geometry-and-trigonometry

[radiant nimbus]

If and when you start Calculus, Khan Academy is great, but I really also like the youtuber Nancy Pi. Sometimes Sal can be a little dry lol

Paul's notes

[idle bobcat]

^^^

that one

lol

thoughts on paul

[radiant nimbus]

Yep that's a good one too

[idle bobcat]

sweet

[radiant nimbus]

I used his practice problems a lot

some are freaking hard, but if you can do the hard ones you can crush the easy ones lol

[idle bobcat]

its a pretty shitty feeling acing all the tests on the khan academy modules but struggling with some of pauls exercises

[radiant nimbus]

lol

[idle bobcat]

which led me to believe even more that I wasn't getting as much out of khan as i thought

[radiant nimbus]

It's a great supplement to learning, but I don't think it should be your only source

you can get a lot of textbooks for free

there are PDFs of a lot of math books that are like a version or 2 behind the "current" one

but usually nothing in an earlier version is any different than a newer version

Calculus is calculus and hasn't really changed since Newton and Leibniz discovered it lol

[idle bobcat]

yeah exactly lool

[radiant nimbus]

I'll probably get in trouble for saying that buuut 🤷

[upper karma]

my only fear if I were learning only on khan acadademy would be getting extremely bored and quitting math

[radiant nimbus]

exactly ^

[idle bobcat]

😮

well maybe its different in the higher levels

[radiant nimbus]

Sal is a good guy and all, but he puts me to sleep lol

[idle bobcat]

but ive enjoyed my time on khan so far

the only thing that irks me is how he repeats himself so many times

even the most trivial things

[upper karma]

I've used it in the past and it was fine for me but it lacks uh...

I don't know, uniqueness and interesting new things?

[idle bobcat]

haven't had an issue with that personally, at least so far

though I don't like how some of the sections have misordered videos

like there would be a lecture on a topic, but some of the prerequisite videos would come AFTER the topic

which is confusing sometimes

[upper karma]

it's mostly like, standard school math presented in a way a good school teacher would present

which is aight and probably the best that a resource made for everyone on the internet can be

[idle bobcat]

what was your go-to resource?

other than just in class lectures and textbooks

[radiant nimbus]

honestly for me, it's my schools tutor lab lolll

but

[idle bobcat]

haha yeah that would be nice to have

[radiant nimbus]

if you grasp the material, the best thing to do is practice, if you don't grasp it. then watch youtube or something until you do, practice it and confirm your answers

nothing is worse than thinking you know something, but you forgot because you didn't commit it to memory via exercise

that's why pauls notes, or some book or something that you can get practice problems for free is important

I've actually heard the manga guides are pretty cool/good too if you like manga/anime

[upper karma]

I had little interest in learning math until after I was done high school, so I didn't use any resources lol

[radiant nimbus]

its a way to learn what you need, and break up the monotony of it with a story in between

[upper karma]

after that a grad student got me into mathematical rigor and showed me lots of cool things

so I guess that was my go to resource

[idle bobcat]

that's cool, so he was kind of like your mentor in a sense?

@radiant nimbus bruh I had no idea that was a thing LOL

[upper karma]

but it's much harder to create a universal plan to get people into cool math because cool math is harder

and yea

[idle bobcat]

I had this huge irrational fear of math since elementary school so I never committed to learning

but ever since picking it up recently everything seemed cool to me lol

maybe that's why I havent gotten bored of khanacademy

maybe I'm still in a "honeymoon phase" lmao

[upper karma]

if you like it then it's not bad

but most people find it too boring for various reasons

I dunno, I guess I am trying to say, if you do find it boring in the future, you don't have to give math up because there is plenty of other fun stuff

since I have an inkling that happens to a fair number of khan academy learners

but I do know someone who did basically all of khan academy and really loves math, so if that is also you then go for it

[idle bobcat]

yeah I don't see myself giving up math any time soon even if it ends up getting boring

thats why I'm glad I know about other learning resources now too

[upper karma]

ah a slave to math already, nice

[idle bobcat]

I'm curious why people find math boring

I'm assuming its the practice problems/homework that they dread?

which is what makes it boring to them?

Because I find it hard to see the actual content as boring

[livid sequoia]

depends on how they get taught

[idle bobcat]

ah yeah overlooked that

since im self learning i have a ton of resources available to me so that isn't an issue for me

[upper karma]

yea math education can get much more boring than khan academy

also, many find no value in anything contained from high school math and beyond (and much of the middle school level) because for most, none of it will find direct application to their day to day lives that they can see

[idle bobcat]

well I guess I don't have too much of a say on what is boring or fun yet considering I'm still learning the most basic of concepts currently

wouldnt be able to gauge the "fun" factor yet really

yeah I can understand that

which is why I think a lot of people start getting more into math after highschool where they start taking on jobs/career paths that warrant a certain level of understanding of the underlying systems of whatever it is they are working on

which typically lends itself to some degree of math

it's why I started learning as well

[upper karma]

I'd argue people who know math can still find math in many of the things people do in their daily lives without realizing it, but it'd be hard to see without first knowing math

[idle bobcat]

also true

[upper karma]

so that also can contribute to their idea that it's not worth it

[idle bobcat]

I also just find it worth it to learn math due to how it forces you to think critically and logically

that in and of itself makes it a worthwhile endeavor for me

[upper karma]

that's where the cool math lies

[idle bobcat]

i guess math really is for the cool kidz

[upper karma]

a lot of education lacks that

in favor of instruction following and plug and chug algorithms and other things that are easy to test

[idle bobcat]

Hi again

Am I doing something wrong or is my calculator pooping itself

[upper karma]

I think you are in radians

[idle bobcat]

this worked

oh i see

[upper karma]

What are the best books to fully learn geometry?

[glacial haven]

like topology?

[upper karma]

The whole subject.

I don't know how large geometry is

[cloud stump]

really large

[upper karma]

What part of geometry

I mean I’m getting some geometry books to review for my geometry class next year, so I’ll check them out

[supple onyx]

I mean "fully learning" a subject cant exactly be done via books

[upper karma]

Why not

One subject I have has a book to teach and a book full of over a thousand problems to do. If I can move onto the next subject without much hardship, I consider myself taught in the previous.

[glacial haven]

what subject

[little osprey]

what book

[livid sequoia]

is this right?

[paper vale]

the first 3 are literallly equivalent lol

[livid sequoia]

the question says which one out of those is congruent

[paper vale]

no it asks what information we need to imply that sas congruence but they all do

[livid sequoia]

I can only pick one tho which is confusing

[paper vale]

well then do the perpendicular one, as it says SAS

but all those 3 work

[uncut pebble]

wdym, A doesnt solve it my bad im an idiot

neither does D

obv

[paper vale]

i said not D

A implies it is isocscles therefore implies B and C

[livid sequoia]

Ill just pick C

[silver fable]

Hi I have an ellipse defined by its center position and it's semi major and minor axis, and a normalized direction vector, what I need is the farthest point on the ellipse bound in the direction of the vector, for circles it was quite easy it was:

(Center) + (Direction) × (Radius)

But is there any easy way to find that out about an ellipse?

[hidden phoenix]

hmm

little bit more complicated than circles

weather you consider it easy is a question mathematics cannot answer

ill send it

okay, lets begin. Im not sure how to use texit but ill try

also you ca prob simplify this further

[little osprey]

Why in bold

[silver fable]

Forget the part I said easy, any answer would be appreciated lol

[hidden phoenix]

FOR X: (Center X coordinate) + sqrt((The x of the vector)^2/(((The x of the vector)/(The A thing in the picture you sent))^2+((The y of the vector)/(The B thing in the picture you sent)^2)))

FOR Y: (Center Y coorninate) + sqrt((The y of the vector)^2/(((The x of the vector)/(The A thing in the picture you sent))^2+((The y of the vector)/(The B thing in the picture you sent)^2)))

@silver fable

That should be the thing you are looking for if I got what you meant and I did not make some mistake

ping me when you see it

[silver fable]

Ohh alright I'll test it out and see how it works thanks for the answer

[hidden phoenix]

ok you dont have to ping me anymore

expect after testing it ping me to tell if it worked

[silver fable]

Only one thing I'm not understanding is, is the whole thing in the square root except the center?

[hidden phoenix]

umm

ill try to use texit

[hidden phoenix]

@silver fable

[somber coyote]

JsMeansJauhesammutin:

[hidden phoenix]

That thing is for x

[somber coyote]

JsMeansJauhesammutin:

[hidden phoenix]

this is for the y

@silver fable ping me if you got it

sorry i was kinda late but i had to learn how to use texit

vector x is x of the vector

centerx coordinate is the x coordinate of the center of the ellipse

the a thing is the ellipse width /2, that is the thing labeled A in your picture

[silver fable]

@hidden phoenix It kinda works there are issues

The first one is that, it's only on the positive side once your direction vector goes to negative side it doesn't give you the right answer for this I actually had a solution and I used the sign(-direction)

The second issue is, it looks like it's giving me the point on a unit ellipse or something like that I multiplied my X and Y by A and B but it didn't seem to give me the point on the ellipse but further

[hidden phoenix]

Oh yes negative side

Sorry umm

Ill try again

[silver fable]

I'm sorry I must've really troubled you

[hidden phoenix]

Btw are you sure the vector is normalised?

I'm sorry I must've really troubled you
Nope

[silver fable]

Yes it's code and I can show you a visualization of it if you want

[hidden phoenix]

Not at all

[silver fable]

I've been tackling the problem of ellipses for quite a while they are really hard to deal with

[hidden phoenix]

Ok

[silver fable]

I did something myself but it wasn't accurate,

What I did was find the angle of the vector and use it to calculate the ellipse radius in that angle but it didn't give me the farthest point, it had offsets

[hidden phoenix]

What programming language are you using btw?

[silver fable]

C#

[hidden phoenix]

Nice

Btw did you square the entire fractions down or only the top parts?

[silver fable]

I first did only the top part but that gave me really bad results like infinity or NAN sometimes

[hidden phoenix]

Yes

Texit did not make it very clear

[silver fable]

This is what I got after multiplying the result with with A and B before adding the center coordinate of the ellipse

Actually multiplying by half of A and B

[hidden phoenix]

Can you make it draw many of those yellow line+red point things at once so that the vector values are based on sin and cos

It may make it easier to debug

Maybe use a for loop

[silver fable]

Yeah I can do that I'll be back after I got it

[hidden phoenix]

Wait i just realised something

I may have been incredibly stupid

@silver fable I got it in my head but I cannot write down yet

[hidden phoenix]

so i think i got it

[silver fable]

@hidden phoenix It's alright take your time and here's what I did, this time I multiplied the radius by 0.2 so it can capture on the screen

[hidden phoenix]

new method.

1. divide the vector's x with the A thing
2. divide the vector's y with the B thing
   3.you get a new vector.
   4.normalise the new vector.
3. You get another vector.
4. multiply the now normalised vector's x with the A thing
5. multiply the now normalised vector's y with the B thing
   6.You get another vector.
   7.Now add the center of the ellipse to the latest vector

as you can see it is that simple if i am correct

if i am unclear ask me again

[silver fable]

It seems pretty straight forward I'll try and let you know

It looks promising, now the only step left is test it out with GJK algorithm if it could detect collision then it's working

[hidden phoenix]

omg collision detection

always messing with your brain

collision detection is the nightmare

[silver fable]

Yeah its a hell when something isn't working and quite mathematical but I enjoy implementing these algorithms

[hidden phoenix]

i feel extremely stupid because it was so simple

[silver fable]

It looks right but unfortunately GJK is still not considering it as collision on cases where they are not deep, in fact we did the same mistake, the problem is we are getting the radius along the direction of the vector but that's not right in ellipse case there are spots where the farthest point is higher or lower, it works when the ellipse is really deep into collision, the big problem is because the minkowski sum of two ellipses it not exactly an ellipse it's something else

[hidden phoenix]

what is GJK

also what is not deep

what same mistake

what is not right in ellipse case

what does ellipse is deep into collision mean

what do minkowski sums have to do with this

[silver fable]

That's alot of questions let me first tell you what GJK is and what it does

[hidden phoenix]

ok

sorry your comment was kinda nonsensical

Gilbert–Johnson–Keerthi distance algorithm?

[silver fable]

Yes

You know it?

[hidden phoenix]

nope

only by name

[silver fable]

GJK is an algorithm, what it does is that it calculates the minkowski difference of two convex shapes and checks if that shape is containing the origin, the way it does it is by randomly selecting a direction vector and find the farthest point along that direction on the convex shape then it does the same for the second shape but in a different direction, that depends on how you implement it, and tries to form triangles into the minkowski difference of the shapes and checks if those triangles contain the origin

Because we know if we had two shapes and calculate their minkowski difference, if those shapes are intersecting, the minkowski difference contains the origin aka (0,0)

[hidden phoenix]

Bruh ill look into this

[silver fable]

You don't have to really, except if your really interested, it doesn't matter if you understand this algorithm, this part about finding the furthest point is known as support mapping

[hidden phoenix]

Nah im interested too

I also like to code stuff

But I have never considered GJK

[silver fable]

GJK is a quite fast algorithm to detect collision, you can also find out the distance between two convex shapes very fast with this algorithm

Or check if a point lies in a convex shape

It's like a black box with many uses

And for what I mean by ellipse, it's just something I found out with experimenting, imagine an ellipse with a diagonal vector, if you draw the shadow of the ellipse on that vector you get two points, now imagine the same vector with the same direction but crossing the center of the ellipse, if you mark the intersecting points with the ellipse bound you find out that there is a difference between the points that originate the shadow and the points that the intersections are located at, which made me come to this conclusion that we are somehow doing the intersection test, there is a margin of error that keeps getting bigger the more you go towards the diagonals of the ellipse

I've tried to do alot of things with ellipses, calculating closest point's, calculating intersections and so far the only thing that I could achieve is, know if a point lies in an ellipse, this shape is so weird calculating anything for it is nearly impossible at least for me

[livid sequoia]

can someone check my answers?

[little osprey]

either the diagram is wrong

or your ray ab thing

idk

[livid sequoia]

the diagram must be wrong since theres no ray in the diagram

[little osprey]

never forget to put 2 decimal places in money

Exactly

[livid sequoia]

so I put 600.00?

[little osprey]

otherwise your geometry thing looks good

Yup

[livid sequoia]

ok, thank you

[little osprey]

no problem

[upper karma]

does anyone know a good way to make a inverse with move able outputs?

[dark sparrow]

?

[upper karma]

like inversec tan will get outputs between -0.5pi and 0.5pi if im not mistaken,
but what if you needed outputs from -0.1 pi and 0.9pi? I've been trying to figure it out for awhile but I can't find a simple way to do it

[umbral snow]

tan^-1(x) + 0.4π

[upper karma]

that doesn't work

that just moves it which is not what I need

[umbral snow]

What did you need?

[upper karma]

I need to fine a f(y) eqasuon that is the inverse of tan between those two lines if that makes sense

[umbral snow]

Oh. Okay, tan(x - 0.4π)

[arctic vortex]

hi guys can anyone help me with some geometry questions??

[upper karma]

No that just shifts it again

[umbral snow]

Well no wait, you want it moved into there? That's more like tan(x - 2)

Oh I see, you want an arctan with a different branch cut

[upper karma]

The main issue that I'm having is I need it get a diffent porsion of the graph
Which is kinda hard to put into words

Yeah

[umbral snow]

Where are those two black dotted lines, anyway?

[upper karma]

Well the thing is I'm trying to make a eqasuon that can move with the lines , the first line is at 'a' and the second is at 'a' plus pi

[umbral snow]

It's fairly easy to do with a piecewise function

[upper karma]

What's a piecewise funcion?

[umbral snow]

It's a function with different definitions depending on which x you're at

[arctic vortex]

can i put a question here pls?

[umbral snow]

That kind of thing

[arctic vortex]

is this one correct?

[paper vale]

yes

[umbral snow]

You could adapt the usual arctan function to "roll back" to the bottom, if it goes above π/2

[upper karma]

Alright, ima try somethings and comebacl

[arctic vortex]

yes
@paper vale are you talkng to me

[paper vale]

yes

[arctic vortex]

ok thank you

can you help me with 1 more pls

[paper vale]

yes

[arctic vortex]

is this correct

[paper vale]

yes

[arctic vortex]

what

say correct or incorrect

i tihnk u are trolling me

[paper vale]

this is trivial

[arctic vortex]

sorry if i am offending you

wait but 1 more

is this lateral or surface

to solve it

oh surface

no wait

i dont udnrstand it

[upper karma]

Add the lateral of the cylinder to the surface of the cone, it will give you the correct result

[arctic vortex]

ohhh

i thought it was asking for everything but it is actually only asking everything except top and bootm

[upper karma]

Wait, it says surface, also there's no top...

[arctic vortex]

o

[upper karma]

You should just exclude the top of the cylinder and the bottom of the cone

The rest is the surface area of the container

Ramanujan himself?

Damn

Just his fan, lol

I suck at math tbh

any idea how to turn this into a f(y)?

[arctic vortex]

sorry u can g o first

[upper karma]

isn't that impossile?

[arctic vortex]

i did not see urs

[upper karma]

wait no its 9 to 25
secse volcume increases by the cube of the scale

[dark sparrow]

yeah except 3^3 is not 9 and 5^3 is not 25

[upper karma]

I might be stupid

23 to 125

[dark sparrow]

3^3 isn't 23 either

[upper karma]

I am def stupid

[arctic vortex]

i think i got mine

nvm i did not

[livid moss]

What is the formula for the volume of a sphere?

[upper karma]

pir^3 I think

[dark sparrow]

no

$V = \frac{4}{3} \pi r^3$

[somber coyote]

Ann:

[livid moss]

It's the ratio of your radii cubed

[upper karma]

Man I'm not as good at math as I thought i was

[livid moss]

Forgetting formulas doesn't make you bad at math, dw

[arctic vortex]

is this 1:4

[upper karma]

no

[livid moss]

Why no? Yes

[upper karma]

wait yes

[livid moss]

Now you do the reverse, before you cubed it, now you take the cube root

[upper karma]

my brain is broken today

[trail osprey]

the answer is B

[marsh river]

Hey guys, can someone tutor me?

Right now, im reviewing Polynomials.

[jaunty sail]

just post quesitons here i guess

[upper karma]

@marsh river ?

[hidden phoenix]

Man I'm not as good at math as I thought i was
Nah, proving that is hard without calculus. You are not bad, you just have not memorised the formula or researched into calculus

[silk crown]

$-3\sin^2 x \cos x + 6\sin x \cos x > 0$

[somber coyote]

CoolShot:

[silk crown]

hewp me uwu

[dark sparrow]

what have you tried

[upper karma]

Double angle formula for sin?

[dark sparrow]

and also are you to solve this on [0,2pi) or on R

[silk crown]

uhh R

[dark sparrow]

right

[silk crown]

actually no

i can get away with

(-pi/2, pi/2)

[dark sparrow]

factor your thing first

as 3 sin(x) cos(x) (2 - sin(x)) > 0

[silk crown]

indeed

[dark sparrow]

and then realize that 2-sin(x) is positive no matter what

and that cos(x) is positive for all x in (-pi/2, pi/2)

[silk crown]

OOOH yeah ok my small brain didnt see that, so sign only depends on sin(x) here

ty

[dark sparrow]

[supple wedge]

i need help again

how do i do the greater than or equal to sign on latex

idk how to latex this qn

[upper karma]

\geq

[supple wedge]

aight

[upper karma]

Wait

[supple wedge]

$\geq$

[somber coyote]

Hmm:

[supple wedge]

$\leq$

[somber coyote]

Hmm:

[supple wedge]

ok thanks

[upper karma]

Np

[supple wedge]

sketch the graph of $y=\abs{cos4x-2}$ for $-\frac{\pi}{2}\leq{x}\leq\pi$ and state the value of $n$ when $\abs{y}=n$ has four solutions.

fuck

wait

[somber coyote]

Hmm:

[supple wedge]

i got the graph alr

let me think i think im onto something

brb

[minor field]

Let's see the graph

[supple wedge]

,w graph abs(cos4x-2)

fuck ill take pic

@minor field

[minor field]

ok

Let's look at x=0

[supple wedge]

yeh

[minor field]

$|\cos(4\times0)-2|$

[somber coyote]

L'Âne 🍐:

[supple wedge]

-1

[minor field]

=|1-2| = |1|

[supple wedge]

1

forgot abs

[minor field]

But for your graph we have 3

[supple wedge]

what do you mean

[minor field]

At x=0 your graph is on 3

[supple wedge]

ohh

and then what

[minor field]

We just said it should be starting at 1?

but yours starts at 3 for some reason

[supple wedge]

oh so something is

fucked

but

,w graph abs(cos4x-2)

[somber coyote]

Results provided by WolframAlpha

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

Requested by Hmm#2554

[supple wedge]

oh

fuck

LOL

[supple wedge]

ok me again

$\le$

[somber coyote]

Hmm:

[supple wedge]

fuck

Given the equation $\frac{2}{sin^2\theta}=5-cot\theta$ for $0<\theta<360$, find the values of $\theta$.

[somber coyote]

Hmm:

[supple wedge]

help

just @ me

[supple onyx]

Im ere

Here*

So, first you want to reformulate the 2 in terms of sines and cosines

At least that's what seems the best step to me

@supple wedge

[supple wedge]

i did i got stuck

[supple onyx]

What did you do with the 2

[supple wedge]

let me latex what i did

$\frac{2}{sin^2\theta}=5-\frac{cos\theta}{sin\theta}$

[somber coyote]

Hmm:

[supple onyx]

Forget about the cot for now

Only focus on the LHS

[supple wedge]

i didnt touch it at all

oh wait i can make it into identity

[supple onyx]

This may seem weird but note that 2 = 2*1

[supple wedge]

yeah

[supple onyx]

Do you see a possible trig identity there

[supple wedge]

yes

i just did

[supple onyx]

Good

So work with that lemme know if you get stuck

[supple wedge]

sure thx

ok im stuck LOL

[supple onyx]

Whatcha got

[supple wedge]

$2cosec^2\theta=5-cot\theta$

[somber coyote]

Hmm:

[supple wedge]

what identity

[supple onyx]

Hmm that's not right

Ill guide you through it

[supple wedge]

sure

[supple onyx]

So not that 2 = 2*1

[supple wedge]

yes

[supple onyx]

We know that sin^2 + cos^2 = 1

(Also from here on ill replace theta by x)

[supple wedge]

understandable

[supple onyx]

So we have 2* (sin^2(x) + cos^2(x))

We can separate the fraction that we get with sin^2(x)

So that we get 2 + cos^2(x)/sin^2(x)

What is cos^2(x)/sin^2(x)?

[supple wedge]

can u latex cuz its hard to see

[supple onyx]

Alright i will

$2 + 2\frac{cos^2(x)}{sin^2(x)}$

[supple wedge]

q.q

yeah

[supple onyx]

Alright what's this second term

[supple wedge]

$cot^2\theta$

[somber coyote]

Hmm:

[supple onyx]

Right

[supple wedge]

form quadratic?

[supple onyx]

Exactly

[supple wedge]

oo

[supple onyx]

You can replace cot with u or smth and solve the quadratic equation

[supple wedge]

yep

thx

[supple onyx]

Np

[supple wedge]

lets see if i got it right

$2+2cot^2\theta=5-cot\theta\0=2cot^2\theta-cot\theta+3$

O SHIT

nothing happened

[supple onyx]

Oh no bot broken

[supple wedge]

no i deleted it

cuz my algebra went dog

[supple onyx]

Oh okay haha

Oh yeah i can see it from the code

[somber coyote]

Little Narwhal:

[supple onyx]

Woops didnt realise it would go down

I was reparing my previous mistake XD

[supple wedge]

hehe

[upper karma]

\cos

[supple onyx]

Woops

[rose tulip]

Ok so im tryna teach myself geo before next semester.

He got 42 which I dont understand.

[pallid sierra]

Missing info.

[rose tulip]

So yeah 2/7 is the scale factor

and fig b 12

[pallid sierra]

What did u try

Okay here’s the deal

[rose tulip]

12*7/2=X

[pallid sierra]

Yeah

You take the inverse if you want to find the larger segment

(2/7)^-1 ⇒ 7/2

It’s easy

[rose tulip]

So flip fraction then mutiply??

[pallid sierra]

You can make an equation given the information:
X*2/7 = 12

When you multiply both sides by the inverse of 2/7, the 2/7 cancels out in the left.

[rose tulip]

yeah

[pallid sierra]

You are left with

X = 12*7/2

Easy

[rose tulip]

mhmm I have that

[pallid sierra]

👍

[rose tulip]

or i flipped it

[pallid sierra]

Yess

Inverse

[rose tulip]

But where is the 42

[pallid sierra]

(12*7)/2 = 42

[rose tulip]

OOOOOO

damn i forgot about dividing to the other side

[pallid sierra]

Oof.

You’ll get there

[rose tulip]

I even had it in the freakin notes

[pallid sierra]

¯_(ツ)_/¯

[rose tulip]

Lemme finish up the intro to geo lessons and ill brb with more questions

Myfriend told me geo isnt that important

Is that true

I mean besides the basic stuff

[pallid sierra]

You’ll take geometry in high school and you need to pass that class.

To graduate

At least in Texas

[rose tulip]

Yea im not just tryna get by

[pallid sierra]

Mmmm

[rose tulip]

That's why i wanna get to where he's at

and not at 9th grade lvl

[pallid sierra]

You may or not use geometry depending on your career

[rose tulip]

well at 10th grade lvl now

[pallid sierra]

I see

[rose tulip]

software engineering?

[pallid sierra]

Idk

[rose tulip]

hmm???

[pallid sierra]

Maybe

[rose tulip]

Yeah ill just get the basics and then some

[pallid sierra]

Sounds good

[rose tulip]

Thanks dude

[pallid sierra]

My pleasure

🙂

[rose tulip]

Like for real once I finnish geo prolly this month ima be goated

Then alg 2 easy

then whatever is after that XD

aight math time

[pallid sierra]

Yeeep

[supple wedge]

hello i need help

let me latex

A piece of wire of length 80 cm is bent into the shape of a trapezium ABCD. AB=CD=$x$cm and angle BAD=angle ADC = $120\degree$. Show that the area of the trapezium $ABCD$ is given by $\frac{\sqrt{3}}{2}x(40-x)cm^2$.

[somber coyote]

Hmm:

A piece of wire of length 80 cm is bent into the shape of a trapezium ABCD. AB=CD=$x$cm and angle BAD=angle ADC = $120\degree$. Show that the area of the trapezium $ABCD$ is given by $\frac{\sqrt{3}}{2}x(40-x)cm^2$.
```Compile error! Output:

! Undefined control sequence.
l.54 ...x$cm and angle BAD=angle ADC = $120\degree
$. Show that the area of t...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., \hobx'), type I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2018/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]

[supple wedge]

whatever i just dk the latex for degree but i need help

wait i think i got it my algebra glitched again

[supple wedge]

thats the question

i copied it off

the worksheet

im almost there anyway idt i need the help

[dire sand]

@supple wedge what are we showing the area of the trapezium to be?

[supple wedge]

er wait

fuck i forgot to put it in my bad

[dire sand]

nvm

[supple wedge]

did i

idk

[dire sand]

I think its in the image

that texit outputted

[supple wedge]

$\frac{\sqrt{3}}{2}x(40-x)cm^2$

[dire sand]

lovely

[supple wedge]

fuck

[somber coyote]

Hmm:

[vapid briar]

What would one call dividing a sine wave into equal steps? partitioning? or is there a better word for this?

[dark sparrow]

what do you even mean

[vapid briar]

Im writing a program for a Q learning crawler. it has two server motors that I have made follow each its own sine wave.
ie. the servo joints sweep back and forth following the sine wave in small steps. the code for partitioning and finding the steps are not self explainable, and it needs some proper comments explaining it.
However since im bad at math and clearly hopeless at explaining stuff I require help.

😄

ugh. sorry in advance

when googling I see people talk about linspacing. is that what I am doing?

[dark sparrow]

i can't tell until i see the code for whatever it is that you're doing

[vapid briar]

it might be easier for you to understand what it does.

// array setup
 // Sinus wave partitioner
    double delta = (double)(2*M_PI / SERVO_STEPS);
    for (i = 0 ; i < SERVO_STEPS; i++){
        radii[i] = i * delta;
    }
  // servo motor iteration
 for (i = 0 ; i < SERVO_STEPS ; i++){
        // S1
        next_angle = (((sin(radii[i]) * sine_states[newgwS1][servo_1]))*100)+angle_delta[servo_1];
        S1.write(next_angle);

        // S2
        next_angle_2 = (((sin(radii[i]+ (M_PI*sine_states[newgw_phase][phase_shift])) * sine_states[newgwS2][servo_2] ))*100)+angle_delta[servo_2];
        // if phase_shifted wait till old S2 angle matches new
        if(phased){
            if(abs(s2_old_angle - next_angle_2) < 4){
                phased = false;
            }
        } else {
            S2.write(next_angle_2);
      }
        delay(servo_delay);
    }
    phased = false;
    s2_old_angle = next_angle_2;

its the variable radii I want to rename to something fitting. as well as explain better how it works in contxt with the servo motor for loop

issue is, my way of doing math is just trying stuff till it works. So I know my code works. My crawler learns to crawl, all servo motors move smooth

but commenting // I got to this code by trying random stuff is not good commenting

[dark sparrow]

hm

so the orientation of each motor follows a sine curve?

[vapid briar]

that can be changed

depending on agents action choice

same with phase shift

the sine curve amplitude can be changed

but yes

they follow the sine curve

larger sine amplitude means bigger angle delta

[dark sparrow]

hm

radii seems to be an array of equally spaced angles from 0 to 2pi

[vapid briar]

yes

or

could I just call it angle_increments?

[dark sparrow]

maybe?

yeah that doesn't sound bad

[vapid briar]

the way the code works now it limmit a servo sweep between lets say 180¤ and20¤ at sine amplitude 0.8

while with sine amp of 0.1 (min) its like 110 to 90

degrees

anyways thanks for listening 😄

🐱

[vale beacon]

How do I go about approaching this problem? I’m not very good at math lol

[upper karma]

How about the definition of sine?

[vale beacon]

Ratio of a triangle from opposite and hypotenuse

[livid moss]

Triangle looks right angled, but they don't say so. You can see by checking that Pythagoras holds that it is in fact right angled. So yes you can use the definition

What is the opposite and hypotenuse respectively?

[vale beacon]

O=15 H=39

[livid moss]

So the sin is 15/39

[vale beacon]

That is not one of the answer choices

[livid moss]

Have you heard of simplifying fractions?

1/2 = 2/4 = 3/6. The same fraction can be written in different ways.

[vale beacon]

Oh yes divide by the highest number that both the numerator and denominator can be divided by

[livid moss]

Indeed

[vale beacon]

Thank you. I’m trying to get a head start before I start high school and as you can see I’m struggling a bit

[livid moss]

Keep at it!

[near trail]

Hey! I'm trying to define a curve using X and Y values. However, this is not enough information to provide a direction for the curve.

What value should I use for defining the curve direction or the from which side of "slope" does the curve bend from? Is there some standardized terminology for this?

[umbral snow]

@near trail
There are standard ways to capture this, they usually stem from calculus or analysis

[near trail]

@umbral snow Ok, any ideas about the terminology for these? Obviously there's going to be 8 different definitions needed, but with X and Y I can only define 4.

[umbral snow]

You can capture it by defining the slope of the tangent line at the start point

So 1. would start with a high slope

More specifically, if f is your function, and g is the secant line, you want the sign of the tangent line at the start of f - g

[upper karma]

@vale beacon if you need a study buddy I’m doing the same thing. Just ping if you need help.

[outer garnet]

,iam adv

[somber coyote]

Your roles have been updated!

[upper karma]

hi, any hints on how I could find the sin of angle C in this diagram?

well you know theta

then you know angle C

I don't think I do? I could use arcsin, but that would only give me an approximation

yeah arcsin

but I'm not supposed to solve with a calculator, since I need to give an exact value.

<@&286206848099549185>

[stone adder]

Uhh, I personally think there's no such exact value that satisfies sin(theta)=1/6?

Can't you just write theta=2n*pi+arcsin(1/6) with n being integers as the final answer?

[upper karma]

I need to find sin(<C)

I don't need to find theta

[stone adder]

Ohh my bad

I'm afraid this might be a long route tho, but how about find the value of AB first by the cosine law?

I mean, you already have that the angle of A is pi/6 and you also have the lengths of AC and BC

So, we try to find the length of AB first

and after got that length, we can apply sine law to find the sin(C)

[upper karma]

okay!

the length of AB is in terms of cos(<C)

[stone adder]

$BC^2=AC^2+AB^2-2(AB)(AC)\cos(A)$

We have the angle of A, so we have to make the use of that

[somber coyote]

Wilston Lynx:

[upper karma]

oh I understand, it will be a quadratic equation in AB

[stone adder]

Yeah

and we find the appropriate value of AB by solving the quadratic equation

[upper karma]

sqrt(3)x/2 ± sqrt(41)x/2

that's what I found for AB

[stone adder]

Can you show your work for that?

[upper karma]

9x^2 = x^2 + AB^2 - 2x * AB * sqrt(3)/2, then I used the quadratic formula to get AB = (sqrt(3)x ± sqrt(41)x)/2

[stone adder]

I think you did some miscalculations there

Especially on the sqrt(41)x part

[upper karma]

my bad! you're right, maybe sqrt(3)x/2 + sqrt(3x^2 + 32x^2)/2 = sqrt(3)x/2 + sqrt(35)x/2?

[stone adder]

Yep!

It's supposed to have 2 solutions right?

[upper karma]

and I think it should be the positive value, since AB cannot be negative

[stone adder]

I was going to type that, but yeah lol

and yeah, now we get the value of AB

We can apply the sine law now

and I think, you should get the value of sin(C) by applying that

Since you have the values of BC, sin(A), and also AB

[upper karma]

yeah! I think it sin(<C) = (sqrt(3)x + sqrt(35)x)/(12x)

which simplifies to (sqrt(3) + sqrt(35))/12

[stone adder]

Yep!

Well done!

[upper karma]

thank you for your help

[stone adder]

You're welcome!

[unborn holly]

can someone help me understand/derive the half angle trig identities? I know what they are but i have no clue where they come from and cant understand from pages on the internet

[stone adder]

Which part you don't understand about the proof?

[unborn holly]

i cant seem to find a proof at all

[stone adder]

By half angle trig identities, you mean $$\cos(x/2)=\pm \sqrt{\frac{1+cos(x)}{2}}$$ and $$\sin(x/2)=\pm\sqrt{\frac{1-cos(x)}{2}}$$ right?

[somber coyote]

Wilston Lynx:

[unborn holly]

yeah

i think i have to use the double angle identities, which i understand, but i dont know how to use them to get to those formulas

[stone adder]

Alright, we actually can derive this formula from the double angle formula of cosine

Yep!

Well, double angle formula of cosine comes in 3 different form

$\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1=1-2\sin^2(x)$

[somber coyote]

Wilston Lynx:

[stone adder]

Let's say we want to derive the cos(x/2) first

[unborn holly]

do we use the last 2 forms for each proof?

[stone adder]

Yep!

If we want to proof the cos(x/2) ones, we can use $$\cos(x)=2\cos^2(x/2)-1$$

[somber coyote]

Wilston Lynx:

[stone adder]

We pick this form because it'll leads us into getting cos(x/2) when we do some algebra

[unborn holly]

oh my god i think i understand how to get to the formulas now. I didn't think to just halve the variables. I'll try to see if I can get both formulas now with this, thank you

[stone adder]

You're welcome! You can mention me if you had some troubles when deriving the formula! (well if you want to lol) 😄

[unborn holly]

i got both of them! it was weird being able to just place 0.5 in front of each angle variable but it makes sense because if the identity works for any angle it should be able to work for half of any angle too @stone adder thanks again

[stone adder]

Well done! 😄

[austere dragon]

why is the term for moving a point translations?? sounds like an odd term compared to rotation and reflection

[upper karma]

There may be no good reason

[austere dragon]

that cant be noooo there has to be a reason

like why x is commonly used

[earnest echo]

We don't really care about etymology in maths

[supple onyx]

Actually there is a reason for x being commonly used

But i forgot it XD

[earnest echo]

That's because, Rene Descartes used it

[upper karma]

It is possible to make decisions that were selected arbitrarily but are still useful to decide on

[earnest echo]

He used it because x was the the word which was used least commonly

[upper karma]

For the sake of everyone using the same word or because something has to be chosen

[earnest echo]

In old type writers you'd have to manually do something and because of that he used x

[supple onyx]

Thank you for the history @earnest echo 😊

[austere dragon]

that sucks, i was expecting a "actually it comes from the latin word transylvania which means to move an object"

[arctic vortex]

okay this is not a question but my final test is coming up soon for geometry so any tips for it?

[upper karma]

Draw

Label

Color

[arctic vortex]

ok

[twilit zenith]

I second what HoboSas says. It could be a good idea to label as many angles and sides as you can with colors and try to find a way to prove that two colors are equal or satisfy some other relation.

One more thing I can suggest is that if you get stuck, try to work backwards.
For example, if you are asked to prove that two sides are equal, then you can try to ask yourself what other conditions can make you conclude that those two sides are equal.

[fading zinc]

im confused on this

[upper karma]

@fading zinc why?

[tired knoll]

well how would you solve for x. I got lost after 3 root 11

[upper karma]

@tired knoll what

why is a different person replying? lmfao

[tired knoll]

Yeah i might have done it wrong.

y not

its math that I though I knew but didn't

[fading zinc]

hmmm

okay so what i dont understand

[tired knoll]

good morning soap

[fading zinc]

ah yes

thank you

anyways

if i find the missing side o ftriangle ABC

im not sure how i find the missing side of BCDE

otherwise known as CE

[upper karma]

@fading zinc http://sketchtoy.com/69284811

does that help

[tired knoll]

ohhhhhh omg I think I grew come brain cells just now

[glacial haven]

similar triangle

[hot patio]

Is a sphere an ellipsoid?

[dark sparrow]

yes

[wise zodiac]

how do i get the radius?

[silent plank]

pythagoras

[wise zodiac]

omg you're the best

thanks man

[ashen stream]

What do I do after this?

[upper karma]

What are you trying to do

[ashen stream]

Should I convert cos^2 /sin^2 to cot ^2?

I am trying to prove this

[upper karma]

What

[ashen stream]

The 15th problem

[upper karma]

Now multiply the 2 parenthesis

[ashen stream]

Okay

Just a sec

[upper karma]

Np

[ashen stream]

Done

After that

I get this

[upper karma]

Now combine fractions

[ashen stream]

Should I make the denominators same?

Ok

Just a sec

I get this

Sorry if my handwriting is shabby....

[upper karma]

On the denominator you get 2 +sin^2cos^2

[ashen stream]

Yes

Just a sec

[upper karma]

$1=(\sin^2+\cos^2)^2=\cos^4+\sin^4+2\cos^2\sin^2$

[somber coyote]

HoboSas:

[upper karma]

So $\cos^4+\sin^4=1-2\cos^2\sin^2$

[somber coyote]

HoboSas:

[ashen stream]

Okay so I should replace it in the numerator?

[silent plank]

one sec, there's an error in the question

denominator should be 2+ not 1+

[upper karma]

Okay so I should replace it in the numerator?
Yep

Then factor a sin^2cos^2 on the other 2 terms

And you are done

[ashen stream]

Yeah just give me a min

[upper karma]

Take your time

[ashen stream]

Yeah am done

But

I suppose there's an error in the question

[upper karma]

Yes

[ashen stream]

Since I got a 2+ in the denominator

[upper karma]

denominator should be 2+ not 1+

[ashen stream]

Umm yeah

Anyways

Thank you @upper karma

[upper karma]

Np

[ashen stream]

:-)

[unborn holly]

i saw a formula for the cos^2 of theta being (1+cos2theta)/2

could someone tell me why it works?

[livid moss]

Do you know the formula for cos(A+B)?

@unborn holly

[unborn holly]

i just figured it out using the double angle formula and one of the identities

but thanks

[livid moss]

Np

Good job on figuring it out

[pearl lava]

https://gyazo.com/f874c4554dbff314db8af6ce3c73930c
where do I start?

[silent plank]

label your diagram with the given properties

that could be proven if you're able to prove triangles STX and VTW are congruent so consider working towards that

[little osprey]

^

[dusky yacht]

Hi, is this OK? The given answer is OA = (0, 50, -75)

Maybe the given answer is wrong, but I don’t think so

[quasi elm]

Hey can someone help me out with this question? Our professor linked a worksheet to practice trig identities but I'm stuck on this one

[midnight totem]

Need help solving this

[polar wave]

@midnight totem you just use the pythagorean theorem: (leg1)^2 +(leg2)^2 = hypotenuse^2

so you do 6^2 + 11^2

then square root the result

[sour jacinth]

Hey can someone help me out with this question? Our professor linked a worksheet to practice trig identities but I'm stuck on this one

@quasi elm Use the pythagorean identity on $$16 \cos^2(\theta)$$ and simplify the expression.

[somber coyote]

Max Hetfield:

[midnight totem]

@aboinpally#8745 so like this

[stone mortar]

does anyone know how to find foci of ellipse

[sour jacinth]

https://discordapp.com/channels/268882317391429632/326138757474680852/740367006418600078
@midnight totem Nope. You have a calculation mistake in the 4th and 5th lines

https://discordapp.com/channels/268882317391429632/326138757474680852/740380457908043800
@stone mortar Yeah, let me check something

[stone mortar]

thank you

a=distance from center to vertex?

b=c^2 -a^2 ?

c=distance from vertex to foci?

i don't know

h,k=centre

[sour jacinth]

Wait a sec

@stone mortar Read this: https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:conics/x9e81a4f98389efdf:ellipse-foci/a/ellipse-foci-review?modal=1

[stone mortar]

thanks this helped me @sour jacinth !

[sour jacinth]

You're welcome, @stone mortar !

[cloud stump]

anyone knows how to draw a perpendicular with ruler and compass

[fossil kestrel]

any recs for a coordinate geometry book?

[pastel anvil]

ya draw a line segment then at one end point as the center draw a big arc with a radius greater than half the length of the line segment then with the same radius draw another big arc with the center as the other endpoint

so then connect the 2 points where the 2 arcs meet and that's the perpendicular line

[sour jacinth]

any recs for a coordinate geometry book?
@fossil kestrel Coordinate geometry?

[upper karma]

@fossil kestrel are you asking about vectors?

[raven arch]

Im doing a review for linear algebrea and the review includes geometry from like 8th grade

i just wanna make sure this is correct

im fairly confident it is

[azure reef]

@raven arch

why trapezoid should have congruent sides?

[raven arch]

because 2 will be parallel

OH but not nesicarrly same length

appologies

thank you

[azure reef]

@raven arch are not sides of rhombus assumed to be equal?

and why rectangle does not havve 90 angles?

also, can parallelogram have opposite sides not equal?

[hard geyser]

and why does the rectangle have none?

(sets of congruent sides)

also, should the trapezoid have any?

well. I assume congruent here means same length, so probably

[azure reef]

also, should the trapezoid have any?
i already pointed that

it should not

[hard geyser]

I mean

a set of sides has to have the same length

wasnt that congruent in geometry?

[azure reef]

if two sides are equal in length they are congruent

but trapezoid can have all sides of diff length

[hard geyser]

a true. Im thinking of a regular trapezoid. im dum

[raven arch]

oh. i forgot that a rhombus had all 4 sides congruent, also i missed some thing on the digital that i wrote on paper sorry

thank you for being so helpful

[hard geyser]

theres also sets of congruent sides on rectangles and parallelograms

[raven arch]

thank you, i have that down on my paper lol but not here

[earnest echo]

The diagonals of rhombus intersect at right angle as well

[fossil kestrel]

@fossil kestrel are you asking about vectors?
@upper karma nope. its geometry on the Cartesian plane and you deal with algebraic expressions of curves, lines and such

[upper karma]

@fossil kestrel so... vectors?

[austere dragon]

can someone explain this to me more thoroughly plz?

i understand that the point will be all the same distance from the center, but why draw a perpendicular line??

and what's with the other points on there excluding the actual p and p' line

[paper vale]

it has to do that otherwise the points wont be same distance from centre

it is like an isosceles triangle

[austere dragon]

what?

[paper vale]

ehh i dont why u are confused

[austere dragon]

it's just, strange, why add those points on the perpendicular bisector?

aren't we just trying to get the center of it

[paper vale]

the centre must be on that line

it could be anywhere on it

[supple onyx]

Any of those points can be a center of rotation

Since all of those points are equidistant to P and P's image

[austere dragon]

owh

but whats the perpendicular line for??

we just need to get the center of between the 2 points

[supple onyx]

To determine where all those points lie

Yes but that's only one center of rotation

Not all possible ones

[paper vale]

bruh

[austere dragon]

🤯

big brain moment

[supple onyx]

If i look at the perpendicular bisector of the line joining P and its image, I can form iscoceles triangles with various base angles

[paper vale]

not really that big brain but ok

[supple onyx]

Because those hypothenuses will be equal on both sides, so the points will still be equidistant to the center of rotation

[midnight totem]

@sour jacinth What is the mistake?

[little osprey]

what

[livid moss]

@midnight totem 12^2 is not 157

[little osprey]

,w calc sqrt 157

[somber coyote]

Results provided by WolframAlpha

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

Requested by JC Denton#9718

[paper vale]

bruh

[livid moss]

Also you took the squareroot of a^2 and just left it as a^2

[little osprey]

^

[livid moss]

That was wack

[little osprey]

Also

No units

[livid moss]

Well, he wasn't given units, to be fair

[little osprey]

?

I was taught

You should always write units

[paper vale]

units are redundent

[little osprey]

even if there are no units

[paper vale]

why would u do that

[livid moss]

I was taught that too, but eh

[little osprey]

why would u do that
@paper vale Big brain stuff dw about it

[livid moss]

When sqrt(a^2) = a^2, you have bigger fish to fry

[paper vale]

um im pretty sure that isnt a thing

[livid moss]

I did it at school too, so it is a thing

[little osprey]

Same

Maybe it wasnt a thing for you kane?

[paper vale]

so u just assume it is like cm^2 wtf

[livid moss]

kane's subjective experience is mathematical fact, duh

No, you don't assume units

[silent plank]

units^2

[livid moss]

You write units^2

[paper vale]

hmm ok

in this instance it isnt area tbh

[little osprey]

^

Just units

[paper vale]

actually that makes sense tbh because that is what defines area

[little osprey]

It seems even Ramonov used to use 'units'

[silent plank]

more important for stuff like area and volume

[paper vale]

yea

[silent plank]

doesn't make much of a difference for length

[midnight totem]

Sorry I mean't 12.5

[silent plank]

12.5 isn't being accurate to 3decimal places

[fossil kestrel]

@fossil kestrel so... vectors?
@upper karma lol yeah sorry i read the definition now and its part of it. the subject is called coordinate geometry or analytic geometry. so the book wouldn't be titled "vectors"

[upper karma]

@fossil kestrel what you're describing is still vectors

@fossil kestrel what you're looking for is a linear algebra book.

I think they are actually looking for an analytic geometry book

@upper karma what would they cover there

I've never read a book on it or learned it explicitly per se

that's because it doesn't exist, or rather, it exists as linear algebra books

but it's not synonymous with linear algebra

sure it is

name something from the field of "analytic geometry"

i.e. a concept

the main concept is geometry over a coordinate system

like what

is "the basics of rectangular, cylindrical, and other coordinate systems" satisfactory

what are basics

define it

I can't be that specific for the reason I mentioned earlier

have you ever done calculus with analytic geometry

like what

i don't know what analytic geometry is, just linear algebra

so

problems like, find the area of triangles which are enclosed by some region which is indicated by a line tangent to a curve

yup, that's linear algebra

my linear algebra book covered exactly that (and a lot more)

oh

to some curve

that's calculus

lol

[livid moss]

Do you really not know or are you being obtuse? Analytic Geometry are things like the distance and midpoint formula that you learn in high school only in the context of R^2 without generalization. It is the geometry of the Cartesian plane.

[upper karma]

vector calculus, in fact

@livid moss a lot of my linear algebra book was dedicated to $\mathbb{R}^2$

[somber coyote]

polynomial:

[upper karma]

since it's the best for building intuition

[livid moss]

Yes, mine too

And a lot of my topology classes was dedicated to concepts from real analysis. Well, shit, I guess we don't need real analysis anymore. Just learn it all in topology.

[upper karma]

not true. topology is a subset of analysis, kind of

it's the part of analysis that most people actually like about analysis

it's analogous to R^3 or something in linear algebra lol

maybe not the best analogy, but yeah

[livid moss]

The point is just because concepts overlap, does not mean the two subjects are the same

[upper karma]

i'm still waiting for you to name some "analytic geometry" concepts

[livid moss]

You never asked me to, but I already did. The midpoint formula.

[earnest echo]

That's what just it is called

[upper karma]

the midpoint formula is definitely covered in my linear algebra book

[livid moss]

I never said it wasn't

[earnest echo]

People call it coordinate geometry everywhere

[upper karma]

and probably all intro linear algebra books

[livid moss]

Name a real analysis concept

[upper karma]

proofs

lol

[earnest echo]

Stop this point less argument

[upper karma]

real numbers

[livid moss]

Proofs were definitely done in my linear algebra class so real analysis is unnecessary

[upper karma]

i agree

[livid moss]

Excellent

[upper karma]

fuck real analysis

i hate it

[livid moss]

Let's all never learn math and just jump to research

[upper karma]

that'd be interesting

@livid moss the point is i doubt you can name an exact concept that you learn in "analytic geometry" but not linear algebra

[livid moss]

I understand your point perfectly, and you seem to be missing mine completely

The same can be said of real analysis. Name a concept you encounter in it, that you don't encounter in a different class.

Sometimes you need to learn something in one setting before you abstract it

[upper karma]

"abstracting" the midpoint formula

lol

https://tenor.com/view/scarecrow-strawman-hobgoblin-scarer-figurehead-gif-13919922

[livid moss]

Sure, why not? From R^2 to R^n or even C^n.

[upper karma]

but he's asking about R^2

so

he just needs to read the R^2 part of a linear alg book

[livid moss]

That's different than what I thought you meant. I thought you meant they should just learn linear algebra immediately. But even so, a linear algebra book will reference the parts that aren't just about R^2. You won't have that problem with notes on Analytic Geometry.

Anyway, I'm done arguing. So if you want to say something else, you may have the last word

[upper karma]

i don't, but i did mean what i said last. you don't have to read the whole book when you read a book, just the parts that you are interested in or are relevant to you

in this case, R^2 to him

[supple onyx]

what's the argument

[upper karma]

is linear algebra in R^2 = "analytic geometry"

(whatever the latter may be)

[supple onyx]

Idk what analytic geometry is, so i cant take part

[upper karma]

that's what i said lol

but from his examples, it's pretty clear he's talking about lin alg in R^2