help-19

[olive needle]

you would have to evaluate the endpoints of each interval to determine which are the upper and lower estimates.

[keen swan]

i see

just this time it was ass as the points were horrible

I now have n = 10, and delta x = 1 as

its [0,10], 10 - 0 / 10 = 1

So in turn, its the same stuff before but instead of multipled by 2, it should be 1?

[olive needle]

Yes.

[keen swan]

I got 7 and 11.6 (12) is wrong

I have 1(5+3+1.8+1.1+.7), and 1(3+1.8+1.1+.7+.4)

giving 7 and 11.6, but is wrong?

[olive needle]

One moment.

[keen swan]

?

[olive needle]

That is the function broken into ten intervals.

[keen swan]

oh

[olive needle]

Each of the points on f(x) is what you need to use for the lower or upper estimate.

[keen swan]

im stupid

i forgot the x's change

i see i see, thank you

gtg, but ill try that later

[odd edge]

@keen swan Has your question been resolved?

[scenic lodge]

#linear-algebra

[odd edge]

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[next swift]

help with 2ii pls, answer to 2i is 2/(2x+3)

[oak crown]

yeah.

$\dv{x}\ln(2x+3)=\frac{2}{2x+3}$.

[clever fjord]

;(

[oak crown]

what does the fundamental theorem of calculus state?

[odd edge]

@next swift Has your question been resolved?

[next swift]

i don’t know we never learnt about it

[oak crown]

it states that, for a function $f$, we have that $\int_a^b f'(x)dx=f(b)-f(a)$.

[clever fjord]

;(

[next swift]

oh yeah i got that

[oak crown]

so, what's your function f here?

[next swift]

i plugged in the limits and i got 1/9-1

so -8/9

[next swift]

f(x) = 1/(2x+3)

[strange aspen]

no you dont plug the limits you need to integrate first ( f'(x)) not (f(x))

[oak crown]

no.

[oak crown]

no.

what did they just tell you to do in part a)?

[oak crown]

this is an introduction towards integration, so there are no known techniques yet; in addition, they just assumed that f was 1/(2x+3)

[strange aspen]

yea I overcomplicated it

[oak crown]

the integrand includes f', so what is in the integrand of the equation in part ii)?

[next swift]

oh

i got my mistake

[oak crown]

👍

[next swift]

ive already done integration my brain was lagging 😭😭

thank you for the help 🙏🏾

[oak crown]

to get full marks, you are probably meant to identify that f'=2/(2x+3), recognize its antiderivative from part a), then plug in the endpoints.

[next swift]

thank you

i’m gonna try again

i’ve ran into a problem

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[next swift]

.reopen

[odd edge]

✅

[next swift]

<@&286206848099549185>

[dim blaze]

?

[tardy pawn]

Its (ln(2x+3))/2

[oak crown]

you can't use power rule when $n=-1$.

[clever fjord]

;(

[next swift]

what should i use instead

[oak crown]

i literally just said, use the result from part a).

[oak crown]

^

[next swift]

but dy/dx is 2/(2x+3)

[oak crown]

mb

[next swift]

the issue is that the function being integrated is different to the one i got in part a

it’s similar tho it’s half

and i know i have to use that answer bc it says hence im just not sure what to do

[odd edge]

@next swift Has your question been resolved?

[next swift]

<@&286206848099549185>

[celest acorn]

You know that integral of a constant times a function is the constant times the integral of the function?

[next swift]

i just called a friend he showed me exactly that

i got it finally

thanks 😊

.close

[odd edge]

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[weary grove]

so am i cooked?

[oak crown]

!status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[weary grove]

1

[oak crown]

oh boy

well, what are the properties of a trapezium?

[weary grove]

parrallel lines?

but those 2 aren't parralell

[oak crown]

well, RQ and OP are parallel.

[weary grove]

yeah but i can't figure what to do with them

[hazy python]

Can you see that OP vector and RQ vectors are parallel

[weary grove]

yes

but what to do with it

[hazy python]

Ok so their direction will be same

[weary grove]

yes

[hazy python]

We are given a relation in their magnitude

Can you use it to find the vector

[weary grove]

but how will we find PQ:OR from OP:RQ

[hazy python]

Find RQ

[weary grove]

5b/3

[hazy python]

Now you got two parallel sides of a trapezoid

[weary grove]

yeah

[hazy python]

In form of vector

[weary grove]

yes

what can we do from here?

[hazy python]

Now do you know concept of similarity

[weary grove]

length scalar factor?

[hazy python]

Yeah for magnitude we are gonna use that

[weary grove]

5/3?

[hazy python]

No I mean do you know similarity in triangles

[weary grove]

yes

but that's in 2 triangles

[hazy python]

See triangles WRQ and WOP

[weary grove]

yeah

OHHHHHHHHHH

[hazy python]

Can you find |OW| from therw

[weary grove]

i think

[hazy python]

Then you are done i think

[weary grove]

done bro

thank u

saved my life

[hazy python]

You're welcome

[weary grove]

u too bruh i was too dumb mb

.close

[odd edge]

Channel closed

Closed by @weary grove

Use .reopen if this was a mistake.

[oak crown]

👍 no, you're not dumb

[weary grove]

yo man thanks for saying that really helped

i got another question shi

now they are saying

write PQ as a column vector

but there cleary is no x and y value given

2nd question is

write 3PQ as a single vector

[odd edge]

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[pastel orbit]

use the chain rule and product rule

[light zodiac]

😔

ITS SO MUCH

[pastel orbit]

sorry, but there's no other option

it's good practice though!

[light zodiac]

I got it wrong

[pastel orbit]

so I'm a fool and forgot that log rules are a thing

it's probably a good idea to use those instead

,, \log(ab) = \log(a) + \log(b)

[clever fjord]

ourfallenstars

[pastel orbit]

,, \log(x^a) = a \log(x)

[clever fjord]

ourfallenstars

[light zodiac]

oh yeahhh

[odd edge]

@light zodiac Has your question been resolved?

[wooden gorge]

use log rules to simplify this function further. what is R(x) simplified

[light zodiac]

Is it possible to do it without it

[wooden gorge]

no, otherwise you will have to differentiate the log

and apply chain rule a bunch of times

I'm saving you from that

[odd edge]

@light zodiac Has your question been resolved?

[odd edge]

@light zodiac Has your question been resolved?

[light zodiac]

can u show how

wat did i do wrong

[wintry kelp]

we don't have the g(x)

[wintry kelp]

oh

[wintry kelp]

,tex $ R(x) = ln\left[ x^{2}\cdot \sqrt{g(x)}\ \right] $

[clever fjord]

salagata

[wintry kelp]

,tex $ R(x) = ln\left[ x^{2} \right]+ln\left[ \sqrt{g(x)} \right]$

[clever fjord]

salagata

[wintry kelp]

,tex $ R(x) = 2 ln\left[ x \right]+\left( \frac{1}{2} \right)ln\left[ g(x) \right] $

[clever fjord]

salagata

[wintry kelp]

lmao i'm solving it cause i'm bored

now you have to derivaate normal

,tex $R'(x) = \frac{d }{dx} 2 ln\left[ x \right]+\frac{d }{dx}\left( \frac{1}{2} \right)ln\left[ g(x) \right] $

can i solve it?

[clever fjord]

salagata

[wintry kelp]

please?

,tex $R'(x) =
2\frac{d }{dx}ln\left[ x \right]+\frac{\frac{d }{dx}ln\left[ g(x) \right] }{2}$

[clever fjord]

salagata

[wintry kelp]

now use chain rule

17/9

[odd edge]

@light zodiac Has your question been resolved?

[odd edge]

@light zodiac Has your question been resolved?

[wary forge]

do u know chain rule

[light zodiac]

i think so

I cant work on it until tmrrw tho

[wary forge]

okay you can dm me the doubt whenever you wanna solve it

[odd edge]

@light zodiac Has your question been resolved?

[light zodiac]

ok i return

[wary forge]

who summoned me

[light zodiac]

Idk what to do from here

@wary forge

[wary forge]

yea im here

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

@light zodiac

[light zodiac]

wat do I do

[wary forge]

do you know this?

[wary forge]

this

do you know $\frac{d}{dx}ln(x)=\frac{1}{x}$ ?

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

do you know how to apply chain rule?

[light zodiac]

Yes

[light zodiac]

Bo

No

I thought I just derivative to both since it just addition

[wary forge]

$\frac{d}{dx}f(g(x))=f'(g(x)) . g'(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

do you know this?

[wary forge]

you have to do that but in the second term you gotta apply chain rule

[light zodiac]

Oh

does the first one become 1/x or 2/x

Or 0

[wary forge]

first one becomes 2/x

cuz we have a 2 multiplying outside

[light zodiac]

I thought we derivative it too

[wary forge]

no

2 is a constant in multiplication

so you can take it outside of d/dx

$2\frac{d}{dx}f(x)=\frac{d}{dx}2f(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

its the same thing

[light zodiac]

but then don't we do the derivative of 2

[wary forge]

you dont need to cuz its a constant

you can do what youre saying but you would have to apply the product rule for that

[light zodiac]

OK LETS DO THE CHAIN RULE ON THE SECOND ONE

idk how to do that with logs in there

[wary forge]

$\frac{d}{dx}f(x)g(x)=g(x)\frac{d}{dx}f(x)+f(x)\frac{d}{dx}g(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

if you treat 2 as g(x) over here

then its the same thing

[alpine sedge]

yo

somone help pls

where do i go for help?

[wary forge]

!occupied

[odd edge]

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

[wary forge]

@light zodiac you there?

[light zodiac]

Mb i had to go to my next class

Anyways

Idk how to do those derivative rule things with logs

[wary forge]

read my last few messages

[wary forge]

ill show you those

this is how you differentiate log functions $\frac{d}{dx} ln(x) =\frac{1}{x}$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[light zodiac]

Oh

[wary forge]

this is how you differentiate log functions $\frac{d}{dx} ln(f(x)) =\frac{1}{x}f'(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

did you understand what im saying?

[light zodiac]

Yes

[wary forge]

okay so what would be $\frac{d}{dx}2ln(x)=?$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[light zodiac]

2/x

[wary forge]

good

you have differentiated the first term now

great

now lets differentiate the second term

do you know this? $\frac{d}{dx}f(g(x))=f'(g(x)) . g'(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

this is the chain rule of differentiation

you need to know this to find the derivative of the 2nd term

[light zodiac]

OK I GOT THE ANSWER

17/9

YAYAYAYA

Ty

[wary forge]

very good

youre welcome

[light zodiac]

Idk what to do abt the g prime already being there

[wary forge]

work from inside to outside

what is f(6)

[light zodiac]

2

[wary forge]

great

g'(f(6))=g'(2)

what is g'(2)

[light zodiac]

idk

[wary forge]

see the graph

at x=2

[light zodiac]

0

[wary forge]

good

[light zodiac]

YAY

[wary forge]

thats your answer

[light zodiac]

I thought I had to do chain rule

[wary forge]

no

[light zodiac]

Wat Abt 22

[wary forge]

if you see the next question, you would have to apply chain rule there

yep thats chain rule

you know h(x)=f(g(x))

[light zodiac]

do i just do the chain rule version of that

[wary forge]

you have to apply chain rule on that

$\frac{d}{dx}H(x)=\frac{d}{dx}f(g(x))=f'(g(x)) . g'(x)$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[wary forge]

did you understand whats done in this step?

[light zodiac]

Wat is g'(6)

Is it like 2

[wary forge]

it looks more like 1

[light zodiac]

Oki

[wary forge]

did you get the answer?

[wary forge]

?

[light zodiac]

yes

[wary forge]

good

[light zodiac]

wat Abt f'(9)

[wary forge]

at 9 f(x) seems to be a straight line

what do you think is the slope of f(x) at 9

[light zodiac]

Like -1/2

[wary forge]

nope

thats incorrect

its a straight line

$m=\frac{y_2-y_1}{x_2-x_1}$

[clever fjord]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[light zodiac]

Oh like 0

NOOO

[wary forge]

nope

use this and tell me what slope youre getting

did you get it?

[light zodiac]

-2

YAYAYA ty

[wary forge]

very good

[wary forge]

are there any more?

[light zodiac]

No

[wary forge]

cool

[light zodiac]

[wary forge]

happy learning, have a nice day

[light zodiac]

.close

[odd edge]

Channel closed

Closed by @light zodiac

Use .reopen if this was a mistake.

[light zodiac]

.reopen

[odd edge]

✅

[brazen saffron]

hmm, what is your question exactly? kinda hard to follow what you have written

[light zodiac]

I messed up😭 😭

It is implicit differentiaiton

@brazen saffron

[brazen saffron]

Yes, I'm still not sure what you've done. There doesn't seem to be any derivatives in the question but it looks like you have a few dy/dx in there

[wet raven]

I think they're meant to differentiate the given curve.

[warped glacier]

yes make sure your working has some structure

so you mean to start off with $x^2 + y^2 = 3xy^2$ first right?

[clever fjord]

south

[warped glacier]

also do you want to find dy/dx or dx/dy? I mean dy/dx = 1/(dx/dy) so it doesn't matter too much in the end

[warped glacier]

then you should realise from this what you have done wrong after differentiating both sides

[odd edge]

@light zodiac Has your question been resolved?

[soft tinsel]

What is the question?

[light zodiac]

Solve with implicit differentiation

[light zodiac]

Oh do i need to

I think when my teacher did she just ended up collapsing it into negative exponents on the left side

<@&286206848099549185>

[light zodiac]

OK I FIGURED IT OUT

[kind yarrow]

Gosh

[light zodiac]

This the answer if anyone wanted it Fr

[light zodiac]

Fr I need to work on my organization 😭

.close

[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[mystic hearth]

hello

[mystic hearth]

i need help with a problem

i’m not sure what to do next

and yes i’m doing it on a napkin lol

[dry fog]

get rid of the ln5 and ln2

and then simplify

[mystic hearth]

how do i get rid of them

[dry fog]

multiply both sides first by ln5, then by ln2

[surreal lichen]

how did you get the second line?

[grizzled bison]

I prefer to do this with ln personally

[silk folio]

property of log ig

[fickle stump]

convert them in a single base

and get all the stuff together

[mystic hearth]

so is there a property where i can simplify the ln’s

[silk folio]

you need to solve for X, thats it right?

[mystic hearth]

yes

[silk folio]

just take log on both sides on the first step itself

then the power will come to the front

and then u can take x to one side

i believe thats the best way

[surreal lichen]

what operation did you do to get the second line?

[grizzled bison]

ln5^(x-1) = ln2^(2x+1)
(x-1)ln5 = (2x+1)ln2
(x-1) / (2x+1) = ln2 / ln5

[mystic hearth]

now they have different bases

[silk folio]

@mystic hearth this

[grizzled bison]

nice I got it, haven't done logs like that in a while 😂

or partially got it

[silk folio]

nah u got it, its easily solvable for x now

[surreal lichen]

$5^{x-1}=2^{2x+1}$, if you take $ln$ of both sides you get a simple equation.

[mystic hearth]

so the property for logs means you can bring the exponent to the coefficient

[grizzled bison]

yup 3rd law of logs

[clever fjord]

Horsi135

[grizzled bison]

you just drag it down

[dry fog]

that's what i was writing

[mystic hearth]

so now i divide?

[dry fog]

open the brackets

[grizzled bison]

divide by 2x+1 on both sides and then divide by ln5 on both sides to get what I got

[dry fog]

then move x to one side and factor them out, divide by what's left

and you get the x

[silk folio]

just leave it at that, let her try on her own bruh

[dry fog]

ok

[grizzled bison]

and then simple (which he says after struggling to do it because he's washed) algebra manipulation should get you the answer

[dry fog]

should have waited for her to ask

[surreal lichen]

also $2^4=4^2$ but $log_2 4$ does not equal to $log_4 2$

[silk folio]

yeah, purpose of this channel is to make others learn, so better make them try on thei rown after a few hints

[mystic hearth]

so this is what the one guy got

[silk folio]

try a bit more to isolate x

think what esle u can do

[clever fjord]

Horsi135

[surreal lichen]

so youre second line in the start was wrong

[dry fog]

try opening the brackets

[mystic hearth]

well with the denominators being there i’m not sure what i can do

what bracket

[dry fog]

multiply (x-1) by ln5

[silk folio]

maybeeeee u can try opening the bracket on x-1 = (2x.... step

[grizzled bison]

^

[mystic hearth]

https://tenor.com/view/sad-hamster-with-bow-violin-hamster-girl-gif-14473982535061953593

i’m sorry i don’t get it

[dry fog]

u get ln(5)x-ln(5)

[silk folio]

LET ANY ONE EXPLAIN

[dry fog]

multiply what's inside

[silk folio]

yall are creating a mess smh

[dry fog]

by ln 5

multiply each term inside the bracket

by ln(5)

[grizzled bison]

no problem, just go over the content again in your own time

[dry fog]

and by ln(2) on the other side

[dry fog]

yes

[surreal lichen]

!noans

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[dry fog]

now open the brackets on the right side

[surreal lichen]

don't tell him what to do

[dry fog]

and see what you can do

[dry fog]

any ideas?

[mystic hearth]

bring the x to the same side?

[dry fog]

yes

[mystic hearth]

is this ok

[dry fog]

here on the right side 2x should not be in brackets

because it wasn't in brackets before, it was (2x+1), not ((2x)+1)

[mystic hearth]

oh ok

[dry fog]

looks right

so it's ok

now i think you can finish it

[dry fog]

yeah it's correct

[dry fog]

you forgot x inside the bracket

it shouldn't be there

[mystic hearth]

i think i’m just going to get this one wrong i’m so confused, thank you for your help though i really appreciate it

[dry fog]

no

u are about to get it

[dry fog]

x(ln5-2ln2)=ln10

you see where?

[mystic hearth]

oh bc it’s not x^2 it’s 2x

[dry fog]

yep

[dry fog]

well done

.close

[mystic hearth]

yay thank you

that was so hard i really appreciate it

[dry fog]

[odd edge]

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

@mystic hearth Has your question been resolved?

Channel closed

Closed by @mystic hearth

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[odd edge]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[steep prawn]

Hi, I need help with combinatorics.

Question: There is figure inside 1x1 square. No two points of this figure are distanced exactly 1/1000. Prove that area of this figure is no greater than 0.29.

[odd edge]

@steep prawn Has your question been resolved?

[open crown]

You can try to convert the square into a grid where each smaller square's length = 1/1000

Since no points of this figure are distanced exactly 1/1000, they must not occupied any adjacent squares

Hmm

https://en.wikipedia.org/wiki/Close-packing_of_equal_spheres

[karmic lotus]

are points vertices or are points every pixel of a line

[open crown]

Then use the sphere-packing theory in 2D

[steep prawn]

Every pixel

[steep prawn]

Nice, I didnt know this theory. Imma look into that

[open crown]

https://en.wikipedia.org/wiki/Circle_packing

[karmic lotus]

do two points within a figure's area constitute for a valid distance measurement(does 1/1000 between such 2 points count)

[steep prawn]

Yeah, all distances count

[open crown]

This is for a sphere with r = 1

Erm...

[steep prawn]

1. Its about hexagons
2. I have to say that density <= 0.29

[wooden gorge]

This is such a physical chemistry question._,

[karmic lotus]

is this legal

[steep prawn]

Note that figure doesnt have to be connected

[karmic lotus]

if the beams and columns are so narrow that green is still under 1/1000

is this legal

[steep prawn]

Yes it is

[karmic lotus]

would the collection of balls or the gap/net of holes be the legal figure

[steep prawn]

Yes if they are enough far apart

[karmic lotus]

which

balls or net

[steep prawn]

Balls

[karmic lotus]

“far enough apart”?

[steep prawn]

I mean: take ANY two points on your figure
Is distance between them 1/1000?
Yes: figure is illegal
No: figure is legal

[karmic lotus]

if a figure doesnt have to be connected and still be one figure, then the entire figure cant exceed 1/1000 diameter or am i lacking imagination

do you mean multiple figures

each abiding by the rule

[open crown]

Honest question: Whats kind of subject is this question comes from?

[steep prawn]

Combinatorics

[steep prawn]

You can assume that figure is defined as set of points

[karmic lotus]

one figure

[steep prawn]

.

[karmic lotus]

not clear enough for me

is there only 1 figure in the box

[steep prawn]

https://www.merriam-webster.com/dictionary/figure

2a from this site

[karmic lotus]

…

[steep prawn]

@karmic lotus

I wrote it down

[karmic lotus]

alr i saw

[tacit elm]

do you need to solve this ?

[steep prawn]

Yes

[tacit elm]

What's A(F) ?

[steep prawn]

Area

[tacit elm]

The way you defined F, isn't that just the whole R^2 plane ?

[steep prawn]

Yes, I didnt know how to nicely put forall condition inside F

[tacit elm]

Aah ok

Do you have the original problem statement ?

[steep prawn]

.

Its combinatorics and my teacher defines figure as set of points

[karmic lotus]

do you agree with this polkadot figure

[steep prawn]

that are dots?

[karmic lotus]

i didnt bother to fill the circles

but yes

[steep prawn]

Wait so circles or points?

[karmic lotus]

points are me being too lazy to draw 289k circles

[tacit elm]

Did you solve it for little balls ?

[karmic lotus]

i solved for balls

[steep prawn]

Doesnt work

I have solution for A(F)<0.34

[karmic lotus]

geometry returned <0.227 for me

far from 0.29 but its in range

[steep prawn]

Ok wait

Area has to be LESS than 0.29

[karmic lotus]

mine is indeed less than that

but its pretty far off

[steep prawn]

But you have to prove that area HAS TO BE, no matter what F you choose

[karmic lotus]

the "no distance of 1/1000" f yes i think i got it

[steep prawn]

Problem with "building" this figure is that you probably wont be able to prove that no bigger figure exists

[karmic lotus]

maybe

but until i learn bigger theorems i think this is it

bottom pit takes 500 balls exact

stacking the lines of 500 balls give 577(might need factchecking) rows

[steep prawn]

Thank you for trying

[karmic lotus]

ok theres space for non circle stuff on this dot figure

[steep prawn]

Proving upper bounds is not that easy ig

@tacit elm u got something?

[tacit elm]

Nah stilll looking

[steep prawn]

For 0.34 solutions looks lie this:

[karmic lotus]

havent learnt

[steep prawn]

We are taking two vectors and transponate F by these vectors

We get F' and F'' that are disjoint

[karmic lotus]

for a certain amount of times until it hits the square?

[steep prawn]

Once

Then we make square bigger like that:

F' and F'' are inside this square

Oh, I didnt mention that length of these vectors is 1/1000

We know that F, F', F'' are disjoit

Area of F+F'+F''=3F
3F <= (1+1/1000)^2
F <= 0.34

[karmic lotus]

looks like F is a hexagonal grid

according to you

[steep prawn]

We dont know anything about F in my prove

We take any F

[karmic lotus]

but arent you assuming that 3 Fs would fill everything

[steep prawn]

No, I imply that 3F is less than this bigger square

[karmic lotus]

my new figure has spikes and stuff on the edges since circles outside the box dont contribute to the 1/1000 space barrier and i aint doing the math for that but this would be closer to the ans

[steep prawn]

.

[karmic lotus]

yeah its just closer im not sure yet

[karmic lotus]

but without a geometric visual i cant be sure that this perfectly accounts for the >1/1000 barrier for every diameter<1/1000 cluster

[steep prawn]

Wait, what

[karmic lotus]

what

[steep prawn]

Nvm

[steep prawn]

I didnt get this message

[karmic lotus]

just saying i prefer seeing the shapes

and that your method doesnt make sure that theyre all legal

[tacit elm]

Is this what you tried ? in terms of packing ?

[karmic lotus]

me yes

[tacit elm]

did you account for the "border effects ?

[karmic lotus]

thinking back i suppose largest figure would have infsmall dots and spacings

[karmic lotus]

spikes on 3 edges no i didnt do the math but ik they exist

[karmic lotus]

how do you trim the ans down from 1/3 from this

[steep prawn]

We know that Fs do not overlap

(because of length of vectors = 1/1000)

We know that Fs are inside bigger square

Sum of Fs areas is 3F (bc they do not overlap)

Area of Fs is less than area of bigger square

3F < (1 + 1/1000)^2

[karmic lotus]

pretty much 1/3

[steep prawn]

Yup

So I know that area of F < ~1/3

I dont know how to prove that this area is < 0.29

[tacit elm]

I found 0.226 without border effects

[karmic lotus]

nice thats what i got

[tacit elm]

Ok so time to get the borders, but I fear they will be negligible

Borders won't cut it

[karmic lotus]

how to shrink the circles and gaps but still making them legal

[steep prawn]

@tacit elm

[tacit elm]

I'm surprised this is actually part of combinatorics

[steep prawn]

Ok, I am going to sleep. I have been thinking about this problem for too long and its already late in my country. Imma bump this problem for anyone intrested in helping me.

[karmic lotus]

start a thread and head to discrete math channel too

[tacit elm]

I'm surprised this is combinatorics

[steep prawn]

Yeah, me too

[steep prawn]

New help thread and ask in discrete math channel for help?

[karmic lotus]

yeah

[steep prawn]

Okok

Thanks a lot again

.close

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[placid sun]

how can i plot geodesics of a surface (I already have the geodesic equations)

[quasi sparrow]

https://www.desmos.com/calculator/t6xayzxbvf

Example with desmos

Google geodesic tutorial and your plotting software

[placid sun]

ok, i will give a look at it

thank you

but the thing is that i have to plot the geodesics from this equation

[odd edge]

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[charred summit]

f(x) i have found to be:

$f(x) = \frac{1}{2} - 6x + \frac1x$

[clever fjord]

theemperor342

[charred summit]

wait let me check this again nvm

nvm i did an oopsie while calculating f(x)

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[wooden gorge]

-f(x) = 1/3x + 2x - 1/2

[charred summit]

yeah idek what i did wrong to get that 💀

fixed it now

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[neon lion]

Explain how to do this task,i tried to use V= 6x3,5x3,5 but i think its incorrect, Its not my homework its just a review before my test

[open crown]

Why 6?

Isnt the tank height just 4?

[neon lion]

I got confused and mistook it for something like this S= 6xAxA

[open crown]

Nope

The general formula for rectangular volume is

[neon lion]

Do i need perhaps to V=3,5x3,5=12,25dm
and then V=4x12,25dm2= 49dm3= 49 liters

[open crown]

Here's the general Formula

[open crown]

This works too

[neon lion]

Ah,i figured it out

[open crown]

think of volume as a 3D extension of an 2D area

When u do 3.5 * 3.5

U get the area of the bottom right?

Then u multiple by 4

Which extend the area into volume

[neon lion]

hm

[oak crown]

it doesn’t say rectangular prism

it says parallelepiped

which makes me think there is a slant height

or maybe i’m just tripping

yeah by that one theorem i don’t remember cross-section volumes are equal to regular volumes (in this case a rectangular prism) so 3.5*3.5*4 works here

but yeah there is some justifications that is needed to be done

[neon lion]

Ahh'

[oak crown]

then again it says 2 decimal places

which makes me think we should use slant height

[neon lion]

it might just be translation error to be honest..

[open crown]

Then the problem doesnt provide the angle?

[oak crown]

no you can use pythagorean

"parallelpiped" and "2 decimal places" is really throwing me off though

[neon lion]

It might be translation error,since english wasnt the language for the task

[oak crown]

oh ok

[neon lion]

But

Atleast i answered correctly i guess

[oak crown]

👍

just wanted to give a heads up

[neon lion]

Thanks

Im gonna close the room right now

.close

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[open crown]

Here's try to solve this using pytha then

[odd edge]

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[wanton bison]

$r = 2(\sin(\theta)+ \cos(\theta))$

[clever fjord]

anti-algebraist 𝔸dωn𝓲²s

[wanton bison]

Why would (r,θ) = (0,3π/4) not correspond to (x,y) = (0,0)

[sullen ferry]

r=0 always corresponds to (x,y)=(0,0)

[wanton bison]

So if I wanted to multiply by r so i can convert it into cartesian i wouldnt introduce a new solution

[wooden gorge]

who said it won't lmao

that is literally a circle

[vernal pilot]

you wont, because r = 0 is already a solution. its a misreading on my part (+ miscommunication)

[wooden gorge]

with (0, 0), (2, 2) as diameter

[wanton bison]

ok I see, was confused ty

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[wooden gorge]

@frigid canopy

[vernal pilot]

his answer was correct, you didnt need to ping him

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[loud grotto]

I know about degrees- but what is a linear function in this context??

[quasi sparrow]

Quite awful wording

I'd interpret it as linear factors

[loud grotto]

factored i get (x−1) (x4+x3+3x2+3x+3)=0

[quasi sparrow]

,w factor x^5+2x^3-3

[clever fjord]

Results provided by WolframAlpha

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

Requested by rie.mann

[quasi sparrow]

Yes then my interpretation is wrong

[loud grotto]

its not just you

[loud grotto]

i guess maybe I could see it as factorization splitting it in to 2.. groups?

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[covert acorn]

Hey Im doing my calc 2 hw and this question has me confused. From my understanding to solve this question I must take the axis of rotation y=4 and subtract 4x to get the radius of the shape. Then take the integral over 0 to 1 of pi(4-4x)^2 to get the area which I calculated to be 16pi/3.

[warped urchin]

is R_1 just a triangle?

[covert acorn]

Ye you could solve it that way but there are other questions like this where it is not a triangle so I would like to be able to solve it with an integral

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[fringe stump]

This was supposed to be a really easy question but idk how the answer is supposed to be a is less than -7/16?

[wooden gorge]

So, for the lines to have two intersections implies that solving the given two equations has exactly two solutions

=> 2(x^2 + a) = x - 1 has exactly two solutions

So, D > 0

[fringe stump]

It just wants the range of values of a that give two solutions and the mark scheme says the answer is “a<-7/16

But idk where I went wrong in my working

[wooden gorge]

Yes dear, and that is what I'm trying to explain

Why this question was supposedly easy

You got till the D > 0 part correctly

[wooden gorge]

or a < -7/16

[fringe stump]

Ohhh ok I see how that works

Can you tell me what’s wrong about the first line after d that I wrote?

Like I thought it was just a different way of getting to the same answer but I got the wrong answer

What was wrong with my equation

[wooden gorge]

the last line lol

[fringe stump]

OHHHHHH OKK

DAMN I THOUGHT I GOT THE STUFF BEFORE THAT WRONG

Thanks

[wooden gorge]

!done ^^"

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[golden granite]

Write an equation in slope-intercept form for the line that is perpendicular to the given line and that passes through the given point, x=4 ; (-3, 1).

[sand horizon]

So x = 2 means the point 2,0 right ?

[golden granite]

I know that the slope of x=4 is undefined

[golden granite]

yes

sorry I edited it was 4

[sand horizon]

Yes you need two points but you have two points

[golden granite]

but like how do you end up with y=mx+b?

[sand horizon]

you have a formula for the slope between two points

A and B be the points, (yb-ya) / (xb - xa)

Gives u the slope of the line AB

So here it will be m

Once you got this

Just put the point you want into the y = mx+b to find b

(I'd pick 4,0)

Easier

[golden granite]

thanks very much

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[hollow flower]

One message removed from a suspended account.

[hollow flower]

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[ancient rampart]

its asking

u to use the definition of integrals in terms of area

integral represents area under the curve

[hollow flower]

One message removed from a suspended account.

[ancient rampart]

so first function is a straight line passing through origin

find its area from x =0 to x=b

[hollow flower]

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[ancient rampart]

its the boudns

[green glen]

draw y = x from x = 0 to x = b

then the area of the triangle

[hollow flower]

One message removed from a suspended account.

[green glen]

some constant

[ancient rampart]

u need to review

what an integral is first

[hollow flower]

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[hollow flower]

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[ionic compass]

Try learning what is derivative and integral

[hollow flower]

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[ionic compass]

The sum of the area below the curve

The number below the integral and the top of the integral

Is the range

[hollow flower]

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[warm aspen]

hey chat can someone help me with this one pls 😭 its applications of integration. i tried average value theorem and got -9 :( the correct answer is 6.0mg, i just dont know how to get there

Suppose the amount of a drug in a patient's bloodstream t hours after intraveneous administration is a(t)=30/((t+1)^2) mg. what is the average amount in the bloodsteam during the first 4 hours?

[wanton bison]

!show

[odd edge]

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

[wanton bison]

You prob did a mistake integrating a(t)

[warm aspen]

Its b9

[wanton bison]

,rccw

[wanton bison]

Did you do a u-sub?

[warm aspen]

yeah

u=(t+1)

[wanton bison]

but then resub

[warm aspen]

yes

[wanton bison]

Ok so I see one mistake

Second term should be +

[warm aspen]

ohh ok

srry my work isn't very clear 💀

oh that was it

lolol

cause doing that and then multiplying by (1/4)(30) gets 6

[wanton bison]

Then you got 30/4 × 3/5

[warm aspen]

yesss thanks so much!! idk how i ended up with a negative 😭

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[vocal flower]

If i multiply it by 2 why is it this solution?

i mean shouldnt be it = (n+2)+(2n+2) ?

math is not mathing

jumping off a bridge would make more sense

https://tenor.com/view/kidding-joke-gif-22204117

[vale vapor]

$\left(\frac12 n+1\right)=\left(\frac12 n +\frac12\cdot 2\right)=\frac12 (n+2)$

[clever fjord]

Bonk

[vocal flower]

I dont understand why the 1/2 just jumps (like me haha) infront of the ()

and why the (n+1) is not effected by 2

[vale vapor]

wdym?

[vocal flower]

this

[vale vapor]

idk what this is supposed to mean

i think it just indicates the question number or smth

it has nothing to do with the equation at least

[odd edge]

@vocal flower Has your question been resolved?

[vocal flower]

wait i show you my math lesson script

😭

[vale vapor]

i dont understand whats confusing here

[wanton bison]

that's probably indicading that you are multiplying now both sides by 2

[vale vapor]

why would you ever do it like this??

[wanton bison]

| basically means commando line

common thing in germany

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[vale vapor]

oh thats a |?

i thought it was a 1

[wanton bison]

yea

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[stiff kettle]

help

[vale vapor]

!da2a

[odd edge]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[stiff kettle]

a)

dont u plug in 2 for x

[stiff kettle]

and it ends up being 0/0

LOL

[vale vapor]

are you allowed to use lhopital?

[stiff kettle]

what

[vale vapor]

okay nevermind

[stiff kettle]

HELP

wait

i dont get what they want

[vale vapor]

the top is the difference of sqaures

[forest sky]

have you tried factoring?

[stiff kettle]

BRUH

OJMG

STO

NO

LITERALY

im so tired

im like

[vale vapor]

u ok?

[stiff kettle]

top of my class in math

[quasi sparrow]

Can you calm down

[stiff kettle]

i swear

no

ok thanks guys

[vale vapor]

a^2-b^2=(a-b)(a+b)

use this

[stiff kettle]

ya

ok thank you

[cerulean flare]

take rest

[stiff kettle]

LOL

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[vagrant turret]

how am i soppoused to turn this into slope intercept format in order to graph it.

[sharp oak]

It is.
f(x) and y may look different, but they're just labels for a value

[vagrant turret]

ohhhhhhh ok i thought it had to be y

[native dove]

slope intercept is Y= MX+b M is the slope

i think if i remember correctly

[vagrant turret]

yeah

so i just graph it normally?

[native dove]

yeah, try that

i hate ALEKS too bro

[sharp oak]

Could also have been
🍞 = -2🧀 - 2

[native dove]

F (x) is just Y btw

noone told me that when i was stressed about this.

[vagrant turret]

i dont think it even shows me what i get wrong since even though its a review quiz

also my teacher linked a text book through it and it broke

alr thanks for the help

[native dove]

good luck!

[vagrant turret]

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[little horizon]

Say you're arranging 4 distinct pencils, 3 distinct markers, and 3 distinct erasers in some order into your pencil case, with the constraint that you cannot put an item in if there are already more of that type of item than another type of item. How many different orders are there for you to put the items in your pencil case?

[little horizon]

my answer + work

did I make any mistakes

if so, could someone point them out

[little horizon]

also pardon the terrible penmanship

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@little horizon Has your question been resolved?

[halcyon stump]

I think you might want to reformulate your answer

Think of it as a ballot counting problem where we must maintain balanced growth of each category

This gives you counting lattice paths from (0,0,0) to (4,3,3)

Apply the lattice theorem

Sum factorial over product each category factorial

That’s no restriction

Add the restriction

m-n/m+n

You get 1/7*4200 =600 possible arrangements

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@little horizon Has your question been resolved?

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[sweet zephyr]

is there a better way to solve this?

[wooden python]

yes there is

find x^2 + y^2 first as (x+y)^2 - 2xy

then find x^4 + 2x^2y^2 + y^4 as (x^2+y^2)^2

then subtract (xy)^2 from that

[sweet zephyr]

ok lemme do that

[blazing elm]

There's also this:
(x^2-xy+y^2)(x^2+xy+y^2)=x^4+x^2y^2+y^4

[sweet zephyr]

yeah ok

both the answers helped me.

thank you everyone for the help

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[white stirrup]

Can I say cot2x=0 where 0 is tan90?

If so why if not why

[wooden python]

well tan(90) isn't 0, for starters.

what is this letter btw?

[white stirrup]

x sorry for the handwriting

I got 3 angles 22.5,45,112.5

As for the 135 idk why this should be counted

Considering tan is positive

[wooden python]

yeah, this doesn't look like x at all lol

[white stirrup]

Cant assume infinity is equivalent to 0 ? Ig no

[wooden python]

no of course you can't...

wait so the original equation is

1/sin^2(2x) - cot(2x) = 0

right?

[white stirrup]

Yes

[wooden python]

with x going from 0 to 180°

[white stirrup]

But I brought cot

[white stirrup]

Yes

[wooden python]

you put a = cot(2x) and got a(a-1)=0

[white stirrup]

Yes

[wooden python]

so you are now solving the two sub-equations cot(2x)=0 and cot(2x)=1

[white stirrup]

Yeah

[wooden python]

cot(2x) = 0 <=> cos(2x)=0 <=> 2x = 90+180k

[white stirrup]

I dont understand that part

[wooden python]

cot = cos/sin, yeah?

[white stirrup]

Yes

[wooden python]

so cot is 0 precisely when cos is 0

[white stirrup]

But if you do cos0/sin0 you get infinity again

[wooden python]

that is not what i am talking about

i am talking about cot(θ)=0 not about cot(0)