help-0

[thin urchin]

oh

got it ty

.close

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[wooden mantle]

A horizontal rectangle of rectangular cross-section
beam in the vertical direction is directly proportional to the
the product of its width and height in cubic form, i.e. it is of the form cxy3,
where c > 0 is a constant. Determine the width and height of the stiffest possible beam when it is
sawn from an R-beam log.

Im not really sure where I should start with solving this.

[wooden mantle]

Its from my calculus class

[lone heart]

@wooden mantle Has your question been resolved?

[wooden mantle]

<@&286206848099549185>

[carmine reef]

is there a picture of the problem

I'm kinda lost

[wooden mantle]

nope, this is the whole question, I translated from Finnish to english

wait It should be (xy^3)^2

lemme fix that

[carmine reef]

oof

[wooden mantle]

thats what I mean, like how do I actually go about solving it 💀

[carmine reef]

I mean I can't understand the problem at all

I think we need to maximize one thing subject to some constraint

Like, maximize cxy³ when xy=R or something except that doesn't even have a solution so idk

[wooden mantle]

Yeah the question is so abstract imo

[carmine reef]

can you convert it to pure math

maximize A under condition B?

[wooden mantle]

Im not sure what you mean

[carmine reef]

make it not a word problem

can you understand what it's asking for

[wooden mantle]

This is prob easier to understand if I draw something gimme a sec

like this

aint they just asking the length and height of the cross sectional beam when at its highest possible stiffness

But like

It can be anything since we dont have actual numerical values right?

[carmine reef]

Yeah I still don't have a clue what's going on

Highest possible stiffness?

[wooden mantle]

Yeah no clue

"Determine the width and height of the stiffest possible beam when it is
sawn from an R-beam log."

[carmine reef]

No clue what an R beam log is either lol

[wooden mantle]

https://math.stackexchange.com/questions/2529525/finding-the-stiffest-wooden-beam-from-inside-a-log-rectangle-inside-circle

I found something similar

[carmine reef]

Ok

So your constraint is x²+y²=R², right

[wooden mantle]

Umm

Probably

But I still dont get what R is

[carmine reef]

Radius of the log

[wooden mantle]

Like is it just a log that we use

To make the beam

Aight

[carmine reef]

Yeah

Actually the constraints is probably (2x)²+(2y)²=R²

[wooden mantle]

I mean I think I understand it better niw

[carmine reef]

ah that's good

[wooden mantle]

Im good for now then, ima look more into it later after school

Thanks

.close

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[carmine reef]

Np!

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[last gulch]

here is a question I got wrong on a test. Can someone show me how to do it properly? Simplify $\frac{x^{2}-16}{x^{2}-3x-4}$

[ocean seal]

PeckyChicken

[solemn juniper]

Can you show what you did on the test?

[last gulch]

$4 - \frac{1}{3x}$

[ocean seal]

PeckyChicken

[solemn juniper]

I'm more talking about the steps leading up to whatever final answer you get

I have 0 idea where you went wrong just by looking at this

[last gulch]

I went wrong by assuming you could just take the 2 x²'s out of the fraction

leaving $\frac{-16}{-3x-4}$

[ocean seal]

PeckyChicken

[last gulch]

I think

[solemn juniper]

Fair enough

Do you know how to factor quadratics? Because ideally, once you factor both sides of the fraction, they'll have a common term you can cancel

[last gulch]

yes 🤦

top is a difference of 2 squares

[solemn juniper]

Ok well then factor them

[last gulch]

so $\frac{(x-4)(x+4)}{(x-4)(x+1)}$

[solemn juniper]

Hey look, a common factor

[last gulch]

$\frac{x+4}{x+1}$

[solemn juniper]

Quite so

[last gulch]

on the test answer it says x +4/x +1. I dont know how to parse that

[solemn juniper]

Oh wait

The denominator isn't factored correctly, my bad

[last gulch]

wait

yeah

[solemn juniper]

I shouldn't have given the go-ahead so fast

[ocean seal]

PeckyChicken

[solemn juniper]

Better

[ocean seal]

PeckyChicken

[last gulch]

that matches the answer on the test

[solemn juniper]

Very nice

[last gulch]

may I close the channel?

[solemn juniper]

You don't need my permission

[keen plinth]

also

[solemn juniper]

If you have nothing else, close whenever

[keen plinth]

this

[keen plinth]

should not have come from this either

[last gulch]

wait I see what i did

[keen plinth]

sums in a denominator cant be split

only in the numerator

[last gulch]

I extracted the -16 and the -4

[last gulch]

yes

my mistake was that

[keen plinth]

[last gulch]

I extracted -16 and -4 which gave me $\frac{x^2}{x^2-3x}+\frac{-16}{-4}$

[ocean seal]

PeckyChicken

[keen plinth]

you totally cant do that

[last gulch]

yes ik now

[keen plinth]

good

[solemn juniper]

They have since found enlightenment

[keen plinth]

amazing

[solemn juniper]

[last gulch]

then my next step was to extract the $x^2$. $\frac{x^2}{x^2} + \frac{1}{-3x} + 4$

[ocean seal]

PeckyChicken

[last gulch]

and you can see how i ended up with my answer there

[keen plinth]

thats extremely cursed

[last gulch]

mathematics if it was different

[keen plinth]

[last gulch]

on another question i gave the answer $\frac{x-1}{2}$. The correct answer was $\frac{1}{2}(x-1)$. They are the same right?

[ocean seal]

PeckyChicken

[solemn juniper]

Indeed

[last gulch]

ok

in that case

.close

[lone heart]

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[radiant karma]

how do I solve
un+1 = 2un - 1, u1=2 for n>= 1

[radiant karma]

Find the term that has value 257

n+1 is in subscript

and n

in the 2un

And also 1 in the u1=2

I tried doing it by substituting values

But i could only find answer by listing all the values manually

Is there a simpler way

and how do I solve this

un+1 = 3un - 2, u1=2 for n>= 1
Find the value of the largest-term that is less rhan 10,000

[gray isle]

try to find the general term

[radiant karma]

Whats the general term?

The gen formula?

[gray isle]

yeh

[radiant karma]

I was thinking of doing that when i realized one was geometric and one was quadratic

But in the tb, they didnt cover general formulas yet

So i thought there was another way

[lone heart]

@radiant karma Has your question been resolved?

[radiant karma]

<@&286206848099549185>

[lone heart]

@radiant karma Has your question been resolved?

[lone heart]

@radiant karma Has your question been resolved?

[lone heart]

@radiant karma Has your question been resolved?

[alpine sable]

<@&286206848099549185> help this dude

literally the last channel 💀

ok, sorry

[radiant karma]

Can i close this

.close

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[arctic birch]

Hello, I'm struggling with this scaling issue in my code.
I thought multiplying C's X,Y by scale would put it in the right spot.

[arctic birch]

Got it. Its Dx = (Cx-Ax)*S%+Ax and the same for y

.close

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[alpine sable]

how would i solve this?
(x+8)^2+(y-4)^2=144

[alpine sable]

get the circumfrence

[gray isle]

are you able to identify the radius?

[alpine sable]

how would i do that

[gray isle]

compare this to the general equation of a circle in standard form

that will tell you where radius is located in that equation

[alpine sable]

ok

[lone heart]

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[sleek furnace]

i just need to know how this problem is named, like, connecting dots with lines, then adding a dot

[soft knot]

I don't understand what you mean. Given a set of points, adding a line between every pair and then adding a new point and connecting it with any of the previous ones?

[weary wyvern]

graph

[lone heart]

@sleek furnace Has your question been resolved?

[sleek furnace]

nvm, i found it, it was the Königsberg and jordan problem

thanks anyway

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[devout zinc]

i need an answer and also an explanation, please

[lone heart]

@devout zinc Has your question been resolved?

[devout zinc]

<@&286206848099549185>

[devout zinc]

bro i cant get a proper answer online

@rustic coral do u have an answer

[rustic coral]

It's one dimensional so it only has length?

[devout zinc]

well, it must have thickness right

if we use a microscope or smth we might be able to see the thickness

[devout zinc]

well, thanks ig

.close

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[edgy crag]

"Determine λ ∈ ℂ such that the quadratic equation has two opposite roots." I've tried multiple times (abc formula) but keep getting wrong answers. Could someone help me out?

[heady pollen]

with opposite you mean a+bi and a-bi?

[plain flame]

yeah the quadratic equation (even the complex version) doesnt really work that well when complex a,b and c

you need to "kwadraat afsplitsen"

dont know the english term

[edgy crag]

no, a+bi and -a-bi

[heady pollen]

ah ok

[edgy crag]

ow, I'll try searching that up. thanks!!

.close

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[delicate ruin]

I need help with question number 2, 3 and 4. I've provided 1 because question 3 requires it. I don't understand what it means by **standard factor ** on question 2. Thanks

[heady pollen]

this table should help

[delicate ruin]

alright what about question 4. could you explain what it means by the question please?

[heady pollen]

question 4 looks weird

seems like they forgot to write down the actual number

[delicate ruin]

alright thank you for your time

.close

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[pure bane]

You are voting in a state election. There are three candidates for governor, five candidates for lieutenant governor, four candidates for the state house of representatives, and two candidates for senator. How many different ways could you fill out the ballot if you vote once for each office?

[pure bane]

this is mine

go to

#help-5

or

#help-14

[patent swallow]

yeah ik

[pure bane]

or #help-21｜아리스킨충1

[patent swallow]

yeah ik

yeah ik

[ornate condor]

i think u just multiply all the numbers

[keen plinth]

i agree for once

[ornate condor]

i feel

quite insulted

[keen plinth]

you've been saying some sus things tonite

[ornate condor]

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[hollow fjord]

This is probably a very easy quotation but how would you find a percentage of a total when the components of that total is weighed differently. So like in my case, 17 of the total 24 is weighed at .8%, but the other 7 is weighed at .3%. How can I find the average percent for the total? I phrased this very poorly but

[hollow fjord]

Basically like the average percent of the total 24 if 17 is multiplied by .8% but 7 is multiplied by .3%

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@hollow fjord Has your question been resolved?

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@hollow fjord Has your question been resolved?

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@hollow fjord Has your question been resolved?

[languid saddle]

@hollow fjord If I interpret your question correctly, I think you're looking for a weighted average. In this case you would need to sum all the amounts multiplied by their corresponding weight factor. Then divide that by the total amount

So you would have $\frac{17 \times .8 + 7 \times .3}{24} = .654...$

[ocean seal]

joshy

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[warped hedge]

hi

[warped hedge]

could i get help in solving this, the answer is -1/2 according to the textbook but i'm not sure how to get there

from my knowledge of limits, i would usually multiply the reciprocal of the denominator on the bottom and top, but in this case theres a second part to the function, that being - 1/t

[abstract fractal]

Try combining the fractions into one

[warped hedge]

would that be the only method?

[royal meadow]

$\frac{1}{t\sqrt{1+t}} - \frac{1}{t}\ = \frac{1}{t\sqrt{1+t}} - \frac{\sqrt{1+t}}{t\sqrt{1+t}}\ = \frac{1-\sqrt{1+t}}{t\sqrt{1+t}}$

[ocean seal]

Kaisheng21

[royal meadow]

now it's easier i feel like

slightly

[warped hedge]

@royal meadow i understand that, but im not sure what to do after

i plugged in zero and got zerro

[royal meadow]

well if you just try and plug t = 0 right in, then you get 0/0 right

[warped hedge]

yeah yeah

[royal meadow]

so it's indeterminate

[warped hedge]

indeterminate

so then you simplify more

[royal meadow]

so you can try l'hopital, or use taylor series if l'hopital isn't allowed

[warped hedge]

my problem is i dont know how to simplify that fraction, after getting 0/0

[royal meadow]

well i don't think you can straight-up cancel any factors

[warped hedge]

yeah

[royal meadow]

you need some calculus to deal with the limit

so try l'hopital, or taylor series if that's not allowed

[abstract fractal]

What if you multiply the top and bottom by √(1 + t)?

[warped hedge]

is thta possible?

[royal meadow]

oh yeah that's slightly simpler

yeah

it doesn't solve it but it helps a bit

$\frac{\sqrt{1+t} - 1 - t}{t(1+t)}$

[warped hedge]

ive been doing that for everything else but since there was a - 1/t i thought it wasnt possible

[ocean seal]

Kaisheng21

[royal meadow]

i wish it wasn't so big

[abstract fractal]

√(1 + t) - (1 + t)

Maybe you can factor out 1 + t

[royal meadow]

can you??

[abstract fractal]

Maybe im just high

Im doing it in my head

[royal meadow]

i mean that gives like

[keen plinth]

[royal meadow]

if you divide top and bottom by 1+t?

[abstract fractal]

Fuck it, let me get some paper

[royal meadow]

is that what you're saying?

[abstract fractal]

Basically, yeah

[royal meadow]

ok

so we just multiplied top and bottom by sqrt(1+t)

and now you want to divide top and bottom by (1+t)

[abstract fractal]

I'm probably high

[royal meadow]

that's the same as just dividing top and bottom by sqrt(1+t) in the first place

i don't think that works

[royal meadow]

my advice is try l'hopital or taylor series on this thing

[keen plinth]

very very bad advice

lets not go there shall we

you know the saying

when you have a hammer

everything starts looking like a nail

[abstract fractal]

There's gotta be some way to get rid of the t in the denominator

[keen plinth]

factor it out

cmon

its just algebra

[abstract fractal]

How? The numerator doesn't have a factor of t

Just a bunch of (1 + t)'s

[keen plinth]

factor out the t from the denominator

[abstract fractal]

But then you have a 1/0

[keen plinth]

this is why we arent finished yet

[royal meadow]

does that actually help

[keen plinth]

yes

ofc it does

[royal meadow]

how

[abstract fractal]

We're back to this

[keen plinth]

where did everyone's algebra skills go

[royal meadow]

explain it then

go on

[abstract fractal]

I'm racking my brain jimbo

[royal meadow]

i don't think it helps

prove me wrong

[warped hedge]

😶

[keen plinth]

the t on the numerator cancels with the t outside the brackets

[abstract fractal]

FUCK

[keen plinth]

then we just plug in 0

voila

[warped hedge]

I got this

[abstract fractal]

I was high all along

[royal meadow]

ah, ok

[keen plinth]

it seems so

i think we all need an algebra refresher

[abstract fractal]

I just helped someone with a problem where you do that exact thing like 5 seconds ago

[royal meadow]

who needs algebra

l'hopital is still faster istg

[keen plinth]

again

i will repeat myself

very very bad advice

[royal meadow]

bluh

[keen plinth]

L'H is a hammer that is almost always not needed

[royal meadow]

disagree lmao but ok

it just depends on what problems you're seeing

[keen plinth]

well you can go and keep using L'H if you want

and ignore being good at algebra

[royal meadow]

some sets of problems are meant to be done without it

[keen plinth]

but here

we solved it without L'H just fine

[abstract fractal]

Using l'h is like using a graphing calculator instead of u sub, integration by parts, trig subs, etc

[keen plinth]

and it was much simpler

[royal meadow]

disagree

it's just another tool in the toolbox

[keen plinth]

its the hammer in the toolbox

its the nuke of all weapons

[royal meadow]

sure

[keen plinth]

you are going around nuking everything

[royal meadow]

yes

it's very fun

[keen plinth]

when a simple knife couldve done

[abstract fractal]

If you use l'h on every limit you come across that's of the form 0/0, you'll never learn the actual methods for limits

Limits

[keen plinth]

mhm

[abstract fractal]

Not integration

[keen plinth]

agree with dio

[royal meadow]

but consider

i know the actual methods for doing limits

[keen plinth]

well

[royal meadow]

and that's why i choose to use L'H's

[keen plinth]

it seems you dont

[royal meadow]

because they're awful

[keen plinth]

because you kinda

failed this one

[royal meadow]

yeah yeah yeah

[keen plinth]

so

:/

[royal meadow]

that's fair

fine, i'll rephrase

i know the other methods for doing limits

so i use L'H's because i suck at them

easy

[abstract fractal]

It's like using the quadratic formula on an easily factorable quadratic

[royal meadow]

perfectly valid

[keen plinth]

you suck at them because you rely on L'H

[royal meadow]

no kill like overkill

[keen plinth]

which i mean

if you want to live life like that

all the power to you

but

[royal meadow]

i do ^-^

[keen plinth]

i would like to give people better advice

[royal meadow]

ok fine

i didn't spot the algebra

it was square roots

i should remember square roots are often algebra

[warped hedge]

Alright I got it now

[royal meadow]

not L'H's

[warped hedge]

thanks guys

[keen plinth]

yw

[warped hedge]

👍

.close

[lone heart]

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[agile blade]

can I get some help?

[agile blade]

So I have a multistep problem, I have done all of it I just need to solve the equation

this is the problem: Lydia inherited a sum of money. She split it into five equal parts. She invested three parts of the money in a high-interest bank account which added 10% to the value. She placed the rest of her inheritance plus $500 in the stock market but lost 20% on that money. If the two accounts end up with exactly the same amount of money in them, how much did she inherit?

I need to solve this but I don't know how

[alpine sable]

You can write left as (3x/5) * 1.1 and right as (2x/5-500) * 0.8

[agile blade]

Like this?

[alpine sable]

Yeah

[agile blade]

how come?

[alpine sable]

well you wrote x/5 * 3 etc

Can you see it's 3x/5?

[agile blade]

because I multiplied it by 3, but I asummed because it's a variable you couldn't do that

[alpine sable]

you wrote 3x/5 * 0.1 + 3x/5 which is the same as 3x/5 * (0.1+1)

Why not

[agile blade]

Just the way my teacher taught me I guess

[alpine sable]

If you have 3 times 1/5 it's 3/5

If you have 3 times x/5 it's 3x/5

[agile blade]

but I did get the initial equation right, correct? I am doing this for extra credit so I really want to get this right.

[alpine sable]

Hmm

Explain your right part

[agile blade]

wdym

[alpine sable]

Well i don't know how you find that

[agile blade]

I have the question above so I tried writing out an equation

[alpine sable]

Alright

Can you explain how you wrote it out?

Only the right part

[alpine sable]

This is what you have right?

[agile blade]

yeah, that should be the equation for this problem

x being the inheritance

I needed to split the it in 5 parts take 3 of those 5 parts and add 10% of its value to it and then for the second parts and $500 in a stock market and lost 20% of it

my teacher said each side should be equal

I think it is right because I asked my teacher for help when I was doing this in class and she never mentioned my equation being wrong

[alpine sable]

Well alright

I don't get it though

The right part is (2x/5+500) * (0,2-1) = (2x/5+500) * (-0.8)

Isn't that negative?

[agile blade]

which part?

[alpine sable]

the right part

[agile blade]

sorry I'm confused its been a long day

[alpine sable]

This won't take long but i think you do not have the right equation

[agile blade]

what did I do wrong

[alpine sable]

It's (2x/5-500) * 0.8 instead of (2x/5-500) * (-0.8)

So it should be -(x/5...)*0.2 + (x/5...)

instead of +(x/5...)*0.2 - (x/5...)

Still right part

[agile blade]

but the problem says that she lost 20% of that money

so why wouldn't it be -

[alpine sable]

she has 100% of the money

• (x/5...)

She loses 20%

-(x/5...)*0.2

If she has 10$ and loses 2$ she has 10-2 not 2-10 can we agree on that?

[agile blade]

yes

so I need to flip it?

[alpine sable]

yes

[agile blade]

give me a second

kinda like this?

[alpine sable]

yes

[agile blade]

that makes sense

[alpine sable]

So can you write it with 3x/5 & 2x/5

Then find (3x/5) * 1.1 = ...

[agile blade]

wait I forgot the x

but x on the fractions right

[alpine sable]

sure

[agile blade]

this?

[alpine sable]

yeah

A little more work to find (3x/5) * 1.1 = ...

[agile blade]

so what would I do next? combined like terms?

[alpine sable]

You need to factorize

[agile blade]

I don't want to sound stupid, but what is that I haven't learned that yet

[alpine sable]

Well i can't explain it very nicely because i'm not an english native so you should check with google

But basically it's the way of writing ax+bx=x(a+b), you find a common factor (this time x) and you can write it like this

2 * 4+2 * 1 = 2 * (4+1)

[agile blade]

okay, so how would that apply here? Would I have to do 6x/5 x 0.1

that would be one side

[alpine sable]

You have (3x/5) * 0.1 + (3x/5) * 1

So 3x/5 is the common factor here

[agile blade]

but the * 1 doesn't do anything so why add it?

[alpine sable]

Why not

In the example 2 * 4+2 * 1 = 2 * (4+1)

You can explain 2 * 4 + 2 = 2 * 5

(3x/5) * 0.1 + (3x/5) * 1 = (3x/5) (1+0.1)= (3x/5) * 1.1

[agile blade]

I think I get it

maybe not as well to do it by myself but I understand what your doing

[alpine sable]

Ok

Can you factorize the right part may be

[agile blade]

yeah I'm not sure of the first step sorry

I have to add a * 1 so I can multiple something to both sides?

then add them?

[alpine sable]

That's right

You have (2x/5+500) * 1 - (2x/5+500) * 0.2

[agile blade]

yes then combined it into (2x/5) * 1.2?

[alpine sable]

you can factor (2x/5+500)

[agile blade]

oh I forgot about the +500

[alpine sable]

(2x/5+500) * 1 - (2x/5+500) * 0.2 = (2x/5+500) * [1-0.2]

[agile blade]

subtract the 1 and the 0.2 because in we needed to subtract for the money in the orginal problem correct?

[alpine sable]

Do you know how to multiply (a+b)*(c+d)?

For example (x+1)(x-1)

[agile blade]

not really

[alpine sable]

Well let's put that aside for now

In the end you have (3x/5) * 1.1 = (2x/5-500) * 0.8

[agile blade]

I have to go in 4 minutes so I don't really think we can do that much right now but this was the other side?

[alpine sable]

yeah

I'll pm you the rest

[agile blade]

thank you so much

[alpine sable]

sure

[sweet breach]

whats 8 + 11

[lone heart]

@agile blade Has your question been resolved?

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[lone heart]

@brittle hedge Has your question been resolved?

[lone heart]

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[stable night]

@keen plinth are u alive? i wanna ask some meth

[keen plinth]

ah crap

possibly not a good time lol

[stable night]

HAI

oh

i will wait

ill just put my question here

[keen plinth]

yeah do that

[stable night]

so a linear transformation is when we transform the entire x/y/z... axis and make sure that

1. origin remains the same
2. all the grids' lines must be straight

so for linear trans, we can imagine it as a function taking in a vector and returning as a vector

so here, vector i (1,0) gets multiplied by -1 so it on the left now
then vector j gets multiplied by 2 which extends it by 2

i seriously dont understand this linear transformation
what made it such that the axis needs to look like this

[keen plinth]

this is just a representation of a vector

so a vector in 2d has 2 components

the i component and the j component

we write it like the column vector

[stable night]

yes

[keen plinth]

that just signifies its components in the i and j directions

[stable night]

yes

understand

[keen plinth]

the thing about linear transformations is

you only need to know where they send i and j

because you can rebuild your vector from knowing how much of the i component and j component it has

[stable night]

hmm ok

[stable night]

but why like this

why turn the whole plane that way

is it because of the values put in

[keen plinth]

thats just a specific example of a linear transformation

nothing particularly special about it

[stable night]

oh so i can rotate the plane to any way i want

[keen plinth]

yes

[stable night]

it doesnt neccessarily need to be this way

o

[keen plinth]

yes

[stable night]

waitbut whats the point of doing that

isnt visualising on the regular plane much more easier

[keen plinth]

what do you mean

[stable night]

like why not use this normal plane

[stable night]

whats the use of doing a linear transofmration

making the plane look so weird

[keen plinth]

well the thing is

we start with the standard cartesian plane

but

we want to apply a transformation to it

if we start with the standard cartesian plane and we change it

it necessarily gets wonky

you cant change something and still stay the same

thats kinda paradoxical

[stable night]

umm

wait why change in the first place

[keen plinth]

well

thats what the study of linear transformations is about

linear transformations are exactly those things that change the cartesian plane in this way

the whole of linear algebra is this

every single matrix youve looked at

does this

[stable night]

wait wut.

wait but i thought the linear transformations are up to us

[keen plinth]

yeah they are

the matrix you write down is up to you

[stable night]

like we can have the most complex matrix and we can still use the original cartesian plane

[keen plinth]

but once you write down the matrix

the linear transformation is decided

[stable night]

wait the linear transformation is the transformation of the plane right

not the resulting vector

[keen plinth]

yeah

[stable night]

how does a matrix determine how the plane will look like

i thought we can choose not to cchange the plane

[keen plinth]

keep watching?

[stable night]

oh

[keen plinth]

has it been explained yet?

[stable night]

im at video 3

midway

[keen plinth]

i cant remember

its explained at some point

[stable night]

oh err

i dont think so

[keen plinth]

he explains at some point that the columns of a matrix correspond exactly to where i and j are sent

[stable night]

wait so we can do this because

[stable night]

ye he did

[keen plinth]

yeah

[stable night]

heh wait wut

[keen plinth]

because v is just 2 numbers

its how much i it has and how much j it has

the linear transformation changes i and j

so v changes along with

[stable night]

wait so that changes the vecotr

not the plane

like if i is (1,0) and j is (0,1)

lets say i multiply i and j by 2

[keen plinth]

the point is that its basically the same thing

[stable night]

then

[keen plinth]

if you move how the plane is

the points get dragged along with

[stable night]

yes

wait then how does the matrix determine how the plane looks like

[keen plinth]

its columns tell you where i and j go

[stable night]

like lets say i change the values of i and j

ok lemme draw it out

is this right?

[trim wagon]

no that’s transformation

[stable night]

then this would be how its represented in the matrix form

[stable night]

wat

im so confused

AAAAAAAAAAAAA

[keen plinth]

yes

except like

you havent changed anything lol

but yes thats right

[stable night]

so the first column stands for the x coords

and second for y

then lets say i wanna multiply the scalar 2 to both x and y

green is aftermath

yea so the matrix will look like

2,2

so 2,2 is the vector of the resulting vector right?

which is the orange line

[keen plinth]

uh

[stable night]

[keen plinth]

what happened

you just got a vector out of nowhere

the matrix would be

[stable night]

umm wym

[keen plinth]

and you're multiplying this against

bot lag

[stable night]

wait why like this

hm

[keen plinth]

that vector

[keen plinth]

yeah

you have to apply a linear transformation to a vector

to get out a new vector

otherwise you just have a linear transformation

[stable night]

wait but he expressed the original vectors as this

[keen plinth]

yeah

[stable night]

HEH

[keen plinth]

but thats a linear transformation

the columns tell you where i and j are sent

[keen plinth]

this is an example of where [x y] is sent

[trim wagon]

https://www.desmos.com/calculator/upooihuy4s

[stable night]

um where did this come from

[trim wagon]

try it for urself

[trim wagon]

what that not correct

[stable night]

nani wat that

[keen plinth]

?

[keen plinth]

im talking about the green vectors here

[stable night]

like this one is the final vector cant we represent it in (2,2)

oh

oh lmao

[keen plinth]

like

your linear transformation turns the red i, j into the green i, j

so its the 2 0 0 2 thing

[trim wagon]

\begin{bmatrix}
0 & 2 \
2 & 0
\end{bmatrix}

[ocean seal]

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

[keen plinth]

this

[trim wagon]

thats what u looking for

[keen plinth]

??

thats reflection

there is no reflection

[trim wagon]

green one

[keen plinth]

the i and j just get scaled

[trim wagon]

no

get better eyes

[keen plinth]

i dont even know what deep is talking about anymore

[trim wagon]

[stable night]

wait how did u get this

[keen plinth]

its just scaling

by 2

[stable night]

ye

[trim wagon]

uhh

[stable night]

o ye that would be reflection

i kinda see it

lol

[keen plinth]

this is a reflection

in the y=x line

and also scaling up by 2

[stable night]

wait so a linear transformation is it the transformation of the ENTIRE PLANE or just the vectors

or do the vectors determine how the plane looks

[keen plinth]

its a function that takes in vectors and spits out vectors

you can think of it as transforming the ENTIRE PLANE

but really you only need to know where i and j are sent

because every other vector can be expressed in terms of i and j

[stable night]

wait so how do i determine how the plane looks after changing i and j?

[keen plinth]

you just

drag the grid lines

along with i and j

so that they remain straight and parallel

he animates it in the video like once every 5 seconds lol

thats how its done

[stable night]

i see how its done but i dont get why its turned to that specific angle thou

[keen plinth]

because the i and j vectors get put into a weird angle with each other

the grid lines need to line up with i and j

so they necessarily follow the same angles

[stable night]

wont this line up with i and j already thou

oh wait

this is the basis one

[keen plinth]

thats before the transformation

probably

after the transformation its gonna get wonky

[keen plinth]

yeah

[stable night]

so here we transformed i by -1

and scaled j by 2

[keen plinth]

no no

thats not the transformation

v is given by -1i + 2j

thats just the definition of v

the transformation is the thing thats sending i and j to where you see they are on the picture

thats the transformation

[stable night]

but why is the i and j that way

like why is i suddenly facing the southeast

wont it face the south

[keen plinth]

this matrix once youve written it down defines a linear transformation

[keen plinth]

its effect is moving i and j like you see in the picture

its not that i and j always move that way or something

its determined by the matrix you are working with

i gets sent to the first column

[1 3]

wait its [1 2]

[stable night]

wait sorry how did u get this lol
-1(1,0) + 2 (0,1)

[keen plinth]

completely wrong

this is right

[stable night]

oh

why dont we express them

[keen plinth]

so the first column is [1 -2]

thats where i goes

the second column is [3 0]

thats where j goes

you have no say about what angle theyre at or how long they are or anything

once the matrix has been written down

everything is decided

[stable night]

and i can "sum" up these two

to get the final vector?

[stable night]

the yellow line here

[keen plinth]

so v = -1i + 2j right

[stable night]

yes

[keen plinth]

and call this matrix T

then

bot lag

[keen plinth]

the transformed v

Tv

is just the sum of the transformed i and the transformed j

Ti and Tj

with the same "amount" of each

thats what it means to be linear

we know what Ti and Tj are

thats just the columns of our matrix

aka where i and j are sent by the matrix

so we can calculate Tv as the sum of those

with the correct amount of each ofc

[stable night]

okay i get this part

i still dont get why the plane transform that way lol

like i get the matrix and why when they add up theyll be some new vector on the plane

and stuff

but i still dont see how this relates to changing the entire plane

we could do all these without changing the plane

i think

[keen plinth]

the thing is that you can visualise the transformation as changing the entire plane

like

the transformation sends the standard cartesian plane

to the new changed plane

[stable night]

this would be without transforming the plane

[keen plinth]

the fact that the grid lines are still parallel and straight and stuff is because each vector is able to be written in terms of i and j

well yes

you can calculate individual vectors and stuff

but the point is like

you want to have the big picture

[stable night]

or like u said we could also change the plane aka linear transformation to get the resulting vector?

[keen plinth]

what does this linear transformation do to the whole plane

you usually dont care about single vectors

[stable night]

umm from the video is like rotating it 80 degreees to the right or something

[keen plinth]

you care about the behaviour of the transformation

stuff like that yeah

[stable night]

but why rotatr right

why not rotate left lol

like wai

[keen plinth]

well i mean that just depends on what matrix you write down

theres infinitely many linear transformations to choose from

the video just has a lot of examples

they arent anything special in particular

[stable night]

how does (-1, 2) tell us to rotate right

[keen plinth]

where does -1 2 come from

the rotation is associated with the matrix

not the vector

you dont care about what happens to individual vectors

you care about how the matrix acts on everything

in order to see how a linear transformation behaves

you care about its behaviour on the entire plane

[stable night]

okay so ill ignore individual vectors entirely
can u tag the matrix that represents this plane again pls

[keen plinth]

this is a reflection in the y axis + scaling the y axis by 2

[stable night]

can i ask where u got the first column from

sorry if im asking dumb qns lol

[keen plinth]

the tip of i is at (-1, 0)

[stable night]

okay then the second column from the transformed j

[keen plinth]

the tip of j is at (0, 2)

[stable night]

okay so we end up with this

how does this matrix

determine how the linear transformation work

[keen plinth]

it tells you that i gets set to (-1, 0)

and j gets sent to (0, 2)

every vector is a combination of i and j

so we know where every vector gets sent

because we just need to write out how much i it has and how much j it has

then recombine it after the transformation

in other words

you just drag all the gridlines along with i and j

and that tells you the entire behaviour of the transformation

[keen plinth]

this is really just saying how many gridlines in each direction the vector crosses

[stable night]

wait "drag all the gridlines along with i and j"

[keen plinth]

which is why we're talking about gridlines

like say here

all those red lines that ive drawn

thats determined by the fact that i has been sent to (1, -2)

so all those red lines must line up

be parallel

and have the same length

because it forms a grid

same thing for j

i and j completely determine the gridlines

[stable night]

wait could i turn it another way tho

[stable night]

like can the red line here

[keen plinth]

well ofc

[stable night]

OH

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[keen plinth]

you can just

change the matrix

[stable night]

ok wait wut

[keen plinth]

it depends on the matrix you start with

this matrix fixes those gridlines

another matrix might fix different gridlines

[stable night]

like in the same example, can the red be facing directly downwards (without changing the matrix)

[keen plinth]

no

[stable night]

we will change y accordingly

o

[keen plinth]

the gridlines are completely determined by where i and j go

they must be parallel to i and j

and be equally spaced according to the lengths of i and j

[stable night]

wait i get that

but why is i

this

and not this

[stable night]

i get that the gridlines must strictly follow i an dj

[keen plinth]

you see that yellow parallelogram

[stable night]

ye

[keen plinth]

thats the parallelogram spanned by i and j

the entire grid is that parallelogram

every single cell of the grid is a copy of that parallelogram

you cant form the grid in any other way

thats how linear transformations work

[stable night]

how did they form that parallelogram in the first place tho

[keen plinth]

its determined by i and j

you know where i and j are

take any two arrows

and you can form a parallelogram with them

[stable night]

sorry but like i get that it depends on i and j

[stable night]

but i dont get why i points in this particular direction

like why southeast

[keen plinth]

because i = (1, -2)

in this particular case

thats just

what the transformation is

its been decided that i = (1, -2)

so thats where it goes

you cant just be like oh but its actually (0, -2)

[stable night]

WAIT IM STARTING TO SEE IT

[keen plinth]

like i give you to random arrows

[stable night]

OH I SEE IT

OMG

I GET IT

OKAY

[stable night]

THIS IS JUST

THE COORDINATES

[keen plinth]

here they form a parallelgram

[stable night]

OF THE TRANSFORMATION

I GET IT

OKAY

[keen plinth]

great

[stable night]

OHHHHHHHHHHHHHHHHHHHHHHHHHH

[stable night]

THIS

this solved it

AAAAAAAAAAAAA

this is amazing

[keen plinth]

bruh

[stable night]

ive never felt this satisfied

thx alot boss

[keen plinth]

now go back to watching your video

i have stuff to do lol

[stable night]

yes sir

i will do anything u want

[keen plinth]

good

[stable night]

[keen plinth]

[stable night]

u r ze best

people would have raged cuz i didnt get wat was happening

but itsok

[keen plinth]

go learn more stuff

idk

[stable night]

[keen plinth]

im off

[stable night]

i worship u

byebye

[keen plinth]

👋

[stable night]

.close

[lone heart]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[sly seal]

Can someone pls explain this step to me

[sly seal]

I get there factoring out an x but why is there a positive 1

[pliant basin]

Because 1 • -2

[ember onyx]

@vagrant moat @surreal mesa @grizzled raptor @static tree

[static tree]

@ember onyx stop posting in other people's help channels

[ember onyx]

A discussion about: The distance vs. time when the carts are moving in uniform motion (first experiment) and uniformly accelerated motion (second experiment) using a data table, and the meaning of the slope in the graph velocity vs. time.
@surreal mesa

[sly seal]

Is that factoring

[ember onyx]

@surreal mesa

[static tree]

STOP @ember onyx

[ember onyx]

where do i post then

[static tree]

I pinged you into the 384 chat

[ember onyx]

ty

[pliant basin]

I think it’s because you factor out an X it becomes (x^2-x-2)

Then you do the funny factoring

[sly seal]

Isn’t it -x^2-2 when factored

[pliant basin]

If you reverse it and multiply (x+1)(x-2)

You should get that

Well you’re taking an x from x^3-x^2-2x

They skipped the part where it’s x(x^2-x-2)

[sly seal]

ohhhhh I got it thx I thought it was x^2 being taken out

.close

[lone heart]

Channel closed

Closed by @sly seal

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Send your question here to claim the channel.

Remember:
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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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stable night]

from the first to second line, what is that property called

[keen plinth]

nyani

[stable night]

hi my lord

what is this LOL

is this some exponential property that i forgot

[keen plinth]

taking 2^(blah) of both sides

[stable night]

u can just do that