help-38

[bright patrol]

$$
\begin{aligned}
2\pi&\int_0^5x^3\sqrt{1+9x^4}:dx \
&u=1+9x^4\qquad \frac{du}{dx}=36x^3 \
&\frac{dx}{du}=\frac1{36x^3} \
&\int_a^bx^3\cdot\frac1{36x^3}\cdot\sqrt{u}:du \
\frac{2\pi}{36}&\int_a^b\sqrt{u}:du
\end{aligned}
$$

[solid kiln]

AWACS Sky Eye

[bright patrol]

after some more work, i'm at the point in the video but now i don't know what to use to recalc limits

[wraith arch]

so since we know u = 1 + 9x^4

we just plug in the original bounds into this equation

to change the bounds so that it follows u

[bright patrol]

...huhn. thanks.

[wraith arch]

the reason why we can’t use the original bounds is because they are what x is equal to

we have to change everything to u when substituting

since we are converting everything to in terms of u

[bright patrol]

$$
\begin{aligned}
2\pi&\int_0^5x^3\sqrt{1+9x^4}:dx \
&u=1+9x^4\qquad \frac{du}{dx}=36x^3 \
&a=1+9(1)^4\qquad a=10 \
&b=1+9(5)^4\qquad b=5626 \
&\frac{dx}{du}=\frac1{36x^3} \
&\int_a^bx^3\cdot\frac1{36x^3}\cdot\sqrt{u}:du \
\frac{2\pi}{36}&\int_a^b\sqrt{u}:du \
&\int_a^bu^{1/2}:du \
\frac{2\pi}{36}\cdot\frac2{3}&\biggr[u^{3/2}\biggr]a^b \
\frac\pi{27}&\biggr[u^{3/2}\biggr]{10}^{5626}
\end{aligned}
$$

[solid kiln]

AWACS Sky Eye

[bright patrol]

this doesn't look right, and it's not right at least when i put it in.....

[wraith arch]

ooh a should be 1

since the lower bound was originally 0

so u=1+9(0)^4

[trim jolt]

@bright patrol Has your question been resolved?

[bright patrol]

OH.

big thanks! @wraith arch

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[raven timber]

hello

[raven timber]

i have a probability question,

why did we add 0 not 6 number

to the universe?

[vagrant prism]

it doesn't matter

you can have 1,2,3,4,5,6 instead

possible explanation is that in programming everything is 0 based, so it might be a habit of your lecturer to start with 0

but in general it is often conventional in math to have a 0 based list

[vagrant prism]

but i digress, in this case it doesn't matter

[raven timber]

oh ok thanks

.close

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[obsidian hornet]

Is it possible to use the thm above with open bracket intervals..

[nimble stone]

what if i had 1/sin(x) on (0,pi/2), its continuous on that region, but it has no maximum

[obsidian hornet]

ahh

[neon dirge]

the closed interval & continuity leads to the existence of minimum/maximum on that interval

which is why it's a requirement for you to use it

[obsidian hornet]

oh ok

I gotta leave for around an hour maybe more..

I'll be back tho!

hopefully bot doesnt close channel

[neon dirge]

sure, occupy as you wish

[neon dirge]

well it will

if you want to ask more about the above images just copy them into your dms

then you can use them later too

[trim jolt]

@obsidian hornet Has your question been resolved?

[trim jolt]

@obsidian hornet Has your question been resolved?

[obsidian hornet]

okk Im finally back

a continuous surjective map doesnt mean its surjective right?

wait no

continuous surjective map means its continous and surjective by the name

[obsidian hornet]

But if we are given the fact that the function is continuous and surjective, then it must have a min/max even if its open bounded

so heres the question

what Im thinking rn is I can apply the Intermediate value theorem, even if its open intervals

because I can find a limit as x approaches 0 from right

and limit as x approaches 2pi from left

limit exists

right so, to prove its not a bijection, I just have to prove its not injective because its already given that its surjective and continous

but for it to be injective, you CANT have two values in the domain to map to the same value in the codomain

so Im thinking then by IVT I can find such c where the limit also goes to 1 or -1

These are just my thoughts for now...

Im still working it out

IVT I can find such c where the limit also goes to 1 or -1 <- this part is just an assumption so Im gonna try finding more info about this

[trim jolt]

@obsidian hornet Has your question been resolved?

[whole coral]

well, IVT is a good idea for sure (noting the continuity of f), and...

[whole coral]

...yes, you do know that f has a maximum (of 1) and minimum (of -1), and must attain those somewhere by the surjectivity

If you haven't gotten an idea yet, hopefully those two statements should do it

[obsidian hornet]

Hello! yup Im working on it now

after sending my last message I left to eat food😅

[whole coral]

Awwww I'm just getting inside now

Food time for me

[obsidian hornet]

if f maps the interval (0, 2pi) onto [-1,1], -1 and 1 would be attained once in (0,2pi) if you suppose injectivity

and surjective just means all y in [-1,1] gets mapped by x in (0,2pi)

[whole coral]

Sure to both, though you may find it easier, rather than assuming injectivity to contradict it, to directly show that there isn't injectivity

[obsidian hornet]

how can I show that more than one x in the domain maps to the same y in the codomain...

id have to show -1 and 1 could be attained more than once in(0, 2pi)

[whole coral]

Not necessarily -1 and/or 1 need to be attained more than once-

As per before, you know that -1 is attained at least once, say, at the point a (between 0 and 2pi exclusive), and that +1 is attained at least once, say, at the point b...

You have either a < b or a > b (you can assume either without loss of generality), but with that, you can make use of those and something else you know

[obsidian hornet]

IVT, so there exist c in (a,b) s.t c=k if f(a)<k<f(b) (or > it doesnt matter)

[whole coral]

Yep, in other words, between a and b, you attain all values between f(a) (= -1) and f(b) (= +1)

But then, there's something slight I had mentioned before that's quite important, see if you can figure it out

[obsidian hornet]

right but you can also find x outside [a,b] wich is a subset of (a,b), and that x also maps to some y in [-1,1]

proving it is not injective

[whole coral]

Yep, that's basically the idea

On the interval [a, b], f attains all values between -1 and 1, but then, you have a > 0 and b < 2pi, so there are some elements in the domain you haven't "used", but you've "used" all the elements in the range already - you basically have to "reuse" some values

[obsidian hornet]

ok

thank you

ahh now I gotta write it down

thats the hard part too

cause its hard to get ideas down on paper

at least for me I think

[whole coral]

Awwww yea it can be hard to write out the ideas properly sometimes, even when you can see how it'd work and have the ideas

[obsidian hornet]

yuppp

I definitely need more practice on proof writing

I take wayyy too long

@whole coral Im wondering if you could help check?

[whole coral]

Of course

[obsidian hornet]

oh no i did smth wrong for sure

[whole coral]

You haven't! The comments I have are mostly "grammatical" more than anything

I would say that to make clear that a is one of those points where f attains the value -1 (and similarly b and +1)
I would also restate the IVT part: basically more phrasing it that "for any k such that f(a) < k < f(b), there exists a c in (a, b) such that f(c) = k"
But you have it

Just tiny things to make it pretty clear what you mean

[whole coral]

(also another really tiny point, I think I would say in the sentence "f attains all values between -1 and 1 on the interval [a, b]" to make it clear that it's within that subset of (0, 2pi) that you've attained the values in the range, making it clearer about the later part that you're not injective, but not super essential I think )

[obsidian hornet]

there!

thank you soo much

I really appreciate the help

and you're also very good at explaining stuff

[whole coral]

Wonderful

[whole coral]

Awww why thank you

Always a pleasure to work with you

[obsidian hornet]

ohhh

wait I missed smth

its f(a)<f(b) cause I stated -1 is attained at a

so I dont need the w.l.o.g stuff in the brackets

[whole coral]

Yeaa you know that f(a) was -1 and f(b) was 1, or at least chose a and b such that those were the case

You could, instead, have "fixed" a and b such that a < b, then let f(a) and f(b) be either -1 or 1, then do the without loss of generality there

[obsidian hornet]

ahhh

i think im gonna take a break from work

[whole coral]

Awwww you definitely should, if you need one

[obsidian hornet]

ahh my friends are asking me to go out with them

do I go do I nottt

i love staying in😂 but also i should socialize too

[whole coral]

Awwww I would say that you should go out though I'm a bit biased maybe cause I wouldn't mind going out myself

It is nice to stay in for sure though

[obsidian hornet]

ye I was gonna just call up a friend and just chill at their place

but they're not picking up

so prob gonna go out LOL

okkkkkk

bbyeeeeeeeeeeeeeeeeeeee

thank yuouuuu

.close

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[whole coral]

.See ya laterrrrr

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[brazen gazelle]

i literally have no idea how to do this

[austere cedar]

Take two points, sub them into the equation.

This gives two equations with two unknowns. You want to simultaneously solve.

[zinc dove]

Find k before you do that

Or if you really want you can sub in (infinity, 4) as a point but you have to be on some crackhead level shit for that

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@brazen gazelle Has your question been resolved?

[brazen gazelle]

oh okay

t.close

mb

mistyped it

.ckise

.close

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[brazen gazelle]

💀

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[normal urchin]

need help with a

[normal urchin]

idk if i’ve done it correctly so far but not sure how to find c

[strange ice]

cuz x-4 is a factor so if u input 4 into the eq it will = 0

[normal urchin]

i understand the factor and remainder part, just unsure of what to do since it’s a factor of f(x) and not f’(x)

[normal urchin]

it says that for f(x) but the given equation is f’(x)

does that change anything

[warm vortex]

factor theorem la

plug x=4

[strange ice]

i think u can work out f(x)'s equation with the 2 factors maybe

[warm vortex]

set qual to 0

[normal urchin]

does it matter that the equation given is f’(x) (derivative) and not f(x)

f’(x) is given but it’s the derivative, no?

[warm vortex]

you can integrate to get f(x)

[normal urchin]

i’m trying to do that as my work is shown in the screenshot but how do i get rid of c or find it

[warm vortex]

then you get one equation in a and c

then

plug x=-1

[normal urchin]

ohh it will make simultaneous equation if i use x=-1 and x=4

[warm vortex]

remainder theorem again

[normal urchin]

and using that i can subtract them to remove c?

[warm vortex]

system of equations in a and c

solve

voila

x+1 is not factor btw

[normal urchin]

i meant x = -1

[warm vortex]

so f(-1)=25

remainder is 25

that's why

[strange ice]

((ax^2)/3 - (7-4a/3)x + (28 - 16a/3)) is when you / f(x) by (x-4)

[warm vortex]

Also is that your dog in your pfp

[normal urchin]

no

and it’s a cat

😭

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[silent knoll]

Given f(x). Find the value of f'(0)

[supple copper]

Did you try differentiating f

[wraith hinge]

differentiate

use u/v

[silent knoll]

Ohhh

[wraith hinge]

and replace x with 0 for f'(0)

[silent knoll]

Yea I'm doing it but

Wait nvm

Imma try it

[wraith hinge]

yeah try

you just need to know u/v forumula

[silent knoll]

But what about that √?

I have to keep it throughout the solving?

[wraith hinge]

convert it to exponent

1/2

[silent knoll]

Ahh

I get it

I've got something like this

,rotate

[silent knoll]

Idk what to do now..

[silent knoll]

Oh and uh /(√(x²+4))²

I forgot that

[wraith hinge]

chain rule

[silent knoll]

What is that

,rotate

[silent knoll]

My handwriting looks terrible

[rough goblet]

,rcw

[rough goblet]

sighs

,rcw

[silent knoll]

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@silent knoll Has your question been resolved?

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[pastel pond]

could anyone help me with this question?

i don't get it

[trim thicket]

fiind surrface area of each of them

[shrewd ridge]

yeah, you don;t need to get it to find the areas and divide

[trim thicket]

then use the formula,
%difference = ((areaBIG - areaSMALL)/(areaBIG)*100)

[pastel pond]

.close

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[candid valve]

I don't know why I'm off by a factor of 3

[steady oracle]

You still need help? @candid valve

[steady oracle]

I guess because int_x^2.dx is x³/3 + c

Hope that helps!

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[candid valve]

I integrated wrt x after integrating wrt y

[steady oracle]

oh

if you integrate x²dxdy wrt y then will it be x²y.dx?

[candid valve]

it's between the limits y=x and y=0, so I get x^3

[steady oracle]

Wait, I haven't learnt double integral

Helpers might help

<@&286206848099549185>

[steady oracle]

Here's the question

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[candid valve]

.reopen

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✅

@candid valve Has your question been resolved?

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[topaz imp]

hello please

[topaz imp]

i need elp

not really specific math hlep but rather on studying the subject any advice

[steady oracle]

what kind of help?

[topaz imp]

when im studying math

i seem to spend a l ong time answering a small amount of questions

when i mean

a long time

i will time myself while im solving the questions

but i will spend a long time wit the teacher

[brave pagoda]

it is completely okay

[steady oracle]

it's okay

[topaz imp]

as i ask a lot of questions about math questions

[brave pagoda]

sometimes ı spent weeks just for 1 question

[steady oracle]

That's a good thing!

[brave pagoda]

if you can't ask you can't learn

[brave pagoda]

ı didn't see you were handling,sorry mate

[steady oracle]

So you're taking more time because you are learning new concepts!

[steady oracle]

It's okay mate, we're both trying to help

[topaz imp]

but my problem is other people can solve 10 20 30 questions without struggling immediately after a class

[steady oracle]

So it's completely fine

[steady oracle]

It comes with practice!

[topaz imp]

and it seems other peole are getting a nice reward for studying less

no right after a class

[steady oracle]

Well you know what?

[brave pagoda]

you cannot know that

[steady oracle]

They might be studying at home

[topaz imp]

i live with them

at dormiitory

[brave pagoda]

god's gift then ı shall say

[steady oracle]

Then what's the problem here?

[topaz imp]

what ive noticed

is when im in class

k

[steady oracle]

It only means that they can apply those ideas more quickly than you

What do you consider as being good at math?

[topaz imp]

but i need to solve more problems

bein able to solve a lot of problems

and be creative in solutions

[steady oracle]

Yes you can say that

[steady oracle]

Exactly!

Solving more questions in less amount of time is cool

[topaz imp]

but i want to go for math olympiads someday

[steady oracle]

But having a solution which is beautiful is ever cooler

[topaz imp]

and i canat go if i spend absurd amount of time solving questions

[brave pagoda]

let me give you an example. ı have a friend that solves questions for real slow. Like when he solves 1 ı solve 5 question hovewer, he always takes the most efficiency from that questions and beats me in any test. So rather than numbers, you should focus on efficiency you get from studying sessions.

[steady oracle]

You can! You just need to find the right mentor and you need the aptitude

[topaz imp]

wait

how

[steady oracle]

For math or olys for any subjects, i would say what's most important is the passion for math

[topaz imp]

yes i have that

[steady oracle]

Then As long as you're passionate about the subject, then you will enjoy the process

[topaz imp]

gokadam can you please share more light on thsi

[brave pagoda]

I just solve questions. but he creates other scenarios from that questions and think about them. he doesn't answer and skip like me

[topaz imp]

but it not only math its physics

[topaz imp]

that what i do

[topaz imp]

but my people just criticise

[brave pagoda]

why do you care

people talk

[steady oracle]

They are almost the same! What I say applies to both or more of them!

As long as you're passionate and don't care about what others say (in a good way), then you're going right!

[topaz imp]

have you guys been to university

[steady oracle]

In my opinion, you're going right

[topaz imp]

when they are teaching you a topic

and when you begin practice

[steady oracle]

You have the passion and you ask questions, so it will be fine for you

[brave pagoda]

no, but ı know unversity lessons

[topaz imp]

do you spend a lot of time

solving questions

[steady oracle]

No, but I know a lot of topics

[brave pagoda]

YES

[steady oracle]

A LOT

[brave pagoda]

we are the same person fr

[topaz imp]

i mean on a question

at first

[brave pagoda]

ofc

it makes me happy

[steady oracle]

Like a say a LOT, I spent like 2 hrs solving 2 questions yesterday

And that of math kangroo of class 8

GRADE 8, when I'm going to grade 12

[topaz imp]

No normal school math

that olympiad math

[steady oracle]

So it doesn't matter to me

[steady oracle]

But they use the same concepts right?

[topaz imp]

yes but olympiad math is more twisted

[steady oracle]

What's the difference? Only the difficulty is different?

[steady oracle]

Not every questions

@topaz imp what I'm trying to say is

You're going in the right path

Yes it does

But it's not "that" necessary

as im not creative

[steady oracle]

If you're passionate about math, then you should spend time solving more questions

[brave pagoda]

just don't hesitate and doubt yourself. keep moving and you will be succesful for sure

[steady oracle]

Be it school math or Olympiad math

[steady oracle]

And you could ask your math teacher to be your mentor

He will definitely help!

Also There is a whole math server dedicated to it

[topaz imp]

ok

[steady oracle]

spending more time on questions is cool btw

it's not something you should be ashamed of

[topaz imp]

alright

.close

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[wraith hinge]

"At the party, there were 8 girls and 13 boys. How many different pairs can be formed for dancing, if all the girls must participate in the dance?"

Note: Dancing pair can only be formed by one girl and one boy. There can’t be two boys or two girls in one pair

[trim jolt]

@wraith hinge Has your question been resolved?

[proven sluice]

13C1×8C1 ?

[wraith hinge]

Well maybe

I looked at it this way and not sure why that would not work

Wait imma send

[proven sluice]

Ok

[wraith hinge]

For first group there are 8 girls avalible and they can pick one of 13 guys (104 possibilities)

In second group there are 7 girls left and they can pick one of 12 guys (84 possibilities)

In third group there are 6 girls left and they can chose one guy from 11 guys left (66 possibilities)

and so on

And then we just do 104 + 84 + 66 + 50….

Why would that not work?

My friend said it’s done like this

We find out in how many ways we can pick 8 guys from 13 guys and just multiply that by 8 girls

[proven sluice]

Well your logic is some what correct

[proven sluice]

Yep I think this is better, you know this is same as ur logic

[proven sluice]

You tried using similar method as we do for finding number of possible numbers formed right?

[proven sluice]

" 8girls available can pick any from 13boys" u are actually pairing them all not just one girl

[wraith hinge]

Thats why when one girl picks a guy, another 7 girls can pick one of 12 boys remaining and so on. Why can’t i pair all the girls that way?

Oh

I got it

Sorry

Never mind

Yeah

[proven sluice]

So u understood?

[wraith hinge]

Yeah

[proven sluice]

Did u get this btw

[proven sluice]

Aight then

[wraith hinge]

Let me try something one moment

[wraith hinge]

I get why this doesn’t work

[wraith hinge]

I don’t get why this works

Could you try to explain?

[proven sluice]

So there is basically two places left to be filled up since we need to pair up

Since number of ways of filling girls is 8 itself it is given the value as 8

But number of ways of filling boy is yet to be found

There are 13b we want just 8, they can be anyone from these 13, therefore 13C8

ways of filling boys space

[wraith hinge]

Also doing it this way, does it matter if let’s say Anna and Mark are in group 2 or group 4.

Does it matter what group they are in or is it just important that they are in 8 different groups

[proven sluice]

Just think of it as numbers forming problem

[proven sluice]

8 diff grp

[wraith hinge]

How would this be done if it doesn’t matter what group someone is in. Sequence of groups is not important

It is just important that they are paired in 8 groups

[proven sluice]

wdym exactly

[wraith hinge]

I mean if Mark and Anna being paired in group 2 is same as Mark and Anna being paired in group 4

[proven sluice]

8 different groups?

[proven sluice]

It's not like this, it is still one group

[wraith hinge]

But the problem would have to be solved differently if sequence of group mattered

[proven sluice]

There would be just 8 possible groups

If they mattered ig

Since u want particular boy and girl in a particular grp

[wraith hinge]

Doing problem this way, is this the same scenario

[wraith hinge]

Same girl matched with same guy everywhere

But in different groups

How would problem if this counted still as same thing and how would problem be solved if this was counted as not same thing

Is my question making sense

[proven sluice]

I'm trying to understand

[proven sluice]

This is not what the question is asking

But that's what I mean there will be only 8 such groups of ur desire

The question is asking:

Any boy + any girl

[wraith hinge]

As someone whose english is their second language i am having barrier in trying to express my self xD

[proven sluice]

So mark can be paired up with cherry

[wraith hinge]

I am not asking that

[proven sluice]

It's okay! I can understand what you are trying to explain though I g?, since it's my second language too it's not easy for me aswell

[wraith hinge]

Okay let’s pretend that group 1 goes to one one room

Group 2 goes to another room

Group 3 goes to third room

And each room is different

That way

[proven sluice]

Okay and

[wraith hinge]

If mark and Jenna got paired INTO group 1

They would to for example room that has tv and cool music

But if Mark and Jenna got paired into group 6

They for example go to room that has vegetables and stinks

[proven sluice]

Lol

[wraith hinge]

So even if they paired same way

It matters if they got paired into group 1 or group 4

But now

[proven sluice]

But I don't think it means anything as grouping them is 1,2 or 3 group numbers

[wraith hinge]

Let’s say that EVERY SINGLE group goes to room with vegetables that stink

[proven sluice]

It's just they have to be paired

[wraith hinge]

So now it doesn’t matter if Mark and Jenna got paired in group 1 or group 4

Because it is SAME thing

[proven sluice]

Correct

[wraith hinge]

It is same thing if they got paired together in group 1 and group 4

But if rooms are different it is not same thing

So how do you solve problem if number which group they are in matters

And how would you solve if it doesn’t matter what group they got sorted into it just matters who paired with who

[proven sluice]

It is , we are not finding 'number of ways to put the pair in a group'
After pairing them it doesn't even matter which grp they are

[wraith hinge]

Okay

[proven sluice]

Well I don't really understand this much

[wraith hinge]

So my problem from beggining

[proven sluice]

But some other helper might help you!

[wraith hinge]

Was that it doesn’t matter what group they are in

Just matters who got paired with who

[proven sluice]

Just matters if they are paired or not

[proven sluice]

Yep I think so

[wraith hinge]

I want to close this chat i am going nuts haha

Thank you

[proven sluice]

Sure lol

[wraith hinge]

I understand better now

[proven sluice]

I'm anyways going to sleep, it's midnight here

[wraith hinge]

How do i close

Chat

[proven sluice]

Hopefully!

use ||.close||

!close

!done

[trim jolt]

If you are done with this channel, please mark your problem as solved by typing .close

[wraith hinge]

.close

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[half lintel]

answer for this ?

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@half lintel Has your question been resolved?

[half lintel]

<@&286206848099549185>

[edgy willow]

most of the configurations are basically the same, so just draw it out and count

or ask how many lines each even one hits

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[faint fulcrum]

How can sum-to-product identities be demonstrated?

[faint fulcrum]

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcRAMd_96p-DeE2IT5OrP2Cn7-1H5_bC90mddoU9i7xN7dvRjelZU5WE7iFiaUwjPMyV2_E&usqp=CAU

[austere cedar]

Take the identities for cos(a+b) and cos(a-b)

solve them simultaneously for cos(a)cos(b) and sin(a)sin(b)

Change the variables around by substituting.

You get the first two identities

These are really just the sum identities in reverse

[faint fulcrum]

bro, go to another channel XD

[faint fulcrum]

You tell me to do something like
cos(x)+cos(y)
x=a+b, y=a-b
cos(a+b)+cos(a-b)?

[austere cedar]

Do you know the identities for cos(a+b) and cos(a-b)?

[faint fulcrum]

yes

?

[austere cedar]

Start from cos(a+b) + cos(a-b)

[austere cedar]

This is already looking suspiciously like the top identity

[faint fulcrum]

then?

[austere cedar]

Let a = (x + y)/2
And b = (x - y)/2

[faint fulcrum]

now?

[faint fulcrum]

I don't understand the why and how of this change of variable

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@faint fulcrum Has your question been resolved?

[austere cedar]

Well, it works, didn't it?

Make sure to put that into your original line too

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[pure oak]

I’m stuck on this problem

[pure oak]

The hint is to use F(r)=c/r^2

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@pure oak Has your question been resolved?

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@pure oak Has your question been resolved?

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@pure oak Has your question been resolved?

[simple haven]

@pure oak the problem is simply about setting up the integral, because it specifies not to evaluate it. Do you recall the integral form of work?

[pure oak]

I think it's just the definition of the definite integral

[simple haven]

If so, then use the integral form of work, you just need to specify the integration bounds, and the function of integration, which you already have

$\int_a^b F(r) , \dd{r}$

[solid kiln]

OmnipotentEntity

[pure oak]

I'm confused how to get the value for c

I understand the radius is 4000

[simple haven]

Ok, so c is going to be related somehow to the mass right?

What is the force due to gravity?

[pure oak]

Mg

[simple haven]

Only if g is constant

$\frac{GmM}{r^2}$

[solid kiln]

OmnipotentEntity

[simple haven]

Where M is the mass of earth, m is the mass of the object, and G is the gravitational constant.

[pure oak]

The way the professor would do it is F(4000)=c/(4000) ^2 and he lets F(4000)=weight of space module but this doesn't make any sense

[simple haven]

That way does make sense

Because you know what g is on the surface

This allows you to find c.

And then you can reuse c

Both methods will work

[pure oak]

How is the way you would approach this particular example

[simple haven]

I would suggest using the professor's way, it seems easier, uses fewer facts.

[pure oak]

I'm okay with using their way but why does F(4000)=10

It's input is 4000 and output is 10

[simple haven]

10 is the weight in metric tons

I guess "metric tons-force"

[pure oak]

For the second part I'm confused at is the bounds of integration

I would've said from 0 to 1000

[simple haven]

4000 to 5000

Because you start at the surface, which is r=4000

[pure oak]

Reason I say that is because it says "How much work is done in propelling the module to a height of 1000 mi above earth"

Honestly I'm not used to seeing integration this way, I'm used to it being area under a curve

[simple haven]

It is here as well

The curve is force vs radius

It's just a little bit more implicit, and you have to go out of your way to graph it

[pure oak]

That does make sense

I guess that's it for this problem even though it makes absolutely no sense. Thanks for all the suggestions

.close

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[steel socket]

how would i write a polynomial function with least degree and leading coefficient of one given the zeros 0, root 5, 2

[steel socket]

is it just f(x)=x(x-root 5)(x-2) then factoroed out, but my worksheet says the factored out function starts with f(x)=x^4....

[marble wharf]

!original

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[marble wharf]

your attempt is correct unless you missed a detail in the question

[soft granite]

I think you’re missing the root part

[soft granite]

When the zero equals a root shouldn’t it be + and - that root or am I wrong

Ohhh I did this in precalc

Hold on I know I’m wrong now

[dull temple]

only if you need the coefficients to be integers / rational numbers

if you don't care then you don't need that

[soft granite]

Oh okay

[steel socket]

so how would i write this function, because from what i know the highest exponent is how many zeros there will be no?, so you would think it starts with f(x)=x^3 but its x^4

[marble wharf]

x^3 is correct

they probably wanted rational coefficients, in which case you would need x^4. but they didnt say that, so whatever

[steel socket]

is what the school says is the answer

[hallow cliff]

They made -√5 also a root to get this

[steel socket]

weird, ill just write it assuming theres only 3 cause they didn't give -root 5

[rough goblet]

that's alright

if you want only rational coefs u need -sqrt 5 too

if ydc, then yea u dont need -sqrt 5

[steel socket]

ya its just homework but im so cooked for my exam tomorrow

thanks yall

.close

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[steel socket]

.reopen

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✅

[steel socket]

next question they say the zeros are 1, -1(multiplicity 3), and 3i

so you would thing it would be like f(x)=x^5 yadadada, but

[soft granite]

Show the actual question

[steel socket]

is the answer apparently

[dull temple]

they (again implicitly) want real coordinates

[steel socket]

so it would be +-3i?

[dull temple]

so if you have a complex root you need its conjugate as well -- again this is them not specifying it. your answer is perfectly fine

[steel socket]

ok give me a minute i wanna write it out

so pre distributed it would be f(x)=(x-1)(x+1)^3(x+3i)(x-3i)?

[dull temple]

yeah

[steel socket]

took a little bit on paint cause i dont have a notebook/whiteboard with me, but i got the answer, so i guess if its a root or imaginery i gotta assume its +-

[dull temple]

if you want integer coefficients yeah

[steel socket]

what other coefficients are there?

like irrational?

[dull temple]

yeah

like $\sqrt5x²+ix + \pi$ is a valid polynomial

[solid kiln]

hay²le

[steel socket]

so is this stuff actually precalc cause that is not the course im taking

[dull temple]

sure, you could call it that

what course are you taking?

[steel socket]

integrated math 3

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[jolly lily]

i need the help:( i feel like im missing something. its top to bottom yeah? so √1-x^2 - 0?

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[modest onyx]

ab and ba numbers(they are two digits not multiplied) and If ab + 24 = ba what is a+b?

[rough goblet]

on the other hand, trial and error could also work

[modest onyx]

so we try every possbility?

[rough goblet]

algebra?

9b = 9a + 24

we know that they're one digit numbers

one possibility

wait no mb

hang on

lemme try again

[modest onyx]

so was 3b=3a+8 wrong

[rough goblet]

yes

smh

mb lmao

[modest onyx]

i gues you didnt multiply it by -1

[rough goblet]

??

[modest onyx]

no wait you did

[rough goblet]

10a + b + 24 = 10b + a

then we get 9a + 24 = 9b

but that's not possible

tf?

[modest onyx]

lol

[rough goblet]

xD

da hell

[shrewd ridge]

it's not possible

[rough goblet]

yea thought so

[modest onyx]

I guess i created the formula incorrectly

[rough goblet]

probably, yes

!xt

i mean

!xy

[trim jolt]

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.

[modest onyx]

In the real question, it says ab is minute part of a digital timer and after 208 minute minute part is ba

[rough goblet]

what?

send a screenshot

[modest onyx]

there is no picture in the question

[shrewd ridge]

so it's like ab+24 = xba

[modest onyx]

i mean you know like a digital clock

like 03:24

[shrewd ridge]

not even like that either

[rough goblet]

oh i see

[modest onyx]

so i thought we need to do that like 208 minute is 3 hours + 24

so we dont care about hour etc.

[shrewd ridge]

no ok it is like that

it's just 28 not 24

[rough goblet]

so smth like

minutes since midnight = 60H_1 + 10A + B
60H_2 + 10B + 10A = that + 208

[modest onyx]

oh wait it was actually 204

not 208 mb

so it is definitely 3 hours + 24 minutes

[shrewd ridge]

got it

[modest onyx]

maybe we need to do it like: ab+24-60=ba ?

or im just making up formulas

*equations

what if ab+24 is higher than 60 then that would be in hour

[shrewd ridge]

it still feels like it's impossible but not as clearly 🤣

i found it

[modest onyx]

how?

[shrewd ridge]

just thought of random numbers

if it went past 60 we could say that the number is something like 36, or 69

b=a+3

because it doens;t go past 60, there are other options

[modest onyx]

If we think it logically and think sum of minutes the equation must look something like: ab+24=60*k + ba

does it make sense to you

k is natural number

[shrewd ridge]

k is 1 or 0

[modest onyx]

yes

i tried k=1

and i got a=4+b

[shrewd ridge]

i don't know how to do it logically

actually wait

there's maybe 2 answers

yeah, there's 2 ways to do a=b+4

they both work

[modest onyx]

then whats a and b?

[shrewd ridge]

we can't tell

[modest onyx]

a+b?

[shrewd ridge]

it could be 40 with sum 4 or 51 with sum 6

[modest onyx]

there must be answer the test says there is 😄

[shrewd ridge]

nah, can't determine a+b

6 is maybe more likely since there's no 0 but not by much

[modest onyx]

what are the numbers you used to make 6

[shrewd ridge]

51

[modest onyx]

correct answer is actually 6

but so what i learned from this question 😄 trying every possbilitiy?

I was supposed to solve this in 1.5 minutes

[shrewd ridge]

well yeah

you got a=4+b from assuming k = 1

and it gave you an answer

and it's sufficiently smart

[modest onyx]

I guess i need to make most simple equation then try every number from 0 to 9 fast

they always ask questions like this in a exam in Turkey called TYT

it's like SAT but its harder

[shrewd ridge]

so wait, there were like 4 options?

[modest onyx]

yeah

no 5

a b c d e

[shrewd ridge]

and there was 6 and there was no 4?

[modest onyx]

options are 7 6 5 4 3

[shrewd ridge]

right

so it's unsolveable

[modest onyx]

but how can we get 4

[shrewd ridge]

40

[modest onyx]

Oh i forgot to say that

question said a and b is numbers except 0

sorry lol

[shrewd ridge]

okay

[modest onyx]

.close

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[shrewd ridge]

.reopen

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✅

[shrewd ridge]

i can see a whole proof kinda

so first we assume a is in {1,2,3} and b is under {1,2,3,4,5}, so nothing rolls over at all

this is a contradiction, both a and b will increase in that case, they can't become equal to ba

[modest onyx]

bro is on his way to prove test is wrong

[shrewd ridge]

no it's fine

nothing wrong in this one

basically you can show that "ab+24=60 + ba" holds, that it rolls over 60, and then it's solveable like you said a=4+b

.close

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[brazen gazelle]

whats happening here

for the reject

i dont understand why they rejected it

wait nvm i get it now

but im stil confused on how they got the blue part

[turbid leaf]

Draw line x=1 on the graph and it should be clear

[brazen gazelle]

huh

i know how to get red

its blue-red area

but idk how to get blue

[brazen gazelle]

oh shit

oh ok

so blue goes to where red's x-int is

from 0 to red's x-int

and then red goes from 0 to 1

@turbid leaf is that right

i dont understand why its addition tho

shouldnt it be subtract

wait no im wrong again

its blue from 0 to 1

add red 1 to its x-int

why are they doing pinks area liike this

is it just easier

@turbid leaf

[turbid leaf]

Yep

[brazen gazelle]

so i can technically do it both ways right

[turbid leaf]

Yeah you can

[brazen gazelle]

alr thanks

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[past sinew]

During the last drought, the water table dropped so low that three villages in the plains ran out of drinking water. The council organised a light duty tanker to carry water from the main reservoir to the villages on a weekly basis.

At the start of each trip the tanker was always full of water; unknown to the driver, just as he started driving from the reservoir, the water tank developed a leak. By the time he reached the first village, the tanker had lost half of the water from the water tank and consumed half of the fuel tank. You can assume that the fuel consumption of the truck is directly proportional to the weight it carries, and that the water flow (in the leakage) and the speed of the truck were constant for the whole journey.

We also know that an empty tanker uses exactly one sixth of the fuel tank when travelling from the first village back to the reservoir.

Disregarding the fuel tank weight, what fraction of the fuel tank would be spent to reach the first village when the truck had no leak?

[past sinew]

i need all the assumptions you made while getting this answer too :3

as long as its logical it works

[trim jolt]

@past sinew Has your question been resolved?

[past sinew]

I think it's 1/3,

as total = 1 + w

c(1+w) = 1/2 + c(1+w/2)

c = 1/2w

1/2w = 1/6 ( empty tank uses one sixth)
w = 1/3

<@&286206848099549185> its been 15 >.<

im just wondering if it looks right

[wraith hinge]

whatas the qns

[past sinew]

.

Disregarding the fuel tank weight, what fraction of the fuel tank would be spent to reach the first village when the truck had no leak?

Assuming proportionality of fuel and water weight, no changes in the conditions that could change the rate of leakage, uniform speed of the truck

[wraith hinge]

woukld it be just 1/2 the tank tho

[past sinew]

wdym?

like the

[wraith hinge]

like the fuel in the tank would have been spent since when it satrts at the resevoir it has a full tank

n when it reaches the first village it has lost half of the water from the water tank

since the fuel consumption of the vehicle is directly proprtional to the weight it carries

if it loses half water of the tank then it loses half of the fuel it contains

assuming the water density remains constant

[past sinew]

yeah but that's half of the fuel and water including the fuel it takes to travel to the first village

which is 1/6

so the total fuel used to get to the first village is 1/2 - 1/6

and if that's 1/3, then the water should also be 1/3

well

[wraith hinge]

yes u are right

if we saying the truck has no leak then f+1/6=1/2

thus f=1/2-1/6=3/6-1/6=1/3

[past sinew]

that should be correct right

it just seems a bit simple

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[cobalt sapphire]

in vector calculus for curvature of a circle to a curve, k, with starting at r(t), and where T=r'(t)/|r'(t)| i have 3 different formulas. are they all interchangable?

k=1/|v| * |dT/dt|

k = |r'(t) X r''(t)| / ( |r'(t)|^3 )

k = |T'| / |r'|

[cobalt sapphire]

i know 1 and 2 are but is 3?

[edgy willow]

yea 3 looks like 1 rearranged a bit

[cobalt sapphire]

alright thanks

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[tame hemlock]

Learned about permutations, at 12:10 they give us the problem of finding the order a 2 cycle, 3 cycle, etc;, but the order of an element x is the power required to get the identity element, but what the hell is the group and identity element in this case? https://youtu.be/MpKG6FmcIHk?si=U-loF1EP97f834jw

[tame hemlock]

Wait the group is the symmetric group and e is (1)(2)(3)…(n), but how do i find the order?

[little garden]

You have to multiply element to itself unless you not get identity and it is that elements order

Here group is one to one and onto function to a set to same set

Like if A={1,2,3,4}

Then all the one-one and onto function from A to A is the symmetric group

And identity element is if
1 goes to 1
2 to 2
3 to 3
4 to 4

[tame hemlock]

You mean until you get the identity right?

[little garden]

Ya

[tame hemlock]

Is there a specific symbol for the identity?

[little garden]

Ya we use epsilon which is not in my keyboard

[tame hemlock]

So for 2 cycles the order is 2 (i did it mentally) and according to google the order of an n cycle is n but how?

[little garden]

So you know what is operation in symmetric group

?

[tame hemlock]

Basically function composition right

[little garden]

Group we working on

Ya

So what the problem you can calculate order of 2 cycle

By function composition

[tame hemlock]

Yh but how do we prove the order of an n cycle is n?

Nvm i did some examples by hand and got it

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[little garden]

I am trying to do by induction method @tame hemlock

[tame hemlock]

Aight i’ll reopen post the proof when you’re done thanks

.reopen

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✅

[little garden]

Sorry I couldn't do it ,i see in books it saying verify yourself no proof

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[urban jungle]

the largest eigenvalue was 3. I understand everything except why did the solution add
0
1
0
even though it's not in the null space? i thought the null space is only:
-2
0
1

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[urban jungle]

When we express the solution set in terms of the free variables, we include a vector for each free variable where that variable is 1 and all other variables are 0. This is a standard way to express the solution set of a system of linear equations.

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[smoky kite]

Don’t rly understand what associated symmetric matrix is

[whole coral]

You should be able to write that as x^t Ax for some matrix A, and letting x = (x1, x2, x3)

They’re basically asking you for that matrix A

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[smoky kite]

Oh

Ok hmm

Yeah I get that theory

just not seeing how you’d find A

[whole coral]

Good question erm…

Well you already know that the matrix A has to be symmetric at least, which makes your life a bit easier

And because you have the x1, x2 and x3 terms all appearing squared, the diagonal entries you know

But then you also have an x1x3 term, so you wanna be able to get that

That would imply the (1, 3) (and by symmetry, the (3, 1) one as well) are nonzero, but then those entries will contribute two factors of x1x3

If any of that seems to make sense? Rather annoyingly I haven't found any resources or anything that explain it better than me @smoky kite

[smoky kite]

Hmm that will definitely help thanks

I’ll try it again in a bit

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[fringe marsh]

can someone explain the induction they did on this

[fringe marsh]

I feel like it was done in a kinda confusing way, but i could be overlooking some things

also, im ngl to prove it is well defined cant you just use the well ordering principle

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[fringe marsh]

<@&286206848099549185>

[edgy willow]

it's tricky because it's both a contradiction and induction proof. for the contradiction they have to search through all of P(N) for a set that has no minimum, and to do that they induct on sets with k in them and work their way up

so like the base case covers sets like {0,5,9}, {0,3,4,6} etc. and for the next step they look at sets with a 1, but none of the numbers less than 1, such as {1,2,3} etc.

and the proof logic is that they try the obvious minimum, k, but the induction hypothesis contradicts this so apparently you can't pick any k in the A and A is empty, for all sets A in P(N)

[fringe marsh]

i hate that

no chance i would have thought of that on an exam

is there a more uh smooth solution

thanks for explaining btw, makes sense.

[edgy willow]

yea I wish there was, but I think the only way to super rigorously prove this is with induction

[fringe marsh]

thank you for explaining, im going to try reconstructing the proof myself (and btw thank you so much for helping me these last few days, really really appreciate it)

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[simple jackal]

I’m trying to understand the concept of a sequence of events happening infinitely often (in the picture shown, please disregard my markings). I’m confused about the set intersection/union definition.

[simple jackal]

image cut off above

I made a toy example, where the sample space is infinite coin tosses and events A_i denote whether there is a Heads in the ith position of each sample point

The set definition of i.o. is A_1 intersect (A_1 union A_2) intersect…

but i can’t really make sense of this

can i get an explanation of how the set definition applies to my example?

[simple jackal]

namely isn’t this long intersection just equal to A_1?

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[night mango]

can anyone explain me this equation?

[sullen ore]

What do u need explained

[sullen ore]

?

[night mango]

nvm

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[vale obsidian]

i got up until ( x + y ) / 5 = xy

but im stuck there

[zinc dove]

Not enough data methinks

[vale obsidian]

anwser key says its option B

just dont know how to get there

[zinc dove]

Oh I think we misunderstood the question

Maybe it’s saying the sum of x and 1/x is 5

[vale obsidian]

i tried that way but couldnt do much either

im kinda lost

<@&286206848099549185>

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[sour talon]

what is this notation and what does it mean

[austere cedar]

Need more context

[sour talon]

how it's used in algebra

[brazen forge]

it looks like a screenshot of a part of text you were reading and you dont know what does it mean

maybe send a pic of all text?

so we get a more context

without context this notation might be used in variety of situations

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[fallen jay]

I need help setting up my triple integral in cylindrical form. I don't know what my bnounds should be on my radius, I've never done an elliptical cylinder before.

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[fallen jay]

<@&286206848099549185> please

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[lost zephyr]

x

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[vapid ruin]

F = mv/t
It doesn't say how long so lets assume 1 second
F = 90*2
F = 180N

since force is preserved it is the same for both of them

180 = 60v
v=180/60
v=3ms-1

yeah

F is force in newtons and t is time in seconds

F=ma a = v/t

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[dapper prairie]

Calculate lnS(x) at x=1.5. With S(x) being the sum of the series ..... know ....

I calculated that the answer to this question is A, someone please help me check the answer

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[dapper prairie]

<@&286206848099549185>

Sorry for pinged

[unreal eagle]

lâu lắm r mới thấy ng Việt

long time ago to se vnese

[dapper prairie]

Sao anh biet em nguoi Viet

[unreal eagle]

*cái đề

[dapper prairie]

đùa thôi

lâu r mới thấy có người Việt ở đây

em chao anh nhe

[unreal eagle]

chao xìn

[dapper prairie]

uồi

anh trong xPho luôn

[unreal eagle]

thực ra tôi cx ko biết giải:)))

[dapper prairie]

:))) à k sao đâu anh

em chỉ cần check kqua thoi ma

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[astral ore]

Hi, could someone help me figure out what are the total possible concatenated numbers?

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<@&286206848099549185>

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[wraith hinge]

THe main ideia is to multiply the 10 numbers than X can choose with the 50 numbers of Y and the 100 numbers of Z to get it

[astral ore]

it's only 10 numbers for X lol