

.open

[livid scarab]

plug x-1 and x^2+4 into the function

[ashen laurel]

Ik i gotta do that i need help after that part

😭

Its giving me an answers that are wrong when i do it

[crisp fossil]

what answers are you getting

[ashen laurel]

I got x^2-x once

💀

[crisp fossil]

yeah i know how

you plugged in x-1 instead of (x-1)

[ashen laurel]

I also got x^2-2-1

😭

[crisp fossil]

try replacing the x's in the first equation with (x-1)

[ashen laurel]

Huh

🤧🤧

[crisp fossil]

dyk how parentheses work in math

[ashen laurel]

Im dumb af when it comes to math ugh

😭

[crisp fossil]

nah you probably arent

[ashen laurel]

Working on it 😔

[crisp fossil]

anyways try this
(x-1)^2
vs
x-1^2

what is the difference?

[ashen laurel]

Well in the first part its being squared with the whole bracket while in the second its being squared with -1 only

[crisp fossil]

yeah so when you are replacing the x's in the function, you have the replace the x with the whole thing and make it parentheses

so the first answer would be:

(x-1)^2 - (x-1)

[ashen laurel]

Yeah i got to that part

We put down the formula there right?

[crisp fossil]

which formula

[ashen laurel]

(a-b)^2

[crisp fossil]

well i wouldnt really think of that as a formula right now

if they make you expand this then FOIL out the first term

[ashen laurel]

🤔

[crisp fossil]

and then distribute the negative in the second term

[ashen laurel]

Foil? 😭

[crisp fossil]

basically x^2 is x*x correct?

[ashen laurel]

Yeah

[crisp fossil]

(x-1)^2 is gonna be (x-1)(x-1)

[ashen laurel]

Ahhh

[crisp fossil]

so you can multiply those two together to make
x^2-2x+1

[ashen laurel]

Is that the final answer?

[crisp fossil]

no

thats the expansion of the first term

[ashen laurel]

🤔

Uhuh

[crisp fossil]

(x^2-2x+1)-(x-1)

solve that and thats your answer

what do you think it is

[ashen laurel]

x^2-3x+2?

[crisp fossil]

correct

[ashen laurel]

Ah

😭😭

And the second part??

[crisp fossil]

do the same method

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[rough walrus]

i know i need to show work but i honestly don’t know what to do here

[quartz scroll]

6444

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[lunar fiber]

Think I missed something in class, how am I supposed to solve this?

[lunar fiber]

I mean I could definitely solve it but it would take a decent bit of calculating and I don't think that's the proper solution

[torn jolt]

wait nvm

if you look at the OBC triangle

is it a right angle triangle?

[lunar fiber]

I can't really tell

[torn jolt]

I don't think I can assume that since there is nothing that claims it is

[lunar fiber]

Yeah

It might be though, lemme solve it assuming it is rq

I have the answer so that may be the case and the question is just stupid

[torn jolt]

I think it should be stated if it was, since then the OB side would just be the radius.

There might be some formula for this though lemme check.

[lunar fiber]

Yeah it isn't a right triangle

[torn jolt]

You might be able to set up a system of equations using this, but no idea if it is the right solution.

[lunar fiber]

Maybe? I'm not completely sure how to do that though

[torn jolt]

is the solution radius = 18 cm?

[lunar fiber]

Should be 10cm

Unless I wrote the wrong number

[torn jolt]

yeah then that formula didn't work.

[lunar fiber]

Was sleeping during class and I just copied down what I thought was the answer

[torn jolt]

Mark the foot of O on the chord AB

and connect it

this would be a perpendicular bisector

[lunar fiber]

Then what?

[torn jolt]

let the foot be D

then

AD would be 13/2

following?

[lunar fiber]

Yep

[torn jolt]

CD would be 13/2 - 4

[lunar fiber]

5/2

I got about that far earlier but didn't know what next

[torn jolt]

yes now you can find OD

since ODC is right angled at D

Now consider the triangle ODB

[lunar fiber]

No idea how I didn't figure that out earlier lol

[torn jolt]

you can now find OB which the radius

[lunar fiber]

I think I get it

Thanks 🙏🏻

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[torn jolt]

is this right

[torn jolt]

and this

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[balmy flicker]

5. Daniel often watches two of his satellite channels on TV. He knows that Kanal A shows 18
   minutes of advertising per hour and Kanal B shows 13 minutes of advertising per hour. He also knows that the two channels show commercials at times that are not dependent on each other.
   a) Daniel turns on channel B. What is the probability that the channel is showing advertising at that moment?
   b) Daniel hits his TV on another occasion. What is the probability that
   at least one of the two channels Kanal A and Kanal B is showing something other than commercials at that time?

[balmy flicker]

I need help with B

[nimble current]

what is the complement of the probability you want to find?

[torn jolt]

is the answer for (a) 13/60 ?

[nimble current]

yes

[torn jolt]

thank you

(b) would be P(A)+P(B)+P(A)*P(B)

right?

@nimble current

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[balmy flicker]

The procentage

Of the B

answer on a) is 22%

[balmy flicker]

that is wrong

a is 13/60

aka 22%

b) answer is 93.5 %

I just need explaintion on how

[nimble current]

so you can use complementary probability @balmy flicker

[balmy flicker]

yes

[nimble current]

the probability that at least one of the channels is not showing comericals is 1 minus the probability that both channels are showing comericials

[balmy flicker]

exactly

[nimble current]

the latter probability is 13/60 * 18/60 = 0.065 = 6.5%, so the answer is 100%-6.5%=93.5%

[balmy flicker]

🤦‍♂️

I am so stupid

thanks for the help

I read the question wrong

I did 47/60 X 42/60

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[torn jolt]

#14

[umbral dome]

what about it?

[torn jolt]

Does the dimension of H equal the dimension of the basis of column A

And why

[umbral dome]

what is H?

[torn jolt]

H is a subspace

[umbral dome]

in this question we are considering the subspace which is the span of the 5 vectors given

[torn jolt]

Do I need to use the basis theorem
I barely understand it

[umbral dome]

since we put them as columns of a matrix, it is also by definition the column space of the matrix

[umbral dome]

so we have that the column pace is span{c1, c2, c3, c4, c5} (where c denotes the columns). so by definition the set {c1, c2, c3, c4, c5} spans the space

however, they are not linearly independent, so they are not a basis.

We are trying to find a subset of {c1, c2, c3, c4, c5} that has the same span, and which is linearly independent

[torn jolt]

So c1 and c2 form the basis because they are pívot columns

OHHHH I GUESS SO

[umbral dome]

yes. they span the same space: span{c1, c2, c3, c4, c5} = span{c1, c2}, and they are linearly independent

[torn jolt]

the rest are just linear combinations of the first two!

OHHHH

omg

It’s making sense

because like a basis generates all the points in the subspace???

and these two vectors generate everything else

[umbral dome]

yes. the original 5 columns did that too, but they were redundant (we didn't need all of them). so a basis is the set of vectors that does that without any redundancy (the bare minimum needed to generate the subspace)

[torn jolt]

Yesssss

I saw that in one YouTube video

What about rank A + dim Nul A = n

So if I wanted dim Nul A

Is it 3

[umbral dome]

yes

[torn jolt]

and I saw in a yt video
This guy said n is like a steak
The meat is the rank and the dim Nul A is the fat

[umbral dome]

the null space of A is the set of solutions to Ax = 0, and each "free column" (non-pivot column) contributes a parameter/free variable

[torn jolt]

Right! That’s why we put Ax = 0 in parametric vector form, because we putting free variables in terms of basic variables

NO

Basic in terms of free

Yeah

[umbral dome]

yes, so then we write the solution as [ t_1 \vb{v}_1 + t_2 \vb{v}_2 + t_3 \vb{v}_3 ] which is exactly the same as $\nullsp A = \vspan{ \vb{v}_1, \vb{v}_2, \vb{v}_3}$ (so the nullspace is 3-dimensional)

[glossy valve]

pnoןɔ

[torn jolt]

I’m stressed I have an exam tomorrow 😭

Yeah

How about 15

How do I do it

There can only be 3 pivot positions

[umbral dome]

so if there are 5 columns and 3 pivots, what is the dimension of the column space, and what is the dimension of the null space?

[torn jolt]

Dim H = 3

#16

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[steep dragon]

I’m assuming I use the fact that a matrix is invertible iff det neq 0

it definitely won’t be diagonal as I will have a number and -a^2 in the first col

[neat aspen]

you could use column 5 as the coefficients for cofactor

it's all 0 except the last entry

then keep cofactoring with the maximum amount of 0's

[steep dragon]

would it be beneficial to row reduce first, or at any step because there’s a lot of nonzero terms

with the first submatrix because we get rid of the a in the first column, we can get rid of all of the entries in that column

[neat aspen]

haven't gone through it myself but if you see any point that you can row reduce for more 0's then it's probably recommended to do so

since it will save you alot of computation time

[steep dragon]

alright thank you

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[violet loom]

how would I go about this integration., i cannot work out how to sub u into this

[neat aspen]

looks like to rewrite x in terms of u and then differentiate

[violet loom]

differentiate?

[neat aspen]

so that you can substitute it in

[violet loom]

hold up ill try that thanks and get back to you!

like this?

[neat aspen]

hmm

looks ugly

lemme think

[violet loom]

but how would i get rid of the 1/(x+9x^3)

like wait ill show

[neat aspen]

well x = 1/sqrt(9(u - 1))

[violet loom]

is that legal

[neat aspen]

so i was saying that you can substitute this in for every place you see an x

but it looks ugly so i am doubting this approach

[violet loom]

lol that would be so clapped but yeah that could work but would it really be easy to integrate

[neat aspen]

and no 1/(a + b) != 1/a + 1/b

[violet loom]

is that a does not equals sign

[neat aspen]

yea

[violet loom]

ahhh

i even asked chat gpt still couldn’t do it

[neat aspen]

oh

it's partial fractions

i think

how did i overlook that

[violet loom]

how would i do partial fractions sorry

[neat aspen]

oh have you not learned that yet

[violet loom]

no maybe its assumed knowledge i must have forgotten

[neat aspen]

it's just if you have a denominator as a product of polynomials, you can rewrite it as a sum that looks like this

$\frac{r(x)}{p(x)q(x)} = \frac{a(x)}{p(x)} + \frac{b(x)}{q(x)}$

[glossy valve]

chebyshev's infinite pee norm

[violet loom]

oh ok i see

how could that be applied here tho

[neat aspen]

have you seen this before?

[violet loom]

yea

[neat aspen]

you have $\frac{1}{x(1 + 9x^2)} = \frac{something}{x} + \frac{something else}{1 + 9x^2}$

[glossy valve]

chebyshev's infinite pee norm

[violet loom]

ahh

yeah

but what is something and something else

[neat aspen]

if you're familiar with the process of partial fractions, then it'll be your job to figure out what those somethings are

[violet loom]

ye okay

[neat aspen]

That's not what the partial fractions should be

how did you get thos

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[steep dragon]

it isn’t when they’re both 1 because we could have i=1 and k=0 and a fine derivative

wait when they’re both zero i+k=0 so can I just have the sun from zero to d?

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[steep dragon]

<@&286206848099549185>

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[steep dragon]

<@&286206848099549185>

[steep dragon]

bruh someone please 😭

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[empty plover]

I have a quick question. How do I solve this?

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[lavish stag]

How would I do this? I'm not 100% sure how do to find the coordinates for the graph when i am just given the derivatives and intervals.

[royal grotto]

I can give a rough idea

Start by marking the given points, then use all of the other information to understand how the funciton behaves between those points

f'(x)=0 signifies that the slope of the graph at that point is 0 in other words the normal at that point would be horizontal
f'(x)>0 means that the function is increasing in that interval
and vice versa for f'(x)<0

f''(x)=0 means that that point is the point of inflection
f''(x)>0 means that the graph has upward concavity in that interval
and vice versa for f''(x)<0

[lavish stag]

okay, I'll try and let you know how i do!

[royal grotto]

sure, gl!

[lavish stag]

I'm still a little confused on the analyzing the derivative part, How do i determine where the function is increasing or decreasing based on the derivatives?

[royal grotto]

for example f'(x)<0 from x=-4 to x=4 so the function would be decreasing in this interval

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[lavish stag]

this lowkey still does not make sense to me 😭 🙏

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[dense edge]

did I solve this correctly? does it want a matrix for the answer or x and y value as the answer?

[quartz gale]

There is a problem here

Did you calculate at any point AB ? Or BA ? @dense edge

[dense edge]

Just did that now

[quartz gale]

Okay so you agree that what you did was wrong ?

[dense edge]

yeah im not sure what i found

A+B

what i need is AB

[quartz gale]

It’s just that you just solved A=B

[dense edge]

oh ok

[quartz gale]

Look at the first photo

[quartz gale]

Where is BA ?

[quartz gale]

It’s right

[dense edge]

is this getting any closer lol

i found y now

[quartz gale]

Too bad

Almost everything is good

Except the first thing you calculate for BA

look at the first coefficient you calculated

[dense edge]

wdym?

i dont see it

[quartz gale]

For BA