help-38

[tardy hemlock]

Have you shown it's a horizontal one?

[vapid token]

yes

[tardy hemlock]

Alright then

[vapid token]

massively appreciate the help, thank you!

[tardy hemlock]

Np

[vapid token]

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[tardy hemlock]

The second point in the yellow box is kinda misleading. You don't just need f''(x) = 0 and f'(x) = 0 but also f'' changes signs at x

Because according to just the second point, f(x) = x^4 has a horizontal point of inflexion at x = 0..

[vapid token]

oh that’s right, i’ll let my teacher know about that

[tardy hemlock]

But he probably meant to include "changes signs" because he did for the first point

[vapid token]

there’s a lot of mistakes on the notes, but we tend to notice a majority of them in class

mainly with the questions jumping from 1 a) to 1 c)

and stuff like that

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[lament igloo]

Yeah I’d just start with abbot

Youll fill in any gaps you have as you go, just like read the proofs they provide carefully and understand what they do and why

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[lethal anvil]

channel is already closed 🙂

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[nimble wind]

I have verified that the provided x values, when inputted into the function, equal eachother

i just have a problem with setting the derivative equal to zero

[fast ginkgo]

what is the problem you have

[nimble wind]

its the picure above

so basically i need to set the first derivative equal to 0

[fast ginkgo]

as in why do you have a problem setting f'(x) = 0

[nimble wind]

Thats what i have on my paper

i just suck at trig

[fast ginkgo]

can you start by finding f'(x)

[nimble wind]

Yeah

its 1/2 cos(x/2)

[fast ginkgo]

ok, we set that to 0

1/2 gets wiped out

[nimble wind]

yeah easy

[fast ginkgo]

do you recall when cos(a) attains 0?

[nimble wind]

isnt it every multiple of pi

im not quite sure

[chrome latch]

Every multiple of pi/2

[fast ginkgo]

try drawing its graph if that helps

[nimble wind]

ohhh okay

[nimble wind]

the graph of the derivative?

[fast ginkgo]

cosine's graph

[nimble wind]

yeah its every multiple of pi/2

i see it

[chrome latch]

So then based off that answer, what values of x would make cos(x/2)=0

those would be where the derivative = 0

[nimble wind]

wait

i see it now

x has to equal pi

becasue then its 0

wowww

i really need to take a trig class again

[chrome latch]

yup, in your case the range only includes one multiple of pi, but if the range was different x may also be any other multiple of pi

[nimble wind]

okay i see

thanks for your help!

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[chrome latch]

sorry correction the answer would be pi+n*2pi, where n is any integer

cuz x=2pi wouldnt equal 0

[nimble wind]

i got the answer right with pi

but i see what your saying

thats every multiple of pi

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[proper shore]

Find the shortest distance from the point with coordinates ( (3; 1; 7) ) to the point, the coordinates of which are positive and satisfy the inequality

[
\frac{3x}{y} + \frac{y}{21z} + \frac{7z}{x} \geq 3(1 - (y + 7z)^4)^{1/4}.
]

[solid kiln]

kitten

[whole sable]

does anyone know year 10 algebra
and is willing to explain it to my class in a call

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[proper shore]

<@&286206848099549185>

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[glacial cipher]

am i right ?

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[noble anchor]

.open

How do i find the surface area of a cube with a space diagonal of 15sqrt3

[dapper swift]

Can you write an equation using Pythagoras in 3D?

For the side length of the cube

[noble anchor]

I don’t know how

[noble anchor]

Thank you

[dapper swift]

No worries

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[latent axle]

The equation of the trucks displacement is 20t

the equation of the cars displacement is where I'm stuck because the acceleration changes

between 0<t<20 the acceleration is 7/4 I think

and then after that the acceleration is 0

how am I supposed to put that into the equation

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[chrome latch]

If displacement = velocity*time, think about what that represents on the graph. As a hint, your equations for the trucks displacement is correct.

So using that knowledge, try figure out what velocity*time represents on the graph

[latent axle]

area

[chrome latch]

👍 now how do you get the area of the graph for the car

[latent axle]

uh

20*35

5*35

[chrome latch]

Close, remember the area of the section where the car is accelerating is a triangle, so make sure to find that area appropriately

[latent axle]

20*35 / 2

mb

[chrome latch]

Just remember you want your area in terms of t

The slightly tricky part is that the rectangle part is only part of the area if t>20

[latent axle]

(delta t * 35) / 2+ (delta t - 15) * 35

is that right

if t = 20 then it looks like it checks out

[chrome latch]

Actually one sec I think I might be over complicating the solution

[latent axle]

Alr

[chrome latch]

Yeah sorry I was being really dumb, I’m used to doing this stuff with integrals so I was making it way more complicated than it needed to be

[latent axle]

I could do it with integrals but I don’t think I’m supposed to

It’s just AP physics

So do you know what to do here

[latent axle]

How is the first part acceleration

1/2tv is acceleration?

[chrome latch]

im just noting that i split the cars total displacement into 2 seperate sections, one where it is accelerating, and the other while it is at uniform speed

so i first find the total displacement of the car before it starts traveling at uniform speed (the acceleration section, everything before t=20)

[latent axle]

What if t is like 15 wouldn’t that make the second part negative

[chrome latch]

yes, but your total displacement would still be positive

[latent axle]

But wouldn’t that affect the displacement

In an unnatural way

Like the accelerating bit of the graph has its own displacement, wouldn’t it be wrong to introduce a negative value not connected to it

[chrome latch]

well if you know beforehand that t<20, you would solve the equation using S=intial displacement + 1/2at^2

[chrome latch]

i skipped showing the calculations but the 1.75 is just velocity/time = acceleration (aka the slope)

[latent axle]

Alr I’ll look at this again with a fresh pair of eyes tomorrow morning it’s 1 am rn

Thanks for the help

[chrome latch]

no prob

[latent axle]

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[young forum]

Can someone double check ans bc when I did it before I was being stupid and I’m not sure if I’m wrong rn or wrong back then

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[foggy depot]

lakalzz

[sacred sapphire]

hmm lemme try first

ive havent done this topic in a long time

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[sacred sapphire]

oh i got it

just use the formula for composite index

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[half bobcat]

how can i prove that this function is

like

its the same when x is - and when x is +

just like x^2

without solving the integral

nvm solved ty

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[elder glade]

how do u solve this:

[orchid wagon]

have you learned about inequality with absolutes

[elder glade]

no

[cosmic meadow]

then you should learn that before

You cannot solve something without the knowledge (usually)

First of, do you know what |x| represents?

[elder glade]

alr

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[quick hound]

what do they mean by r<0

[quick hound]

how can it be less then 0

isnt r the circle

i thought it meant the line was going backwards through 0

am i dumb?

its like a circle right even if it goes to -x axis it should still be positive r

[dapper swift]

Yeah so it's a weird convention

r < 0 means that you take an angle clockwise from the positive x-axis

You're going backwards through the origin

[quick hound]

yeah i got the angle right

but why is my radius wrong

isnt it the same

wait

negative radius is correct

why the hell

man whatever

[dapper swift]

Ah it should be -4 then

It's the same radius but negative

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[earnest flume]

..

[earnest flume]

Not able to understand soln pls send someone

[dapper swift]

https://math.stackexchange.com/questions/3496870/number-of-one-to-one-functions-f-1-2-3-4-5-to-0-1-2-3-4-5-s

[dapper swift]

There are two answers; which one are you referring to?

[earnest flume]

2nd one

[dapper swift]

Ah

Yeah so there are 5 options for the image of 2, 4 options for the image of 3 etc, so 5 * 4 * 3 * 2

[dapper swift]

I guess you're confused about the inclusion-exclusion part right?

[earnest flume]

Yes

Like how they wrote 4c5*4!

And after that how they wrote about dearrangemrnts

[dapper swift]

Wait I think we're talking about different answers

Wait so the one by aryan bansal oh

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

[dapper swift]

A derangement is just the number of ways that no number ends up in its original spot

[earnest flume]

This one

[dapper swift]

Ah okay so that doesn't have any derangeemnts

Yeah ok now we're talking about the same one

[earnest flume]

So how did they use inclusion exclusion here

[dapper swift]

Give me a moment

I think I'm lost sorry

[earnest flume]

Np

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[viscid cairn]

I can't figure out thus dervative, please help

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[dusty hound]

Friendly reminder that advanced questions belong in the advanced channels too

[viscid cairn]

I don't know if this is considered advanced or not.

It's just a dervative right?

(one I can't seem to do)

😂

Would optimization be the channel for tikhonav regulation?

[dusty hound]

Idk, I never studied that stuff

[viscid cairn]

I'll ask there 🙂

[dusty hound]

Also, by late undergrad most things can fit in advanced channels, and will fit unless people redirect you
But here they won't, from experience

Though idk about the relevant advanced channels, I don't go there

[viscid cairn]

Nah I get it. No issue. Thanks man 🙂

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[lucid torrent]

If I had (2n+1)/n, the value is 2 as n approaches infinity

[lucid torrent]

But will it ever actually be 2?

[marble wharf]

it doesnt approach anything

check what you wrote again

[lucid torrent]

Oh I mean is 2

[marble wharf]

do you mean (2n+1)/(2n) ?

[lucid torrent]

No

Ohhh

Over n

My bad

[marble wharf]

that indeed approaches 2

but it never equals it

[latent meteor]

huh

oh he edited it

[lucid torrent]

Okay thanks

[latent meteor]

, frac{2n+1}{2n}

does that work

[lucid torrent]

I was just macking sure

[latent meteor]

,w \frac{2n+1}{2n}

[solid kiln]

Results provided by WolframAlpha

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

Requested by yuk.um

[latent meteor]

$$\frac{2n+1}{n}$$

epic

[lucid torrent]

Yeah fr

I meant over n tho

[solid kiln]

UsingApp

[lucid torrent]

$$\frac{2n+1}{n}$$

[solid kiln]

itsnotabbas

[lucid torrent]

This is kinda cool

[marble wharf]

you could also write it as 2+1/n which is easier to deal with imo

[lucid torrent]

Yeah

[latent meteor]

yeah just say 1/n approaches 0

[latent meteor]

i mean its kinda 1 step either way

or just say to ppl like "ignore the constant"

2n/n ==> 2n

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[thorn geyser]

how many real answers does this equation have?

[dapper swift]

One: both sqrt(x) and x^(1/4) are monotonically increasing functions

So sqrt(x) + x^(1/4) is monotonically increasing

i.e there are no turning points or saddle points and it never goes down

[thorn geyser]

Oh thank you I thought I needed to calculate it somehow

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[stiff locust]

Hello is anyone here

[ruby hound]

depends on what kinda help u need

im in helpers, but im 14 so idk much

idk like calc and allay

[marsh forum]

ask, if anyone can help, they;ll help

[ruby hound]

ik trig if u need

[stiff locust]

My internet speed is so bad rn

Need help with sinx = sin2x

[marsh forum]

sin(x)=sin(2x)

or sin(x)=sin(2z)

[stiff locust]

Is the only viable answer x = 0?

[marsh forum]

sin(x)-2sin(x)cos(x)=0

[ruby hound]

sinx-2sinxcosx=0

sinx(1-2cosx)=0

sinx=0 v 1-2cosx=0

[stiff locust]

How does this help

I don’t have -2sinxcosx

[ruby hound]

factor out sinx

and u get sinx(1-2cosx)=0

so either sinx=0

or 1-2cosx=0

2cosx=1

[stiff locust]

my problem is

[ruby hound]

cosx=0.5

[stiff locust]

-2sin(x) + 2cos(2x) = 0

[ruby hound]

[0,2pi] is the range?

[stiff locust]

I can’t factor sin out

[ruby hound]

ohh

i thought u were solving a diff problem

u can divide both sides by 2

-sinx+cos2x=0

cos(2x)=sinx

[stiff locust]

Thats what I do

look at pic

[ruby hound]

1-2sin^2x=sinx

solve quadratic for sinx

u get 2 solutions

a=-2 b=-1 c=1

plug those into the quadratic formula

u get

sinx=1 or sinx=-0.5

[stiff locust]

how do u get cos(2x) = 1-2sin^2x

[ruby hound]

pretty simple

[ruby hound]

its an identity

idk how to write latex so ill try in paint

[stiff locust]

no its alright

-2sin^(2x) -sinx + 1 = 0

[ruby hound]

proof is pretty simple

[ruby hound]

its -2sin^2(x)

[ruby hound]

you know cos2x=cos^2(x)-sin^2(x)

and cos^2(x)=1-sin^2(x)

you then substitute the 2nd into the 1st

[stiff locust]

I see

[stiff locust]

how would u solve this w/o quadratic formula

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[wraith hinge]

To solve this do we need to turn this into a quadratic equation?

[wraith arch]

yeah i would put it into standard form first

we basically just need to find the vertex of the parabola

[split chasm]

converting the form isn't really needed

you could apply symmetry of the parabola,
vertex will be located at the average of roots (which can be easily identifiable from factored from)

[wraith arch]

true

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[wraith hinge]

so just take the average of the roots?

[split chasm]

not just

still more stuff to do after that,
but doing that will give you the x-coord of the vertex

[wraith hinge]

ooh

I think I'll probably get confused in that one so I'll just convert it to standard form

I had a doubt regarding how do I convert it

[split chasm]

what's your issue?

are you aware of the required method? and/or have an issue applying it

btw, converting is more tedious, will take a few more steps than what i'm recommending

[wraith hinge]

nope its my first time I encountered to convert an equation to its standarad form

[split chasm]

have you been introduced to completing the square yet?

[wraith hinge]

I don't quite remember

[split chasm]

,tex .cts

[solid kiln]

ℝαμΩℕωⅤ

[split chasm]

that will most likely lead to more confusion

[wraith hinge]

oh

[wraith hinge]

yeah this is a bit lengthier

how do I do with the root average method?

[split chasm]

do you know factor theorem and/or are able to identify the roots/x-intercepts for the given equation?

[wraith hinge]

yeah I think so

oh the factor tree thing

yeah ik that

[split chasm]

no

there shouldn't be any reference of trees here

[wraith hinge]

no

I mean

[split chasm]

state what you think the roots/x-intercepts are here

[wraith hinge]

this tree thing

[split chasm]

like i said,

there shouldn't be any reference of trees here

factor theorem refere to something entirely different

[wraith hinge]

they would be -2 and -6

[split chasm]

yes

factor theorem is the theorem that links the factored form of a polynomial with its roots

i used multiple different wordings to ask the question, knowing any of them would've been enough

anyway, what's the average of -2 and -6?

no

[wraith hinge]

-4

[split chasm]

yes

that will be the x-coordinate of the vertex

now plug that value into the given equation

[wraith hinge]

substitute -4 for x?

[split chasm]

yes

[wraith hinge]

y=2

this gives us the vertex?

(2,-4)

[split chasm]

you have your coordinates swapped

x-coord comes first

(-4,2)

[wraith hinge]

oh yeah mb

R = {yER/y is less than or equal to 2}

[split chasm]

yeh

[wraith hinge]

thanks finally

.close

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[safe apex]

mathematical gods

[safe apex]

can y'all check my ans

hol up

Or check if my steps are aight

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[halcyon onyx]

why can you not deduce if a point is a minimum or maximum or saddle if one of the eigenvalues of the hessian matrix is 0?

[halcyon onyx]

surely if one is 0 but the rest are positive would it nost still imply that the stationary point is a minimum?

or maximum for the opposite argument

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[kind forum]

The determinant of the hessian is what determines the concavity

if one of the eigenvalues is 0, the determinant of the hessian becomes 0

Therefore, the test is inconclusive.

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[tardy jewel]

(Lagrange interpolation method) (Math IA) (desperate) (did not procrastinate just suck at maths)
long story short, I don't really know how this method works - I know the formula yes, it worked for the first green curve on the left but not so much for later parts of this image - its a chess piece.
Struggling to know what points I'm really supposed to be using to be able to plot out the curves with this method.
Any help would be appreciated -- esp since deadline is in a couple of hours and I'm basically not done with my IA.

Examples -- from point I to V (sorry for the weird naming convention this isn't very organised)
And point V-J, can't seem to find curves that fit

the current ones in use were previously failed functions with limits/ whole other sorts of functions that don't involve the lagrange interpolation
X-J is a translated sine curve

[tardy jewel]

t(x)=37.32768 x^(8)-887.00935 x^(7)+9184.83775 x^(6)-54131.43403 x^(5)+198603.87176 x^(4)-464510.78606 x^(3)+676373.09784 x^(2)-560604.87805 x+202509.25832

example of a failed function that I used points along the part i wanted to plot which didnt work

[strong zinc]

!original

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

[tardy jewel]

what am i supposed to show though

[strong zinc]

You’re trying to fit a curve to these points yes?

[tardy jewel]

yeah

[strong zinc]

And it has to be Lagrange interpolation?

[tardy jewel]

yup

there isnt like a question for it

its my ib math ia

part of it at least

[strong zinc]

Idk what that means but okay

[tardy jewel]

my bad

its for the internal assessment for some fancy worded program

[strong zinc]

Were these points given to you or did you choose them?

[tardy jewel]

i chose them

thats the issue
I dont know what points im supposed to be choosing for the curve to fit that specific segment

[strong zinc]

Ah okay this is your problem

https://www.youtube.com/watch?v=F_43oTnTXiw

[tardy jewel]

yeah, sorry im not very clear rn

[strong zinc]

https://en.wikipedia.org/wiki/Runge's_phenomenon

[tardy jewel]

oh yeah i saw that jn, just didnt really read much into it
i'll read up a bit, tysm

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[slate stream]

i need help filling out the frequency distrbution table

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[stiff locust]

yo

[stiff locust]

I'm confused about the factoring out

at the very bottom they took out x^4/5 (x-2)

Specifically the x^4/5

f'(x) has two terms that have an x

x^-1/5 and x^4/5

how does x^(4/5) factor out of x^(-1/5)?

[balmy spear]

What is $x^{\frac{4}{5}} \cdot x^{-1}$

[solid kiln]

TayBee

[stiff locust]

x^4/5 * 1/x

[balmy spear]

To answer the question a bit more directly though, it doesn't inherently factor out, but you can factor it out if you want, and clearly they decided it makes it look a bit nicer

Use index rules

$x^a \cdot x^b = x^{a+b}$

[solid kiln]

TayBee

[stiff locust]

ok

-1/5

x^

[balmy spear]

Yup, so there's your x^-1/5 that you had at the beginning

So that's how they factored out 4/5

[stiff locust]

ok

[balmy spear]

They just worked out what you add to 4/5 to get -1/5 so it's equivalent

[stiff locust]

How would I think of dividing mathematically

lets see

x^-1 / x^(4/5)

because we decided we need to take out x^4/5

when we're dividng we subtract exponents

[balmy spear]

They basically wanted to factor it to get rid of one of the x's

So they decided to factor out the 4/5

Could have factored out the x^-1/5 if you wanted and then had x on the inside as -1/5 + 1 = 4/5

[balmy spear]

And yes true

So you can do -1/5 - 4/5

[stiff locust]

-1/5 -4/5 = -1

alright perfect

[balmy spear]

[stiff locust]

so index rules are for same base

what if its diff base do those rules have a name

[balmy spear]

As in same variable like x? Yes

$x^a \cdot y^b$

[solid kiln]

TayBee

[balmy spear]

Would just remain as it is, you cant combine those

They're separate variables and separate powers

$x^a \cdot y^a = (xy)^a$

[solid kiln]

TayBee

[balmy spear]

That occasionally helps but not really applicable here

[stiff locust]

awesome thats a good summarization ty

[balmy spear]

Np

[stiff locust]

Wouldn't f(0) be 0 and not -1/8

??

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[wispy sparrow]

could be a mistake

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[misty cosmos]

For part C) is there a reason why: 4^8 to find the total numer of Bauman string doesnt work bc i used it in part B) which worked. I tried that since the first string is C and last string D, then C - - - - - - - - D, where there are four possible options within the string?

[misty cosmos]

Like for the second string I believe there is A/B/D/E as the options and in the third string, if A is chosen in the second string, B/C/D/E, etc.?

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[wraith hinge]

Hi

[wraith hinge]

I need help

[unique pine]

shoot

[rare oak]

Get some help

[wraith hinge]

:D if my step patterns

are

positive does it go up?

[wraith hinge]

bruh

[turbid leaf]

Wdym

[wraith hinge]

hold on

[nimble stone]

[wraith hinge]

I am having hard time to plot the step patterns

Idk I might have made mistakes

[wraith hinge]

help me pls

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[halcyon onyx]

how is it possible to integrate this

[halcyon onyx]

F(x) and r are both vectors of R^3 btw

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[jaunty pagoda]

guys is this equation correct my mom kepe asking me if its correct or not 4x = y7 + 6 (4x - 5y)?

[halcyon onyx]

.close

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[unborn solstice]

How to do?

[twilit fable]

Which one

!status

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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

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[livid dune]

Is this correct?

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[livid dune]

<@&286206848099549185>

[livid dune]

.close

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[rapid horizon]

What

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[abstract thorn]

Please help me solve q 16

[abstract thorn]

Tomorrow is my maths retest , i need to solve the exam sheet before hand , so atleast i am somewhat prepared

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[velvet bolt]

draw all the lines

so you can visualize the problem

[abstract thorn]

That's what I am cunfused in , should I find the distance of each and then check weather if 2 ate same or not ?

[velvet bolt]

did you draw the lines

[abstract thorn]

No

Idk what type and how to draw the line according to the question

[velvet bolt]

do you know how to draw a line which is in the form y=mx+b ?

[abstract thorn]

No bro

[velvet bolt]

https://www.youtube.com/watch?v=AYCElcAQ3Es&t

[abstract thorn]

Hey @velvet bolt I got the question. Can u help me here ,I am stuck at one point

[velvet bolt]

?

did you watch the video

[abstract thorn]

Watching it , previously I was doing another question while waiting for a reply

[abstract thorn]

I am still not able to relate it to my question fully

[velvet bolt]

are you trolling or what

[abstract thorn]

I am serious, I am weak in maths , i can't understand until someone teaches me

I am really sorry for that

[velvet bolt]

what is your question

@abstract thorn

[abstract thorn]

How to solve 16 th question

[velvet bolt]

good

watch the video then come back

[abstract thorn]

Full or from where u said to start from , the link starts from a time

[velvet bolt]

from the start

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[dull narwhal]

How do i Find the length of the small gap in the center, like the orsnge line

[slender vale]

if both of the circles had diamteter 4.5, then there would be no gap

each circle overlaps ths centre of the rectangle by 0.5, as 4.5 + 0.5 = 5

hence together, they overlap each other by 1cm (the orangle length)

[dull narwhal]

.close

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[cinder prawn]

hello

[cinder prawn]

what does an R group mean?

in chemistry

actually nvm

.close

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[ionic pendant]

you can solve for v in that equation

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[short trellis]

3 of the same cubes are on too of each other what is the volume of the given right parallelepiped if the surfice area is 224

[nimble stone]

what do you think

[tepid hamlet]

think about how many total cube faces are visible, then you can find the SA of a single one

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[short trellis]

224/16

Is the side sqrt14

Of the cube

@tepid hamlet

[tepid hamlet]

you counted 16 sides? count again

[short trellis]

Faces

16 faces

[tepid hamlet]

that are visible?

how are you getting that

[short trellis]

Well I don’t counf the top one

[tepid hamlet]

yes facs is what i meant sorry

[short trellis]

What is facs?

[tepid hamlet]

faces

count the # of faces visible carefully, it is not 16

[short trellis]

18?

[tepid hamlet]

...

theres a top, a bottom, then 4 on each row, how many is that total?

[short trellis]

Like visble from the front?

14

[tepid hamlet]

yes

were you subtracting 2 from 18 to get 16 maybe? remember where the cubes meet, each one has a face that becomes hidden. two meeting places = 2 x 2 that you don't count out of the 18. 18 - 4 = 14 also, so you can think of it that way if you wish

[short trellis]

Ok now what?

@tepid hamlet

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[orchid wagon]

How many integer pair $(x,y)$ satisfying $0\le x\le2023$ and $\log_5(5x+5)-3y=125^y-x$

[solid kiln]

lemonsaurus aficionado

[orchid wagon]

i dont really know where to start

i notice 125^y-=5^(3y)

[wispy sparrow]

try simplifying log_5(5x+5)
also collect like terms

[orchid wagon]

$\log_5(x+1)+1-3y=5^{3y}-x$

[solid kiln]

lemonsaurus aficionado

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[orchid wagon]

<@&286206848099549185>

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[trim jolt]

Please don't occupy multiple help channels.

[serene sleet]

what algorithm can i use

to solve this

[edgy willow]

you'll have to try a bunch of different ways to place 2,6,3,a into the 4 boxes, and once you make a choice, you can use the ratio in one row to apply to the other, like the setup
3 a
2 6
has ratio 2:6 on the bottom, so a would be 3*6/2=9

[serene sleet]

what do you mean by
3 a
2 6

[neon dirge]

the layout for where to place the areas

[serene sleet]

oh

[neon dirge]

3 topleft

a topright

etc.

[serene sleet]

so then i need to try all possible combinations

and find all a

wouldnt that be many many possible ways

[edgy willow]

yes, but can you figure out how to try all the meaningfully different ways

[neon dirge]

it's not how you think it works .-.

[serene sleet]

how would i go about this. any idea

[neon dirge]

<@&268886789983436800>

[obsidian ferry]

ty

[neon dirge]

consider it this way:

since we can place them arbitrarily, we need to start with a choice

so let's put a in the bottom right area

meaning a is fixed there

[serene sleet]

yes

[neon dirge]

if a were somewhere else, we could just rotate the rectangle

and it'd be on the bottom right again

meaning you now have three remaining areas

and 3 values to put into

so you just have 3 * 2 * 1 = 6 possible combinations

[serene sleet]

oh because we only need to swap the positions three numbers

[neon dirge]

yop

[serene sleet]

and a is fixed

alright

[neon dirge]

yes

[serene sleet]

that makes it easier

thank you

[neon dirge]

and if you want to reduce more

if you put one number in the top left for instance

notice that it's irrelevant in which order you place the other two

the area of a remains constant

meaning you only have 3 cases

which are for the 3 values

1.case: 2 is in top left

2.case: 3 is in top left

3.case: 6 is in top left

[serene sleet]

damn

you figured that out quickly

what branch of maths does this fall under

[neon dirge]

idk

but if you go through the 3 cases

you'll get 9, 4, 1

[neon dirge]

maybe stochastics

and geometry

[serene sleet]

2 3
a 6

3 6
a 2

6 2
a 3

all possible cases covered?

[neon dirge]

yop

@serene sleet and now comes a sick part ;)

multiply all the three areas

2 * 3 * 6

[serene sleet]

yeah

[neon dirge]

and divide by the square of one of them

so 2 * 3 * 6 = 36

36 / 2² = 9

36 / 3³ = 4

36 / 6² = 1

:O

magic!

heresy!

witchcraft!

[serene sleet]

amazing

[neon dirge]

ban this lunatic for his forbidden technique

[serene sleet]

wait what is this called

like is there an actual name to this technique

[neon dirge]

hm don't think so no

but it works in higher dimensions too

e.g. if you had a cube

and split it into 8 volumes

or a hypercube

and split it into 16 hypervolumes

[serene sleet]

damn

well thanks for the help!

[neon dirge]

🦇 np

[serene sleet]

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[latent axle]

Just learning Riemann sums and my book is pointing to this equation, I can’t find it online and I’m confused what I’m looking at

[latent axle]

I can only find the f(x)dx one

[trim jolt]

@latent axle Has your question been resolved?

[pseudo grotto]

you are summing rectangles, if you partition into n rectangles, a particular rectangle will have f(value) as the height, and (b-a)/n width.
f(a + k * width) represents the height of the k-th rectangle to the right of a

[latent axle]

what is the a+ representing

[latent axle]

do you mean rectangle or do you actually mean triangle here

because the book is just doing rectangles rn

[pseudo grotto]

a is the starting value, since you are summing over the interval [a, b], and (a + k * width) is which rectangle you sum over

[pseudo grotto]

i mean rectangle, where are triangles showing up?

[pseudo grotto]

ah typo

[latent axle]

alr

good

okay so

a is the left end of the partition?

[pseudo grotto]

yes

[latent axle]

and whats the right end or is that not its own constant in there

[pseudo grotto]

so how it's normally taught in calculus is you just choose a consistent point to use as the height of the rectangle

it could be the top left corner, it could be the top right corner

it could be the exact middle

[latent axle]

yeah

[pseudo grotto]

whichever scenario your questions refer to, you just want to think it's to compute the area's of the rectangle, then sum over all the rectangles

it might help you to do a practice problem to visualize

[latent axle]

wait a minute how does f(a+k*width) represent height?

[pseudo grotto]

f(x) is height

[latent axle]

yeah

[pseudo grotto]

the input is a + k * width is which rectangle you are taking the height of

[latent axle]

where does the width come in

[pseudo grotto]

you have the interval over [a, b]

lets say i want 2 rectangles over this interval

[latent axle]

ok

[pseudo grotto]

then the first rectangle goes from [a, (a + b) / 2] and the second rectangle goes from [(a + b) / 2, b]

sorry lemme rewrite that

first interval is [a, a + (b - a) / 2] and second is [a + (b - a) / 2, b]

[latent axle]

im confused

whats the a+(b-a)/2

i get its the partition

[pseudo grotto]

ok

let's say my interval is [1, 5]

[latent axle]

okay

[pseudo grotto]

i want 2 rectangles

[latent axle]

yeah

[pseudo grotto]

where would they be

[latent axle]

[1,3] and [3, 5]

[pseudo grotto]

ok

what is 1 + (5 - 1) / 2

[latent axle]

i gort that with 1+5/2

[latent axle]

5/2

[pseudo grotto]

no

[latent axle]

?

[pseudo grotto]

1 + [(5 - 1) / 2]

order of operations

[latent axle]

mm

3

[pseudo grotto]

yes

now in general form

[latent axle]

why not just do what i did though?

(a+b)/2

[pseudo grotto]

verify for yourself that
a + (b - a) / 2 = (a + b) / 2

then in general, for n rectangles instead of 2,
a + (b - a) / n,
a + 2 * (b - a) / n,
...
a + (n - 1) * (b - a) / n
a + n * (b - a) / n
lists out all the endpoints

where the last line, a + n * (b - a) / n = b

[latent axle]

Okay, first off, if they are equal why can’t I use either, second, I’m not understanding the multiple part

[pseudo grotto]

you're free to express them however you want to since they are equal. However, this way in particular is intended to give insight as to what's going on when you generalize to taking a limit as you approach infinitely many rectangles having diminishing width

as for the multiples part

consider the example [1, 5] again

what would the end points be if i asked you to partition it into 4 rectangles? 8 rectangles? 20 rectangles?

if i asked you to partition them into 20 rectangles, a particular rectangle should have width (5 - 1) / 20

[latent axle]

6/20 would be the width

[pseudo grotto]

no,

the line segment is 4 units long

not 6

[latent axle]

Then how did I do (1+5)/2 earlier

[pseudo grotto]

.

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

[latent axle]

Yes

[pseudo grotto]

but a + (b - a) / n is not equal to (a + b) / n

for every value of n

[latent axle]

Only for /2?

[pseudo grotto]

yes

[latent axle]

Okay

[pseudo grotto]

anyway you have width (5 - 1) / 20

the first endpoint would be at 1

where would the second one be?

1 + (5 - 1) / 20

where would the third one be?
1 + 2 * (5 - 1)/ 20

[latent axle]

Oh

[pseudo grotto]

where would the k-th endpoint be?
1 + k * (5 - 1)/ 20

[latent axle]

I think I see

[pseudo grotto]

let me reindex so it aligns with the book

[latent axle]

So you are just getting multiples of the width to find different partitions with a base starting point at a=1

[pseudo grotto]

yes

[latent axle]

That’s where the a comes in

[pseudo grotto]

yes

[latent axle]

Kinda sorta?

Okay

[pseudo grotto]

if i wanted to know about an interval [100, 150]

i can't just begin at 0

i need to shift a to the right

[latent axle]

Right

So that gives us the x value of the end points of partitions

Subinyervals

Whatever

[pseudo grotto]

yes

a + k * width

[latent axle]

And the f(of that) gives us the height of the function at that point

[pseudo grotto]

yes

then you sum over all the rectangles

[latent axle]

Times what

[pseudo grotto]

the width of a rectangle

(b - a) / n

[latent axle]

Ohh

[pseudo grotto]

area of rectangle = height * width

[latent axle]

Right

Okay

Okay

I see

So another question if you don’t mind

[pseudo grotto]

ok

[latent axle]

The book talked about unequal widths being fine to use

Why would I want that

If I’m approaching infinite rectangles anyways

What’s it matter

[pseudo grotto]

so

the limit, should it exist, is called the integral

[latent axle]

And that’s the area right?

The area is the limit

[pseudo grotto]

yes it represents the area under a curve

[latent axle]

Okay

[pseudo grotto]

in a more formal setting, you're allowed to take any partition of an interval, you're even allowed to integrate over sets that aren't intervals themselves

and how you take a limit in those situations is called taking a refinement, basically you take the partition you have and put more endpoints into it

in the formal setting, there is no requirement that you need equal width's of these rectangles

[latent axle]

I assume there’s a seperate formula for it though right?

[pseudo grotto]

but in traditional calculus, they'll give you questions like
compute the partial riemann sum over the interval [1, 5] with endpoints being [1, e, pi, 4.234982725, 4.95486, 5]

and you need to know how to answer a question like that

[latent axle]

Is this a calc 1 thing or more advanced

[pseudo grotto]

calc 1

[latent axle]

So my book should go over it then

[pseudo grotto]

yes

the idea is the identical

find the height of the endpoints

multiply by the width of the rectangles

add them all up

[latent axle]

Is 41 and 42 what you are describing

Idk what norm is

[pseudo grotto]

the technique is still sum (height * width)

[pseudo grotto]

?

[latent axle]

Yeah, but in terms of like a formula, some the widths will differ

[latent axle]

41

[supple saffron]

yes

[pseudo grotto]

idk but according to a quick google

[latent axle]

Uh

[pseudo grotto]

which would be 1.2

[latent axle]

So do I just

Is that it

[pseudo grotto]

unless your book uses norm in a different way

[latent axle]

Yeah book says 1.2

Okay

I think that covers everything

Thanks

[pseudo grotto]

np

[latent axle]

Are you available later if I ping you for certain practice problems?

[pseudo grotto]

maybe idk

[latent axle]

Alr

[pseudo grotto]

i'll answer if i answer

[latent axle]

Got it

Thanks

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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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[hybrid bluff]

I don’t understand this question. Can anyone help pls 😭

[velvet vector]

hmmm

[near vessel]

whats have you tried

[velvet vector]

lets plug some values in

[hybrid bluff]

I don’t really understand what I’m doing. I tried to plugging numbers into the system of equations but that didn’t work.

[near vessel]

notice the table of values

its not yet complete for the 2nd equation

[velvet vector]

-9=A(-1)+B, -7=A(1)+B

[near vessel]

you should probably complete it first

[velvet vector]

lets solve for A and B first

[near vessel]

you are given 2 known ordered pairs for the second equation, meaning you can easily determine its slope

[hybrid bluff]

2 ordered pairs?

In the table?

[velvet vector]

(-1,-9) and (1,-7)

slope is rise over run

[hybrid bluff]

How did you get that

[velvet vector]

table

[velvet vector]

it says when x=-1, y1 is -9

[near vessel]

use these for solving the slope for the 1st equation

[velvet vector]

yes

[near vessel]

and use these for the 2nd

[hybrid bluff]

Ohh wait let me try this

[hybrid bluff]

I calculated the slope I got 1

[hybrid bluff]

I got -2 for this

What do I do now 😭

[near vessel]

correct

so the first equation has a slope of 1

and the second have a slope of -2

does any equation come to mind that involves x, y, a slope, and a y-intercept?

[hybrid bluff]

slope y-int form?

y=mx+b

But I don’t have y-int

[near vessel]

correct

you are going to use y=mx+b to find the y-intercept

choose a point in equation 1

any point you want

[hybrid bluff]

(7,-1)

[near vessel]

alright

we have
x = 7
y = -1
and the slope of equation 1:
m = 1

substitute those values in y=mx+b,then solve for b

[hybrid bluff]

Oh ok

b=-8

[near vessel]

alright

so the first equation will be

y=mx+b
y=(1)x+(-8)

which simplifies to
y=x-8

thats your first equation