help-38

[proper kernel]

I didnt say that

hey are we doing with the actual explanation or this one

you gotta choose one, remember

[torn shale]

Tbh if what youre abt to continue explaining will make me better at math then heck yeah but ill waste your time and annoy you with my questions and stupidity. So its really your choice

[proper kernel]

then its faster if we start with the initial explanation

[torn shale]

Go ahead. I’ll let u write everything out then ill ask (if i have questions)

[proper kernel]

you know r(x) = x^2
it squares numbers

now say the domain is [0, √3]
you can figure out that the range is [0, 3]
that is what f([0, √3]) = [0, 3] means, I thought you would know that

any questions so far

[torn shale]

Oh it means range

Ok no questions

[proper kernel]

oh awit

oh no theres a typo in there

you know r(x) = x^2
it squares numbers

now say the domain of r is [0, √3]
you can figure out that the range of r is [0, 3]
that is what r([0, √3]) = [0, 3] means

[torn shale]

Oh

[torn shale]

(So name of the)(domain of the function} = (range of the function) is what that formula means

Ok.

[proper kernel]

thats not a formula

[torn shale]

Wrong phrasing

[proper kernel]

you keep calling it a formula, I dont think you know what that word means

also you are misreading what the pattern is

[torn shale]

English isnt my native

[proper kernel]

the pattern is f(this) = that

and I am using the pattern on intervals instead of numbers

[torn shale]

Oooh so its like a normal function

Yess okay

I understand

[proper kernel]

you gotta tell me how you didnt see that earlier

[torn shale]

R(the donain) = (the range) yes that makes sense

[proper kernel]

I have no idea how you missed that, you could be missing a lot more if I keep leaving holes in the explanation like that

no extra parenthese on the range

r(domain) = range

[torn shale]

Just

Yes

[torn shale]

Im dumbbbbb

Okay? 😭

[proper kernel]

thats not a very helpful description for you is it

surely you should get to the bottom of that

its easier than you think but I cant do it for you

[torn shale]

What are you even talking about? My dumbness?

[proper kernel]

let me repeat

I have no idea how you missed that, you could be missing a lot more if I keep leaving holes in the explanation like that
just saying "Im dumb" is not a very helpful description for you, is it
surely you should get to the bottom of this and be more specific
its easier than you think but I cant do it for you

[torn shale]

Oh

Ok ok

I underatand

Continue if you want

[proper kernel]

you know r(x) = x^2
it squares numbers

now say the domain of r is [0, √3]
you can figure out that the range of r is [0, 3]
that is what " r( [0, √3] ) = [0, 3] " means

you know F(x) = x + 1
it adds 1 to numbers

now say the domain of F is [0, 3]
you can figure out that the range of F is [1, 4]
that is what " F( [0, 3] ) = [1, 4] " means

[torn shale]

Yes okay

[proper kernel]

now you can see that r( [0, √3] ) = [0, 3]

and F( [0, 3] ) = [1, 4]

so F( r( [0, √3] ) ) = F( [0, 3] ) = [1, 4]

[torn shale]

Ooooh

Yes

[proper kernel]

and there you can see that the pattern has emerged again

the domain of F(r(x)) is [0, √3]

the range of F(r(x)) is [1, 4]

and its directly from considering how the domain would pass through the functions in order

it cant get more direct than that

the rules/formulas youre learning are procedures to help work with that

so theyre not as direct but they answer the question you had on finding the domain of F(r(x))

[torn shale]

Can i ask now?

[proper kernel]

yea I left a pause

I shouldve asked you for questions earlier

[torn shale]

I understand this

[torn shale]

But this? How?

[proper kernel]

consider the following

you said earlier that f(domain) = range

this is a general pattern just like with say F(r(0)) = 1

that pattern is how you write the input and output of a function

to be sure, do you get that F(r(0)) = 1

[torn shale]

I mean- when you plug in x=r(x) in f(x) it’ll give you the equation for f(r(x)) which you can plug in x = 0 and you’ll get 1 ig

F(r(x)) is x^2 +1

Yes.

[torn shale]

Isnt that pattern = equation of the function

X^2 +1 for this case

[proper kernel]

back, thats roughly the idea

youre forgetting though that F and r each have specific domains

so F(r(x)) isnt just x^2 + 1

[torn shale]

I wont ask you to explain, say what f(r(x)) other than x^2+1 and perhaps ill understand since i think the problem here is the phrasing

F(r(x)) is x^2+1 considering that x is a part of f(r(x))’s domain

Of thats what you mean

If*

[torn shale]

I mean wait

Isnt the domain of f(x) is what values x can be

So the domain of f(r(x)) is what values r(x) can be which is the domain of r(x) whicg is indeed [0,sqrt3]

Ok

I understand

@proper kernel so sorry for the ping but its 7am and if my parents wake up theyll kill me

[proper kernel]

sorry I gtg

[proper kernel]

which is the range

[torn shale]

So yes the domain of f(r(x)) is [0,sqrt3]

[torn shale]

Yea sorry

Values for r(x) is range

Yes

Okay… we’re done yeah

So you made me understand how the entire thing works

Tysm

[torn shale]

Wish you luck in whatever you’ll do

[proper kernel]

you too

[torn shale]

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[prime shoal]

is there an easy way to find x?

[prime shoal]

actually I plotted it on desmos and found the value for when y=0

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[dense breach]

there's no other way then a calculator for this btw

no exact solution

[prime shoal]

yeah thanks

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[rapid horizon]

Question 3rd is linearly dependent

The author says it should be LI how?

[hallow kite]

author must be wrong, you can't have four linearly independent vectors in R^3

[native radish]

Jai hind

Glad to see a Indian here

[brazen pebble]

I'm Indian too

[native radish]

Nice I am also Indian , hindustani

[brazen pebble]

Yes

Jai hind

[rapid horizon]

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[native radish]

Can I use Hindi here or not?

[rapid horizon]

I closed the chat

Hope next time

[native radish]

Ok

[brazen pebble]

You didnt

[native radish]

Wah bhaiya wah

Bete moj kardi

[hallow kite]

she did close it, it just won't disappear if everyone keeps talking in here

[native radish]

... Cairo is typing...

[rapid horizon]

I am also waiting for cairo mesage

Cairo is typing...

[native radish]

Bhai india mai kaha se ho?

[lilac bloom]

Rajasthan 👿

[rapid horizon]

I am not Indian

[native radish]

It seems to anant

[native radish]

Oh ok

[frail heron]

yes

[hallow kite]

where else speaks hindi?

[native radish]

Ok

[rapid horizon]

Kota

[native radish]

In india

[brazen pebble]

I'm tamil

[hallow kite]

i know but she said she wasn't indian

[native radish]

Oh

[hallow kite]

i didn't know if there were other countries

[native radish]

Ya

[brazen pebble]

Nepal?

[hallow kite]

as far as i know pakistan is mainly urdu and sri lanka is sinhala

[lilac bloom]

India mai kaha se ho?

[native radish]

Bangal

Mai Bengali hu

[hallow kite]

gujarat 🙋

[native radish]

I am bengali

Joy bangla

[brazen pebble]

Jai shri ram

[lilac bloom]

Jai shri ram

[native radish]

Jai Vishnu Brahma Mahesh

Sorry for it

[native radish]

Gujarat is in india

[brazen pebble]

@native radish Your good at maths?

[native radish]

My dad is a maths teacher ( msc first class) I little good at maths as his son

[brazen pebble]

If your I need helpers because I'm preparing for imo and ap level for Cambridge University

[brazen pebble]

You wanna be my friend

[native radish]

I am preparing for ssc cgl

[native radish]

Yeah why not

Mitra

[brazen pebble]

?

[lilac bloom]

Shrimad ramayana youtube pe aani lag gyi finally!!

[lilac bloom]

it means friend

[native radish]

I mean friend in Hindi

Ramanujan was a Tamil so I like Tamil Nadu and it's a good state for scientific research

[brazen pebble]

Yes

[native radish]

Bodhi dharman who was the founder of kunfu is also from tamil

[brazen pebble]

Yes

He was from pallava dynasty

Which is on tamil nadu

[native radish]

Yeah

What about sn bose

[brazen pebble]

sn Bose was born on Kolkata I think

[native radish]

Yes a Bengali who worked with Einstein and founder of god particle it's a honour to us cus we are also bose

We are in same group

🫡

[brazen pebble]

You mean Subash Chandra Bose?

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[native radish]

No I mean satyendra nath Bose
A well known mathematician

[frail heron]

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[stable ore]

If you want to talk about non-math things, try to keep it to social channels like #discussion

[native radish]

Ok sorry

[stable ore]

.close

Can I not? rip

[stable ore]

Ok can you close it

[native radish]

Yeah sure

.close

[native radish]

Nothing happened

Lol

[frail heron]

I did close it

[native radish]

I am sorry for discussing non maths things

You can close it

[shy nacelle]

@native radish You have to close it now

[native radish]

Ok I do

. close

Nothing happened

[native radish]

Fairyrose >

Wait a min

What you are saying to close

About

I can't

I have not the control

Then why saying

Lmao

[native radish]

Mean see rose what he is saying

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[stable wing]

can a regular language set have infinite elements in it?

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[gentle sleet]

I dont understand how they got the bound for y2 to be 0 < y2 < 2

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[gentle sleet]

wtvr

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[rapid horizon]

I guess here a is -2?

[wraith hinge]

how can it be negative it's on positive y axis

what does the area under probability density function represent ?

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[shy nacelle]

,calc 1-1/230.125

[solid kiln]

Result:

0.8125

[rapid horizon]

What is that cairo?

[shy nacelle]

1/2×(0.25+a)×1 = 0.8125

,calc 2*0.8125 - 0.25

[solid kiln]

Result:

1.375

[shy nacelle]

Isnt the total area = 1 ?

Is 11/8 in option ?

[rapid horizon]

Of course 1

[shy nacelle]

?

[rapid horizon]

Let me check

How to calculated?@shy nacelle

[shy nacelle]

Hey here i subtracted the area of triangle

1/2×b×h , b= 3 h = 0.125

Next i equated the area of trapezium to remaining part

Area of trapezium = 1/2(sum of parallel sides) × base

Ok

[rapid horizon]

Yes i know you find the area but i am cofused what area you find and what you subtracted

[rapid horizon]

How three?

What triangle is this?

[shy nacelle]

This one

-2 to 1 so base 3

[rapid horizon]

Should we not ignore negative area?

[shy nacelle]

I think we cant

[rapid horizon]

You added these area right?

[shy nacelle]

Yeah

[rapid horizon]

Fine thanks

[shy nacelle]

Does it gives correct ans?

[shy nacelle]

Welcome !

[rapid horizon]

Yes correct

[shy nacelle]

Ok

[rapid horizon]

I was confused by the calculation 🤣🙈

[shy nacelle]

I used * they wrote in some weird way

[rapid horizon]

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[rapid horizon]

Thanks

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[gleaming ermine]

need help in part a
says that 180 per foot across
so i always need to get 800 ft across so 800x 180
plus the rest 100x
but i think im interpreting this wrong

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[gleaming ermine]

<@&286206848099549185>

[gleaming ermine]

<@&286206848099549185>

[cursive pulsar]

Uve been here for so long 😭

So first convert the 2mi to ft to make everything easier

[cyan nebula]

Shi mb I would of helped but it’s in miles and feet

[cursive pulsar]

You need to find the amount of ft for across the river (the diagonal blue line) then multiply by 180
Just use the pythag theorem.

Then for the distance across the land (the straight blue line) it will be whatever 2mi is in ft - x, multiply that by 100!

[cursive pulsar]

Right ☠️

When I first saw it I was like “what the heck is mi”

[cyan nebula]

I can’t think in miles and feet lmfao

[gleaming ermine]

wait

why is it 2-x

[cursive pulsar]

Not 2 - x

2mi

[gleaming ermine]

shouldnt it just be x

[cursive pulsar]

But you need to convert it to ft first

Look at the picture

[gleaming ermine]

oh right

no but still

[cursive pulsar]

You’re tryna find that blue line

The blue line will be 2 mi - x

The cable is the blue lines

[gleaming ermine]

OH

THERES ANOTHER BLUE LINE

I COULDNT SEE

[cursive pulsar]

HAHA

[gleaming ermine]

i just saw the hypotenuse

and im like

how am i supposed to do both

[cursive pulsar]

Loll

Do you get it now ?

[cyan nebula]

Should be light now

[gleaming ermine]

so its hypotenuse x 180 + (2mi - x) x 100

right

[cursive pulsar]

Yes!

Convert the mi tho

[gleaming ermine]

thank you thank you

mhm

no worries

💀

[cyan nebula]

!close

[gleaming ermine]

.close

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[quartz arrow]

can anyone help me understand why this integral becomes ucos(u)du

[quartz arrow]

instead of it becoming cos(u)du

wait dumb question

nvm

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[hexed shard]

How can I determine if A spans all of R4?

[hexed shard]

I know this is the same as the system of equations encoded by A having infintely many solutions

but how can I determine such a thing?

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[loud trellis]

if i'm trying to find the area between these two curves, how should i split them up? or are they both correct?

[celest heath]

The second picture is easier, but it doesn't really matter what you do

[wanton rune]

when integrating, if going along the x-axis, keep each as functions of x (solve for y), if y, turn into functions of y (solve for x)

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[limpid dawn]

try to find left angle

trignometry

For example

oh ok sorry

that's smart btw

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[slim wedge]

Is it possible to change the graph for a cosine function from blue to green?

[slim wedge]

My current function is

[barren tartan]

use area

[wanton rune]

the blue plot doesn't look like it matches this function

[slim wedge]

Oh... I actually pulled up the wrong pic. The blue graph is f(x) plotted with values 0, 5, 10, 15, 20, ...

My bad

I don't know if my question makes any sense now.. I want the plot to look like the green graph.

[old tiger]

square it?

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[hidden sleet]

This one is giving me algebra difficulty

Here is how I have been solving these problems thus far:

1. I take the derivatives partial to x and y.
   In this case, fx = 3x^2 - 3 + 3y^2, and fy = 6xy
2. I set these equations to 0 to find the coordinates of the critical points.
   Here is where I am having trouble. x = sqrt(3)*sqrt(1-y^2), -sqrt(3)*sqrt(1-y^2)
   and 0 = 6xy

So I do not know how to proceed further. Plugging x into 0 = 6xy has proven a bit beyond me.

then later on... past the point where I am stuck:
3) take fxx, fyy, and fxy. solve for the hessian determinant and do the 2nd partial test. then plug these coordinates into the function to get the local min/max and/or saddle point

I guess we can tell that x = 0 and y = 0 from 0 = 6xy and plug that into the fx = 0

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[warm nymph]

im just trying to understand how wolfram alpha get's this result since that's not the result I got.

[lusty delta]

what result do you get?

[warm nymph]

the normal log(|x-1|) - log(|x|)

• C

but I was trying to solve a bigger problem and using that solution for that integral actually yields an answer i would expect in that bigger problem so I was trying to understand what i got wrong and came to this weird thing

[mighty prism]

logarithm only accepts positive values, so these are essentially the same

[warm nymph]

I know but then why the -(x-1) i dont understand

thats the part of |x-1| that's in 0,1

[kindred pier]

Depends on your domain

!xy

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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.

[warm nymph]

sure

[mighty prism]

yeah wolfram might be doing some weird formatting

but what you got is correct

[warm nymph]

let me send it

[midnight frigate]

Ohh actually x(x-1) = x^2-x

so that is always positive except on [0,1]

[warm nymph]

this is the original problem as you can see its a separable differential equation that shouldn't be too hard to solve and at some point in the solution you have to solve that integral

idk if this really helps because its just more complicated but sure

[warm nymph]

yeah

so what does that mean

[lusty delta]

do you see that part where it says "assuming a complex-valued logarithm"

wolfram's answer is correct since it doesn't care if the log has a negative input

[warm nymph]

hm

[lusty delta]

though normally on paper you write it with the absolute value signs

[warm nymph]

but i still dont understand why multiply by -1

if it didnt care then it would have just not used the absolute value

[kindred pier]

for complex logs, yes, but you want to stay in the real domain

[lusty delta]

it's the same thing

you can think of it as ln(-x) = ln(x) + ln(-1) and ln(-1) is just a constant (even though it's imaginary), meaning it's added into the +C when integrating

[warm nymph]

i still dont completely get it how would that "stay in the real domain" maybe a visual explanation could help

[warm nymph]

yes but that would yield a different solution overall and not a good one for the differential equation

my question isn't really why would wolfram alpha do that its more like why is that necessary for the solution to the differential

[lusty delta]

then add in the absolute value signs

[warm nymph]

that sounds awful ngl

for a solution

well

[lusty delta]

if you assume a complex-valued logarithm (as wolfram has showed you) you don't need to

but that's normally how it's done

[warm nymph]

this is a graph of 1/(1-e^(|x|)) vs 1/(e^(|x|)+1)

when i made the differential i had the green solution in mind

that's if you do that little step in the logarithm

blue one is if you just plain ignore the absolute value

[lusty delta]

i have no idea what this is

[warm nymph]

solutions to ^

[lusty delta]

alright

what is r?

[warm nymph]

mb read the functions again i misstyped a sign

[warm nymph]

constant

i assumed r=1 on those graphs

and c=0

[lusty delta]

for a differential equation involving sgn you can solve it piecewise i guess

i.e. if x > 0, solution is [this]; if x < 0, solution is [this]

[warm nymph]

I did that and i wanted to try just using |x| since if you integrate sgn(x) you get |x|

i mean yesterday i was trying to solve the same thing but without the (1-y) term

and its just ce^(-r|x|)

hm

maybe thats it

it just assumed x was in between 0 and 1

idk why but ig thats what it did

so then what is the actual solution to this

[warm nymph]

ill try doing this and looking at the graph

shit you need to do the exact same integral

@lusty delta if i wanted to be rigorous how would i go about solving ln|y-1| - ln|y| = -rx+C

[trim jolt]

@warm nymph Has your question been resolved?

[warm nymph]

hm

this is beyond me i might just accept that wolfram alpha is so good it transcends any understanding

[trim jolt]

@warm nymph Has your question been resolved?

[delicate pagoda]

there is a big hint in the "assuming a complex-valued logarithm" note

you use partial fractions to get

[delicate pagoda]

$=\int (\frac{1}{x-1}-\frac{1}{x})dx$

[solid kiln]

Bob Goldham

[delicate pagoda]

then split the integral like so

$=\int \frac{1}{x-1}dx - \int \frac{1}{x}dx$

[solid kiln]

Bob Goldham

[delicate pagoda]

the integral of 1/x is ln(x) so

$=\int\frac{1}{x-1}dx-\ln(x)$

[solid kiln]

Bob Goldham

[delicate pagoda]

if you want the step-by-step for the left integral we do integration by substitution, define u:=x-1 and du=a integrate du

which yields ln(u)

[delicate pagoda]

undo the substitutions and you end up with the result you got here

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[blazing geode]

The length

What

What is it referring to, what it means

God

Mickey lord

What is the length

It means I have to straighten the curve in [0,1] and measure the length in the span

I do not have a steam iron at my hand

What should I do

[hallow kite]

they're referring to arc length

there's a formula

arc length of function y(x) from x = a to x = b is integral from a to b of sqrt(1 + (dy/dx)^2)

[blazing geode]

Tell me Neil

Should I memorize it

Or there’s a way of interpretation

[hallow kite]

yeah it's important enough to memorize, but there's a logical way to understand it too

[blazing geode]

Neil

[hallow kite]

it comes from the pythagorean theorem

[blazing geode]

Neil is such a good name

[hallow kite]

oh uh thank you lol

[blazing geode]

Neil nailed it

[blazing geode]

I see

[hallow kite]

so the integrand becomes sqrt(1 + (dy/dx)^2) dx

imagine that dy and dx were actual variables for a moment

then dx = sqrt(dx^2)

[blazing geode]

Is there a name for this

[hallow kite]

arc length formula

[blazing geode]

Ohh

[hallow kite]

https://tutorial.math.lamar.edu/classes/calcii/arclength.aspx

here is a super simple derivation

they chop the curve up into a bunch of little line segments which have a length that can be found using the change in x and change in y

[blazing geode]

Thank you Neil, you’re so kind

[hallow kite]

happy to help

also i like the kamikaze profile

i'm an eminem fan as well

[blazing geode]

Neil is a hiphop head

I didn’t expect that

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[sick gale]

Is this the same as (p and not q) implies r?

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@sick gale Has your question been resolved?

[sick gale]

<@&286206848099549185>

[junior saffron]

yes, it is the same. Apparently it varies from litterature to litterature, I commonly see $\lnot$, but ~ is also regarded as 'not'.

[solid kiln]

Crystopher

[sick gale]

Alright, thank you. Is there anywhere I can find rules regarding parentheses?

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[red scarab]

what happened here? At first I thought they were multiplying both sides by -3, but on the bottom the 3 stayed as a 3 and didn't turn into a 9.

[livid thunder]

they multiplied just the top by 3

and then took the negative to the denominator

[red scarab]

oh like they just picked a side to have the negative?

[livid thunder]

yeah, it wont matter whether the denominator or the numerator recieves the negative

[red scarab]

makes sense, thank you!

[livid thunder]

all g

[red scarab]

where did this 2 come from?

[split chasm]

chain rule

derivative of inner function (2t+1)

[red scarab]

ah

this test ain't gonna go well xD

ty <3

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[devout prawn]

Hello! I'm recreating this question because the channel with it is hidden (can't reopen). I need help proving a fact. Conditions: There are containers of 20 or 40 tonnes, 60% of all containers are filled with sugar. Can the mass of the containers with sugar be 40% of the total mass?

Containers can weigh 20 tonnes or 40 tonnes so there are two options for containers with sugar - there are containers with 20 tonnes of sugar and/or with 40 tonnes of sugar. The same with containers with not-sugar. So we can take 2 sugar containers with 40 tonnes and 1 sugar container with 20 tonnes but the amount of sugar containers (in this case the amount of sugar containers is 3) must be 60% of total amount of containers (so in this case there must be 5 containers in total and 2 of them are not-sugar containers and the amount ratio of not-sugar containers is 40%).

For example, question: can the mass ratio be 50%?
Yes, if we take 4 not-sugar container with 40 tonnes, 2 sugar containers with 40 tonnes and 4 sugar containers with 20 tonnes. So it will be 160/320 which is sugar container mass ratio and 6/(4+6)=6/10 which is sugar container amount ratio (one of possible combinations, but this one shows how we can choose containers to get the right mass ratio (in this question right mass ratio is 50%)).

I know about a way of proving it by finding minimum of mass ratio, which is 3/7 (which is greater than 40%). However, I need to explain why the proof I compiled is correct (not the one with the minimum).

I knew that the answer is 'No' but couldn't normally prove my position when I was solving it for the first time. I have written (on the screenshot):

So I am asking for help in showing a proof by which I can get exactly the same ratio ( 4[20x+40y] = 6[20x+40y] ) by which I have shown that 40% is impossible. In the proof, I had to use only those ideas that were given in the condition or were written by me in the proof (on the screenshot). So mass ratio and amount ratio can be used.

[trim jolt]

@devout prawn Has your question been resolved?

[warm nymph]

Not true (it was 3 am I went to sleep) because the integral evaluates to ln|x-1|

Wolfram alpha considers the alternative ln(1-x) because it's positive in ]0,1[

Idk why that would give such a more appropriate solution to the differential

[trim jolt]

@devout prawn Has your question been resolved?

[trim jolt]

@devout prawn Has your question been resolved?

[old tiger]

lets say we have 4 types of containers:
With sugar & 20 tonnes, this will be a
Without sugar & 20 tonnes, this will be b
With sugar & 40 tonnes, this will be c
Without sugar & 40 tonnes, this will be d

We will write a system of equations to prove there is no solution:

[solid kiln]

clonesolopros

clonesolopros

clonesolopros

clonesolopros

clonesolopros

clonesolopros

[old tiger]

$10b+2d+12c=0 \implies (b<0) \lor (c<0) \lor (d<0)$

[solid kiln]

clonesolopros

[devout prawn]

Thank you for the solution, I understand it. I have also solved this system, got a slightly different equation but still be the same proof of impossibility. Is there a way to get a final equation:

4(20x+40y)=6(20x + 40y) or 4(a+c)=6(a+c)

I have transformed this system of equations but I can’t get rid of b and d and several moments with c. Here what I got (can’t be sure, maybe did something wrong)

[solid kiln]

𝚃𝚛𝚒𝚜𝚝𝚎

𝚃𝚛𝚒𝚜𝚝𝚎

[devout prawn]

As I can see, it hasnt got any solutions in natural numbers. It should be an answer but I need to transform it to 4(20x+40y)=6(20x + 40y) or 4(a+c)=6(a+c). Is there a way to do so?

[old tiger]

it simplifies to:
\begin{gather*}
2(20x+40y)=0\
2(a+c)=0
\end{gather*}

[solid kiln]

clonesolopros

[old tiger]

tldr:

\begin{gather*}
20x =-40y\
2a = -2c
\end{gather*}

[solid kiln]

clonesolopros

[devout prawn]

I mean is there a way to get exactly

4(20x+40y)=6(20x + 40y) or 4(a+c)=6(a+c)

It can be not a final result, I just need to get it somewhere in the solution. Doesn’t matter if it will be in the end or in the middle, just somewhere, but exactly one of those two equations.

I might not understand something and one of those equation has already been given, please tell me if so

[old tiger]

those equations dont make sense, well you just subtract the terms from both sides

[devout prawn]

yeah they dont make sense but I need to get them somehow. As I said I know how to solve this task with normal ways but I will have to proof that my method is right so I have to get this equation through some conversions. Could you show how to subtract to get what I need? I haven't got any ideas how to get rid of 0,4d and 0,4b

[old tiger]

my idea is this:

[old tiger]

$\frac{12a+24c}{4} = \frac{8b+16d}{4}$

[solid kiln]

clonesolopros

[old tiger]

$= 3a+6c=2b+4d$

[solid kiln]

clonesolopros

[old tiger]

and since we have $2a +2c = 3b + 3d$

[solid kiln]

clonesolopros

[old tiger]

$\implies c=\frac{2b+4d-3a}{6}$

[solid kiln]

clonesolopros

[old tiger]

and $2a+2(\frac{2b+4d-3a}{6}) = 3b+3d$

[solid kiln]

clonesolopros

[old tiger]

$=a=3b+3d-\frac{2b+4d}{3}$

[solid kiln]

clonesolopros

[old tiger]

$=\frac{7b+5d}{3}$

[solid kiln]

clonesolopros

[old tiger]

so: $a=\frac{7b+5d}{3}$ and $c=\frac{2b+4d-3a}{6}$

[solid kiln]

clonesolopros

[devout prawn]

so b and/or d will remain under any conversions? And there is no way to get that equation that I have written?

[old tiger]

1. can you explain it? xd, 2.I dont think so

[devout prawn]

explain the equation that I have written?

[old tiger]

no the remaining under any conversion

[devout prawn]

Can't really explain, I just cant imagine the way of getting 4(a+c)=6(a+c) through these formulas for 'a' and 'c' that you have shown above. I might be wrong tho

[old tiger]

well bc we "know" a, we can further simplify c, by putting the formula for a into c

[devout prawn]

$c=\frac{-5b - d}{6}$

[solid kiln]

𝚃𝚛𝚒𝚜𝚝𝚎

[old tiger]

yea

also its interesting if b=d, then a and c are integers

then $d = -6c - 5b$

[solid kiln]

clonesolopros

[old tiger]

but we have a problem

again

[old tiger]

C is negative

[devout prawn]

so no natural solutions

[old tiger]

nope

[devout prawn]

so we cant get 4(a+c)=6(a+c)

[old tiger]

that is like saying 1=0

your formula doesnt have to do anything with this

I think

but if the mass was a bit closer it could have a solution maybe

[devout prawn]

yep, it was like a random equation that I wrote on exam two weeks ago and I was trying to proof that it will work but I think it was just a random thing that has nothing with real solution

[old tiger]

I hope you can do something with this 😂

[devout prawn]

sorry for wasting your time and thank you for trying to help me! I'll close that

[devout prawn]

I wont speak about it XD

[old tiger]

its always an investment of time XD

[devout prawn]

I think this problem was a real time-waster. On exam and here :D

Thank you)

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[trim jolt]

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[stable ore]

What have you tried

[frail heron]

What seems to be the problem?

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@sacred basin Has your question been resolved?

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[umbral river]

I just want to check that I am starting this question correctly? I am on question 6., and I am getting x=x and y=-y, which seems a bit weird, but I guess is possible if y=0?

[celest heath]

You didn’t put the name sake of the method, the Lagrange multiplier, into any of your equations

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@umbral river Has your question been resolved?

[umbral river]

Oh true

But when I do I get that it equals -1 from the y equation, and 1 from the x equation

Meaning -1=1

Which isn't true obviously

[umbral river]

<@&286206848099549185>

[slender shard]

That means you have two possible cases for lambda. Try solving for each one.

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@umbral river Has your question been resolved?

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[wraith hinge]

In a physics experiment, a group of students showed that the position equation of a particle in uniformly accelerated rectilinear motion can be written in the form x(t)=10(3t-5)(2t-1), with x in cm and t in s. The values of the initial velocity v0 and the acceleration a of the particle are

[wraith arch]

!status

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[wraith hinge]

4

[stable ore]

Can you show the work?

[stable ore]

Looks right

[wraith hinge]

okok, thanksssss

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[umbral river]

Could I get some help on this question? I am supposed to solve it with lagrange multiples. (I solved using the distance formula already and got 9/sqrt5). It is getting a lot more complex than any of the example questions and I feel like I may have done something wrong, but regardless I don't really know where to go from here.

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@umbral river Has your question been resolved?

[balmy spear]

I think you have gone down a bit of a rabbit hole trying to find the gradient of the distance function here

Start with $$\mathcal{L} (x,y,\lambda) = f(x,y) + \lambda g(x,y)$$

[solid kiln]

TayBee

[balmy spear]

f(x,y) is the objective function - it is much easier to work with the square of the distance rather than the distance itself

[umbral river]

Sure, but how do I get g(x,y)?

[balmy spear]

$f(x,y) = (x-4)^2 + (y-2)^2$

[solid kiln]

TayBee

[balmy spear]

g(x,y) is the constraint

So it's what you've defined as f(x,y)

[umbral river]

Oh, I see. So I can just square it?

[balmy spear]

Yeah, if you minimise the square of the distance you will get the same result as minimising the distance

[umbral river]

Oh true

That makes sense

Thank you

[balmy spear]

Np, so ye plug those in, set the partial derivatives to 0 and solve the system of eqns, you'll arrive at the answer

[umbral river]

Gotcha

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[cinder lark]

just want a second pair of eyes im super sleep deprived did i plot these right

[cinder lark]

!status

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[cinder lark]

Four

[hallow kite]

everything is right except (6, -2)

you plotted (6,0)

you also plotted (0,-2) for some reason

i mean visually this can't be correct since you have five points listed and six points plotted

[cinder lark]

ah

ok let me redo

i was thinking visually it didnt look rght

okay @hallow kite

should look a little better 😅

[hallow kite]

isn't this the exact same as the first one?

[cinder lark]

wrong ss

[hallow kite]

yeah that looks right now

[cinder lark]

.close

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[wraith hinge]

👍

[sudden mortar]

whats the problem?

[cinder lark]

I was uploading my work I wanted to make sure it was right.

[kindred pier]

yes

[cinder lark]

thank you

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[bright pine]

How to solve $$ \int{x cos 4x dx} $$ I keep messing it up

[solid kiln]

Nerdy_Coder

[unique minnow]

We could point out where things go wrong in your work. I suppose you're doing integration by parts?

[kindred pier]

!status

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@bright pine Has your question been resolved?

[bright pine]

Yea

[split chasm]

the work is too poorly structured to see what's going on

stop doing everything on a single line

it's hard to see where equations start and stop

can you try doing it again

(also respond to the bot so the channel doesn't close)

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[kindred pier]

simplify and then rational root theorem. That's my guess.

You can also just try RRT by calculating only 2x3x8x12

But, assuming RRT does work, it doesn't guarantee you find every solution

[marsh forum]

,w (x+2)(x+3)(x+8)(x+12)=4x^2

[kindred pier]

[marsh forum]

ok, so cancelling the 4x^2 doesn't help

[kindred pier]

I'm putting money on RRT

[rapid horizon]

I bet on SWR

20¥

Hope we will win

[kindred pier]

!nosols

[trim jolt]

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[kindred pier]

@rapid horizon okay, try RRT.

[kindred pier]

And I think this will be good shortcut

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[rapid horizon]

After simplifying ingot x^4+25x^3+152x^2+552x+576=0@kindred pier

So should I factor 576 now?

[kindred pier]

That's cool. You didn't need to do the whole thing, but sure

You could have just done 2x3x8x12 to get 576

and you don't even need 576, you just need its factors, which are the factors of 2,3,8 and 12

[rapid horizon]

Yes

2^6,3^2

[kindred pier]

Looks good

Now you just have the painful task of testing which work

idk who gave you this problem, but they mean

[rapid horizon]

Can we use calculus?

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[drifting rose]

can someone show me how I would do this

[woven nova]

You wanna use gauss

[drifting rose]

ok i'm on the right track at least

I already found it's gauss elim

since i found it, now what

[woven nova]

Instead of like this, I would write it like an augmented matrix

[drifting rose]

?

[woven nova]

Where B is just 0 0 0

[drifting rose]

oh with the line?

[woven nova]

Yeah

[drifting rose]

yea the program can't do that

you just have to imagine it's there

[woven nova]

OK haha

[drifting rose]

yea lol soz

[woven nova]

The apply gauss algorithm

If you manage to get a row of zeros, then you have one equation like this:
0=0
And the next one would be:
ay+bz=0

In these cases we say we have a degree of freedom and chose z to be a variable

Which means that we found solutions that are not 0

[drifting rose]

wait z?

where did z come from

[woven nova]

You could say a vector v out of V has 3 components

And you can call them x y z

[drifting rose]

hmmm

so how would I figure this out?

[woven nova]

Using gauss elim

[drifting rose]

ok i did the gauss elim alreadt

I got this

now we could say we have 2x+2y-6z=0 and -1y+2z=0

i think

now idk what to do

[woven nova]

By choosing g z=t, we can find solutions for y

And then we can find solutions for x

So in total, we find solutions for the coefficients where they are not all 0

Here z=t, y=2t, ...

[drifting rose]

so for -1y+2z, would it become -1y+2t?

[woven nova]

Yes

[drifting rose]

and find for t?

[woven nova]

Ah wait

I did a mistake sry

[drifting rose]

oh

[woven nova]

We are not solving for x y z, we are solving for r s t
But the result is the same, it's just names

So we have
2r+2s-6t=0
-1s+2t=0
0 =0

And here we can say t=t
And then s=2t

And then find r

Thus we find the set of solutions

Sry for the confusion there

[drifting rose]

r=s+3t then righ?

[drifting rose]

yea it was a bit confusing but I realsied what you were slowly refering to

then you just sub everything in?

[woven nova]

r=3t-s

[drifting rose]

oops my mistake

so r=t then?

and what does that mean? r=2?

[woven nova]

It means we found infinitely many solutions

[drifting rose]

So how would I find values where r s and t make 1 of the infinite solutions

[woven nova]

Generally we found
(r,s,t)=(t,2t,t)

To find on of these solutions, just choose any t

Except for t=0

[drifting rose]

Any number?

[woven nova]

Yes

[drifting rose]

Like 2

[woven nova]

Yes

[drifting rose]

Woah.

[woven nova]

You can also check if with f=2 so r=2 and s=4, the scaled vectors add to 0

Btw, visually, this means that if we have n vectors and they are linearly dependent, they span in a structure that is at most n-1 dimensional

[drifting rose]

ah

[woven nova]

For example here we have 3 dependant vectors which means they don't span a 3 dimensional space

When they are independent, then they do span a n dimensional space

Easiest example are the vectors (1,0,0), (0,1,0) and (0,0,1)
They are clearly independent and also span the 3d space

[drifting rose]

ah I understand

thanks for the help martin, that was really good!

.close

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[woven nova]

Btw @drifting rose for more intuition I highly highly recommend the linear algebra series on youtube by 3blue1brown

[drifting rose]

oh I heard of that guy, will definately look into him more, his videos seem amazingly good from observation

But ty for the help, it really made me understnd

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[livid inlet]

so I have a very simple question i just need confirmation: i am aware it is in dutch here is the translation:

'The arc-sin function is the invese function of the sin function with limited domain [-pi/2, pi/2]' now my question is: isnt this definition wrong?

[livid inlet]

isnt it limited domain [-1,1] ?

[hallow kite]

what they mean is that arcsin comes from the sine function where you've restricted the domain of the sine function to [-pi/2, pi/2]

they're not referring to [-pi/2, pi/2] as the domain of arcsin

for two functions to be inverses, the domain of one is the range of the other and vice-versa

[livid inlet]

oh

well then i suppose i just interpreted it weird

i suppose if i rotate it

like this

i get something that resembles the sin(x) with limited domein [-pi/1,pi/2]

okay

thanks for the help

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[hasty idol]

Help me to derive an integral expression for this volume

[hasty idol]

Using (r, theta, z) i get that theta is in [0, pi]

With gives me this on the rz plane

Now I need to convert this to spherical coordinates, and I was thinking about moving the z axis to r = 4 that way it would be much easier but idk how to consider the aditional area

Maybe there's a easier approach idk

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[slate bolt]

<@&286206848099549185>

[slate bolt]

How do you do this 😭

[wraith hinge]

any other info given?

hmmm i feel like we will make some triangles and use congruence rules

[split chasm]

you have two intersecting chords

you can apply intersecting chords theorem

(part of power of a point)

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[eternal sky]

Hi can someone please let me know if my answers are correct.

[junior saffron]

1. is good, in 2) using sine is wrong since $\sin T = \frac{RS}{RT}$. So even if you computed it correctly you would solve for $RS$, not $ST$. Otherwise the method was alright.

[solid kiln]

Crystopher

[flat creek]

Can someone help me solve this questions? I'm supposed to use linear equations ro solve it and idk where to start 😭😭😭

The total cost after 3 months includes a 10% tax and is $825. Tax increases to 13% at month 6 and the total cost after 6 months with 13% tax is $1356. What is the monthly maintenance fee and building fee

[kindred pier]

!occupied

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[flat creek]

Oh, sorry 😅

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[rustic arrow]

How is this not undetermined??????

[river willow]

can u show the whole thing

[wooden plover]

Do you know anything else about a ?

@rustic arrow

[rustic arrow]

Its physics but the problem is maths

@wooden plover

@river willow

[wooden plover]

This doesn't answer my question

Can you post the problem statement instead of the solution ?

[rustic arrow]

A is a constant

It’s Fourier transform of the sgn (t) signal

[river willow]

as t approaches 0 or -inf respectively u get 1 and 0 as the value of the expression

[rustic arrow]

Yes

But in last one it approaches +inf

And somehow the result is zero

…

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[rustic arrow]

No 🥲

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[still plinth]

This would be 1000 possible permutations, not combinations, right?

[lusty delta]

selecting the digits to use is combinations
arranging the digits in a specific order is permutations

for language purposes like in that message you posted, writing combinations makes more sense

[still plinth]

combinations is where order doesn't matter

so 475 would be equal to 745

which is obviously not that case here

[elder rivet]

permutation is definitely the correct word to use mathematically but that reddit comment used the language loosely ig and used the word "combination" like we use in everyday life

[still plinth]

oh, fair enough

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[magic sinew]

find X in number 5

[magic sinew]

my work is written in pencil, the original figure is in ink

i am stuck, what do i do

<@&286206848099549185>

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[native raptor]

@magic sinew did you get 9

or is it unsolved

nvm

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[stone comet]

Hi.

[stone comet]

$[
\lim_{m \to \infty} \frac{\sum_{n=0}^{m} \left( \int_{0}^{n} -\frac{1}{x^2} , dx \right)}{\ln(\ln(m))}
]$

[solid kiln]

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

[stone comet]

Does the harmonic 1/x series grow faster or slower than ln(ln(m))?

[placid radish]

you can do an integral test on 1/x that approximate the growth rate of the sum of 1/x

note that the integral from 1 to n of 1/x is larger than the sum of 1/m from m goes from 2 to n

[ionic pendant]

i don't think the integral converges

[stone comet]

No, the integral diverges, i am talking about the limit comparing to ln(ln(m))

[placid radish]

i think they mean the integral of 1/x^2

integral of 1/x^2 from 0 to 1 is divergent

[stone comet]

-1/x^2 and yes it is divergent

[placid radish]

so integral of -1/x^2 from 0 to n will always be undefined

and the limit wouldnt make sense

[stone comet]

wait what

[placid radish]

you probably meant to integrate from 1 to n

in which case you get someting* 1/x +something

[stone comet]

Oh yeah, that was a typo sorry

[placid radish]

no worries

[placid radish]

but anyways for your question

[placid radish]

you can do this

hopefully the method is clear to you

[stone comet]

Yes it is

thank you

[placid radish]

u get get a lower bound and an upper bound for the sum of 1/x

in this case lower bound is useful

[stone comet]

Enough info, ill do it from here on my own

thx

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[marsh forum]

Find the inverse of the function $f(x)=log_a(x+\sqrt{x^2+1})$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

so to start

$y=log_a(x+\sqrt{x^2+1})$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

so $a^y= x+\sqrt{x^2+1}$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

or $(a^y-x)^2 =x^2+1$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

$\implies a^2y+x^2-2a^yx=x^2+1$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

or

$a^{2y}-2a^yx=1$

[solid kiln]

ƒ(Why am. I here)=I don't know

[marsh forum]

now what?

nvm, got it

.close

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[pseudo grotto]

[marsh forum]

how do I get this emoji

Can't steal it

[pseudo grotto]

what prevents you from stealing it

[marsh forum]

carl bot

[pseudo grotto]

you can download this link then reupload it into discord emoji server https://cdn.discordapp.com/emojis/1129577505771225118.gif?size=96&quality=lossless

[marsh forum]

thanks!

[pseudo grotto]

alternatively

https://discord.gg/gNEkg8YV

[marsh forum]

It's fine, I downloaded it

thanks

[frozen plover]

please don't promote your server in help channels!

to ask for help check out #❓how-to-get-help

[pseudo grotto]

it's not my server

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[gaunt temple]

Hello can somebody please explain why lambda = 2 has algebraic multeplicity = 2

(I calculated the determinant using Sarrus)

[marble wharf]

well this char poly isnt fully factored yet

as a note, it would be imo easier to expand the det along the second row

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[gaunt temple]

in that case, using laplace the determinant would be (2-lambda)[(3-lambda)^2 - 1)] right?

and if i factor everything i can find the zeros of the polynomial

but how can i find the algebraic multeplicity?

if it's something like (10-lambda)^2 * (6-lambda) it's easy to see that 10 is an eigenvalue of algebraic multeplicity = 2 and 6 is an eigenvalue of algebraic multeplicity = 1

[gaunt temple]

but here it doesnt seem very obvious to me